Samplebook-Tools of organic chemistry

(i)

By

AJNISH GUPTA & BHARTI GUPTAProfessor of Organic Chemistry

OrganicFor world of competitions

Tools of

Chemistry

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All rights reserved.No part of this publication may be reproduced, storedin a database or retrieval system, or transmitted in anyform or by any means, electronic, mechanical, photo-copying, recording, or otherwise, without the priorwritten Permission of the writter.

First Indian Reprint, 2015

This edition is manufactured in Indian and isauthorized for sale only in India, Bangladesh, Pakistan,Nepal, Sri Lanka & Maldeves.

Printed & Distributed by:Udaan Classes Pvt. Ltd.Rainbow building, Patna &Madhur SatyapushpaShubhash Nagar, Kota9122057123

Price: Rs. 499/-

Copyright © 2015 by Ajnish Gupta

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PrefaceThe guiding principle in writing this book was to createa textbook for students- a textbook that presents thematerial in a way that they learn to solve all thequestions along with the strategy to approach theproblems.In this book we mixed all our teaching experience of 11years along with theoratical and experimentalknowledge to generate a hand book for all students toreason their way to a solution rather than memorize amultitude of facts, hoping they don’t run out of memory.This book covers mainly 4 units with 61 sections whichare real tools of Organic chemistry, which a studentsmust know before dealing any chemical reactions.Organic chemistry is very easy and conceptual subjectand need proper understanding of the basics andstretegy to solve the questions in corret manner.This book will prepare your right mindset for learningOrganic Chemistry. This mindset is essentially the onethat focuses you on a small number of straight forward,repeated, fundamental concepts and helps you toapply them in different ways to solve the variety ofproblems you face in organic chemistry.This book is complete as it not only covers theory inproper sequence but also provide varieties of questionsalong with 10 test papers to judge your knowledge be-fore going to start chemical reactions.This book also covers all the questions of AIEEE,IIT-Mains, IIT-Advanced, AIPMT & other medical ex-ams from 2000 to 2014.In this book balance has to be achieved between thenumber of questions and the quality of the questions,

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Organic Chemistry is easyby

www. co.in

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especially because it is relatively easy to frame a verylarge number of multiple-choice questions and theoryof the subject.The questions in this book have been selected keepingthree things in mind.First- the questions are such that they really test theunderstanding of the subject.Second- the questions cover all concepts.Third- the number of questions has been kept largeenough to offer meaningful practice to the students.We would like to thank the editors and production stafffor their efforts in bringing out the book. Suggestionsfrom readers for the improvement of the book are wel-come.

Ajnish Kumar Gupta & Bharti [email protected]

9122057123

mailto:[email protected]

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AcknowledgementWe are thankful to all the teachers who taught usduring the concept building session of our life speciallyDr. Nizamuddin sir, senior Chandra Vijay Rao and Dr.Vijay Pratima Mittal madam.We have written this book to remove the fever oforganic chemistry from mind of students.We particularly want to thank many wonderful andtalented students whom we have taught over the yearswho in turn taught us how to be a good teacher andhow we can help others.We want to make this book as user friendly as possible,and we will appreciate any comments that will help usto achieve this goal in future editions.Finally, this edition has been presented before you withefforts to make it error free. Any that remain are ourresponsibility; if you find any, please let us know so theycan be corrected in future printing.

Ajnish Gupta & Bharti GuptaProfessors of Organic [email protected]

09122057123

mailto:[email protected]

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Table of ContentsPage No.

Unit 1 : Basic understanding of organic chemistry 01Section 01 : Introduction of organic chemistry 02Section 02 : What is organic chemistry 06Section 03 : Representation of organic compounds 09Section 04 : Reason for the formation of large

number of organic compounds 15Section 05 : Functional groups 18Section 06 : Homologue & Homologous series 25Section 07 : Nature of C, H & functional groups 29Section 08 : Saturated & Unsaturated molecules

with important points 34Section 09 : Hybridization 38Section 10 : Classification of organic compounds 42Section 11 : Baeyer’s strain theory 47Exercise 1 : Subjective approach 51Answer : Subjective approach 63Exercise 1 : Objective approach 79

Single correct questions (SCQ) 79Multiple correct questions (MCQ) 90Assertion/reason type questions (A/R) 95Match the column type ques. (MTC) 97Comprehension type questions 98Integer type questions 99

Answer : Objective approach 100

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Unit 2 : Nomenclature of organic compounds 101Section 01 : Alkyl groups 102Section 02 : Basic concept of IUPAC nomenclature 106Section 03 : Points to write IUPAC nomenclature 111Section 04 : Points to write IUPAC nomenclature

for multiple bonds & rings 116Section 05 : Points to write IUPAC nomenclature

of compounds with functional groups 122Section 06 : Nomenclature of Bicyclic &

Spiro compounds 131Section 07 : Some common names commonly used

in organic chemistry 135Section 08 : Degree of unsaturation 157Section 09 : Chemically different hydrogen 163Exercise 2 : Subjective approach 168Answer : Subjective approach 194Exercise 2 : Objective approach 214


Answer : Objective approach 240Test 1 : Basic understanding of organic

chemistry & nomenclature 241Test 2 : Basic understanding of organic

chemistry & nomenclature 250Test 3 : Basic understanding of organic

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chemistry & nomenclature 259Test- 1, 2, 3 : Answer 268

Unit 3 : Study of isomerism 269Section 01 : Basic concept of isomerism 270Section 02 : Chain isomerism 273Section 03 : Positional isomerism 277Section 04 : Functional isomerism 281Section 05 : Metamerism 285Section 06 : Basic concept of tautomerism 292Section 07 : % enol content 298Section 08 : Types of tautomerism 303Section 09 : Basic concept of geometrical isomerism 305Section 10 : Nomenclature used in geometrical

isomerism 310Section 11 : Basic concept of optical isomerism 315Section 12 : Nomenclature used in optical isomerism 321Section 13 : Terms used in optical isomerism 326Section 14 : Compound showing optical isomerism

along with calculation of number ofstereoisomers 331

Section 15 : Conformational isomerism 336Section 16 : Conformation of cyclohexane 344Section 17 : Interconversion of one form of

molecule into another 351Section 18 : Finding relationship between

pair of compounds 354Section 19 : Physical properties of organic

compounds 360Exercise 3 : Subjective approach 366

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Answer : Subjective approach 384Exercise 3 : Objective approach 401


Answer : Objective approach 446Test 4 : Concept of isomerism 449Test 5 : Concept of isomerism 459Test 6 : Concept of isomerism 468Test- 4, 5, 6 : Answer 478

Unit 4 : Electronic effect & it’s application 479Section 01 : General concept of organic reactions 480Section 02 : Inductive effect 487Section 03 : Basic concept of resonance 491Section 04 : Points to draw resonating structures 495Section 05 : points to check stability of resonating

structures 499Section 06 : Mesomeric effect 504Section 07 : Hyperconjugation 508Section 08 : Electromeric effect 511Section 09 : Organic reaction intermediates 514Section 10 : Carbocations 517Section 11 : Stability of carbocation 520Section 12 : Rearrangement of carbocations 524Section 13 : Classification of carbocation 532Section 14 : Carbanions 535

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Section 15 : Free radical 540Section 16 : Carbenes 546Section 17 : Nitrene 550Section 18 : Benzyne 554Section 19 : Ajnish rule & It’s application in

organic chemistry 558Section 20 : General concept of Acid & Base 567Section 21 : Concept to decide relative acidic

strength 572Section 22 : Concept to decide relative basic

strength 577Exercise 4 : Subjective approach 581Answer : Subjective approach 607Exercise 4 : Objective approach 639


Answer : Objective approach 726Test 7 : Electronic effect & it’s application 729Test 8 : Electronic effect & it’s application 739Test 9 : Electronic effect & it’s application 749Test- 7, 8, 9 : Answer 759Test 10 : Tools of organic chemistry 760Test 10 : Answer 769

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Dedicated to all those students whoare in fever of organic chemistry.

Dedication

1

Basic understanding of organic chemistry

Organic Chemistry is easy bywww.CoachMe.co.in

This section is designed to lay the foundation for your organicchemistry. Here you will learn what is organic chemistry, howorganic compounds are represented, what is the reason forformation of large number of organic compounds, functionalgroups, homologues, nature of carbon & hydrogen, saturation andunsaturation, hybridization, classification of organic compoundsand Baeyer’s strain theory.

1Unit

Basic understanding ofOrganic Chemistry

http://www.CoachMe.co.in



2

Have you ever thought that when you read anything from “anybook” your eyes use an organic compound (retinal) to convert visiblelight into nerve impulses. When you pick up this book from anyplace, your muscles do chemical reactions on sugars to give youenergy. As you read the words and sentences of this book, gapsbetween your brain cells are being bridged by simple organicmolecules (neurotransmitter amines) so that nerve impulse can bepassed around your brain and you can understand all matterscorrectly and you did all that without consciously thinking aboutit.

You do not yet understand these processes in your mind butyou are regularly carrying them out in your brain & body.

3CH3H C

O3CH

11-Cis retinal3H C

H

3CH

(Absorbs light when we see)

HO

NH

Serotonin2NH

(Human neurotransmitter)You are not alone there.........No organic chemist, is so brilliant who can understand the

1Section

Introduction ofOrganic Chemistry


3



detailed chemical working of human mind or body very well.So don’t be sad and give a smile to yourself so that your

journey should be cheerful....You do at least one thing after reading this page “Think the

importance of organic chemistry and try to search at least one thingwhich makes your life easy but not made of CARBON”.

??

Are you thinking of this book, which you are readingpresently - Made up of cellulose and indeed carbon

Notice board, Black or white board of schools - Made upof wood and indeed carbon.

Cloths which you are wearing - Polyester, tericot, silk etc.- Polymer of organic compounds

Perfumes, which you are using for good smell - Againorganic molecule such as cis-jasmone from jasmine flowerand damascenone in smell of rose.

Medicines at the time of your sickness - Made up of organiccompounds such as paracetamol and aspirin

Sugars to make your life sweet - Made up of glucose,sucrose and indeed carbon.

Are you thinking for opposite sex attraction- Pheromones &hormones are responsible for it, which are again made up ofcarbon.

Are you now thinking for drugs - Then they are againorganic and made up of carbon such as cocaine, nicotine,etc.

Polythene which are used to carry things - made up ofethene which indeed is made up of carbon

Blue jeans, which you are wearing - made up of indigodyes, made up of carbon.

Petrol or diesel- Again made up of carbon




4

Candles for your birthdays and parties for light - Againmade up of carbon

Are you thinking for bigger molecules such as Vitamins,Proteins, DNA, RNA – they are again organic and madeup of carbon.

Trees, plants, flowers - They too are made up of carbon You, I or any human being along with all animals &

plants- has some sort of carbon, in one form or another.

3H C

O

OCocaine

MeOOCN

3H C 3CH

3CHDamascenone-The smell of rose

O

3CH

HOO

HO OH

OHO

Vitamin C

3CH

O 3CH

Cis Jasmone of jasmine

OHHO O

HO OHOH

Glucose

3CHOCOOH

O

Aspirin

3CH 3CH

OH

Vitamin A

3H C 3CH

3CH

N2NH

HNN

NAdenine


5



N

2NHHNNH

N

O

Guanine O

HN

NH

O

Indigo dye for blue colour

ON

N

2NH

HCytocine

ONH

NH

Uracil

O

ONH

NH

Thymine

O3H C

Anything more which you want to suggest....Examples may varies from yours to our’s but from these

example. We want to give a clear cut picture of importance of organicchemistry and its applications in daily life i.e. everything aroundus are organic in nature so

Love organic, feel organic and give a smile for organic....




6

What is Organic chemistry?Why do so many people study it?And why should I study it?The answer to these questions is all around you. Every living

organism is made up of organic compounds. The proteins that makeyour hair, skin and muscles; the DNA that controls your geneticheritage; the food that nourish you; the clothes that keep you warmand the medicines that heal you, are all organic chemicals. Anyonewith a curiosity about life and living things must have a basicunderstanding of organic chemistry.

The foundation of organic chemistry dates from the mid 1700s,when chemistry was evolving from alchemist’s art into a modernscience. At that time unexplainable differences were noted betweensubstances obtained from living sources and those obtained frommineral sources. Compounds obtained from plants and animalswere often difficult to isolate, purify and even difficult to work.The Swedish chemist Torbern Bergman in 1770 was the first todifferentiate “organic” and “inorganic” substances and the termorganic chemistry soon came to mean the chemistry of compounds

2Section

What is Organic Chemistry


7



found in living organism.To many chemists at that time the only explanation for the

differences in behavior of organic and inorganic compound wasthat the organic compounds must contain a peculiar “vital force”as a result of their origin in living substance. Due to this vital forcetheory chemist believed that organic compounds could not beprepared and manipulated in the laboratory, as could inorganiccompounds.

In 1816, however this vital force theory received a heavy blowwhen Michel Chevreul found that soap, prepared by the reactionof alkali with animal fatcould be separated into several pure organiccompounds,which he termed “fatty acid”. For the first time, oneorganic substance (fat) was converted into other organic substances(fatty acids and glycerin).

Animal fat Soap + Glycerin2H O

NaOH

soap Fatty acid3H O+

In 1828, vital force theory again got a heavy blow when FriedrichWöhler prepared organic compound urea, of human urine frominorganic compound ammonium cyanate.

–4 2 2

HeatNH OCN NH CONHAmmonium Cyanate Urea

By the mid of 1800s, the weight of evidence was clearly againstthe vital force theory which led William Brande in 1848 to write“No definite line can be drawn between organic and inorganicchemistry......... Any distinction........ Must for the present be merelyconsidered as matters of practical convenience calculated to furtherthe progress of students”.

Chemistry today is unified. The same principles that explainthe simplest inorganic compounds also explain the most complexorganic ones. The only distinguishing characteristic of organicchemistry is that all contains the element carbon. Nevertheless, thedivision between organic and inorganic chemistry, which began




8

for historical reasons, maintains its practical convenience... to furtherthe progress of students. So, in general

“Organic chemistry is a branch of chemistry which deals withcarbon and its compounds”

C6

12

Atomic number

Atomic massSymbol

–number of electrons–number of protons

From the periodic table we can easily find that it comprises ofseven periods and eighteen groups. First and second groupselements are called as s block elements, thirteen to eighteen groupelements are called p block elements, third to twelve group elementsare d block while La to Lr which are downside are called as f blockelements.

When we carefully analyze the periodic table, it is clear thatposition of carbon is in second period and fourteenth group andcarbon is a non-metal.

Carbon is the basic element for life but it forms its compoundsby combination with other elements such as H, N, O, F, Cl, Br, I, S,P etc.

In Organic chemistry we mainly deal with Structural determination – How to find out the structures

of new compounds, even if they are available only ininvisibly small amount.

Theoretical organic chemistry – How to understand thosestructures in terms of atoms and the electrons that bindsthem together.

Reaction mechanism – How to find out how thosemolecules reacts with each other and to predict theirreactions.

Biological chemistry – How to find out what nature doesand how the structures of biologically active molecules arerelated to what they do.


9



Organic chemistry is the study of compounds that containcarbon but there are some other elements too associated with it suchas H, O, and N. We know that carbon is tetravalent in nature andform covalent compounds by sharing of electrons by generally H,F, Cl, Br, I, N, O, S and P. For sake of simplicity all halogens aregenerally represented by X. In the study of molecules, you will noticethat oxygen has two covalent bonds, nitrogen has three covalentbonds and carbon has four covalent bonds when they are neutral.Neutral hydrogen & halogens each have one covalent bond. If atomshave more bonds or fewer bonds than the number required forneutral molecule then they would have either a formal charge or anunpaired electron. These numbers are very much important toremember when you are first drawing structures of organiccompounds because they provide a quick way to recognize whenyou have made a mistake.

C N O H F Cl Br I X

This covalent bond may be of sigma ( )or pi ( ). Sigma bondis formed by lateral overlapping of orbitals. It may be of s-s, s-p or

3Section

Representation of OrganicCompounds




10

p-p but pi bonds are formed by sidewise overlapping of p-p orbitalsin terms of organic chemistry.

+s s s-s overlap

s p s-p overlap

y x+

y x

p orbital p orbitalzz

p-p overlap alongthe orbital axis

Py

y

Py

y

+

p-p sideways overlap

y y

As sigma bonds are strong as compared to pi bonds so generallypi bonds are involved in chemical reaction with reagent, if presentin molecule. When we compare the bond strength of bonds formedby s-p & p-p overlapping; p-p bond is stronger so generally s-pbond break on treatment with suitable reagent, such as in case ofalkanes.

Organic chemistry concerns itself with the way in which theseatoms are bonded together into stable molecular structures and theway in which these structures change during chemical reactions.

For this organic molecules are represented by several ways1. Lewis structure or dot structure –Here valence electrons

in a molecule are represented as dots. When you drawLewis structure you must make sure that hydrogen atomis surrounded by no more than two electrons, and C, O, Nand halogen (F, Cl, Br, I) atoms are surrounded by no morethan eight electrons. In other words, they must obey theoctet rule. The valence electrons not used in bonding are


11



called as non-bonding electrons or lone pair of electrons.Lewis structures are useful because they show us whichatoms are bonded together and tell us whether any atompossess lone pair of electrons or have a formal charge.

H C HH

H

.. .. H C OH

H

.. .. H.2. Dash structure or Kekule structure–In this representation,

dash represents the bonds. Single dash represents a singlebond, double dash is used for double bond of any type suchas of C=C, C=N, C=S etc. and triple dash represent triplebond of any type such as C C or C N, etc. In thisrepresentation lone pair of electrons of hetero atoms (O, N,S, and P etc.) may or may not be shown. This representationgives complete structural formula of any molecule.

H–C–C–HHH

HH C=C

H

H H

H H–C C–H

3. Condensed s tructure – In condensed structuralrepresentation, dash structures can further abbreviated byomitting some or all of the dashes representing covalentbonds and by indicating the number of identical groupsattached to an atom by a subscript. The resulting expressionof the compound is now called a condensed structuralformula. For example, CH3CH3, CH2 =CH2, CH CH,CH3OH. Similarly, CH3CH2CH2CH2CH2CH2CH2CH3 can befurther condensed to CH3(CH2)6CH3.

4. Bond line structure – In this representation of organicmolecules, carbon and hydrogen atoms bonded to carbonare not shown and the lines representing carbon-carbonbonds are drawn in zigzag fashion.

Even without writing the hydrogen atoms we know that theyare there and we assume that any carbon atom that does not appearto have its potential for four bonds satisfied is also attached to theappropriate number of hydrogen atoms.




12

We also rub out all the C representing carbon atoms and writeonly zig-zag line assuming that every kink in the line representscarbon atom.

Pentane Cyclohexane

Benzene

As the functional groups are the key to the organic chemistryso they are specifically shown if present so the only atomsspecifically written are heteroatoms such as oxygen, nitrogen,halogens, which makes the functional groups, which we will discusslater.

O

OH

This H is shown becauseit is attached to an atom

other than C

all 4 bonds are shown to this c atom so no Hatoms are implied

these C atoms must also carry1 H atom because only 3 bonds

are shown for each atom

end C have 3 H atoms

these C atoms must also carry2 H atoms because only 2 bonds

are shown for each atom

every kink in thechain represents

a C atom

Organic molecules should be drawn to be realistic, economicaland clear so here are 3 guidelines, which help you make structuremore realistic. Guideline 1- Draw the chain of atoms as zigzag. Guideline 2 – Miss out the hydrogen attach to carbon

atoms, along with the C-H bonds. Guideline 3 – Miss out the capital C representing carbon

atoms.5. 3D structure – Of course, all the structures we have drawn

give only an idea of the real structure of the molecules. For example,the central carbon atom of CH3CH(NH2)COOH has tetrahedralarrangement around it but so far we have completely ignored it.Now we want to emphasize this fact, so for this 3D structure of


13



molecules are written using wedge-dashed projection and Fischerprojection.

COOH

3CH

H

2H N

A. Wedge-dash projection or perspective formula:-In this wedge-dash projection, the solid wedge ( ) isused to indicate a bond projecting above the plane of paper,towards the observer and dashed wedge ( ) is used todepict the bond projecting below the plane of the paperand away from the observer. The bonds lying in the planeof the paper are depicted by using a normal line ( ).

CH

H

H

H

bonds are awayfrom the observer(Dashed wedge)

bonds are towards the observer(Solid wedge)

bonds are inthe plane of paper

(Normal line)

Sometime we could miss out the hydrogen atom and drawsomething bit neater though slightly less realistic. Here we assumethe missing hydrogen atom is below or above the plane by lookingthe 3 attachments. If two bonds are in the plane and one is abovethe plane then hydrogen must be certainly below the plane andsimilarly if two bonds are in the plane and one is below the planethen hydrogen must be above the plane.

COOH

H is assumed to be below the plane3CH

2NH2NHHCOOH

3CHor




14

2NHHCOOH

3CHH is assumed to be above the plane

2NH

COOH3CH

or

B. Fischer projection:–In this projection molecule is represented by horizontal andvertical lines. Groups on horizontal line are above the plane(towards the observer) and groups on vertical line arebehind the plane (away from observer) and point ofintersection of horizontal and vertical line is carbon.

C

H

Perspective formula of methane

H H

H 109.5°


15



What makes the carbon so special?What is it that sets carbon apart from all other elements in theperiodic table?Why are there so many organic compounds?The answer lies in carbon’s position in the periodic table. Carbon

is in the centre of second row elements

Li

Elements of the second row of the periodic table

Be B C N O F

First think why molecules are formed from atoms? It is becauseof the reason that atoms combines with same or with other atomsto form molecule so as to complete its octet and attain lower energystate and hence become stable. That is the reason why noble gasesare considered as inert gases, they generally do not combine withitself or with other atoms because they have complete octet. Butwhat about other atoms? They have incomplete octet, so they mustcombines with same or other atoms to form molecule for betterstability.

4Section

Reason for the formation of largenumber of organic compounds




16

Elements on the left hand side of carbon have less than 4electrons in the valence shell (Li-1, Be-2, B-3) so they have moretendencies to loose electron to attain noble gas configuration forstability. That’s why they generally forms compounds with Li+, Be2+,B3+ by losing 1, 2 , 3 electrons respectively. Elements presentdownside in the same group too have similar tendency as that ofLi, Be and B, hence form compounds in the following states; Na+,K+, Rb+, Cs+, Mg2+, Ca2+, Sr2+, Ba2+, Al3+, Ga3+,etc.

Elements on the right hand side of carbon have more than 4electrons in the valence shell (N-5, O-6, F-7). To complete their octet,valence electrons must be subtracted from 8 that’s why the valencyof N is (8-5) i.e. 3, O is (8-6) i.e. 2 and that of F is (8-7) i.e. 1. It ismuch easier to gain 3, 2, 1 electrons to complete their octet ascompared to loosing 5,6,7 electrons to complete their octet. So theseelements have more tendency to gain electrons and form compoundsin the following states; N3-, P3-, As3-, Sb3-, Bi3-, O2-, S2-, Se2-, Te2-, Po2-,F-, Cl-, Br-, I- .

As elements present on the left hand side of carbon looseelectrons to form compounds and elements of right hand side gainelectrons to form compounds so compounds formed are ionic innature.

But think about carbon and the elements present down side,which are present in the middle of each period and have equaltendency to loose or gain electrons as they have 4 electrons in theiroctet. This led carbon and other elements of this group (Si, Ge, Sn &Pb) to share electrons with itself and other elements of periodictable to complete octet. As these compounds are formed by sharingof electrons so they are considered to be covalently bonded.

Carbon by sharing its electrons with other carbon atoms leadsto formation of long chain carbon compounds which may be single,double or triple bonded, cyclic or acyclic, linear or branched. Thisself-linking property of carbon is called catenation. All the atomsof 14th group show the property of catenation but it decreases downthe group because of weak overlapping due to large size and followsorder:

C > Si >> Ge > Sn > Pb


17



Carbon may also form multiple bonds with N, P, O, S etc.forming large number of functional group, which we will discusslater.

This is not the end of compound formation. Carbon forms manyabnormal compounds with elements of s, p & d blocks. So for sakeof simplicity we are constructing an organic chemist’s periodic tablewith the most important elements emphasized.

Elements, which are in dark box, are generally involved inmaking organic compounds along with deuterium (D), which is anisotope of hydrogen (H).

18

K Ca Sc Ti Cr Mn Fe Co Ni Cu Zn Ga Ge As Se BrRb Sr Pd Ag Cd In Sn Sb Te I

Cs Ba Os Ir Pt Au HgFr Ra Hs Mt Ds Rg

3 4 5 6 7 8 9 10 11 12Na Mg

Li

H1

2

Al Si P S Cl

B C N O F13 14 15 16 17

D

T

Be

Kr

Xe

Tl Pb Bi Po At Rn

Ar

NeHe

La Ce Pr Nd Pm Sm Eu Gd Dy Ho Er Tm Yb Lu

Ac Th Pa U Np Pu Am Cm Cf Es Fm Md No Lr

Tb

Bk

Y Zr Nb Mo Tc

Hf Ta W Re

Rf Db Sg Bh

VRu Rh

At. mass = 12.01Valency = 4EN = 2.5

CarbonAt. radius = 67 pm

E. Con. = [He]2s pStructure = Hexagonal

2 2

As there are large number of atoms in periodic table which havevalence electrons, atomic orbital of carbon may overlap with themand share its electron to form large number of compounds. But forthat many other factors such as size, activation energy,electronegativity, electron affinity, catenation etc. are responsiblew hich all come u nder one w ord “Position” i.e. position of carbonin the periodic table. This word “position” include everythingrelated with molecule formation therefore the main reason behindlarge number of organic compound is the position of carbon inthe periodic table.




18

A functional group is an atom such as halogen (-X) or group ofatoms such as carboxylic acid (-COOH) in a molecule that gives themolecule its characteristic chemical properties. They are the actiongroup or reactive site in a chemical reaction and the remaininghydrocarbon part remains inert.Each functional group undergoescharacteristic chemical reaction so by recognizing them it is possibleto predict the reaction, which that molecule undergoes.

O

3CHHO 3CH

Carboxylic acidInert

hydrocarbon part

The concept of functional group is important for organicchemistry due to 3 reasons: Each functional group shows its characteristic chemical

reaction i.e. a particular functional group shows samenature of chemical reaction when present in any compound.

5Section

Functional Groups


19



Functional group helps in nomenclature of organiccompounds. Each functional group have a secondary suffixfor it such as oic acid for carboxylic acid, ol for alcohol, etc.

Functional group serves to classify organic compounds intodifferent classes or families i.e. Compounds with samefunctional group belong to same class.

A molecule may have more than one functional group and arecalled as poly functional group compounds or simply polyfunctional compounds and properties of each functional group maybe modified by the presence of other.

O

OH

O

HO

Carboxylic acid

Ketone

Alcohol

(Polyfunctional compound)As carbon has 4 valencies therefore compounds with all single

bond, one double bond or with one triple bond may be formed withany atoms, which satisfy octet rule. Functional group with bond only-

Carbon may combine with those atoms which needs 1electron for completing their octet (such as hydrogen orhalogen) or with some groups which require one electronfor completing octet such as alkyl group (–CH3, –CH2CH3,etc.), hydroxyl group (–OH), alkoxy group (–OCH3 ,–OCH2CH3), amino group ( –NH2), thiol group ( –SH) orthioether (–SCH3) to form corresponding functional groups.

| |

| |– – –C H/C Alkane

|

|– –C X Alkyl halide

|

|– –C OH Alcohol

|

|– –C SH Thiol




20

–C O – C Ether C – S –C Thioether

|C – N–H 1 amine

H

|C – N– C 2 amine

H

|C – N– C 3 amine

C

Functional group with one bond-Carbon forms 1 bond with many atoms such as O, N, S,

P and even itself leading to formation of different functionalgroups. Along C=O bond

OH/C – C–H Aldehyde

O

H/C – C– X Acid halide

O Carboxylic

acidH/C–C –OH

O O AcidanhydrideH/C–C –O–C –H/C

|

OH/C – C– N– H 1 amide

H

|

OH/C – C– N– C 2 amide

H

|

OH/C – C– N– C 3 amide

C

O

C – C–C Ketone

Along C=N bond

C=N–HC/H

C/HImine C=N–R

C/H

C/H

Alkyl imine(Schiff base)

C=N–OHR/H

R/HOxime 2C=N–NH

R/H

R/HHydrazone


21



C=N–N–PhC/H

C/HPhenylhydrazone

H

C=N–NC/H

C/H2,4-Dimitrophenyl hydrazone

H2NO

2NO

2C=N–N–C–NHC/H

C/HSemicarbazone

H

O

Along C=C, C=S and C=P bond

Carbon disulphideS=C=S C=C Alkene

Methylene triphenylphospheneC=P–PhPh

Ph

Functional group with two bond-

AlkynesC/H H/C

C/H C N Nitrile (Cyanide)

H/C N C Isonitriles (Isocyanide)

Carbon do not form 3 bonds with any atomAs the size of carbon is so small it do not form compounds

which have 3 bonds. This does not mean that no compound with3– bonds exists. There are some d block elements, which form 3bonds and total four bonds. In that case there are 1 , 2 & 1 bonds. This bond indicates the 3rd bond.

Stable quadruple bonds are most common among thetransition metals such as rhenium, tungsten, molybdenum &chromium. Some common examples are




22

Chromium (II) acetateCr2 ( -O2CMe) 4(H2O) 2,

Potassium octachlorodirhenate (III), K2 [Re2Cl8].2H2O &Potassium octachlorodimolybdate K4 [Mo2Cl8]

Re Re

Cl Cl

ClClCl

Cl2-

Cl

Cl

Along with these functional groups there are some otherfunctional groups which are also encountered in organic chemistry.So learn these too with very carefully so that your understandingabout functional group should be crystal clear. We have used R foralkyl group, which means simply the hydrocarbon part of themolecule. It may be acyclic or cyclic but attachments are same asgiven above or below. For more simplicity treat R as C so that youshould get a better observation in molecules. These additionalfunctional groups are

Benzene

O

OORR

Carbonate ester Carboxylate

O

OR –

Hydroperoxy

OOR

HPeroxy

OOR

RCarbonothionyl

H

S

R

Imide

O O

NH

R R

Hemiacetal

O–RHO

HRAcetal

O–R

HR

R–O

Hemiketal

O–R

R

HO

RKetal

O–R

R R

R–O

Orthoester

O–RR

R–O

O–R


23



Orthocarbonate ester

O–RR–O

O–RR–OR–N

H

H1° Amine

R–NR

H2° Amine

R–NR

3° AmineR

R–NR

4° Ammonium saltRR

+ –HN=N =NAzide

N=NR

RAzo

R O NCyanate

R–N=C=OIsocyanate

O–NO

RNitrate

O

–

+ R N CIsonitrile

+ –O

Nitroso oxy

R NO

R–NO

NitroO

–+ N

RNitroso

OR–S–OH

O

OSulphonic acid

SDisulfide

R SR R

Sulfinile

SR

OR–S–R

O

OSulfonyl

R–S–RO

Sulfino

R S NThiocyanate

While identifying a functional group, look at its attachmentsvery carefully. Same group with different attachements may becomedifferent functional group. You will be surprised to note that same




24

group such as –OH when attached to single bonded carbon isalcohol, with benzene ring it is phenol, with C=N it is oxime andwith C=O it is called as carboxylic acid.

3H C

3H C

OH

3CHAlcohol

OH

Phenol

3H CN OH

3H COxime

O

Carboxylic acidOH3H C

Similarly many other differences are observed just on the basisof attachment such as

3H C

3H C

Cl

3CHAlkyl halide

2H CCl

Vinyl halide 2H C Cl

Allyl halide

Cl

Aryl halide

O

Acid halideCl3H C O 3CH3H C

Ether

O

O3CH

Ester3H C

OO

O3CH

Acid anhydride3H C

Amine2NHR

O

Amide2NHR

N

HydrazoneR R

2NH

As whole organic chemistry is based on functional groups andtheir interconversion so revise these functional groups regularlyfor best picture of organic chemistry in mind.


25



A series of compounds in which members have same functionalgroup but differ by one or more 2–CH – units (molecular weight 14or multiple of 14) is called Homologous series (homos is Greekword which means the same as). Members of homologous seriesare called homologues. Homologues are the compounds with samegeneral formula and possess similar chemical properties due topresence of same functional group (if any) i.e. all carboxylic acidsare homologues to each other starting from formic acid (HCOOH)to any carboxylic acid with formula RCOOH where R is any acyclicsaturated alkyl group. All alkanes starting from one carbon toinfinite carbon are homologues.

Example: 14

3 3

3 2 3

3 2 2 3

(a) CH(b) CH – CH Homologous sereis(c) CH – CH –CH of alkanes(d) CH – CH –CH –CH

Example: 2

(a)3CH –C–OH

O(b)

3 2CH –CH –C–OHO

6Section

Homologue & Homologous series




26

(c)3 2 2CH –CH –CH –C–OH

O(d)

3 2 3CH –CH –C–O–CHO

Here (a), (b), (c) are homologues but (d) is not homologue of (a),(b), (c) as it has different functional group. One point should always kept in mind that homologues

have same functional groups or in other word have sametype of chemical reactions so never i nsert CH 2 unit to anybond which create a compound with different functionalgroup.

O

O

HH

3H C O

OH

Homologues2Insert CH unit

O

OH H 3H C

Homologues2Insert CH unit

2CH O

OH

O

OH

3H C

Not Homologues

2Insert CH unit

O

O

3H C 3CH

Carboxylic acid Ester

As the compounds within a homologous series havethe same general molecular formula and the samefunctional group (amines, alcohol, carboxyl acid, ester,alkane, alkene, alkyne etc.), so they can be preparedusing similar methods. Such as

O

3CHRCu/Heat

OH

3CHRWe can prepare any methyl ketone from any alcohol having

CH3CH(OH)– (by changing the number of carbon in R) as


27



Cu/heat is used to convert secondary alcohol into ketone,which we will study later. Similarly we can prepare anycarboxylic acid from primary alcohol (by changing the numberof carbon in R) using KMnO4/heat

O4KMnO /Heat

RR OHOHCompounds within a homologous series show gradual

change in physical properties due to increased molecular sizeand mass, caused by the longer carbon chains. For example,ethane (CH3CH3), has a higher boiling point than methane(CH4). It is because of the reason that ethane moleculeexperiences intermolecular attraction force than methane, asin a large molecule, the electron cloud tends to be distorted atrandom to a greater extent. Thus, the London Dispersion Forcesbetween ethane molecules are higher than that betweenmethane molecules, resulting in stronger forces ofintermolecular attraction, raising the boiling point.

By observing the relative number of carbon and hydrogenatoms in acyclic alkanes it has general formula CnH2n+2 wheren is any integer. So if alkane has 1 carbon atom it must have 4hydrogen atoms; if it has 2 carbon atoms then it must have 6hydrogen atoms and so on. As we know that carbon formsfour covalent bonds and hydrogen forms only one so there isonly one possible structure for an alkane with molecularformula CH4 (methane) and only one structure for an alkanewith formula C2H6 (ethane). As the number of carbon increasesthe number of possible structures also increases. For exampleC4H10 have two, C5H12 have three, C6H14 have 5, and C7H16 have9 possible structures. This number increases very rapidly asthe number of carbon increases such as C10H22 have 75 andC15H32 have 4347 possible structures which we will call later asconstitutional isomers.

As C15H32 have 4347 possible structures but all are havingsame molecular formula so they are simply constitutional




28

isomers which we will study later but starting from CH4, C2H6,C3H8 to C15H32 they differ (CH2)n unit, so they all are consideredas homologues and have same nature of chemical reactionsand method of preparation. We use only the letter R for generalalkyl group (alkane –1H) starting from 1 to infinite.

Here is a table, which helps you to understand homologuebetter with general formula

Homologous General formula Exampleseries Functional group if anyAlkane CnH2n+2 (n=1) CH4, n=1Alkene CnH2n (n=2) C2H4, n=2Alkyne CnH2n-2 (n=2) C2H2, n=2Alcohol CnH2n+1OH (n=1) CH3OH, n=1Carboxylic acid CnH2n+1COOH (n=0) CH2O2, n=0Carbohydrate Cn(H2O)y (n=3) C6H12O6 n=6There are some direct chemical reactions from which we can

convert one member of a homologous series to the next memberand such reactions are called as homologation reactions that wewill study in chemical reactions

O

R OHCarboxylic acid

2SOClO

R Cl2 2CH N

OR

OHNext higher homologue of

carboxylic acid

2moist Ag O


29



When you carefully look over the organic compounds then youwill find that carbon atoms are bonded with other carbon atom orsome other atoms such as H, N, S, P, O, X etc. to form molecule.This difference in attachment leads to difference in rate of chemicalreactions. So it is important to analyze carefully the atoms directlyattached to carbon in any organic molecule. On the basis ofattachment of carbon with other carbon atoms, the carbon atoms inorganic compounds are classified into four types Primary carbon (10C) – Carbon attached to none or one

carbon is called as primary carbon. In case when no othercarbon is attached then such carbon is sometime called assuper primary carbon such as CH4.

Secondary carbon (20C) –Carbon attached to two carbonsis called as secondary carbon.

Tertiary carbon (30C) – Carbon attached to three carbon iscalled as tertiary carbon.

Quaternary carbon (40C) – Carbon attached to four carbonis called as quaternary carbon.

These carbon atoms may be in open chain or in ring; they may

7Section

Nature of C, H & Functional groups




30

be neutral, charged or in radical form. But the important part is theattachment with other atoms and for simplicity they are generallywritten with single bonded forms. These forms may be derived fromalkane, alkyl halide, alcohol, ether, amines, thioethers, thiols, etc.

3CH3H C

3CH

1°2°

3°1°

1°

3H C 3CH

3CH2°

2°

2°

2°

1° 1°

1°

3°4°

3H C

3CH

3CH

2°

1°

3CH

3H C3CH

1°

1°

1°1°

1°

3°4° 3°

Hydrogen’s attached at primary, secondary, tertiarycarbons are called primary, secondary & tertiary hydrogenrespectively. It is important to note that carbons are of fourtypes but hydrogen’s are only of three types, as quaternarycarbon does not have any hydrogen.

Only in case of methane primary carbon have 4 H but in allother alkanes and cycloalkanes each 10, 20, 30 carbon have3, 2, 1 H respectively. To find out total number of 10, 20,30,40 carbon and 10, 20, 30 hydrogen’s you have to count allthe carbons or hydrogen of similar types.

In organic compounds other than alkane or cycloalkanesyou must be more careful in reporting the number H asindividual carbon may have any number of H ranging from3 to 0 as these hydrogen are already replaced with someother atoms or groups.

O3CH

3CH3CHN3H C

3CH

1°

1°

1°

1°

1°2° 2°

1°H=152°H=53°H=0

OS

Cl

Br

Cl

1°H=52°H=83°H=0

1°

1°

2°2°

2°2°2°

3°

3H C1°

NHNH

HS1°

2°

3H C

2NO2°

1°

1°H=42°H=23°H=0

In organic chemistry some functional groups are formed


31



just by replacement of one hydrogen, with an atom orgroups such as alkyl halide, alcohol, primary amine, thiol,nitro compounds, sulphonic acid, etc.

3 3H C–CHOH

3H C–H+OH

Alkane Alcohol

3 3H C–CH 3H C–H

Alkane Primary amine

2NH

2+NH


Alkane Thioalcohol

SH

+SH


Alkane Nitro alkane

2NO

2+NO


Alkane Alkane sulphonic acid

3SO H

3+SO H

So alkyl halide formed from replacement of primary H,secondary H and tertiary H are called as primary alkyl halide,secondary alkyl halide and tertiary alkyl halide respectively.

3H C Cl1° Alkyl halide

3H C2° Alkyl halide

Cl3CH

3H C3° Alkyl halide

3CH

Cl3CH

Similarly alcohol formed from replacement of primary H,secondary H and tertiary H is called as primary alcohol, secondaryalcohol and tertiary alcohol respectively.




32

3H C1° Alcohol

OH 3H C2° Alcohol

3CHOH

3H C3° Alcohol

3CHOH3CH

OH

1° Alcohol OH

2° Alcohol

3CH

OH

3° Alcohol

3CH

But some functional groups are formed by chemical reactionsof two same or different functional groups such as ether, secondaryamine, tertiary amine, thioether, ester, amide, acid anhydride, etc.

3CH3H CAlcohol

AlcoholO H + H O

3H C

EtherO

3CH

2–H O

O

Acid anhydride3H C

3CH3H C

Carboxylic acid

O H + H O2–H O O

O

3CH

O O

Carboxylic acid

O

Ester3H C

3CH3H C

Alcohol

O H + H O2–H O O 3CH

O

Carboxylic acid

O

Primary amide3H C

3CH3H C

Primary amine

O H + H NH2–H O NH 3CH

O

Carboxylic acid


33



3CH3H C

Primary amineNH

3–NHPrimary amine

2NH + H3H C

NH3CH

Secondary amine

3CH3H C

Secondary amine

N3–NH

Primary amine2NH + H

3H CN

3CH

Tertiary amine3CH

3H C

So be careful in reporting your answer for amines or amides.Secondary alcohol is that in which –OH is attached tosecondary carbon but secondary amine is that in whichnitrogen is directly attached to two carbon and onehydrogen. Tertiary amine is one in which N is directlybonded to three carbon. So secondary or tertiary amine maybe present over primary, secondary or tertiary carbon.

3H C 2NH1° Amine

3H C1° Amine

3CH2NH

3H C1° Amine

3CH

3CH2NH

2° Amine

NH

3° Amine

N3CH

2° Amine

NH

NH

2° Amine

NH

2° Amine

N

3° Amine

N

3° Amine

N

3° Amine




34

In organic chemistry when we look over the molecules we willfind that some molecules are only single bonded but some moleculesare multiple bonded and that multiple bond may be of many typessuch as double bonds of C=C, C=N, C=O, C=S or triple bonds ofC N or C C. To differentiate these compounds word saturatedand unsaturated is used.

To understand these terms, take a glass full of water and add aspoon of sugar in it and stir it with spoon. You will find that wholesugar dissolves in it. But when you continuously go on addingspoons of sugar in it and stirring it then after some time you willfind that dissolution stop and sugar settles at the bottom of it. Atthat time you will say that no more sugar is dissolving in it i.e. it issaturated. Similarly when no further bond can be added to amolecule at room temperature then such compounds are called assaturated compounds and if further bonds can be added in it thenthey are called as unsaturated compounds. As we know that singlebond has no tendency to add anything at room temperature andhave only tendency to substitute some atom or group so they areconsidered as saturated compounds and multiple bondedcompounds have tendency to add bond so they are considered asunsaturated compounds.

8Section

Saturated & unsaturated moleculeswith important positions


35



| |Substitution reaction Addition reactionA – B C A –C B A B C –D A– B

DC

Saturated compounds may be cyclic or acyclic; may be only ofcarbon and hydrogen or may have some hetero atoms such ashalogens, nitrogen, oxygen or sulphur. The multiple bond may beof any form such as C=C, C=N, C=O, C=S, C N or C C with orwithout rings.

Example of saturated compounds are-

3H C 3CH 3H C 3CH

Cl

3H C 3CH

OH

3H C 3CH

2NH

3H C 3CHO

3H C 3CH

S

3H C 3CH

SH

NH O

OH

NH

S

Example of unsaturated compounds are-

3H C 2CH CH3H C O

H3H C

O

3H C 3CH O

3H C OH

O

3H C 2NH

O

3H C OMe NH

O

N S

OH

If compounds are saturated with presence of only carbon &hydrogen then such compounds are called as saturated




36

hydrocarbon, such as alkane and cycloalkane but if any hydrogenin it is replaced with any other atom or group such as –Cl, –OH,–NH2, –SH then they are considered as derivatives of saturatedhydrocarbons. Similarly if compounds are unsaturated and haveonly carbon & hydrogen, they are called as unsaturatedhydrocarbon such as alkene, alkyne, cycloalkenes & cycloalkynesbut if hydrogen in it is replaced with any atom or group then theyare considered as derivatives of unsaturated hydrocarbon.

3 2 3 3 2 2

Alkane Derivative of alkaneCH CH CH CH CH CH Cl

2 3 3

Alkene Derivative of alkeneCH =CHCH Cl–CH=CHCH

When you again carefully analyze the functional groups youwill find that some functional groups have presence of carbon withinit such as nitrile (–CN), aldehyde (-CHO), ketone (-CO-), carboxylicacid (-COOH), acid halide (-COX), acid anhydride (-COOCO-), ester(-COOC), amide (-CONH2) etc but some functional groups do nothave carbons such as amine (–NH2), alcohol (-OH), thiol (-SH), nitro(-NO2), Sulphonic acid (-SO3H), ether (-O-), thioether (-S-), etc.

In organic chemistry some common positions such as zero (0),alpha ( ) , beta ( ) , gamma ( ) , delta ( ) , and epsilon ( ) havebeen commonly encountered so we must have a clear cut picture ofit too.

If any functional group do not have carbon in it then the carbondirectly attach to it is given the position alpha followed by beta,gamma, delta and epsilon to next carbons in sequence such as

3H C αβδε Cl

3H C αβδε OH

3H C αβδε 2NH

3CH3H C α

βδ

β

2NO

3CH3H C α

βδ

β

3SO H

3CH3H C α

βδ

β

O3CH

α β


37



N+3H C 3CH

3CH

3CH

α

α

α

αβ

N

+

3CH3H C

βα α

β

αα

S+

3CH

βα α

β

α 3CHβ

But if any functional groups have carbon in it then that carbonis give position zero followed by alpha, beta, gamma, delta andepsilon to next carbons in sequence such as

3H C αβδε

O

H03H C α

βδεO

OH0

3H C αβδε

O

Cl03CH

3H C αββ

3COOCH0

δ

3CH3H C α

ββ

3COOCOCH0

δ

α

3CHO

ββα

β0

δ

O

Oαβ

δ0

ε

This zero (0), alpha ( ) , beta ( ) , gamma ( ) , delta ( ) , and

epsilon ( ) have very much importance in organic chemistry asmany named reactions such as Aldol reaction, Cannizzaro reaction,Claisen rearrangement, HVZ reaction etc involve these terms, whichwe will study later. Aldol condensation- When aldehyde having alpha

hydrogen is treated with base, it first give beta hydroxy aldehydewhich later looses a water molecule to form alpha,beta-unsaturatedaldehyde as a product.

3CH –C–HO

3 2CH –CH–CH –C–H 3CH –CH=CH–C–HΔBaseOOHO

α ,β-unsaturatedaldehyde

β-Hydroxyaldehydeα

Aldehydehaving –H

000α α αββ




38

When we carefully look over the organic molecules we will findthat some carbon atom is single bonded while some others aremultiple bonded. These single or multiple bonds create some fixedshape or geometry in molecule such as linear, planner, tetrahedral,etc. Molecular geometry or molecular structure is the 3-dimensionalarrangement of the atoms that constitute a molecule. It determinesseveral properties of a substance including its reactivity, polarity,phase of matter, color, magnetic properties, and even biologicalactivities. The molecular geometry can be determined by variousspectroscopic methods & diffraction methods such as Infrared (IR),microwave & Raman spectroscopy can give information about themolecule geometry. X-ray crystallography, Neutron diffraction andElectron diffraction can give molecular structure for crystallinesolids based on the distance between nuclei and concentration ofelectron density. Gas electron diffraction can be used for smallmolecules in the gas phase. Nuclear magnetic resonance (NMR)& Förster (Fluorescence) resonance energy transfer (FRET)methods can be used to determine complementary informationincluding relative distances, dihedral angles, angles, andconnectivity. Molecular geometries are best determined at low

9Section

Hybridization


39



temperature and can be different as a solid, in solution, and as agas.

As you learned earlier that these single or multiple bonds areformed by overlapping of atomic orbitals either by axial overlappingor by lateral overlapping, molecular geometries can be specified interms of bond lengths, bond angles and torsional angles.The bondlength is defined to be the average distance between the centers oftwo atoms bonded together in any given molecule. A bond angle isthe angle formed between three atoms across at least two bonds.For four atoms bonded together in a chain, the torsional angle isthe angle between the plane formed by the first three atoms andthe plane formed by the last three atoms.

To explain the shape of any molecule new concept ofhybridization was proposed which is mixing of atomic orbitals ofsame or nearly same energy and redistribution of energy to formnew hybrid orbitals of equal energy and same shape. Sohybridization is simply a mathematical approach to explain theshape of any molecule i.e. if a molecule is tetrahedral then itstetrahedral shape is explained by sp3 hybridization and similarly ifa molecule is planner then to explain it we have to use the conceptof sp2 hybridization and for linear molecules sp hybridization. Inorganic chemistry when we have to find out hybridization of anyatom always count the number of sigma bond and lone pair ofelectrons present on it (if any).

ATetrahedral shape

109.5°

A

Planner shape

120°

180°

Linear shape

A

As the shape of atomic orbitals is derived from Schrodinger waveequation so it is purely a mathematical term. Some shapes commonlyencounted in chemistry are




40

(A) Linear shape (B) Trigonal planner(C) Bent shape (D) Tetrahedral shape(E) Trigonal pyramidal (F) Trigonal bipyramidal(G) Seesaw (H) T shaped(I) Octahedral (J) Square pyramidal(K) Square planar (L) Pentagonal bipyramidal(M) Pentagonal pyramidal (N) Planner pentagonal(O) Square antiprismatic (P) Tricapped trigonal prismaticAs organic compounds are carbon containing compounds and

carbon has only s and p orbitals so the hybridization commonlyobserved in organic chemistry are sp, sp² & sp³. To calculatehybridization of any atom always counts the number of sigma bondsand lone pair of electrons present on it. If the sum of number of sigma bonds and lone pair of

electrons = 2, atom have sp hybridization. If the sum of number of sigma bonds and lone pair of

electrons = 3, atom have sp² hybridization. If the sum of number of sigma bond and lone pair of

electrons = 4, atom have sp³ hybridization.There is no role of positive charge in calculating the

hybridization of any atom as that contains no electrons or bonds.Negative charge is also considered as 2 electrons system similar tolone pair.

H

HHH

3sp

3sp3H C – OH

2sp

2H C2CH2sp

2sp

2sp

2H CCH2sp

2sp

sp

sp

sp2H C 2CHC

2sp 2sp 3CH

O3sp

3sp3sp

3sp 2sp

3sp

2sp2sp

2sp 2sp2sp

2sp

2sp2sp

2sp

As we told you earlier that it is simply a mathematical approach


41



to explain the shape of molecule so there may be some violation ofit. These violations are commonly of three types: If system have A=B-C, where C have lone pair of electrons

then one of the lone pairs present is not counted forhybridization and geometry along C is planner & itshybridization will be either sp or sp²

2H CO–H....

2sp 2H C

2NH2sp.. 2H C

S–H....

2sp

If system have A=B-C, where C have negative charge then–ve charge is not counted for hybridization and geometryalong C is planner & its hybridization will be either sp orsp²

2H C–2CH

2sp 2H C

O.... –2sp 2H C

–NH

2sp.. You must remember these two structures where all atoms

are sp² but molecule is not planner at all.

Cyclooctatetraene = COT

or (Non Planner)

Overall Tub shaped

10 Annulene

or

Hydrogens present in the central carbon disturbe the plane

HH (Non Planner)




42

Up to this point you are familiar with the common functionalgroups, saturated and unsaturated compounds with theirhybridization and shape so now we can classify the organiccompounds into different categories on the basis of their shape andstability. Organic compounds are divided into two classes- Acyclicor aliphatic & cyclic compound. Acyclic compounds or aliphaticcompounds are those which do not have any ring. The wordaliphatic comes from aleiphar which mean fat or oil. Here carbonatoms can be joined together in straight chains or branched chains.They can be saturated, joined by single bonds (alkanes), orunsaturated, with double bonds (alkenes) or triple bonds (alkynes).Besides hydrogen, other elements can be bound to the carbon chain,the most common being oxygen, nitrogen, sulfur and halogen. Thesimplest aliphatic compound is methane (CH4). If aliphaticcompounds are cyclic in nature then such compounds are calledas alicyclic (Aliphatic + cyclic) such as cycloalkanes, cycloalkenes,cycloalkynes, cyclic esters, cyclic ketones etc. Aromatic compounds are the most important class of

organic molecules due to their extra stability. Many naturalproducts such as DNA, RNA, hemoglobin, chlorophyll,medicines, and polymers are aromatic in nature. For a

10Section

Classification of organic compounds


43



compound to show aromatic nature it must follow thesefour conditions.

(I) Compounds should have at least one ring.(II) Each atom of the ring should have sp² or sp hybridization,

no matter it is made up of C, N, S, O or any other atom.(III) The ring should be planner or nearly planner, so that

effective overlapping of pi bond is possible.(IV) The ring should follow Huckel’s rule i.e. the ring must have

(4n + 2) electrons, where n is any number starting from 0to . If the value of n=0, ring must have 2 electrons; forn=1, 6 electrons must be present and so on.

If any compound violates any of the above conditions then itwill not be considered as aromatic and not get extra stability. Forexample, Hexa-1,3,5-triene (CH2=CH–CH=CH–CH=CH2) followslast three conditions but violates the first one so is not an aromaticcompounds. Aromatic compounds are further of two types – one with

attachment of benzene nucleus and other with absence ofbenzene nucleus. Those which contain benzene ring arecalled benzenoid while those which do not have benzenering are considered as non-benzenoid.

Cyclic system may be homocyclic (carbocyclic) where eachatom of the ring is made up of carbon or may be heterocyclicwhere at least one atom of the ring is hetero atom such asN, S or O.

Examples of 2 Pi electrons aromatic system:

+

+

O

HO

HO

O

O

Ph

Ph

Ph

Ph

++

O

OO

O

–

–

NC CN

NC CN

3CH3H C

O

Ph Ph

++

O–

O–

O–

O–




44

Examples of 6 pi electronsaromatic system:

3H C 3CH

Dimethyl fulvene Calicene

N

N+

–

Fe

Ferrocene

–

+

Tropylium ion

2CH

+

+

2CH+

Group

SubstitutedBenzene

Examples of heterocyclic 6 pi electron aromatic system:

NH

Pyrole

Furan

O

Thiphene

S

NH

NImidazole

NPyridine

Pyrylium ion

O+

NN

3CH

Nicotine

Pyrimidine

NN

NNiacin

COOH

N

2NHN

OH

N

2NHN

HO

Cytosine


45



NOH

NN

HO

Uracil

HNO OH

NOH

NN

HO

Thymine

HNO OH

3CH 3CH

Examples of 10 pi electronsaromatic system:

NaphthaleneN

Quinoline Iso quinoleneN

Indole

NH

Benzofuran

O

Benzothiophene

S

Indenyl anion Purine

NHNN N

Indole

Azulene Adenine

NHNN N

2NH

NHN

NNH

+Guanine

NHNN N

NHN

NOH

2H N 2H N

O

Examples of 14 pi electron aromatic system:

Anthracene

Phenanthrene

Phenalenyl anion




46

(14) Annulene

+

Anti- Aromatic compounds are very unstable compoundsat room temperature so they either dimerize or trimerise tobecome alicyclic one.

Theoretically Antiaromatic compounds follow these 4conditions: Compounds should have at least one ring. All atoms of ring should have sp² or sp hybridization, no

matter it is made up of C, N, S, O or any other atom. The ring should be planner or nearly planner, but it is

important to note that there no effective overlapping of pibond is observed.

Compound will not follow Huckel’s rule and ring musthave (4n) electrons, where n = 1 to HSome common examples of anti aromatic compounds are:

–

Cycloprop-2-enide ion

Cyclobuta-1, 3-diene

Cyclopenta-2,4-dienylium ion

+


47



When we carefully look over the cyclic saturated compoundswe find that each atom is sp3 hybridized so it must have bond angle109 0 28´ but in cycloalkanes this angle is mathematically180-(360/n) where n is the number of atoms making ring i.e. inCyclopropane this angle is 600; in Cyclobutane it is 900and so on.This difference in desired bond angle and real bond angle causesstrain in bond which affects the reactivity as well as stability ofmolecule. Greater is the deviation from the theoretical angle greateris the strain. To calculate the distortion or strain in ring we assumethe atoms of ring in a plane, such as in cyclopropane, all the 3 carbonatoms occupy one corner of an equilateral triangle with bond angle600. As two corners bent themselves to form bond so strain too isdivided equally. So strain in cyclopropane will be ½ (109028´ – 600)= +24044 .́

109°28'24°44'

60°24°44'

Deviation of bond angle in cyclopropanefrom normal tetrahedral angle

11Section

Baeyer’s strain theory




48

Distortion or strain = ½ (109028 ́– bond angle of ring). So anglestrains in some cycloalkanes are listed in the table below.

Compound No. of C Angle between Distortion

in the ring the Carbon atoms or strainCyclopropane 3 600 24044Ćyclobutane 4 900 9044Ćyclopentane 5 1080 0044Ćyclohexane 6 1200 -5016Ćycloheptane 7 128034´ -9033Ćyclooctane 8 1350 -12062´From the table it is clear that cyclopropane has the maximum

distortion, so it is highly strained molecule and consequently morereactive than any of the monocyclic alkanes, which is clear fromthe reaction that ring can be opened very easily to relieve strain onreaction with Br2, HBr or H2/Ni at high temperature. In contrast,cyclopentane & cyclohexane have least strain so they are found morereadily and are very stable as compared to cyclopropane.

Baeyer strain theory satisfactorily explains the typical reactivityand stability of smaller rings (from C3 to C5) i.e.

Cyclopropane< Cyclobutane < Cyclopentane(stability order)

but not valid for cyclohexane onwards because the strain againincreases with the increase in number of carbon atom but actuallylarge rings are more stable. So molecular orbital theory is alsoconsidered according to which covalent bond is formed by coaxialoverlapping of atomic orbitals. The greater is the extent of overlapthe stronger is the bond formed. In case of sp3 carbon, C-C bondwill have maximum strength if the C-C-C bonds have the angle109028´. If cyclopropane is an equilateral triangle then the bond angleof each C-C-C bond would be 600. Therefore it was proposed byCouson that in cyclopropane the sp3 hybridized orbitals are notpresent exactly in one straight line due to mutual repulsion of orbitalof these bonds resulting thereby loss of overlap. This loss of overlap


49



weakens the bond and is responsible for its instability and strain inmolecule. Similarly, in case of cyclobutane, there is also loss ofoverlap but the loss is less than in cyclopropane, so cyclobutane ismore stable than cyclopropane. Overlapping of orbitals in large ringcompound (5 or more carbon atoms) is however much better whichaccounts for the greater stability of such compounds.

HH

HHH H

CC C

It is natural that when a molecule has strain within it, it willaffect the stability of molecule. The stability of molecule can becalculated easily by measuring heat of combustion which will givethe measure of total strain and thermochemical stability which canbe calculated mathematically.

Total strain = (No of carbon atom in the ring × observed heatof combustion/CH 2) - observed heat ofcombustion/CH2 for n alkane.

Experimental data of total strain for different cycloalkanesNo of C Heat of combustion Total

atom in ring KJ/CH2 strain in KJ3 697 1204 685 1125 664 356 659 127 662 35

8-11 661-665 32-8812-onwards 657-661 0-48

From the data above it is clear that strain decreases from C3 toC6 i.e. stability increases, but stability again deteriorates from C7 toC11 ring system but interestingly increases from C12and attain the




50

same stability of six membered ring. According to this theory, thecarbon atoms in 5 membered and smaller rings can lie in one planeas explained by Baeyer but Sachse suggested that in six memberedand higher rings the carbon atoms are present in different plane i.e.the ring is puckered. In this way the normal valency angle 109028´is retained and the ring produced is completely strainless. Thus heproposed that cyclohexane exist in two puckered forms as boat andchair form in which chair form is more stable. These forms arereadily interconvertible through half chair and twist boat formssimply by rotation about the single bonds which we will study laterin conformation.

Chair Half chair Twist boat Boat

These forms of cyclohexane are to give a real picture of it but onpaper we commonly make its planner form.


51



You can’t mug-up organic chemistry because there’s toomuch of it. You can remember trivial things like name of compoundsbut that doesn’t help you to understand the principle behind thesubject. You have to understand the principle because the onlyway to tackle organic chemistry is to learn to work it out. That iswhy we are first providing you some questions so that yourunderstanding about the topic should be increased. These problemswill set you on your way but they are not the end of the journey, ashere you are beginning your journey to understand organicchemistry.

The problem would be of little use to you when you couldnot check your answers. For the maximum benefit, you need to solveall the problems without looking the answers. Then you shouldcompare your answers or suggestions with ours before going tonext chapter. If any answer is mismatch then darken those questionnumbers with red pen and again have a microscopic look over itand its theory and correct it before proceeding further. Here startsthe first set of 60 questions01. What is organic chemistry? Write at least 4 characters which

distinguish organic compounds from inorganic compounds.

01Exercise

Subjective Approach




52

02. What is the reason behind the formation of large number oforganic compounds?

03. Which organic compound was first prepared in laboratory?04. Which atom is the centre of attraction in whole organic

chemistry?05. What is the position of carbon in the periodic table?06. What is Octet Rule?07. What is electronegativity and electron affinity?08. What is the electronegativity value of C, H, N, O and F in Pauling

scale?09. What is the valency of C, H, N, O and F?10. What is the symbol for the representation of all halogens?11. Write C-F, C-Cl, C-Br and C-I bonds in decreasing order of their

bond strength.12. Draw structures of single bonded hydrocarbons with 6 carbon

atoms having linear, branched and cyclic frameworks.13. What is wrong in given structure? Suggest better way to

represent the molecules?

H

HH

ONH

HHN

3H C HOH

2NH

14. What is the carbon carbon bond length in ethane, ethene, ethyne& benzene?

15. Which bond is weaker in C-H & C-C bond of alkane?16. Write the decreasing order of bond strength of carbon- carbon

single, double and triple bond?17. How many electrons are involved in formation of single, double

and triple bond?


53



18. What are isotopes? Comment on the isotopes of carbon &hydrogen?

19. Oxygen has atomic number 8 and have 3 isotopes with molecularmass 16, 17 & 18 respectively. How many protons, electronsand neutrons does each of these isotopes have?

20. What is Aufbau Principle?21. What is Pauli Exclusion Principle?22. What is Hund’s Rule?23. What is the basic difference between Principle and Rule?24. Potassium has atomic number 19 and atomic mass 39, with one

unpaired electron in its valence shell. Which orbital does theunpaired electron occupy?

25. Write electronic configuration of F (Atomic no. 9), Cl (Atomicno. 17), Br (Atomic no. 35) & I (Atomic no. 53)?

26. What are ionic, covalent, co-ordinate & polar covalent bonds?27. Which of the following among given compound have most and

least polar bond? NaI, LiBr, Cl2‚ , KCl28. What is dipole moment?29. What is formal charge and how it is calculated in molecules

such as H3O+?

30. Write the Lewis structures of formic acid, formaldehyde &

methanol?

31. Determine the partial positive charge on oxygen atom in a C=Obond if it has bond length 1.22Å and bond dipole moment 2.30D?

32. Predict the relative length of HF, HCl, HBr & HI bonds?

33. Use the symbol and – to show the direction of polarity of thebonds shown in given compounds?CH3-Cl, CH3-NH2, HO-Br, I-Cl, CH3-OH, CH3-MgBr & NH2-OH

34. Name at least 10 organic compounds which make your life easy?




54

35. What is functional groups and why they are important in organicchemistry?

36. Name the compound in which -OH group is attach to tetrahedralcarbon, C=C, aromatic ring and C=O group?

37. Convert these condensed form of molecules into bond linestructures.(A) C6H5CH(OH).(CH2)4COC2H5

(B) O(CH2CH2)2O(C) (CH3O) 2CHCH=CHCH(OMe)2

38. Draw bond line structures for these compounds showing thehydrocarbon framework clearly and showing all the bondspresent in functional groups(A) AcO(CH2)3NO2 (B) MeO2.CH2.OCOEt(C) CH2=CH.CO.NH(CH2) 2CN

39. What are the probable formulas for the following compoundsGeCl?, AlH?, CH?Cl2, SiF?, CH3NH?, AlCl?, CF2Cl?, NI? & PH?.

40. Why can’t organic molecule having formula C2H7, C2H7N &C3H5Br2 exist at room temperature?

41. Fill the non bonding valence electrons if present that are missingfrom the following bond line structure.

3H C 3CHO

O

3H C 3CH O

3H C Cl

3H C N O

3H C O 3CH

42. Convert the following molecular formulas into bond linestructures leaving lone pair electrons as they are assumed to bepresent there(A) C3H8 (B) CH5N (C) C2H6O (2 possibilities)(D) C2H4O (3 possibilities) (E) C3H9N (4 possibilities)


55



43. Sodium methoxide (NaOCH3) contains both covalent and ionicbonds. Indicate the linkage which gives a clear cut picture of it.

44. What are homologues? Write at least 4 homologues of methane,formaldehyde, acetone and formic acid each.

45. How are molecules with a single functional group represented?Write a general formula of alkyl chloride, alkyl alcohol, and alkylcarboxylic acid?

46. How do the compounds in homologous series differ in molecularformula, physical & chemical properties?

47. Name the word used for sulphur analogs of alcohol & ether?48. Find out the functional groups present in following compounds

3H C3CH

SH

SHCompound with worst smell

I

3H C

3CH

SH

Compound with worst smellII

O

3CH

OO

Olean sex pheromone of olive fly(III)

2H C

Vinyl chlorideCl

(IV)

N

N

3CH

3CHCompound from coffee

(V)

Compound from cakes & biscuits(VI)

O

O

OHO

3H C

Coryloneceramel roasted protein

(VII)

O NHOH

3CH

3CHZeneca's tenormin

For treatment & prevention of heart deasese

(VIII)




56

CHOMeO

HOVanillin of vanilla

(IX)

Pyridine(X)

N Phenol.

(XI)

OH

Aniline(XII)

2NH

Diazonium salt(XIII)

–2N Cl+

2CH

Styrene(XIV)

Cummene

(XV)

3H C 3CH

Cummene peroxide

(XVI)

3H C 3CHO–O–H

N 3CH

3CHS

O

HNH

R

OCOOH

Penicilin(XVII)

N3CHN

Nicotine(XVIII)

COOH

O 3CH

O

Aspirin(XIX)

HO

N

O

3CHH

Paracetamole(XX)

S

NN

3CH

EtOO

OO

NN NHN

3H C

3H C

Pfizer's sildenafil (Viagra)(XXI)

OTestosteron (Hormone)

(XXII)

3CH3CH OH

H H

H

3H C

O

OCocaine

N O 3CHO

(XXIII)


57



NQuinine(XXIV)

2H CN

HO3H C

O

3CHHO

NHN

OI

OH O

Fialuridine (Anti viral compound)(XXV)

3CH

OH

Cl3H C

Chloroxylenol(XXVI)

ONH

HNO

Indigo dye(XXVII)

O

SO

2NH

ClO

NH

OOH

Furosemide (Sulpha drug)(XXVIII)

49. Give the structural formula for a 3 carbon compound containingeach of the following functional groups(a) C=C (b) C C (c) Cl (d) OH(e) CHO (f) C=O (g) C N

50. Draw one possible structure for each of these molecules selectingany group of your choice for the “wild card” substituents for

1R2R

A 3Ar

O

1Ar

2Ar




58

51. Find the total number of primary, secondary, tertiary andquaternary carbon in given molecule.

3CH

3H C

3H CI

3H C

3H C

3CH

3CH

3CH3CH

3H CII

3H C

3CH

3CH3H CIII

OH Cl2H N

3CH3CH

3CHIV

3CH

3CHVO

3CH

HN

3CH

3CH3OCH

VI

3CH

VII

Br N OH

N

3CH

NO

O O

3H CVIII

52. Find the total number of primary, secondary and tertiaryhydrogen present in given molecule.

3CH

3H C

3H CI

3CH

3H C

3H CII

3CH

3CH

3H C

3H CIII

Cl3CH

OH

3CH3CH

3CHIV

3CH

3CHV

ClO

53. What are alicyclic compounds? Write at list 10 compounds withdifferent chemical nature.

54. Write the structure of smallest cyclic(a) Alkane (b) Alkene(c) Alkyne (d) Ether


59



(e) sec-Amine (f) t-Amine(g) Thioether (h) Ester(i) Acid anhydride (j) sec-Amide(k) t-Amide (l) Aromatic compound(m) Anti-aromatic compound.

55. Write the structure of open chain molecule that meet thefollowing descriptionsA. Contains 2 sp² hybridized carbon and 2 sp³ hybridized C.B. Contains only 4 carbons, all sp² hybridized.C. Contains 2 sp hybridized carbons and 2 sp² hybridized C.

56. What bond angles do you expect for each of the following andwhat kind of hybridization do you expect for the central atomin each case?A. The C-O-C angle in CH3-O-CH3

B. C-N-C angle in CH3-NH-CH3

C. The C-N-H angle in CH3-NH-CH3

D. The O=C-O angle in acetic acid57. Write the hybridization of each carbon in the given molecule?

3H C2CH

I

II

CH

H

H

H

IIIOH

3H C

2H C

IV

2H N

Cl

COOH

3CH V

VI

VII

VIII

IX

X

2NH

XI

O

2CH.

XII

2CH

SHN

–

2CH+

XIII

N




60

3CH3CH3CH

3CH

3H C3H C

XIV

3CH

3CH HHH

HXV

58. State whether the given molecules are planner or non-planner.

I II III IV

VC

H

H

H

H VI

3H C2CH

VII

2H C+

2CHVIII

–2H C

2CH

IX

3H C2CH

.

X

2H N2CH

XI

XII

H

H H

XIII

O

XIV

S

NH

NXV

59. Classify the following molecules into saturated and unsaturatedcompounds?

I

3H C3CH

3CH3H C

II

Cl

3CH

3H C

IIIHO

IV

Br

3H C

V

VI

VII

3H C

O3CH

VIII

3H C

O3CH

IX

3H C

O3CH


61



X

3H C

CN

3H C

3H CXICOOH

CN

XII

NH

XIII

XIV

XV

Cl

60. Classify the following molecules into aromatic, anti aromaticand non aromatic compounds.

I

II

III

IV

V

VI

VII

VIII

IX

X

XI

XII

.

XIII

XIV

XV

XVI

XVII

XVIII

XIX

XX

XXI

XXII

XXIII

XXIV

XXV

O

XXVI

S

NH

XXVII

NH

XXVIII

N

XXIX

XXX

XXXI

XXXII

XXXIII XXXIV

2N Cl

HO OH

OO

Squaric acidXXXV

OO

+

–

SydnoneXXXVI

N

N

Ph

O

TroponeXXXVII

O

TropoloneXXXVIII

OH




62

O+

Pyrylium ionXXXIX

Phenalenyl aniorXL

PentaleneXLI

HeptaleneXLII

XLIII

N+N

XLIV

3CH3H C

XLV

XLVI

XLVII

O

Ph PhXLVIII

XLIX

L


63



01. Organic chemistry is the chemistry of carbon and itscompounds with some exception of CO3

2– , CO, CO2 etc whoseproperties resembles more with inorganic compounds. Ingeneral organic compounds(A) React more slowly and required higher temperature for

reaction.(B) Undergoes more complex reactions and produce side

products.(C) Has lower melting and boiling points with low solubility

in water.(D) Are classified into families of compounds such as

alcohols, ethers, carboxylic acid etc which have samereactive groups so have similar chemical reactions.

02. Position of carbon in the periodic table.03. Urea04. Carbon05. 2nd period & 14th group.06. In trying to explain why atoms form bonds, Lewis proposed

that an atom is most stable if it has a filled outermost shell oran outer shell of eight electrons, which is called as Octet rule.According to it, an atom will give up, accept, or shareelectrons in order to achieve a filled shell or an outer shellthat contains 8 electrons.

AnswersSubjective Approach

01




64

07. Electronegativity is a chemical property that describes thetendency of an atom or a functional group to attract the bondpair electrons towards itself. An atom’s electronegativity isaffected by both its atomic number and the distance that itsvalence electrons reside from the charged nucleus. The higherthe associated electronegativity number, the more an elementattracts electrons towards it. There are many scales to measureit but Pauling scale is most common among these.Electron affinity of an atom or molecule is the amount ofenergy released when an electron is added to a neutral atomor molecule to form a negative ion.

– –X e X

This property is measured for atoms and molecules in thegaseous state only, since in the solid or liquid states theirenergy levels would be changed by contact with other atomsor molecules.

08. 2.5 , 2.1, 3.0, 3.5, 4 .0 respectively09. 4, 1, 3, 2 & 1 respectively10. X11. C-F > C-Cl > C-Br > C-I12. Linear, branched and cyclic frameworks of 6 carbon

hydrocarbons are

3CH3H C

3CH3H C

3CH 3CH3H C

3CH

3CH3H C

3CH 3CH 3CH

3CH

3CH

3H C

3CH

3CH 3H C

3H C 3H C 3CH

3H C

3CH


65



3CH

3CH3H C

3CH3H C

3CH

3CH

3CH

3CH3H C

3CH

3CH

3H C13. Those structures are not according to geometry and may

create confusions so better representations are

N

N 3CH

O

H

2NH

OH

3H C

14. 1.54Å, 1.34Å, 1.20Å and 1.39Å respectively15. C-H bond is weak16. Bond strength follows C C > C=C > C-C17. 2, 4 & 6 respectively18. Isotopes are variants of a particular element such as C or H.

All isotopes of a given element share the same number ofprotons but differs in its number of neutrons. The term isotopeis formed from the Greek words “isos” means “equal” and“topos” means “place”, hence: “the same place,” meaningthat different isotopes of a single element occupy the sameposition on the periodic table. The number of protons withinthe atom’s nucleus uniquely identifies an element, but a givenelement may in principle have any number of neutrons. Thenumber of nucleons (protons and neutrons) in the nucleus isthe mass number, and each isotope of a given element has adifferent mass number. For example, C-12, C-13 & C-14 arethree isotopes of the element carbon with mass numbers 12,13 and 14 respectively. The atomic number of carbon is 6which means that every carbon atom has 6 protons, so thatthe neutron number of these isotopes are 6,7 & 8 respectively.




66

H have three naturally occurring isotopes, sometimes denotedas 1H (Hydrogen), 2H (Deuterium, D), and 3H(Tritium, T)with 1 proton and 1, 2, 3 neutron respectively.

19. Each isotopes of oxygen have 8 protons & 8 electrons but have8,9,10 neutrons respectively.

20. Aufba u princi ple (German wor d, Aufbau mea ning“building up, construction”) is used to determine theelectronic configuration of an atom, molecule or ion. Theprinciple postulates a hypothetical process in which an atomis “built up” by progressively adding electrons. As they areadded, they assume their most stable conditions (electronorbital) with respect to the nucleus and those electronsalready there. According to the principle, electrons fill orbitalsstarting at the lowest available (possible) energy levels beforefilling higher levels (e.g. 1s before 2s). The number of electronsthat can occupy each orbital is limited by the Pauli ExclusionPrinciple& Hund’s rule.

21. Pauli Exclusion Principle is the quantum mechanicalprinciple that states that no two identical fermions (particleswith half-integer spin) may occupy the same quantum statesimultaneously. A more rigorous statement is that the totalwave function for two identical fermions is anti symmetricwith respect to exchange of the particles. For example, no twoelectrons in a single atom can have the same four quantumnumbers; if n,l, and ml are the same, ms must be differentsuch that the electrons have opposite spins, and so on.

22. Hund’s rules refer to a set of rules which are used todetermine the term symbol that corresponds to the groundstate of a multi electrons atom. In chemistry, the first rule isespecially important and is often referred to as simply Hund’srule. For a given electronic configuration, the term with

maximum multiplicity has the lowest energy. Themultiplicity is equal to2S + 1, where S is the total spinangular momentum for all electrons. The term withlowest energy is also the term with maximum S.


67



For a given multiplicity, the term with the largest valueof the orbital angular momentum number L has thelowest energy.

For a given term, in an atom with outermost sub shellhalf-filled or less, the level with the lowest value of thetotal angular momentum quantum number J (for theoperator J = L + S) lies lowest in energy. If the outermostshell is more than half-filled, the level with the highestvalue of is lowest in energy.

These rules specify in a simple way how the usual energyinteractions dictate the ground state term. The rules assumethat the repulsion between the outer electrons is very muchgreater than the spin–orbit interaction which is in turnstronger than any other remaining interactions. This isreferred to as the LS coupling regime. Full shells and subshells do not contribute to the quantum numbers for total S,the total spin angular momentum and for L, the total orbitalangular momentum. It can be shown that for full orbitals andsub orbitals both the residual electrostatic term (repulsionbetween electrons) and the spin–orbit interaction can onlyshift all the energy levels together. Thus when determiningthe ordering of energy levels in general only the outer valenceelectrons need to be considered.

23. Rules can be violated but Principle can’t be violated and haveno exceptions at all.

24. 4S1

25. F 1s²2s²2p5 Cl [Ne]3s²,3p5Br [Ar]4s²3d104p5 I [Kr]5s²4d105p5

26. Ionic bonds are formed by either loss or gain of electronssuch as NaCl which is originally Na+Cl–. Covalent bonds areformed by sharing of electrons between two atoms. Thissharing may be from same atoms or from different atoms suchas H or O. If atoms are same then bonds are purely covalentbut if atoms are different then such bonds are considered aspolar covalent. Coordinate bond is also a type of covalent




68

bond but here sharing of electrons takes place from a singleat om such as comp oun d of N H 3 BF3 in which N shares itselectron with B.

27. Most polar is KCl & least polar is Cl2.28. The negative and positive ends of a polar bond makes it a

dipole. The polarity of the dipole is indicated by the dipolemoment ( ). The dipole moment of a bond is the product ofmagnitude of the charge (e) on the atom (either the partialpositive charge or the partial negative charge because theyhave the same magnitude) and the distance between the twocharges (d). Dipole moment is measured in a unit calledDebye (D).

= e × d29. Formal charge is the difference between the number of

valence electrons an atom has when it is not bonded to anyatom and the number of electrons it actually “owns” when itis bonded. An atom “owns” all of its nonbonding electronsand half of its bonding electrons.

Formal charge = No of valence electrons – (No of nonbonding electrons + ½ No of bonding electrons)

Oxygen has 6 valence electrons but “own” electrons of oxygenin H3O+ is 5 (2 non bonding plus 3, half of six bonding). Asthe number of number of “own” electron is 1 less than itsvalence electron so its formal charge is 1.

30. Lewis structures of formic acid, formaldehyde and methanolare respectively

O

CO

HH

Formic acid

O

CHH

Formaldehyde

C

Methanol

OHH

H H

31. If there was a full negative charge on the oxygen atom, thedipole moment would be(4.80 × 10-10 esu)×(1.22 × 10-8 cm) = 5.86 × 10-18 esu.cm = 5.86 D


69



Knowing that the dipole moment is 2.30 D, partial positivecharge on oxygen atom will be 2.30/5.86 = 0.39. From thiswe can conclude that oxygen has an excess of about 0 .4electron and carbon atom has a deficiency of 0.4 electrons.

32. Bond length of HX followsHI (160 pm) > HBr (141 pm) > HCl (127 pm) > HF (92 pm)

33. –3CH Cl , –

3 2CH NH ‚ –HO Br , –I Cl ,

–3CH OH , –

3CH Mg Br and –2NH OH

34. Ten organic compounds areSN Compound Use

01

3CH

3CH3H COH

Menthol

Flavoring compound from

the essential oil of spearmint

023CH

O 3CH


Perfume distilled from jasmine

03Aniline

2NH

Basis for the Dyestuffs industry

04Phenol

OH

Antiseptic in surgery




70

053CH

3CH3CH

3CH3H C

Iso-octane

Major constituent of petrol

06

3CH 3CH

OH

Vitamin A

3H C 3CH

3CH

For treatment ofblindness

07

OHHO O

HO OHOH

Glucose

Energy source for living organism

08 OH

H

H

OH

HO H

2CH OH

HO O

(Sucrose)

H O

H

2HOH CHO

H

2CH OHH

OH

Ordinary sugarisolated fromsugarcane or

sugar beet

09N

Quinine

2H CN

HO3H C

Medicine used to treat malaria

10 3H C

SH

3CHPropanedithiol

HS

Worst smell in the world


71



35. Functional group is an atom or group of atoms in a moleculethat gives the molecule its characteristic chemical properties.The importance of functional groups lies in the fact that it isthe site of attack of reagent, makes families for organiccompounds and helps in nomenclature.

36. Alcohol, enol, phenol and carboxylic acid respectively37. Bond line structures are

(A)

OHO

(B) O

O(C)

3OCH

3OCH

3H CO

3H CO

38. Bond line structures are

(A)

O

OON+

O–(B) O

O O

O

(C)

NH

N

O

39. Probable formulas are GeCl4 , AlH3 , CH2Cl2 , SiF4 & CH3NH2,AlCl3 , CF2Cl2 , NI3 , PH3 .

40. Carbon has only four valencies and donot have any vacantorbital so cannot expand its octet. That’s why molecules withsuch structures are not possible at room temperature.

41. Non bonding electrons are

3H C3CH

O××

×× 3H C

O××××

3CH 3H C

O

Cl××

××××

××××

3H C××N

3H C

O

O××

××

×××× 3CH




72

42. Bond line structures are(A)(B) 3 2H C – NH

(C) OH Oand

(D) andO

H OHO

,

(E) and,2NH

2NH, NN

H

43. 3H C –O Na+

Ionic bond

Covalent bond

44. Homologues are the compounds with same general formulaand possess similar chemical properties due to presence ofsame functional group (if any) i.e. all carboxylic acids arehomologues of each other starting from formic acid (HCOOH)to any carboxylic acid with formula RCOOH where R is anyacyclic saturated alkyl group. All alkanes starting from 1carbon to infinite carbon are homologues. Homologues differby (CH2)n units. Homologues of Methane are Ethane, Propane, Butane &Pentane; Homologues of Formaldehyde (Methanal) are Ethanal,Propanal, Butanal and Pentanal; Homologues of Acetone are Butanone, Pentanone,Hexanone & Heptanone; Homologues of Formic acid are Acetic acid, Propanoicacid, Butanoic acid and Pentanoic acid.

45. Molecules with single functional groups are considerd to bederived from a hydrocarbon by replacing one of its H atomswith a functional group. If the hydrocarbon is aliphatic then


73



then the alkyl group is represented by R, and the moleculecontaining functional group may be represented as RG whereG is functional group such as RCl for alkyl chloride, ROH foralcohol and RCOOH for carboxylic acid.

46. Compounds in homologous series differ by (CH2)n unitswhich increases the size of alkyl group, so physical propertieschanges gradually such as general increase in melting point,boiling point is observed with decrease in solubility in water.They undergo similar chemical reactions whose rate dependson the size of alkyl group and shape of molecule.

47. The word thio is usually used to indicate the presence of S.So sulphur analog of alcohol is thiol & ether is thioether.

48. The functional groups present are(I) Thiol(II) Thiol & Ketone(III) Acetal(IV) Alkene & Halide(V) Imine(VI) Ether & Ketone(VII) Ketone & Enol(VIII) Ether, Alcohol & Amine(IX) Ether, Phenol & Aldehyde(X) Imine(XI) Phenol(XII) Aniline(XIII) Diazo(XIV) Alkene & Benzene(XV) Alkyl Benzene(XVI) Peroxide(XVII) Amide, Thioether & Carboxylic acid(XVIII) Amine & Imine(XIX) Carboxylic acid & Ester




74

(XX) Phenol & Amide(XI) Amine, Amide, Ether & Sulphonamide(XXII) Alcohol, Alkene & Ketone(XXIII) Amine & Ester(XXIV) Amine & Alcohol(XXV) Alcohol, Ether & Amide(XXVI) Phenol & Halide(XXVII) Amine, Ketone & Alkene(XXVIII) Ether, Amine, Carboxylic acid & Sulphonamide

49. Structural formulas for compound with such skeletons are

(A)2CH

3H C (B) CH3H C

(C) 3H CCl

Cl

3H C3CHor

(D) 3H C 3H C3CHor

HOOH

(E)3H C

OH (F)

3H C

O3CH (G)

3H CN

50. There are of course many possible structures. A could beheteroatom or a structural fragment while Ar could be anyof a very large number of substituted benzene rings or evenother types of aromatic rings. Four membered rings couldhave A = O, NH, CO, SO2 or even alkene while the R1 & R2could be same or different. To make such structure alwaysbe careful that molecule must follow octet rule.

O3CH

3CH 3CH

HN

Ph

2NH

COOH

2H C


75



3OCHPh

O

3OCH3OCH

SO O

The three aryl groups in second example, all might bedifferent or some might be same.

O

OH

COOHOH

CHO

O

CHO OH

Cl

ON

O

O

O

51. Total number of primary, secondary, tertiary and quaternarycarbons areCom. Primary C Secondary C Tertiary C Quaternary C

I 3 1 1 0II 7 0 3 1III 4 2 2 0IV 3 4 1 1V 4 2 2 0VI 4 3 1 0VII 3 7 1 0VIII 5 10 2 0




76

52. Total number of primary, secondary, tertiary hydrogens areCompound Primary H Secondary H Tertiary H

I 9 2 1II 15 0 1III 9 7 0IV 9 8 1V 8 4 1

53. Alicyclic compounds a re simply the cyc lic alip haticcompounds (Cyclic + Aliphatic), which are commonlycycloalkane, cycloalkene, cycloalkyne, cyclic ether, cyclicsecondary amines, cyclic tertiary amine, cyclic ester, cyclicketones, cyclic amide, cyclic acid anhydride.

54. Structures are

(a)

(b)

(c)

(d)

O

(e)

NH

(f)

N3CH

(g)

S

O(h)

O

O(i)O

O

O(j)

NH

O(k)

N3CH

(l)

+

–

(m)

55. A. 3H C 3CH B. 2CH2H C C. 2CH

2H C

56. A. 109° 28´, sp3 B. 109° 28´, sp3 C. 109° 28´, sp3 D. 120° , sp2

57.3H C

2CH2sp

2sp

3sp

I

2sp

II

C

spH

H

H

H2sp


77



2sp

III

3sp

OH

3H C

2H C

3sp

2sp3sp

2sp

IV

2H N

3sp

Cl

COOH

3CH

2sp2sp

2All spV

2All spVI

2All spVII

2All spVIII

2All spIX

2All sp(X)

2NH

XI

O

2CH.

2sp 2sp3sp

3sp

2sp

3sp

XII

2CH2sp 2sp

3sp

2sp

S3sp

HN

–

2CH+

XIII2All sp

N

3CH3CH3CH

3CH

3H C3H C

XIV

3All sp 3CH

3CH HHH

H

XIV

3All sp

58. Overall shape of molecules areCom. Shape Com. ShapeI Planner II Non-PlannerIII Non-Planner IV PlannerV Non-Planner VI Non-PlannerVII Planner VIII PlannerIX Planner X PlannerXI Planner XII PlannerXIII Planner XIV PlannerXV Planner




78

59. Saturated and Unsaturated compound areComp. Nature Comp. NatureI Saturated II SaturatedIII Saturated IV UnsaturatedV Saturated VI UnsaturatedVII Unsaturated VIII UnsaturatedIX Unsaturated X UnsaturatedXI Unsaturated XII UnsaturatedXIII Unsaturated XIV SaturatedXV Saturated

60. Classification of compounds areCom. Classification Com. ClassificationI Non-aromatic II AromaticIII Anti-aromatic IV Non-aromaticV Anti-aromatic VI AromaticVII Anti-aromatic VIII Non-aromaticIX Non-aromatic X Anti-aromaticXI Aromatic XII Non-aromaticXIII Aromatic XIV AromaticXV Aromatic XVI Non-aromaticXVII Aromatic XVIII Anti-aromaticXIX Non-aromatic XX AromaticXXI Non-aromatic XXII AromaticXXIII Aromatic XXIV AromaticXXV Aromatic XXVI AromaticXXVII Aromatic XXVIII Non-aromaticXXIX Aromatic XXX AromaticXXXI Aromatic XXXII AromaticXXXIII Aromatic XXXIV AromaticXXXV Aromatic XXXVI AromaticXXXVII Aromatic XXXVIII AromaticXXXIX Aromatic XL AromaticXLI Non-aromatic XLII Non-aromaticXLIII Aromatic XLIV AromaticXLV Aromatic XLVI AromaticXLVII Non-aromatic XLVIII AromaticXLIX Aromatic L Aromatic


79



01Exercise

Objective Approach

Single Correct Questions (SCQ) :1. First organic compound generated in 1828 by Friedrich wohler

is

(A) 2CO (B)+ –

4N H OCN

(C) 2 2

OH N - C- NH

(D) 2 2

SH N - C- NH

2. Formation of large number of organic compound is due to(A) size of carbon(B) Electronegativity of carbon(C) Catenation property of carbon(D) Position of carbon in the periodic table

3. How many electrons are used to make a single, double & triplebond respectively.(A) 1,2,3 (B) 2,4,6(C) 3,6,9 (D) 4,6,8




80

4. Electronegativity value on Pauling scale for C, H, N, O, F arerespectively(A) 2.1, 2.5, 3.5, 4.0 (B) 2.5, 2.1, 3.0, 3.5, 4.0(C) 2.5, 2.1, 3.5, 3.0, 4.0 (D) 2.1, 2.5, 3.5, 3.0, 4.0

5. Total number of bonds in the given compound2 2H C = C = CH - CH = C = CH is :

(A) 9 (B) 15(C) 14 (D) 16

6. Molecular formula of Phenantharene is :-

(A) 14 14C H (B) 14 10C H

(C) 14 12C H (D) 12 12C H

7. Number of sigma bonds in the given compound3 2CH - CH - CH = CH - CN is :-

(A) 11 (B) 15(C) 12 (D) 13

8. Find the functional group which is absent in penicillin

N

SR – C – N

H

O

PenicillinO COOH

(A) Amine (B) Amide(C) Thio ether (D) Carboxylic acid

9. Homologue of 3CH COOH is

(A) 3 3CH COOCH (B) 3 2CH COCH OH

(C) 3 2CH CH COOH (D) 2 3HCOCH OCH


81



10. Number of 1°, 2°, 3° & 4° carbons present in given compound

respectively is

(A) 5, 2, 1, 0 (B) 5, 1, 1, 1(C) 5, 1, 0, 1 (D) 5, 1, 1, 0

11. Number of 1°, 2°, 3° Hydrogens present in given compound respectively is

(A) 15, 2, 1 (B) 20, 2, 1(C) 15, 4, 0 (D) 15, 4, 1

12. Number of 3° & 2° carbon atoms respectively in the following

compound are -

(A) 5, 6 (B) 6, 6(C) 5, 7 (D) 4, 7

13. Number of 2° H atoms in the following compound OH

is

(A) 7 (B) 5(C) 6 (D) 4

14. Which molecule is considered as saturated one.(A) Alkane (B) Alkene(C) Alkyne (D) Aromatic comp

15. Which molecule is considered as unsaturated one.

(A) (B)O

(C) NH

(D)




82

16. Central carbon in 2 2CH = C = CH will have which type ofhybridization.

(A) sp (B) 2sp

(C) 3sp (D) None

17. The given compound is

HN 3CH

(A) Alicyclic heterocyclic (B) Unsaturated homocyclic(C) Aromatic heterocyclic (D) Saturated heterocyclic

18. Which of the following is an alicyclic compound ?

(A) (B)

(C)O

(D) O

19. The saturated heterocyclic compound is –

(A) NH

(B)

(C)O

(D)

20. Which of the following do not have bridge head carbon in bicyclocompound.

(A) (B)

(C) (D)


83



21. Which of the following is an unsaturated hydrocarbon.

(A) 3 2CH - CH - C N (B)3 2 3CH – C – CH – CH

O

(C) (D

22. Which molecule is not planner in shape.

(A) (B)

(C) (D)

23. Which molecule is heterocyclic in nature.

(A)2NH

(B)N

(C)

OH

(D)

O3O – C – CH

COOH

24. Which molecule is anti-aromatic in nature.

(A)+

(B)

(C) (D)

25. Which molecule is aromatic in nature.

(A)+

(B)+

(C) + (D)




84

26. Which of the following alkane do not have any 2° carbon.(A) 2, 2-Dimethyl pentane (B) Pentane(C) 2-Methyl butane (D) 2, 3-Dimethylbutane.

27. Maximum angle strain is observed in(A) Cyclopropane (B) Cyclobutane(C) Cyclopentane (D) Cyclohexane

28. In which of the following species is the underlined carbonhaving 3sp hybridisation ? [AIEEE-2002](A) 3CH COOH (B) 3 2CH CH OH(C) 3 3CH COCH (D) 2 3CH = CH - CH

29. The general formula n 2n 2C H O could be for open chain [AIEEE-2003]

(A) carboxylic acids (B) diols(C) dialdehydes (D) diketones

30. Which one of the following does not have 2sp hybridizedcarbon ? [AIEEE-2004](A) Acetonitrile (B) Acetic acid(C) Acetone (D) Acetamide

31. Which of the following represents the given mode ofhybridisation 2 2sp - sp - sp - sp from left to right ? [IIT-2003]

(A) 2H C = CH - C N (B) HC C - C CH

(C) 2 2H C = C = C = CH (D) 2 2H C = CH - CH = CH32. Which of the following molecules, in pure form, is (are) unstable

at room temperature ? [IIT-2012]

(A) (B)

(C)O

(D)O


85



33. The number of loan pair of electrons in melamine is[IIT-2013 (Advanced)]

(A) 4 (B) 6(C) 6 (D) 8

34. In allene (C3H4), the type(s) of hybridization of the carbon atomsis (are): [IIT-2014 (mains)](A) sp and sp3 (B) sp2 and sp(C) only sp2 (D) sp2 and sp3

35. The number and type of bonds in 2–2C ion in CaC2 are:

[IIT-2014 (mains)](A) One bond and one bond(B) One bond and two bond(C) Two bond and two bond(D) Two bond and one bond

36. For the compounds 3 3 3 3CH Cl, CH Br, CH I and CH F , the correct

order of increasing C-halogen bond length is : [IIT-2014 (mains)]

(A) 3 3 3 3CH F CH Cl CH Br CH I

(B) 3 3 3 3CH F CH Br CH Cl CH I

(C) 3 3 3 3CH F CH I CH Br CH Cl

(D) 3 3 3 3CH Cl CH Br CH F CH I

37. Match the orbital overlap figures shown in List-I with thedescription given in List-II and select the correct answer usingthe code given below the lists. [IIT-2014 (Advanced)]

List-I List-II

(P) 1. p-d antibonding




86

(Q) 2. d-d bonding

(R) 3. p-d bonding

(S) 4. d-d antibonding

Codes:P Q R S

(A) 2 1 3 4(B) 4 3 1 2(C) 2 3 1 4(D) 4 1 3 2

38. In the compound 2 3H C = C = CH – CH1 2 3 4

, the hybridization of 1st

and 2nd carbon atom is: [BHU 2003]

(A) sp3-sp (B) sp3-sp3

(C) sp2-sp (D) sp2-sp2

39. Allyl cyanide contains and -bonds: [PMT (MP) 2004](A) 5 , 7 (B) 9 , 3

(C) 3 , 4 (D) 9 , 9

40. The C–H bond distance is longest in: [UGET (Med.) 2006]

(A) 2 2C H (B) 2 4C H

(C) 2 6C H (D) 2 2 2C H Br

41. The correct order regarding the electronegativity of hybridorbitals of carbon is: [AIPMT 2006]

(A) 2 3> spsp < sp (B) 2 3< spsp < sp

(C) 2 3< spsp > sp (D) 2 3> spsp > sp


87



42. Number of and π bonds in 6 5H COOHC is: [DPMT 2007]

(A) 13 , 4 (B) 14 , 4

(C) 15 , 4 (D) 16 , 4

43. What is the percentage of p-character of hybrid orbits of carbonin methane, ethene and ethyne respectively?

[SCRA (Med.) 2007](A) 75, 66, 50 (B) 50, 66, 75(C) 25, 33, 50 (D) 50, 33, 25

44. Which one of the follow ing does not have sp2-hybridized carbon?[JCECE (Med.) 2008]

(A) Acetone (B) Acetic acid(C) Acetonitrile (D) Acetamide

45. In the hydrocarbon 3 2H C – CH = CH – CH – C CH123456

The state of hybridization of carbons 1, 3, 5 are in the followingsequence: [CPMT 2008]

(A) 2 3, spsp, sp (B) 3 2, sp , spsp

(C) 2 3, sp, spsp (D) 3 2, spsp, sp

46. Hybridization of nitrogen atom in pyridine N

is:

[BHU (Mains) 2008](A) 3sp (B) 2sp(C) sp (D) 3sp d

47. The number of π bonds in the following compound2 2O N – C C – NO is : [DPMT 2008]

(A) 2 (B) 3(C) 4 (D) 1




88

48. Compound where underlined carbon use 3sp hybrid orbitals forbond formation is: [BCECE (Med.) 2008](A) 3CH COOH (B) 3 2CH CONH(C) 3 2CH CH OH (D) 3 2CH CH = CH

49. The state of hybridization of 2 3 5 6C , C , C and C of thehydrocarbon,

3CH – C – CH = CH – CH – C CH3CH

6 1234573CH

3CHis in the following sequence

[AIPMT 2009]

(A) 3 2 3sp, sp , sp and sp (B) 2 2 3sp, sp , sp and sp

(C) 2 3 2sp, sp , sp and sp (D) 3 2 2sp , sp , sp and sp

50. Increasing order of carbon-carbn bond length for the followingis : [CET (karnataka) 2011]

2 4 2 2 6 6 2 6C H C H C H C H(A) (B) (C) (D)

(A) B < C < A < D (B) C < B < A < D(C) D < C < A < B (D) B < A < C < D

51. Considering state of hybridization of carbon atoms, find out themolecule among the following which is linear?

[AIPMT (Prelims) 2011]

(A) 3 3CH - CH = CH - CH (B) 3 3CH - C C - CH

(C) 2 2H C = CH - CH - C CH (D) 3 2 2 3CH - CH - CH - CH

52.Which one of the following is a non-benzenoid aromaticcompound? [PMT (Kerala) 2011]

(A) Anthracene (B) Tropolone

(C) Aniline (D) Benzoic acid


89



53. Only sp and sp2 hybrid orbitals are involved in the formation of[PMT (Kerala) 2012]

(A) 3 2CH - CH = CH (B) 3 3CH - CH

(C) 3CH - C CH (D) 2 2H C = C = CH

54. The C–H bond and C–C bond in ethane are formed by which ofthe following types of overlap? [CET (Karnataka) 2012](A) sp3-s and sp3-sp3 (B) sp2-s and sp2-sp2

(C) sp–s and sp–sp (D) p–s and p–p

55. The radical 2CH.

is aromatic because it has [NEET 2013]

(A) 6 p orbitals & 6 unpaired electrons.

(B) 7 p orbitals & 6 unpaired electrons.

(C) 7 p orbitals & 7 unpaired electrons.

(D) 6 p orbitals & 7 unpaired electrons.

56. Which of the following compound has same hybridization asits combusion product CO2 ? [AIPMT 2014]

(A) Ethane (B) Ethyne

(C) Ethene (D) Ethanol




90

Multiple Correct Questions (MCQ) :57. Which statement is correct for element with single bond.

(A) Nitrogen have one lone pair of electron.(B) Oxygen have two lone pair of electron.(C) Fluorine have three lone pair of electron.(D) Carbon have four lone pair of electron.

58. Which statement is correct for bond.(A) Sigma bond is formed by axial overlapping of orbitals.(B) Pi bond is formed by sidewise overlapping of p–p orbitals.(C) Sigma & Pi bonds are a type of ionic bond.(D) Co-ordinate bond is a type of covalent bond.

59. Which statement is correct for functional group.(A) Functional group shows its characteristic chemical reaction.(B) Functional groups helps in nomenclature of organic

compounds.(C) Functional groups serves to classify organic compound into

different classes or families.(D) All functional groups have same physical properites.

60. Functional group is present in vitamin C O

HO

O

OHVitamin C

HOOH

is:

(A) Alcohol (B) Enol(C) Ester (D) Ether

61. Which functional group have absence bond ?(A) Alcohol (B) Aldehyde(C) Amine (D) Alkyl halide


91



62. Which functional group have bond ?(A) Ketone (B) Carboxylic acid(C) 3° amine (D) 2° amide

63. Which compound have C=N linkage ?(A) Imine (B) Amine(C) Oxime (D) Hydrazone

64. Which functional group is present is Viagra

NN

HO

NSO

ON 3CHEtO

NN (Viagra)

3CH(A) Amide (B) Sulphonamide(C) Amine (C) Ester

65. Which functional group is present in sulpha drug

SO

O

2NH

Cl

NH

COOH

O

(A) Sulphonamide (B) Carboxylic acid(C) Amine (D) Alcohol

66. Which compound(s) is/are homologue of Butanal ?(A) Formaldehyde (B) Acetaldehyde(C) Propanal (D) Pentanone

67. Which statement is correct for homologues ?

(A) They have same type of chemical reaction.

(B) They differ in 2 n(CH ) unit.

(C) They have same physical properties.(D) They are prepared by same general method.




92

68. Esters are formed by chemical reaction of(A) Carboxylic acid & alcohol(B) Acid halide & alcohol(C) Acid anhydride & alcohol(D) Aldehyde & alcohol

69. Which compound is unsaturated in nature ?

(A) 2,3-Dimethylbutane (B) 2,3-Dimethylbut-2-ene

(C) 3,3-Dimethylbutyne (D) 3,3-Dimethylbut-1-ene

70. Which compound is saturated in nature.

(A) Butan-2-ol (B) Butan-2-amine

(C) Butan-2-one (D) 2-Chlorobutane

71. Which molecule have correct number of 2°C present.

(A) (3) (B) O

O O (0)

(C) (3)

OH

OHHO(D)

NN N (3)

H

HH

72. Which molecule have correct number of 2 H present ?

(A)OH

(10) (B)2NH

(9)

(C) (8)Cl

(D) (8)

73. Which molecule will have 3Sp hybridized carbon ?

(A) (B)Cl+


93



(C)O+

(D)

74. Which molecule is/are planner in shape ?

(A) (B)

(C) (D)

75. Which molecules will exist at room temperature ?

(A)O

(B)O

(C)

O

(D)

76. Which molecule is/are polar in nature ?

(A) (B)

(C) (D)

77. Which molecule is/are aromatic in nature ?

(A) O

(B)O

O

HO

HO

(C)CNNC

(D)O




94

78. Which molecule is/are aromatic in nature ?

(A) (B)

(C) (D)

79. Which molecule have ( ) gamma position ?

(A) N+ (B) N+

(C) N+ (D) S+

80. Which molecule with correct shape is mentioned ?

(A) Tub shaped (B) non-planner

(C) Planner (D) HC CH Planner

81. The hybridizations of carbon atoms present in cumene is/are

(A) sp (B) 2sp

(C) 3sp (D) 2dsp


95



Assertion / Reason type Questions (A/R) :Each question has 5 choice (A), (B), (C), (D) & (E) out of whichonly one is correct.(A) Statement-1 is true, Statement-2 is true and statement-2 is

a correct explanation for statement 1.(B) Statement-1 is true, Statement-2 is true and statement-2 is

not correct explanation for statement 1.(C) Statement-1 is true and Statement-2 is false.(D) Statement-1 is false, Statement-2 is true(E) Both Statement-1 and Statement-2 is false.

82. Statement-1: The concept of ‘vital force theory’ was objectedby Wohler by preparing urea from ammoniumcyanate.

Statement-2: ‘Vital force theory’ state that we can not prepareorganic molecule in laboratory.

83. Statement-1: Sec-butyl amine is a secondary amine.

Statement-2: In secondary amine, 2–NH group attached withsecondary carbon.

84. Statement-1: tert-butyl alcohol is a tertiary alcohol.Statement-2: In tertiary alcohol, –OH group is attached with

tertiary carbon.85. Statement-1: Oxalic acid & formic acid are homologue.

Statement-2: Homologues are those compounds which havesame nature of chemical reaction and differ in

2 n(CH ) unit.

86. Statement-1: have 1°, 2°, 3° & 4° carbon.

Statement-2: 1°, 2°, 3° & 4° carbon is that carbon which isattached with 1,2,3 & 4 other carbon respectively.




96

87. Statement-1: Cyclohexane is saturated while cyclohexene isunsaturated.

Statement-2: Saturated compound is that which have onlysingle bond while unsaturated compound is thatwhich have multiple bond of any form.

88. Statement-1: Hybridization of oxygen atom in furan is sp2.Statement-2: Hybridization is a mathematical approach to

explain the shape of any atom in molecule.

89. Statement-1:+

is a aromatic compound.

Statement-2: Cyclic planner molecule with (4n + 2) electronsare called aromatic compounds.

90. Statement-1: Toluene is an alicyclic compound.Statement-2: Alicyclic compounds are simply the cyclic

aliphatic compounds.91. Statement-1: Angle strain in decreasing order follows-

cyclopropane > Cyclobutane > Cyclopentane >Cyclohexane.

Statement-2: Angle strain is the half of angle differencebetween desired angle of sp3 atom and real angleof the atom in cycloalkane.

92. Statement-1: Loss of one hydrogen from alkane leads to alkylgroup.

Statement-2: Butane have four types of alkyl groups.


97



Match the Column type Questions (MTC) :93. Match the compounds written in Column-I with its nature

in Column-II :-Column - I Column - II

(A) (P) Aliphatic compound

(B)O

(Q) Aromatic compound

(C)3OCH

(R) Saturated compound

(D)N

(S) Unsaturated compound

(T) Heterocyclic compound94. Match the compounds written in Column-I with its nature

in Column-II :-Column - I Column - II

(A) (P) Aliphatic

(B) (Q) Aromatic

(C) (R) Alicyclic

(D)O

(S) Heterocyclic

(T) Homocyclic




98

Comprehension type Questions :Comprehension-01 :

Paracetamol is widely used as analgesic (pain reliever) &antipyretic (fever reducer) medicine. Based on its structure, selectthe correct answer among this

3CH

O

N

H

HOParacetamol

95. Functional group present in paracetamol is(A) Alcohol (B) Amine (C) Ketone (D) 2° amide

96. Which statement is incorrect for paracetamol is(A) Paracetamol is unsaturated compound.(B) Paracetamol is derivative of hydrocarbon.(C) Paracetamol is heterocyclic compound.(D) Paracetamol is carbocyclic compound.

Comprehension-02 :OH

I

2H /NiΔ

OH

II

Cu

O

III

2NH –OH

H +

N–OH

IV2H O

H+

PolymerizationNylon-6

ON

HV

Δ

Based on this, answer the following questions.97. Incorrect statement about above compound is

(A) A functional group present in II is alcohol.(B) A functional group present in III is ketone.


99



(C) A functional group present in IV is oxime.(D) A functional group present in V is lactone.

98. Incorrect statement about above compound is(A) Compound I is saturated(B) Compound II is saturated(C) Compound III is unsaturated(D) Compound IV is unsaturated

Integer type Questions :99. Total number of primary carbon atoms present in is

100. Total number of aromatic compounds among the givenmolecule is

+

++

101. Total number of sigma bond ( ) present in benzene is

102. Total number of homocyclic compounds among the givenmolecule is

O

N

S

O

OH

OHHO

OO




100

AnswersObjective Approach

0101. (C) 02. (D) 03. (B) 04. (B) 05. (B)06. (B) 07. (C) 08. (A) 09. (C) 10. (B)11. (D) 12. (A) 13. (B) 14. (A) 15. (D)16. (A) 17. (A) 18. (D) 19. (C) 20. (C)21. (C) 22. (C) 23. (B) 24. (B) 25. (B)26. (C) 27. (A) 28. (B) 29. (A) 30. (A)31. (A) 32. (B,C) 33. (B) 34. (B) 35. (B)36. (A) 37. (C) 38. (C) 39. (B) 40. (C)41. (C) 42. (C) 43. (A) 44. (C) 45. (D)46. (B) 47. (C) 48. (C) 49. (A) 50. (D)51. (B) 52. (B) 53. (D) 54. (A) 55. (A)56. (B) 57. (A,B,C) 58. (A,B,D) 59. (A,B,C) 60. (A,B,C)61. (A,C,D) 62. (A,B,D) 63. (A,C,D) 64. (A,B,C) 65. (A,B,C)66. (A,B,C) 67. (A,B,D) 68. (A,B,C) 69. (B,C,D) 70. (A,B,D)71. (A,B) 72. (A,B,D) 73. (A,B,D) 74. (A,B,D) 75. (A,C,D)76. (B,C,D) 77. (A,B,C) 78.(A,B,C,D)79. (B,C,D) 80. (A,B,C,D)81. (B,C) 82. (A) 83. (E) 84. (A) 85. (B)86. (A) 87. (A) 88. (B) 89. (A) 90. (D)91. (A) 92. (B) 93. (A-P,S ; B-P,R,T ; C-P,R ; D-Q,S,T)94. (A-Q,T ; B-P,R,T ; C-P ; D-Q,S) 95. (D) 96. (C)97. (D) 98. (A) 99. (4) 100. (6) 101. (12)102. (6)


101



2Unit

Nomenclature of organiccompounds

In this section you are going to learn the basic concepts of IUPACnomenclature. Here you are going to learn how organic moleculesare named along with some common names which you willregularly encounter in due course of organic chemistry




102

1Section

Alkyl groups

The concept of alkyl group is very much important in organicchemistry as it is the backbone of carbon skeleton of organicmolecule. As every organic molecule is derived from alkane, so thealkyl group is too derived from alkane itself. If a hydrogen atom isremoved from an alkane then the partial structure that remains iscalled as alkyl group.

Alkane - H = Alkyl groupIt is important to note that, as hydrogen is removed from alkane

i.e. all valency of carbon is not fulfilled, alkyl groups are not stablecompounds but they are simply a part of large compound.

Alkyl groups are named by replacing the -ane ending of theparent alkane with an -yl ending. For example, the removal of ahydrogen from methane (CH4) generate a methyl group (-CH3) andremoval of hydrogen from ethane (CH3CH3) generate ethyl group(-CH2CH3). Similarly removal of a hydrogen atom from the endcarbon of any n-alkane gives corresponding n- alkyl group such as

Methane - H = Methyl group (-CH3 or -Me)Ethane - H = Ethyl group (-CH2CH3 or -C2H5or -Et)n- Propane - H = n-Propyl group (-CH2CH2CH3 or -Pr)n- Butane - H = n-Butyl group (-CH2CH2CH2CH3 or -Bu)


103



Just as straight chain alkyl groups are generated by removinghydrogen from an end carbon, branched alkyl groups can begenerated by removing a hydrogen atom from an internal carbon.It is possible to remove hydrogen from primary, secondary or eventertiary carbon. So, on this basis, there are few alkyl groups whichare commonly used in organic chemistry. n-alkyl group- If a hydrogen is removed from first carbon

of straight chain alkane then we use the word n-alkyl forthat. For example

3H Cn-propyl 3H C

n-butyl

iso-alkyl group- If a hydrogen is removed from first carbonof straight chain alkane in which second last carbon haveone methyl group then we use the word iso-alkyl for that.For example

3H C

3CH

Iso-butyl 3H C

Iso-pentyl

3CH

neo-alkyl group-If a hydrogen is removed from first carbonof straight chain alkane in which second last carbon havetwo methyl groups then we use the word neo-alkyl for that.For example

3H C

3CH3H C

neo-pentyl

3H C

3CH3H C

neo-hexyl sec-alkyl group-If a hydrogen is removed from secondary

carbon then we use the word sec representing forsecondary. For example

3H C3CH

sec-butyl




104

tert-alkyl group- If a hydrogen is removed from tertiarycarbon then we use the word tert representing for tertiary.For example

3H C 3CH

3CH

tert-butyl 3H C

3CH

3CH

tert-pentylIt is important to note that n, iso & neo alkyl group are generally

used for removing hydrogen from primary carbon, sec for secondarycarbon and tert for tertiary carbon with an exception in iso-propylwhere hydrogen is removed from secondary carbon. It is becauseof the reason that, here there is a tie between iso and sec and asalphabetically ‘i’ comes first before ‘s’, iso is considered.

3CH3H Ciso-propyl

By combining these alkyl groups with different functionalgroups we can generate many of the organic compounds. Forexample methane can form methyl alcohol, methyl amine, methylchloride etc.

3 3 2 3H C –OH H C – NH H C –ClMethyl alcohol Methyl amine Methyl chloride

To make organic chemistry easy the symbol R is used for allsuch alkyl groups. The R group can be methyl, ethyl, propyl, butylor any alkyl group of any form i.e. n, iso, neo, sec or tert. You mighttake R representing the Rest of the molecule, which we are notbothering to specify because it is not important as chemical reactiondo not occur at that part.

The term primary, secondary and tertiary are routinely used inorganic chemistry along with these common alkyl groups so theymust be crystal clear. For example if it is said that “Tertiary alkylhalide undergoes solvolysis reaction faster than secondary andprimary alkyl halide’’ then it must be clear to you that we are talking


105



about those compound in which halogen is attached with tertiarycarbon.

3CH3H C

3H C

Cl

tertiary butyl chloride

Similarly, if it is said that “Tertiary alcohol reacts faster thansecondary and primary alcohol with Lucas reagent’’ then it mustbe clear to you that we are talking about those compound in whichhydroxyl group is attached with tertiary carbon.

3CH3H C

3H C

OH

tertiary butyl alcohol




106

2Section

Basic concept of IUPACNomenclature

There are billions of organic compounds synthesized & yetanother are in lab to be synthesized. To remember their namesindividually is very very difficult for chemist. So, InternationalUnion of Pure & Applied Chemistry (IUPAC) has given a generalmethod to name organic compounds. This method is applied to allthe organic molecules whether it is aliphatic, alicyclic, bicyclic, spiro,aromatic which may contain one or more functional groups.

The IUPAC name of any compound may consists of 5components which must be written in a sequence so that a correctname is given to a molecule. All of these have a characteristicmeaning. If in any compound, any component is missing then leavethat component & move on to next one.

These components are as follow:2º prefix + 1º prefix + word root + 1º suffix + 2º suffix Secondary (2°) prefix: It tells about the nature and position

of substituent if any. Substituent is any atom or group ofatoms (other than functional group for monofunctionalgroup containing compounds), which replaces one or morehydrogen from the main skeleton of carbon. To find out


107



the substituent in organic molecule encircle the parentcarbon chain along with functional group. If you do notfind any cut means molecule have no substitution but ifyou find any cut at any position that means it hassubstitution. The number of cuts give the number ofsubstituent in the molecule.

Compound with0 substituents

3H C 3CH

Compound with 2 substituents

3CH3H C

3CH Cl

Compound with 2 substituents

FBr

Compound with

4 substituents

FBr

Cl

3H C

Some atoms or groups are always considered as substituentsirrespective of their chemical nature. Some common substituentsare given below

Substituent 20 Prefix-F Fluoro-Cl Chloro-Br Bromo-I Iodo-NO2 Nitro-NO Nitroso-R Alkyl (Example- methyl, ethyl)-OR Alkoxy (Example- methoxy, ethoxy)-Ph Phenyl

Sometimes functional groups may also be considered assubstituent. If compound has only one functional group then it will




108

be treated as secondary suffix. But if compound have more thanone functional group then priority of functional group is considered.High priority functional group is considered as main functionalgroup & for it “secondary suffix” is used while low priorityfunctional group is considered as substituent for which “secondaryprefix” is used. Primary (1°) Prefix: It is used to specify cyclic, bicyclic or

spiro nature of main chain. If main skeleton of carbon iscyclic then primary prefix ‘cyclo’ is used before the wordroot; for bicyclic compounds primary prefix ‘bicyclo’ isused; for spiro cyclic compounds primary prefix ‘spiro’ isused. But if it is open chain compound then no primaryprefix is used.

3H C3CH

No 1° Prefix

1° Prefix-cyclo

No 1° Prefix3H C

3CH

1° Prefix-cyclo

3CH

1° Prefix-Bicyclo

1° Prefix-Spiro

1° Prefix-Bicyclo

1° Prefix-Bicyclo

When two or more atoms are common between the two ringsthen such compounds are called as bicyclic compounds but whenone atom is common between the two rings then it is called as spirocyclic compound. Bicyclic compound has 2 bridge head carbonconnected to each other.

Spiro cycliccompound

Bridge head C

Bridge head C

Bridges

Bridge head C

Bridge head C


109



Word root: It tells about the longest continuous carbonchain which includes functional group, carbon carbonmultiple bond (double or triple bond) and substituent (ifpresent) at minimum position. It is called as parent chain.

For parent chain selection, the decreasing order of priority forassigning minimum position is

Functional group > Multiple bond > SubstituentIt means minimum position is given to functional group

(if present) followed by multiple bond (if present) then substituent(if present). The longest chain of carbon may contain any numberof carbon starting from 1 to and for each of them specific wordroot is assigned;

Some commonly used word roots are given below-

Carbon no. Word root

C1 Meth

C2 Eth

C3 Prop

C4 But

C5 Pent

C6 Hex

C7 Hept

C8 Oct

C9 Non

C10 Dec

For more than 10 carbons some special word such as Undec forC11 , Dodec for C12 and so on is used which is less common. Primary suffix: It tells about saturated or unsaturated

nature of carbon-carbon bonds in the parent carbon chain.If in the parent chain all carbon carbon bonds are singlebond then 10 suffix ‘ane’ is used, if C=C bond is present in




110

parent chain then 10 suffix ‘ene’ is used and for C C bond10 suffix ‘yne’ is used.

Secondary suffix: It tells about the main functional groupspresent in the compound. As we discussed earlier that ifmore than one functional group is present in any compoundthen priority of functional group is taken into consideration.High priority functional group is considered as mainfunctional group and for it secondary suffix is written. Theother functional groups are then considered as substituentand for them 2° prefix is written. For each functional groupa fixed secondary suffix is given which is tabulated belowin decreasing order of priorityFunctional group Secondary suffixCarboxylic acid (–COOH) oic acidSulphonic acid (–SO3H) Sulphonic acidEster (–COOR) oateAcid halide (–COX) oyl halideAmide (–CONH2) amideNitrile (–CN) nitrileAldehyde (–CHO) alKetone (–CO–) oneAlcohol (–OH) olAmine (–NH2) amine

So any organic nomenclature is written by using these 5components in sequential order such as

543213CH3H C

PentaneWR 1°S

54321

3CH3H C

3-Methylpentane

2°P

3CH

WR 1°S

3CH

Methylcyclohexane2°P WR1°P 1°S


111



3Section

Points to write IUPACNomenclature

Now we are going to discuss the important points to writeIUPAC nomenclature of any organic molecule. As organiccompounds are infinite in number so use this 5-component system(in a sequence) to name any organic molecule so that you may writecorrect name. Never forget that all IUPAC names are derived fromthe basic five components only which we discussed earlier. So usethose five components in sequential order keeping the underlyingpoints in your mind. First select the parent chain, which is denoted by word root.

It is the longest continuous chain of carbon containingfunctional group, multiple bond & substituent at minimumposition, following the priority order: Functional group >multiple bond > substituent.

3CH

3CH

3CH3H C

321

4

Incorrect

3CH

3CH

3CH3H C 34

Incorrect

5

21




112

3H C3CH

3CH

3CH1

2 34

Incorrect

3CH3H C

3CH

3CH

23 4 5

6

1 7

Correct

If number of carbon atoms are same in more than onelongest chain then parent chain will be that, which havemore number of substituent.

3

45

67 3CH

8

3CH

3H C3CH2

1

Incorrect8 C in parent chain & 1 subst.

3

45

67 3CH

8

3CH

3H C3CH

correct8 C in parent chain & 1 subst.

21

If numbers of substituent is also same then parent chainmust have substituent at lower number.

3H C 3CH

3H C2 134

56

73CH

8

Incorrect3CH

3H C 3CH

3H C4

56

73CH

8

3CH

321

CorrectIn first case parent chain is of 8 carbons with 2 substituents and

first substituent is at position 3 but in second case parent chain is of8 carbon and 2 substituent but first substituent is at position 2. Numbering is done from that side of the parent chain so as

to give minimum number to the first substituent (lowestset of locants). Locant is the number specifying the positionof functional group, multiple bond or substituent in parentchain. Lowest set of locant means minimum number at firstpoint of difference.


113



4 653213H C

Cl

ClIncorrect

3 124563H C

Cl

ClCorrect

In first case the locant is 2,6 but in second case locant is 1,5. If the position of substituents are identical from both the

ends of the parent chain, then numbering in done inalphabetical order, i.e. substituent coming alphabeticallyfirst must gets minimum number. For example:.

4 6532

13H C

Incorrect

3CH

3CH7 8

3CH

4 653

213H C

Correct

3CH

3CH7 8

3CHEthyl (E)

Methyl (M)Here ethyl comes alphabetically before methyl so, ethyl must

get minimum number. If alphabets are also same then numbering is done from

that side of the parent chain having substituent of thesubstituent at lower number

4 6532

13H C

Incorrect

3CH7 8

Cl

Cl3H C 5 3

467

83H C3CH

2 1Cl

Cl3H C1-Chloroethyl

2-Chloroethyl

Correct Substituents are of two types- simple and complex. Simple

substituents are those in which no further branching isobserved in substituents but complex substituents are thosein which further branching is observed i.e. substituents onsubstituent are observed.

5 346

783H C 2 1

Cl

Cl

3H C

Complex substituent

Complex substituent 3CH Simple substituent

For assigning the name to the complex substituents, the




114

carbon directly attached to the main skeleton of carbon isgiven number 1 & so on. First substituent of substituent iswritten followed by main substituent.

5 346

783H C 2 1

Cl

Cl

3H C3CH1-Methylethyl

212-Chloroethyl

12

Here for the substituent at position-5, the main substituent isethyl on which another substituent Cl is present (Cl is substituentof substituent). So it is named as 2-Chloroethyl. For substituent atposition-4, the main substituent is again ethyl on which methyl issubstituted at position 1. So it is named as 1-Methylethyl. ‘ – ’ is used between number & alphabets while comma (,)

is used between two numbers, there is no gap in writingthe IUPAC name and first alphabet must be written withcapital letter.

54321

63H C3H C

3CH

3CH

2,2-Dimethylhexane If a compound has more than one substituent then

substituents are written in alphabetical order irrespectiveof their locant. For example

3H C 3CH

3-Bromo-2-chloropentane

Cl

Br 3H C 3CH

2-Bromo-3-chloropentaneCl

Br

Word di, tri or tetra is used for 2, 3 or 4 times presence ofsame substituent.

54321

63H C3H C

3CH

2,2,4-Trimethylhexane

3CH 3CH


115



In IUPAC name of simple molecules which do not havecomplex substituent, ‘d’ of di and ‘t’ of tri is not consideredfor alphabetization.

54

32-Fluoro-4, 4-dimethylhexane

2 61

3H C3CH

3H C 3CHF

In IUPAC name of molecules which have complexsubstituent, if ‘di’ or ‘tri’ are the part of the name of complexsubstituent then ‘d’ of di and ‘t’ of tri is considered foralphabetization and the name of complex substituent iswritten in bracket.

1-(1,1-Dimethylethyl)-4-ethylcyclohexane

1,1-DimethylethylEthyl 123

4

If same complex substituents present two, three or fourtimes in the main chain then word bis, tris or tetrakisrespectively is used instead of di, tri, or tetra.

1,4-Bis-(1,1-dimethylethyl)cyclohexane

1,1-Dimethylethyl123

41,1-Dimethylethyl




116

4Section

Points to write IUPAC nomenclaturefor multiple bonds & rings

Earlier we had discussed about the points to write compoundswhich have single bonds and may or may not have substituents.Now we are going to learn nomenclature of those compounds whichhave multiple bonds (such as alkene or alkyne) or rings (such ascycloalkane). Multiple bond is given more priority as compared to

substituents or no. of carbon. The multiple bonds may bedouble (C=C) or triple (C C) in nature.

If multiple bond C=C or C C is present then parent chainis one which includes maximum number of multiple bonds.While selecting parent chain we may leave another chainwhich have more carbon but lesser number of multiplebond. Numbering is done from that side of the parent chain,so that multiple bond gets minimum number (if functionalgroup is absent).

Pent-1-ene 4 532

4-Methylhex-2-ene1

3H C 63CH

3CH

5

6

3-Propyl hepta-1,6-diene

7

1 23

4


117



When e of alkane, alkene or alkyne is followed by anothera, e, i, o or y of primary or secondary suffix then e of theseprimary suffix is not written.

2 34

CH12H C

5

Pent-1-en-4-yne When primary suffix other than ene or yne (i.e. diene, diyne,

triene, triyne etc), is used then at the end of word root an‘a’ is added.

2 3 413H C

Buta-1,3-diene2CH

When double bond & triple bond are at identical positionthen double bond is given high priority & compound isnamed as enyne as e of ene comes first then y of yne inalphabets.

2 3 4CH

13H C 5Hex-2-en-4-yne

6

3-Ethynyl hepta-1,6-diene12

3 4 5 67

When more than two multiple bonds are present, thenalways take the lowest number for second multiple bond ifposition of first multiple bond is identical.

Deca-4,9-dien-1-yne

2CHHC

1 2 3 4 5 6 7 8 9 10

To write the IUPAC name of cyclic compounds we use thesame process i.e. secondary prefix, primary prefix, wordroot, primary suffix and secondary suffix. For example

Cyclohexane

Word root 1° suffix1° Prefix




118

Substituted cycloalkanes are also named by rules similarto those used in open chain alkanes.

In case of monosubstitued cycloalkanes the carbon of thering to which the substituent is attached is given no. 1.

Methylcyclohexane

324

56 1

In case of disubstituted cycloalkane, the carbon to whichsubstituent coming alphabetically first is attached is givennumber one and numbering continues in such direction,so as to give minimum number to second substituent.

1,5(Incorrect)

564

32 1

1,3(Correct)

324

56 1

Locants:

In case of cycloalkanes with more than two substituentnumbering is done in such a way that lowest set of locantis obtained.

1,3,6

654

3

1

Locants:

Br

2

1,2,5

234

5

1 Br

6

1,2,4(Lowest set of locant)

165 4

2 Br

3

It there is more than one way in which identical locants areobtained then substituent which comes alphabetically firstmust get minimum number.

1-Bromo-5-chloro-4-ethyl-2-methylcyclohexane

56

3

2 1

Cl

Br

4


119



If ring & open chain both is present then to select parentchain priority order is :

No. of multiple bond > No. of carbon > No. of substituentIf no. of multiple bond, no. of carbon & no. of substituent, all

are same in ring and open chain then ring is parent chain.

3CH

3 carbons 4 carbons1-Cyclopropylbutane

3CH

Methylcyclopropene

3H C

1-Cyclopropylpropene

3CH3H C

1-Ethyl-5-methylcyclopentene

3-Bromo-1-iodocyclohexa-1,4-diene

I

Br Cyclohexene

12

1,3-Dimethylcyclohexene

3CH

3CH

1

23

21

Cyclooctyne 1

2

3-Methylcyclooctyne

3CH3

Ethylcyclohexane

3CH

3CH

1-Cyclopropylpentane

3H C

1-Chloro-2-cyclopropylpropane

Cl132

As we had told you earlier that all organic compounds arenamed by same five parts of IUPAC nomenclature i.e.Secondary prefix, primary prefix, word root, primary suffixand secondary suffix, no matter compound is aliphatic oraromatic. For simplicity, we will discuss the IUPAC nameof benzene and their derivative.




120

When you will carefully observe benzene you will find that it iscyclic with 6 carbon & alternate single and double bond i.e. primaryprefix, word root and primary suffix is fix for benzene, so all thesethree parts comes together to form the name “Benzene”.

1° Prefix + Word rood + 1° Suffix=BenzeneBe careful that benzene is not Cyclohexa-1,3,5-triene as it gives

the picture of alkene derivative but we know that all carbon carbonbond length in benzene is same and it have extra stability so it is notaliphatic but it is aromatic.

If there is any substituent present at benzene ring then it iswritten as secondary prefix before the benzene, if some functionalgroup is present then secondary suffix for that functional group iswritten at the last. For example:

Methylbenzene

3CH

Benzenecarboxylic acid

COOH

2-Methylbenzenecarboxylic acid

COOH3CH

It is also important to know that all carbon carbon bond lengthof benzene is equal and lies in between carbon carbon single anddouble bond so we can theoretically think the bond length as (one& half) which is greater than single bond but less than multiplebond. If aliphatic or alicyclic carbon chain without multiple bond

& functional group attached to benzene ring, then benzeneis considered as parent chain. But if aliphatic or alicycliccarbon chain with multiple bond or functional group isattached to benzene ring then benzene ring is consideredas substituent. When benzene is placed as a substituent thenwe use the 1° prefix phenyl for it.

Benzene

Methylbenzene

3CH

Cyclohexylbenzene


121



Phenylethene

2CH

Phenylethyne

CH

If aliphatic or alicyclic carbon chain attached to benzenering (without functional group) has more number ofsubstituents, then benzene ring is consider as substituent.

Cl

3CH43 2

1

2-Chloro-1-methyl-4-phenylcyclohexane

1-Ethyl-2-Methylbenzene

3CH3CH

12




122

5Section

Points to write IUPAC nomenclatureof compounds with functional groups

Functional groups are very much important in organic chemistryas they are the site of chemical reactions. The same functional groupwill undergo the same or similar chemical reaction(s) regardless ofthe size of the molecule however, its relative reactivity can bemodified by nearby functional groups.

For naming mono functional group containing compounds weuse specific secondary suffix for each functional groups such as oicacid for carboxylic acid.

3H C

O

Butanoic acidOH

Carboxylic acid123

4

but for poly functional groups containing compounds we haveto consider the priority of functional groups. For higher priorityfunctional groups we use secondary suffix but for lower priorityfunctional groups we use secondary prefix. Let us take an examplewhere high priority carboxylic acid along with low priority alcoholic


123



group is present. For this secondary suffix ‘oic acid’ is used forcarboxylic acid and secondary prefix ‘hydroxy’ is used for alcoholicgroup.

3H C

O

3-Hydroxybutanoic acid

OH

OHCarboxylic acidAlcohol

1234

The priority of functional groups in decreasing order along withsecondary prefix and secondary suffix are given below which mustbe remembered by heart without any confusion.

SN Group Common Name 2°Prefix 2°Suffix1. –COOH Carboxylic acid Carboxy oic acid2. –SO3H Sulphonic acid Sulpho sulphonic acid3. -COOCO- Acid anhydride - oic anhydride4. -COOC Ester Alkoxycarbonyl oate

& Alkanoyloxy5. -COX Acid halide Halocarbonyl oyl halide6. -CONH2 Amide Carbamoyl amide7. -CN Nitrile Cyano nitrile8. -CHO Aldehyde Formyl, aldo al

or Oxo9. -CO- Ketone Oxo one10. -OH Alcohol Hydroxy ol11. -NH2 Amine Amino amineHere are some example of compounds with one functional

group:

3H COH

O

Propanoic acid

3H C

Butane-2-sulphonic acid3CH

3SO H

3H C

Propanamide2NH

O

3H C

Cl

O

Propanoyl chloride

3H COH

O

Propanoic acid

3H CN

Propane nitrile




124

3H CH

O

Propanal

3H CO

Butan-2-one3CH 3H C

Butan-2-amine3CH

2NH

Some functional groups such as acid anhydride, ester andamides are formed by chemical reaction of two similar ordifferent functional groups. So in those cases it is importantto know the attachments with positions. Such as

O

Acid anhydride3H C

3CH3H C

Carboxylic acid

O H + H O2–H O O

O

3CH

O O

Carboxylic acid

O

Ester3H C

3CH3H C

Alcohol

O H + H O2–H O O 3CH

O

Carboxylic acid

O

Amide3H C

3CH3H C

Primary amine

O H + H NH2–H O NH 3CH

O

Carboxylic acid

3CH3H CAlcohol

AlcoholO H + H O

3H C

EtherO

3CH

2–H O

3CH3H C

Primary amineNH

3–NHPrimary amine

2NH + H3H C

NH3CH

Secondary amine


125



3CH3H C

Secondary amine

N3–NH

Primary amine2NH + H

3H CN

3CH

Tertiary amine3CH

3H C

Acid anhydride may be symmetrical (formed by chemicalreaction of two similar carboxylic acids) or dissymmetrical(formed by chemical reaction of two dissimilar carboxylicacids). To name symmetrical acid anhydride we replace ‘oicacid’ of carboxylic acid (from which it is formed) by ‘oicanhydride’ and it is named as alkanoic anhydride. Butdissymmetrical acid anhydrides are named as Alkanoicalkanoic anhydride, taking alphabetization of carboxylicacids.

3CHO

3H CO

O

Propanoic anhydride3CH

O3H C

O

O

Ethanoic propanoic anhydride An ester is formed from chemical reaction of two functional

groups – carboxylic acid and alcohol and both have carbon.To name an ester alcoholic part is written first as ‘alkyl’followed by the acidic part as ‘alkanoate’.

O

EsterEthylethanoate

3H C3CH3H C

Alcohol

O H + H O O 3CH

O

Carboxylic acid

12

Secondary amides are named as N-Alkyl alkanamide andtertiary amide as N,N-Dialkyl alkanamide (for amidederived from same amine) or N-Alkyl N-alkyl alkanamide(for amide derived from two different amines).

N-Ethylethanamide

13CHNH

23H C

O

N,N-Diethylethanamide

13CHN

23H C

3CH

O




126

N-Ethyl-N-methylethanamide

13CHN

23H C

3CH

O

In case of secondary or tertiary amines the alkyl group withlongest carbon chain is parent chain & the other alkylgroups attached to nitrogen are considered as substituenton nitrogen. Secondary amines are named as N-Alkylalkanamine and tertiar y amine as N,N-Dialkylalkanamine, if both alkyl substituent on nitrogen are sameand N-Alkyl N-alkyl alkanamine for tertiary amines withtwo different alkyl substituents on N.

N-Ethylethanamine

13CHNH

23H C

N,N-Diethylethanamine

13CHN

23H C

3CH

N-Ethyl-N-methylethanamine

13CHN

23H C

3CH

Ethers are always written as Alkoxyalkane as –OR is alwaysconsidered as secondary prefix.

Ethoxyethane

13CHO

23H C

Some functional groups are always present at the terminalpositions so called as chain terminating functional groupswhile some functional groups may be present at anyposition, hence called as non-chain terminating functionalgroups. Example of chain terminating functional groupsare Carboxylic acid (-COOH), Acid halide (-COX), Amide(CONH2), Nitrile (-CN), Aldehyde (-CHO) while exampleof non terminating functional groups are Sulphonic acid(SO3H), ketone (-CO-), alcohol (-OH), amine (-NH2)


127



When chain terminating functional group is directlyattached with ring (may be of aliphatic or aromatic), thenspecial suffix is used for these functional groups.

S.N. Functional Groups Secondary suffix1 Carboxylic acid (-COOH) Carboxylic acid2 Ester (-COOC) Carboxylate3 Acid halide (-COX) Carbonyl halide4 Amide (-CONH2) Carboxamide5 Nitrile (-CN) Carbonitrile6 Aldehyde (-CHO) CarbaldehydeSome common examples are

Cyclohexanecarboxylic acid

COOH

Cyclohexanecarbaldehyde

CHO

Cyclohexanecarbonitrile

CN

Cyclohexanecarboxamide

2CONH

Cyclohexanecarbonylchloride

COCl

Methylcyclohexanecarboxylate

COOMe


COOH

Benzenecarbaldehyde

CHO

Benzene

carbonitrile

CN

Benzenecarboxamide

2CONH

Benzenecarbonyl chloride

COCl

Methylbenzenecarboxylate

COOMe




128

Phenyl group is commonly written as C6H5 or Ph. When itis directly attached with any functional group containingcarbon then it is considered as parent chain and word‘benzene’ is used for it but if even one carbon is presentbetween phenyl and functional group then benzene isconsidered as substituent and word ‘phenyl’ is used.


COOH

2-Phenylethanoicacid

COOH2 1

In writing the name of compounds with more than onefunctional group the senior functional group is selected asthe principle functional group and for it secondary suffixis considered while other functional groups are treated assubstituents and for them secondary prefix is used.

OH

3 1245

O O

3H C6

OH

3-Hydroxy-5-oxohexanoic acid

When ester is present as a substituent then two differenttypes of secondary prefixes are used – alkoxycarbonyl(when CO is directly attach to main chain) and alkanoyloxy(when O is directly attached with main chain).

3-Ethanoyloxy-5-methoxycarbonylcyclohexane carboxylic acid

COOH2

3 4 5

61Ethanoyloxy Methoxycarbonyl

O3CH3H C O

O

O

In case of aldehydes, two different secondary prefixes areused- Aldo or formyl & oxo. Secondary prefix formyl or


129



aldo is used when -CHO is directly attached to mainskeleton but secondary prefix oxo is used when alkyl groupwith -CHO is attached to main skeleton.

3-Formyl-5-oxoethylcyclohexanecarbonitrileOHC

CN2

3 4 5

61

CHOformyl 2-Oxoethyl

Secondary prefix oxo is used for aldehydes as well asketones when CO is part of main or side skeleton and thecarbon of CO is counted in skeleton.

3 1245

O O

3H C6

OH

5-Oxohexanoic acid 3 1245

O O

OH5-Oxopentanoic acidH

3-(1-Oxoethyl)-5-(2-oxoethyl)benzenecarboxylic acid

H3CH

O

O

COOH2

345

61

When ester, acid halide or amide is present as a substituentthen carbon of CO is not counted in main chain.

3-(Methoxycarbonyl)propanoic acid

3 12

O

3H CO

OOH

3-(Chlorocarbonyl)propanoic acid

3 12

O

OOH

Cl

3-(N-Methylcarbamoyl)propanoic acid

3 12

O

OOH

N3CH

H




130

Some aromatic compounds are given a fixed name and isretained in IUPAC nomenclature. Such as

Benzene

Phenol

OH

Aniline

2NH


131



6Section

Nomenclature of Bicyclic &Spiro compounds

A bicyclic molecule is a molecule that fuses two rings. Thesefusion of the rings can occur in three ways:

1. If two rings shares one carbon atoms then such compoundsare called as spirocyclic or simply spiro compound

2. If the two rings shares two adjacent carbons then suchcompounds are called as fused bicyclic compounds

3. If the rings shares two non adjacent carbons then suchcompounds are called as bridged bicyclic compounds.

Spiro compounds

Fused bicyclic

or

Bridged bicyclic

As we told you that all IUPAC nomenclature consist of fiveparts written in sequential order i.e.

20 prefix + 10 prefix + Word root + 10 suffix + 20 suffix,IUPAC name of bicyclic compounds and spirocyclic compounds

can also be written in the same way as the other molecules, the only




132

difference occur in primary prefix where “Bicyclo” or “Spiro” isused so as to differentiate from acyclic and cyclic compounds.

Bicyclic compounds:(1) Prefix ‘Bicyclo’ is used followed by the name of alkane

corresponding to the total number of carbon atoms in bothring.

(2) The number of C atoms in each of the three bridgesconnecting the two tertiary carbon atoms is indicated inbracket in descending order after ‘bicyclo’.

(3) Numbering is started from one of the bridgehead carbonand proceeds by longest possible path to the secondbridgehead carbon & continuous through next longer pathback to first bridgehead and is completed by the shortestpath.

1 23

45

6 12

3456

7

or7

Bicyclo[2,2,1]heptane

Here the carbon frame contains a total of 7 atoms, hence theparent name heptane and this molecule has three bridges having 2,2 and 1 carbon atoms, hence the name becomes Bicyclo [2, 2, 1]heptane.

Bicyclo[4,4,0]decane

1 234

56789

10

.

In case of writing IUPAC name of substituted compoundsor compounds with functional groups we use the concept ofsecondary prefix and secondary suffix as we had used earlier.

It is also important to note that in case of bicyclo compoundsbridge head is given number 1, no matter it has substitution or


133



functional groups, then follow the sequence in decreasing orderso that all atoms are taken in decreasing way.

3-Fluorobicyclo[2,2,1]heptan-2-one

1

345

6 1

3456

7

or7O

2FF

O2

1-Methylbicyclo[2,2,1]heptane

1 23

456

12

3456

7

or7

3CH3CH

1 23

456

3H C

3H C1

2

3456

7

or

7-Methylbicyclo[2,2,1]heptane

4-Iodobicyclo[4,4,0]dec-1-en-3-one

O

I

1 2 34

56789

10

Spiro compounds:(1) Prefix ‘spiro’ is used followed by alkane corresponding to

the total number of carbon atoms in both ring.(2) The number of carbon is both the rings, leaving the spiro

atom (atom joining the rings) is written in bracket inascending order after ‘spiro’.

(3) The numbering starts from the atom next to spiro atom inthe smaller ring & proceeds to the larger ring via spiroatom.




134

12

3 487 5

6Spiro[2,5]octane

Here the carbon frame contains a total of 8 atoms, hence theparent name become octane and this molecule has two bridgeshaving 2 and 5 carbon atoms which we have to write in increasingorder, hence the name becomes Spiro[2,5]octane.

5 123

4

8 769

Spiro[4,4]nonane

Cl

3CH12

3 487 5

61-MethylSpiro[2,5]octane

123 48

7 56

1-Chlorospiro[2,5]oct-4-ene

Cl 12

3 487 56

Spiro[2,5]octan-4-one

O


135



7Section

Some common names commonlyused in organic chemistry

There are lot of organic compounds whose common names arealso important along with IUPAC name and they are frequentlyencountered in organic chemistry. So remember these commonnames along with a quick look of their IUPAC name

SN Com. Common name IUPAC name

1. CH4

Marsh gas, damp fire Methane

2.3H C 3CH

3CHIso-butane 2-Methylpropane

3.3H C

3CH3CH

Iso-pentane 2-Methylbutane

4.3H C 3CH

3CH

3CHNeo-pentane

2,2-Dimethylpropane




136

5. 3H C 3CH

3CHIso-hexane 2-Methylpentane

6.3H C

3CH

3CH3CH

Neo-hexane 2,2-Dimethylbutane

7. Benzene Benzene

8. Naphthalene Naphthalene

9. Anthracene Anthracene

10. Phenanthrene Phenanthrene

11. 2H C 2CH Ethylene Ethene

12. 3H C2CH Propylene Propene

13.3H C

3H C2CH Iso-butylene 2-Methylpropene

14. 2CH C 2CH Allene Propa-1,2-diene15. HC CH Acetylene Ethyne16. 3H C–C CH Methylacetylene Propyne

17. 2 3CH CH3H C CCEthylmethyl

acetylene Pent-2-yne

18. 3H C Cl Methyl chloride Chloromethane

19. 2 2CH Cl Methylene chloride Dichloromethane20. 3CHCl Chloroform Trichloromethane


137



21. 4CCl Methanetetrachloride Tetrachloromethane

22. Cl3H C Ethylchloride Chloroethane

23.3H C

3H CCl Iso-Propylchloride 2-Chloropropane

24.3CH3H C

ClSec-Butyl chloride 2-Chlorobutane

25.3CH

3CHCl Iso-Butyl chloride

1-Chloro-2-methylpropane

26.3CH

3CHCl

3H C Tert-Butyl chloride2-Chloro-2-methyl

propane

27. 3H C

3CH

3CH

Cl Neo-Pentyl chloride

1-Chloro-2,2-dimethylpropane

28. ClCl Ethylene dichloride 1,2-Dichloroethane

29. 3H CCl

Cl Ethyledine dichloride 1,1-Dichloroethane

30.2H C

ClVinyl chloride 1-Chloroethene

31. Cl2H C Allyl chloride 3-Chloropropene




138

32.Cl

Phenylchloride Chlorobenzene

33.

Cl

Benzyl chloride 1-Chloro-1-phenylmethane

34.

ClCl

Benzal chloride 1,1-Dichloro-1-phenylmethane

35.

ClClCl Benzo chloride 1,1,1-Trichloro-1-phenyl

methane

36.

ClCl

ClCl

Cl

Cl

Gammexane, Lindane,Benzenehexa

chloride(BHC) Hexachlorocyclo

hexane

37.OH

3H CEthyl alcohol

(Methyl carbinol) Ethanol

38.3H C

OHPropyl alcohol Propan-1-ol

39. 3H COH

3CHIso-Propyl alcohol Propan-2-ol

40.3CH

OH3H C Sec-Butyl alcohol Butan-2-ol

41.3CH

HO3CH

Iso-Butyl alcohol 2-Methylpropan-1-ol


139



42.3CH

3CHOH

3H C Tert-Butyl alcohol 2-Methylpropan-2-ol

43.3H C

3CH

3CH

OHNeo-Pentyl alcohol

2,2-Dimethylpropan-1-ol

44. OH2H C Allyl alcohol Prop-2-en-1-ol

45.

OH

Benzyl alcohol 1-Phenylmethanol

46. OHHC Propargyl alcohol Prop-2-yn-1-ol

47.OH

OH

Ethylene glycol(Glycol) Ethane-1,2-diol

48.3H C

OH

OHPropylene glycol Propane-1,2-diol

49. OHHO Trimethylene

glycol Propane-1,3-diol

50.OH

OHOH Glycerol

(Glycerine) Propane-1,2,3-triol

51.H

O

HFormaldehyde Methanal

52.3H C

O

H Acetaldehyde Ethanal

53.H

O3H C Propionaldehyde Prapanal




140

54.O

H3H C n-Butyraldehyde Butanal

55.

O3H C

H3CH

Iso-Butyraldehyde 2-Methylpropanal

56.O

2H CH

Acrolein Prop-2-enal

57.O

H3H C Crotonaldehyde But-2-enal

58.O

3CH3H CDimethyl ketone

(Acetone) Propanone

59.O

3CHPh Acetophenone Phenylethanone

60.O

Ph PhBenzophenone Diphenyl

methanone

61.O

3H C3CH

Ethyl methylketone Butanone

62.O

3CH3H C Methyl propylketone Pentan-2-one

63.O

3CH3H C Diethyl ketone Pentan-3-one

64.O

2H C3CH Methyl vinyl

ketone But-3-en-2-one


141



65.O

H OHFormic acid Methanoic acid

66.O

3H C OH Acetic acid Ethanoic acid

67. 3H CO

OHPropionic acid Propanoic acid

68.3H C

O

OH n-Butyric acid Butanoic acid

69. 3H CO

OH n-Valeric acid Pentanoic acid

70. OH

O3H C

3CHIso-butyric acid 2-Methylpropanoic

acid

71. 2H CO

OHAcrylic acid Prop-2-enoic acid

72.O

OH3H C Crotonic acid But-2-enoic acid

73.COOH

COOHOxalic acid Ethane-1,2-dioic acid

74.COOH

COOHMalonic acid Propane-1,3-dioic

acid

75.COOH

COOHSuccinic acid Butane-1,4-dioic acid

76.COOHCOOH

Glutaric acid Pentane-1,5-dioicacid




142

77. COOHCOOH Adipic acid Hexane-1,6-dioic

acid

78.

COOHH OH

3CHLactic acid

2-Hydroxypropanoic acid

79.

COOH

COOH

H OHOHH Meso-Tartaric acid

(2R,3S)2,3-Dihydroxybutane-1,4-dioic acid

80.

COOH

COOH

HHOH

HOd or l

Tartaric acid Or

(2S,3S)2,3-Dihydroxybutane-1,4-dioic acid

(2R,3R)2,3-Dihydroxybutane-1,4-dioic acid

81.HO

COOH

COOHMalic acid

2-Hydroxybutane-1,4-dioic acid

82. HOCOOH

COOHCOOH Citric acid

2-Hydroxypropane-1,2,3-

tricarboxylic acid

83.O

3H C COOH Pyruvic acid 2-Oxopropanoic acid

84.COOH

COOHMaleic acid

Cis-But-2-ene-1,4-dioic acid

85.COOH

HOOC Fumaric acid

Trans-But-2-ene-1,4-dioic acid

86.COOH

Benzoic acid Benzene carboxylicacid


143



87. OHCOOH

Salicylic acid2-Hydroxybenzene

carboxylic acid

88.

COOHO

O

3CH Acetyl salicylic acid(Aspirin)

2-(Ethanoyloxy)benzene

carboxylic acid

89.3COOCH

OH Oil of winter green(Methyl salicylate)

Methyl-2-hydroxybenzenecarboxylate

90.COOPh

OH Phenyl salicylate(Salol)

Phenyl-2-hydroxybenzenecarboxylate

91.2NO

OH2O N

2NOPicric acid 2,4,6-Trinitrophenol

92.CH=CH–COOH

Cinnamic acid3-Phenyl

prop-2-enoic acid

93.O

H 3OCH Methylformate Methylmethanoate

94.O

H 2 3OCH CH Ethylformate Ethylmethanoate

95.O

2 3OCH CH3H C Ethyl acetate Ethylethanoate

96.O

ClHFormylchloride

(Unstable) Methanoyl chloride




144

97.

ClO

Benzoyl chloride Benzene carbonylchloride

98.O

Cl3H C Acetyl chloride Ethanoyl chloride

99.

O

O3H C

O

3CH3 2or (CH CO) O

Acetic

anhydrideEthanoic

anhydride

100.

O

O3H C

O

3CH3 2 2or (CH CH CO) O

Propionicanhydride

Propanoicanhydride

101.O

2NHH Formamide Methanamide

102.O

2NH3H C Acetamide Ethanamide

103. 3H C2NH

OPropionamide Propanamide

104.3NHCOCH

Acetanilide N-Phenylethanamide

105. 3H C–O–N=O Methyl nitrite Nitrito methane

106.3H C

O–N=OEthyl nitrite Nitrito ethane

107. 3 2H C–NH Methyl amine Methanamine


145



108. 3H C

2NH Ethyl amine Ethanamine

109.3H C–NH

3CH Dimethyl amine N-Methylmethanamine

110. 3H C 3CHNH

Diethyl amine N-Ethyl ethanamine

111. 3H C–N3CH

3CHTrimethyl amine N,N-Dimethyl

methanamine

112.N

3CH

3H C

3H C

Triethyl amine N,N-Diethylethanamine

113. 2 3H N–SO H Sulphamic acid Amino sulphonicacid

114. 3 3H C–SO H Methylsulphonicacid

Methanesulphonicacid

115.3SO H

Phenyl sulphonicacid

Benzene sulphonicacid

116. 3H C–CN Methyl cyanideor Acetonitrile Ethane nitrile

117.CN

Phenyl cyanideor Benzonitrile Benzene carbonitrile

118. 3H C–NC Methyl iso-cyanide orMethylcarbyl amine Methane iso-nitrile

119. 3H CNC

Ethyl iso-cyanide orEthyl carbylamine Ethane iso-nitrile




146

120.NC

Phenyl iso-cyanide orPhenylcarbyl amine Benzene iso-nitrile

121. O epoxide Oxirane

122. O Trimethyleneoxide Oxetane

123.O Tetrahydrofuran

(THF) Oxolane

124.O

ODioxane 1,4-Dioxacyclo

hexane

125.O

OO Trioxane 1,3,5-Trioxacyclohexane

126.3CH

Toluene Methylbenzene

127.3CH

Ethylbenzene Ethylbenzene

128.3CH3H C

Iso-propylbenzene(Cumene)

(1-Methylethyl)Benzene

129.3CH3H C

3H C Tert-Butylbenzene

(1,1-dimethylethyl)benzene

130.2CH=CH

Styrene Phenylethene


147



131. 3CH3CH

o-Xylene 1,2-Dimethylbenzene

132.3CH

3CH

m-Xylene 1,3-Dimethylbenzene

133.

3CH

3CH

p-Xylene 1,4-Dimethylbenzene

134.3CH

3CH

3H CMesitylene 1,3,5-Trimethyl

benzene

135.OH

Carbolic acid Phenol

136.OH

OH Catechol Benzene-1,2-diol

137.

OH

OHResorcinol Benzene-1,3-diol

138.

OH

OHHydroquinone Benzene-1,4-diol




148

139.OH

- Naphthol 1- Naphthol

140.OH

- Naphthol 2-Naphthol

141.CHO

BenzaldehydeBenzene

carbaldehyde

142. OHCHO

Salicyldehyde2-Hydroxybenzene

carbaldehyde

143.2NO

Oil of mirbane Nitrobenzene

144.3OCH

Anisole Methoxybenzene

145.2NH

Aniline Aniline

146. 3CH2NH

o-Toluidine 2-Methylaniline

147.3CH

2NH

m-Toluidine 3-Methylaniline


149



148.3CH

2NH

p-Toluidine 4-Methylaniline

149. 3CHOH

o-Cresol 2-Methylphenol

150.3CH

OH

m-Cresol 3-Methylphenol

151.3CH

OH

p-Cresol 4-Methylphenol

152. Iso-octane2,2,4-Trimethyl

pentane

153.2 2CH – S – CH2CH Cl 2CH Cl

Mustard gasBis-(2-Chloroethyl)

sulphide

154.2CHCl

2CHCl Westron 1,1,2,2-Tetrachloroethane

155. 2ClCH CH Westrosol 1,1,2-Trichloroethene

156. 3 2Cl C –NOChloropicrin

tear gas1,1,1-Trichloro-1-

nitromethane

157. 3 3CH –C–CH3CCl

OHChloretone

1,1,1-Trichloro-2-methylpropan-2-ol




150

158. 3CH –C 3CH–CH

3CH

3CH

3CH Triptane 2,2,3-Trimethyl

butane

159.2 2H C=C–CH=CH

3CHIsoprene

2-Methylbuta-1,3-diene

160.Cl

2 2CH –C=CH–CH Chloroprene2-Chloro

buta-1,3-diene

161.Cl–C–H

2H–C–AsCl Lewisite 2-chloroethenylarsonousdichloride

162. 3CH –C 3C–CH3CH

OH

3CH

OHPinacol

2,3-Dimethylbutane-2,3-diol

163. 3CH –CO

3C–CH3CH

3CHPinacolone 3,3-Dimethyl

butan-2-one

164.CHOCOOH

Glyoxalic acid 2-oxoethanoic acid

165. CHOC3CH

3CH3CH Pivaldehyde

2,2-Dimethylpropanal

166.3 3CH –C–C–CH

OODimethylglyoxal Butan-2,3-dione

167. Ph–C–C–PhOO

Benzil1,2-Diphenyl

ethane-1,2-dione


151



168.

CHOH OH

2CH OHGlyceraldehyde

2,3-Dihydroxypropanal

169.CHO

CHOGlyoxal Ethanedial

170.

CHO

3C–CH

O

Methylglycol(Pyruvic aldehyde) 2-oxopropanal

171. C=CH–C–CH=C

O3H C

3H C 3CH

3CH Phorone 2,5-Dimethyl-4-en-3-one

172. 3C=CH–C–CH

O3H C

3H C Mesityloxide 4-Methylpent

-3-en-2-one

173. 2H C=C=O Ketene Ethenone

174. HO–C–OH

OCarbonic acid Carbonic acid

175.3CH –C–COOH

OPyruvic acid 2-oxopropanoic acid

176. Ph–CH–COOH

OHMendalic acid 2-Hydroxy-2-Phenyl

ethanoic acid

177. 2H N–CH–COOH Glycine Aminoethanoic acid

178.2H N–CH–COOH

3CH Alanine

2-Aminopropanoic acid

179. 2H N – COOH Carbamic acid Carbamic acid




152

180.COOH

OHGlycolic acid 2-Hydroxy

ethanoic acid

181. Cl–C–C–Cl

O OOxalyl chloride Ethanedioyl

dichloride

182.2 4H N–C–ONH

O Ammoniumcarbonate

Ammoniumcarbonate

183.3 2CH –C–CH –C–OEt

O O Acetoacetic

esterEthyl-3-oxobutanoate

184. Cl – C – Cl

OPhosgene Carbonyldichloride

185.2 2H N–C–NH

OUrea Carbamide

186. NH

Indole Indole

187. N Pyridine Pyridine

188. NH

Pyrrole Pyrrole

189. S Thiophene Thiophene

190. O Furan Furan

191. AzuleneBicyclo[5,3,0]

decapendaene

192. 2H N 3SO H Sulphanilicacid

4-Aminobenzenesulphonic acid


153



193.O

TroponeCyclohepta-2,4,6-trienone

194.O

OHTropolone

2-Hydroxycyclohepta-2,4,6-trienone

195. O

O

HO OH

Phenol

phthalin

3,3-Bis-(4-Hydroxyphenyl) Isobenzo

furan-1-one

196.

Cl

ClClH

ClCl DPT

1,1,1-Trichloro-2,2-Bis-(4-chlorophenyl)ethane

197. N N

H H

Hydrazobenzene

1,2-Diphenylhydrazine

198.

OH

2NO

2NO2O NPicric acid 2,4,6-Trinitrophenol

199.

COOH

Cl

O

mcpba3-chloroperbenzene

carboxylic acid

200.COOH

3CH O-toluic acid2-Methylbenzene

carboxylic acid

201.

COOH

3CHm-toluic acid





154

202.

COOH

3CHp-toluic acid


203.COOH

COOHPhthalic acid

Benzene-1,2-dicarboxylic acid

204.COOH

COOH

Isophthalic acidBenzene-1,3-dicarboxylic acid

205.

COOH

COOHTerephthalic acid

Benzene-1,4-dicarboxylic acid

206.COOH

2NH Anthranilic acid2-Aminobenzenecarboxylic acid

207. NH

Pyrrolidine Pyrrolidine

208. N

HPiperidine Piperidine

209.

O

NH

Morpholine Morpholine

210. NH

Aziridine Aziridine


155



211. Ph–CH=CH–Ph Stibene 1,2-Diphenylethene

212.N

NN

N

Urotropine1,3,5,7-Tetraazatricyclo [3,3,1,1]

decane

213.2 2H N–C–NH

NHGuanidine Guanidine

214.3 2CH –C–NH

NHAmidine Acetamidine

215.

O

O

O Phthalicanhydride

Benzene-1,2-dicarboxylicanhydride

216. N–H

O

OPhthalimide Isoindole-1,3-dione

217.CH=CH–CHO

Cinnamaldehyde 3-Phenylprop-2-enal

218. 3 3CH –S–CHDimethylsulphide

(DMS) Dimethysulphide

219.3 3CH –S–CH

O Dimethylsulphoxide

(DMSO)

Dimethylsulphoxide

220.3CH

3H–C–N–CH

ODimethyl

formamide(DMF)

N,N-Dimethylmethanamide




156

221.

O

OON

HN

H Barbituric acid

Pyrimidine-2,4,6-trione

222.OO

OHHO

HOOH

Ascorbic acid (1,2-Dihydroxyethyl)

-3,4-dihydroxyfuran-2-one

222. Ph–C–CH–Ph

O OH Benzoin

2-Hydroxy-1,2-diphenylethanone


157



08Section

Degree of unsaturation

Degree of unsaturation (DU) is very important in organicchemistry as it helps to predict different structures for any molecularformula. It is also called Double bond equivalent (DBE) or Indexof hydrogen deficiency (IHD). Degree of unsaturation is simplythe hydrogen deficiency index from any acyclic saturatedhydrocarbon.We know that formation of a bond occur by loss of two hydrogenatoms from adjacent position, for the formation of a triple bondfour hydrogen atoms are removed from adjacent position, loss oftwo hydrogen atoms also lead to the formation of a cyclic system ifthey are not removed from adjacent position. So one bond or onering is equal to 1 DU. So DU is equal to sum of bonds and ringsin the structures. Such as

Structure DU Structure DU2 3

2 2

4O

1




158

7 HO OH

OH OH0

2O

5

As the bond or ring is formed by loss of hydrogen from acyclicsaturated hydrocarbon (i.e. alkane having general formula (CnH2n+2),mathematically it can be calculated from known molecular formula.

(2C + 2) – H – X + NDU =2

Where C is the total number of carbon atoms, H is the total numberof hydrogen atoms, X is the total number of halogen atoms and N isthe total number of nitrogen atoms.

It is important to note that there is no role of oxygen atom orany atom of 16th group such as S in calculation of DU. If the value of DU comes to be zero, it means that molecule

is acyclic and saturated with no double or triple bond andalso without any ring.

If DU is equal to one then molecule must have a bond ora ring. The bond may be between carbon atoms orbetween carbon and others atoms (like N, O etc) if theseatoms are present in the molecular formula. Similarly, thering may contain only carbon or ring with hetero atomsmay be formed.

If the value of DU comes to be two that means moleculemust have a triple bond, two double bonds, one doublebond and a ring or two rings.

Example 1: Write all the possible structures with molecular formulaC5H12

To solve such questions first calculate DU. It will give a clearcut picture to think about acyclic saturated molecule as DU comes


159



to be zero. Then draw all possible structures starting from maximumcarbon in main chain and then go on decreasing one carbon in mainchain and placing the remaining carbon in the form of side chain atpossible positions. So the possible structures are:

3H C 3CHPentane

3H C

3CH3CH

2-Methylbutane

3H C

3CH

3CH

2,2-Dimethylpropane

3CH

Example 2: Write all the possible structures with molecular formulaC6H14.

Use the same process i.e. first calculate DU and then write thestructures in same way. The possible structures are :

3H C3CH

Hexane 3H C

3CH2-Methylpentane

3CH

3H C 3CH

3-Methylpentane

3CH

2,2-Dimethylbutane

3H C3CH

3CH3CH

3H C3CH

2,3-Dimethylbutane3CH

3CH

Example 3: Write all the possible structures with molecular formulaC5H10.Here the value of DU comes to be one so write all the structures

containing one double bond or one ring. To write structures withone bond first draw all possible skeleton of carbon then findchemically different positions (positions which leads to differentIUPAC name) where bonds can be placed. To write structureswith one ring first draw ring with maximum number of carbon andthen go on reducing the number of carbon in the ring and place theremaining carbon in the form of substituents at possible positions.In this way, all possible structures with molecular formula C5H10are:




160

2H C 3CH

Pent-1-ene1

2H C 3CH

Pent-2-ene2

2H C3CH

2-Methylbut-1-ene3

3CH

3H C3CH

2-Methylbut-2-ene4

3CH

3H C2CH

3-Methylbut-1-ene5

3CH Cyclo

pentane6

3CH

Methylcyclobutane

7

3CH

Ethylcyclopropane

8

3CH3H C

1,1-Dimethylcyclopropane

9

3CH

1,2-Dimethylcyclopropane

10

3CH

Example 4: Write all the possible structures with molecular formulaC4H6.Here value of DU comes to be two so write all the structures

containing one triple bond, two double bonds, one double bondand one ring or compounds with two rings. To write such structuresfollow the same concept as discussed above i.e. first draw theskeleton and then find the chemically different positions and finallywrite the structures. So all possible structures of compounds withmolecular formula C4H6 are :Example 5: Write all the possible structures with molecular formula

C5H10O.When some hetero atom is present in molecular formula then

structures must contain functional group with that hetero atom.Here, DU is zero so alcohol and ether functional groups areconsidered. The possible structures are:

HO 3CHButan-1-ol

OH3CH

3H CButan-2-ol

3CHHO

3CH2-Methyl

propan-1-ol

2-Methylpropane-2-ol

3CH

3CHOH3H C


161



3H C 3CH

1-MethoxypropaneO 3H C 3CH

Ethoxyethane

O 3H C

3CH

3CHO

2-Methoxypropane

Example 6: Write all the possible structures with molecular formulaC3H6O.Here DU is one which can be in the form of one bond (C=C,

C=O) or one ring (homocyclic or heterocyclic) along with thefunctional groups. In this way nine compounds with molecularformula C3H6O are possible with different IUPAC names .

Prop-1-en-1-ol1

3CHHO

2H C 3CH

Prop-1-en-2-ol2

OH OH

2H C

Prop-2-en-1-ol3

O

Propanal4

3CHH

O

Propanone5

3H C 3CH Cyclopropan-1-ol

6

OH

2-methyloxirane7

3CH

O

O

Oxetane8

3H C

O 2CHMethoxyethene

9

Limitations of DU:Calculation of DU is very much important to draw structure of

any compound with known molecular formula but it has somelimitations such as – It is used to draw all possible structures of a molecular

formula but we cannot reach the final structure only byDU calculation .




162

It does not tell about the position of double bond incompound with known molecular formula.

It does not tell anything about the nature of loss of hydrogeni.e. it does not tell whether DU is in the form of ring or bond.

It is generally applicable to those organic molecules in whichring or bond is with carbon. S=O, S=S, N=O bonds also haveunsaturation but it can not be found by DU calculation. For example,Methanesulphonic acid (CH3SO3H) have two S=O bonds but bycalculation DU comes to be zero.


163



09Section

Chemically different hydrogen

Earlier we have discussed about the types of carbon andhydrogen (i.e. carbon are of four types 10, 20, 30, 40 and hydrogen areof three types 10, 20, 30) on the basis of number of carbon to which acarbon is directly bonded.

3H C

3H C

3CH

3CH3CH

1°4°

1° 2°3°

1°

1°

1°

3CH

3H C1° 2° 2°

2°2°

1°

Example 1:In pentane there are two types of carbon (two 10 and three 20

carbon). But when we carefully analyze this molecule we find thatreplacement of one H by any atom or group such as –Cl from thecarbon skeleton leads to three different products.

3CH3H C

Pentane–H+Cl 3CH3H C

2-Chloropentane

3CH3H C3-Chloropentane

Cl

Cl3CH

1-ChloropentaneCl




164

On the basis of chemical reaction it is seen that replacement ofsecondary H from 2nd and 3rd positions leads to different products,it means that they are of different chemical nature so are consideredas chemically different. Therefore by definition chemically differenthydrogens are those hydrogens which give different products onchemical reactions. These types of H atoms are also known as non-identical H, non-equivalent H.As we have not discussed about thechemical reactions so far, the best way to find the chemicallydifferent H is to see the attachments (atoms or groups) on carbon. Ifattachments (atoms or groups) are different on carbon thensuch carbons are chemically different and hydrogens attached tosuch carbons are chemically different hydrogen. For example, incase of Pentane – 1st carbon and last carbon is attached with three Hand one butyl group, so they are of same chemical nature; 2nd carbonand 2nd last carbon are attached with two H, one methyl and onepropyl groups so they are of same chemical nature but 3 rd carbon isattached with two H and two ethyl groups. So H at 1st, 2nd and 3rd

carbon are of different chemical nature.Alternatively, to find chemically different hydrogen, write the

IUPAC name of products by replacing H with any atom or groupsuch as Cl. The hydrogen which on replacement with any atom orgroup gives product with different IUPAC name is chemicallydifferent.

3CH3H C

Pentane–H+Cl

3CH3H C2-Chloropentane

3CH

3H C

3-Chloropentane

Cl

Cl

3CH

1-ChloropentaneCl

Cl

3H C

or

3CH3H C

Cl

or

The knowledge of chemically different hydrogen is of muchimportance in organic chemistry as it gives idea about all possiblestructures which can be obtained on replacement of hydrogen withany atom or group.In case of saturated hydrocarbons such as alkane,


165



cycloalkanes, bicyclic alkanes we can finds chemically differenthydrogen by replacing H with Cl or Br by monohalogenation whichis carried out in presence of light.

Example 2:

Write the total number of chemically different hydrogen in2-Methylbutane.

3CH3H C

2-Methylbutane

3CH2Cl /light

3CH1-Chloro-3-methyl

butane

3CH3CH3H C

3CH

3CH3H C3CH

3H C3CH

2-Chloro-3-methylbutane



Cl Cl

ClCl

+

+

Here, four different products are obtained on monochlorinationso four types of chemically different hydrogens are present.Example 3:

Write the total number of chemically different hydrogen inMethylcyclopropane.

Methylcyclopropane

3CH2Br

Br

+Br

+

3CH 3CH

BrBromomethylcyclopropane

1-Bromo-1-methylcyclopropane

1-Bromo-2-methylcyclopropane

light

Here, three different products are obtained on monobrominationso three types of chemically different hydrogens are present.Example 4:

Write the total number of chemically different hydrogen inBicyclo[1.1.0] butane.

Here, two different products are obtained on monobrominationso two types of chemically different hydrogens are present.




166

To find chemically different hydrogen in aromatic compoundswe generally use nitration reaction in which aromatic compound istreated with nitrating mixture (conc. H2SO4/HNO3) in which H ofaromatic ring is replaced with NO2 group.

2NO

2 4 3Conc.H SO /HNO

Benzene Nitrobenzene

1-Chloro-2-nitrobenzene

Cl Cl

2 4Conc. H SO 2NO+

Cl

2NO+

Cl

2NO1-Chloro-3-nitrobenzene 1-Chloro-4-nitro

benzene

Chlorobenzene

3HNO

Benzene on nitration gives only one product so all hydrogensin benzene are identical whereas chlorobenzene on nitration givesthree different products so three types of chemically differenthydrogens are present in chlorobenzene.

In case of aromatic compounds having alkyl substituentschemically different hydrogen depends on the nature of reaction.For example, hydrogen of alkyl part is replaced with chlorine inpresence of light but that of aromatic part is replaced with nitrogroup in presence of conc.H2SO4/HNO3 .

2 4Conc.H SO 2NO+

2NO+

2NO3HNO

3CH3CH3CH3CH

2Cl /Light2CH Cl3CH

Chemically different position is used to decide position ofmultiple bond ( C=C or CC) in the skeleton of carbon. For example,to know the total number of alkene possible in skeleton of n-butaneplace double bond in that carbon skeleton at different positions and


167



write their IUPAC name. If IUPAC name are different then positionsare also different.Example 5:

Write the total number of chemically different positions ofdouble bond in skeleton of n-Hexane.

To solve such questions first draw the skeleton of carbon andplace double bond at different positions and write their IUPACnames. It can be seen that double bond between 1-2 and 6-5 givesame name as Hex-1-ene so they are chemically same positions;similarly double bond between 2-3 and 5-4 give same compound,Hex-2-ene so are also same but double bond between 3-4 bond giveproduct Hex-3-ene, so skeleton of n-Hexane have three chemicallydifferent positions which are a=e, b=d and c

3H C3CH1 2 4 6

53a

bc d e

Example 6:Write the total number of chemically different positions for

ketones in skeleton of n-pentane.As in above question, first draw the skeleton and then try to

place =O group in the carbon skeleton. Both terminal positions aresame and leads to aldehyde. Position 2, 3 and 4 forms ketone ofwhich ketone at position 2 and 4 are identical and are different fromketone at position 3. So total ketones possible in skeleton ofn-pentane are two.

3H C

O

3CH3H C

O

3CH& are same

3H C 3CH3H C

O

3CH& are different

O




168

02Exercise

Subjective Approach

01. What are the five basic components which are always used towrite the IUPAC nomenclature of any organic molecules andalso write the role which they play in naming organiccompounds?

02. Name the prefix which is used for following substituents- F, Cl,Br, I, OCH3, OC2H5 OR, NO, NO2, C6H5, CH3, C2H5

03. Name the suffix which is used for following functional groups-carboxylic acid, sulphonic acid, ester, acid halide, amide, nitrile,aldehyde, ketone, alcohol, amine.

04. What is difference between phenyl and benzene? When shouldwe use phenyl and when should we use benzene in writing thenomenclature?

05. What is correct IUPAC nomenclature given for n propyl, isopropyl, n butyl, iso butyl, sec butyl, tert-butyl and neo-pentylgroup?

06. Write the IUPAC nomenclature of these common compounds-n butane, iso butane, neo-pentane, acetone, iso butene,formaldehyde, acetic acid.


169



07. Name the functional groups for which we use some othersecondary suffix such as addition of the word carb when directlyattached to aliphatic or aromatic rings?

08. Write the structure of following compoundsA. 1,1-Dibromo-3-ethyl-4-flourohexaneB. 4-Ethyl-2-methyl-1-propylcyclohexaneC. 6-Chloro-4-ethyl-5-methylhept-5-en-1-yneD. 3-Methylcyclohex-1-ene

E. 2-Methylpentane-2,4-diol

09. Select the main chain which is used as a principal chain or wordroot for naming compounds?

3CH

3H C3CH

3CH

I

3CH

3H C

II

3CH3CH3H C

3H C3CH

10. How many substituents are present on the main chain

3CH

3H C3CH

3CH

I

3CH

3H C

II

3CH3CH3H C

3H C3CH

11. Name the substituents present over the main chain

3CH

3H C

3CH

I

Br

2NO

3CH

3H C

II

3CH

3H C

MeO

Cl




170

12. Write the position of substituent over the main chain

3H C3CH

3CHI

3H C

3CH

3CHII

Cl

Br

3CH

3CHIII

Cl

2O N

IV3CHCl

3H C

Br

MeO

3CH

V

Br3CH

2NOCl

Br

I

ON

3CH

VI

Br3CH

2NOCl

13. Write the IUPAC nomenclature of following compounds

Br 3CH

3CH

3H C

2NOI

3CH

3CH

3H C

II3H C Cl

3CH O

3CH

3CH

3H C

III5 2H C Br

3CH O

3CH

3H C

IV

3CH

3H C3CH

V

3CH3H C

VI

Me

Pr

3CH3H C

VII

Et

Pr

Et

VIII

Et

Pr

Et

Me Bu


171



14. Write the name of complex substituents

3CH

3CH

3H C

3H C

3H C 3H C

IIII II

3CH

3CH

3H C

3H C

IV

3CH

3CH

3CH

3CHV

VI

15. Write the IUPAC name of following compounds?

3CH 3CH 3CH

3CH2CH

I3CH

3H C

3H C

II 2CHCH

3CH

III

3H C

IV

3CH

3CH

3CHV

16. What is the general priority of functional groups in decreasingorder in IUPAC nomenclature?

17. When the prefix formyl and oxo is used for aldehyde?18. Why two different prefixes - alkoxycarbonyl and alkanoyloxy

is used for ester. How they are different. Explain with the helpof an example?

19. Which multiple bond is given priority when double and triplebond is attached in same molecules in different positions.

20. What are the two functional groups which on chemical reactionlosses a water molecule to form ester as a product. Which alkylgroup is written first in IUPAC nomenclature followed by alkylgroup of another?

21. When the word bis is used in IUPAC nomenclature. Explainwith the help of an example.




172

22. What are the two functional groups which on chemical reactionlosses a water molecule to form acid anhydride as a product.Write the IUPAC name of smallest anhydride and its next higherhomologues?

23. When will we use the prefix methylene, ethylidene?24. What are the structures of vinyl, allyl, benzyl halides.25. Write the structure of benzene, benzyne, benzil and also give

the structure of benzyl, benzoyl, benzal & benzo groups?26. Write the IUPAC names and also the more logical common

names which sounds more easy to be pronounced.

3CH3CH

3CH

3H C

I3H C

3CH3CH

3CH

3CH

3CH3H C

II27. Write the IUPAC name of following compounds.

HCOHI

OH

Cl

3H CII

O3CH

III

3H C 2NHIV

O

3CH 3CH 3CH2CH

3CH2CH

V28. Write the IUPAC name of following compounds.

O

O3H C 3CH

3CH

I

O

O3H C

3CH

3CH

3CHII

O

O3H C 3CHIII

3CH3CH

O3H C

O

IV

3CHO

3CH Cl


173



29. Write the IUPAC name of following compounds.

O

O3H C

O

3CHI

O

O3H C

O

3CHII3CH

O

O3H C

O

3CHIII

3CH O

O3H C

O

IV3CH

O

30. Write the IUPAC name of following compounds.

3CH3CH

3CH

3CH

3CH3H C

I

ClCl

Cl

Cl

Cl

II

3CH3H C ClCl Cl ClIII

31. Name the following halides according to IUPAC system andclassify them as alkyl, allyl, benzyl (primary, secondary, tertiary),vinyl or aryl halides:

(A) 3 2 3(CH ) CHCH(Cl)CH

(B) 3 2 3 2 5CH CH CH(CH )CH(C H )Cl

(C) 3 2 3 2 2CH CH C(CH ) CH I

(D) 3 3 2 6 5(CH ) CCH CH(Br)C H

(E) 3 3 3CH CH(CH )CH(Br)CH

(F) 3 2 5 2 2CH C(C H ) CH Br

(G)3 2 5 2 3CH C(Cl)(C H )CH CH




174

(H) 3 2 3 2CH CH = C(Cl)CH CH(CH )

(I) 3 3 2CH CH = CHC(Br)(CH )

(J) 6 4 2 3 2p - ClC H CH CH(CH )

(K) 2 6 4 2 3 3m - ClCH C H CH C(CH )

(L) 6 4 3 2 3o - Br - C H CH(CH )CH CH

(M)3 2 3CH CH CHCH

Cl

(N)3 2 2 2 2CH CHCH CH CH CH Cl

3CH

(O)Br

(P) 3 3CH CHCHF

(Q) 3 2 2 2 2CH CH CH CH CH Cl

(R)3CH

Cl

(S)3CH

3 2 2 2CH CCH CH CH ClOH

(T) 3 2 2 3CH CHCH CHCH CHCl3CH

(U) 3 2 2 2 3CH CHCH CHCH CHCH CHCl3CH 3CH


175



(V)OH

Cl

3 2CH CH(W) 2NO

Br

(X)2NO

Br

(Y)

F

CHO

(Z)

COOH

F

32. Give the IUPAC names of the following compounds:

(A) 3 3CH CH(Cl)CH(Br)CH

(B) 2CHF CBrClF

(C) 2 2CICH C CCH Br

(D) 3 3(CCl ) CCl

(E) 3 6 4 2 3CH C(p - ClC H ) CH(Br)CH

(F) 3 3 6 4(CH ) CCH = C(Cl)C H I - p

(G). 3 2 3CH OCH CH

(H) 3 2 2 3CH CH OCH CH

(I) 3 2 2 2 2 2 3CH CH CH CH CHCH CH CH

3OCH

(J) 3 2 2 2 2 2 3CH CH CH OCH CH CH CH




176

(K)3 2 2 3CH CHOCHCH CH CH

3CH

3CH

(L) 3 2 2 3CH CHOCH CH CHCH

3CH 3CH

(M) O

(N) OH(O) OH

(P)

(Q)Cl

(R)HO

(S)

(T) 3 2 2 2 3CH CH CHCH CH CHCH

3NHCH 3CH

(U)3 2 2 2 3CH CH CHCH CHCH CH

3CH2 3CH CH

2 3CH CH

3CH


177



(V) 3 2 2 2CH CHCHCH CH CH Cl2 3CH CH

Cl(W) 3 2 2 2 2 2 2 3CH CH CH CH CHCH CH CH CH

3 2 3CH CCH CH

3CH

(X) 3 2 2 2 2 2 2 3CH CH CH CH CH CHCH CHCH CH

2CH

3 3CH CCH

2 3CH CH

2 3CH CH

(Y)OH

2H C

(Z)

Br

3H CCl

OH

33. Write IUPAC names of the following compounds:

(A) 3 3CH – CH – CH – C – CHOH

3CH

3CH3CH

(B) 3 2 2 3H C – CH – CH – CH –CH – CH – CH2 5C HOHOH

(C) 3 3CH – CH –CH – CHOHOH

(D) 2 2HO – CH – CH – CH – OHOH

(E)3CHOH (F)

3CH

OH




178

(G) OH3CH

3CH

(H)

3CHOH

3CH

(I)3 2 3CH – O – CH – CH – CH

3CH

(J) 6 5 2 5C H – O –C H

(K) 6 5 7 15C H – O –C H (n-)

(L) 3 2 2 3CH – CH – O – CH – CH – CH3CH

(M) 3 2 2 3CH CH CHCH CCH3CH 3CH

3CH

(N) 3 2 3 3CH CH C(CH )

(O)

2 3CH CH3 2 2 3CH CHCH CH CHCH

3CH

(P) 3 2 2 3 2 2 2 3CH CH C(CH CH ) CH CH CH

(Q) 3 2 2 3 2 3 2 2 3 2CH CH C(CH CH ) CH(CH )CH(CH CH CH )

(R)

3 2 3CH CHCH CH3 2 2 2 2 3CH CH CH CHCH CH CH

(S)2 2 3CH CH CH

3 2 3CH C – CHCH CH3CH 2 2 3CH CH CH


179



(T)

3 2CH(CH )3 2 2 2 2 2 3CH CH CH CH CHCH CH CH

(U)2 3CH CH

3CH (V) 2 3CH CH

(W)3H C

2 3CH CH

3H C(X)

(Y)3 2 2 3CH CHCH CH CH

(Z)

2 3CH CH

2 3CH CHCH

3CH

34. Name the following compounds according to IUPAC system ofnomenclature:

(A) 3 3 2 2CH CH(CH )CH CH CHO

(B) 3 2 2 5 2 2CH CH COCH(C H )CH CH Cl

(C) 3CH CH = CHCHO

(D) 3 2 3CH COCH COCH

(E) 3 3 2 3 2 3CH CH(CH )CH C(CH ) COCH

(F) 3 3 2(CH ) CCH COOH

(G) 6 4OHCC H CHO - p

(H) 3 2 2 2 2 3CH CHCH CH CHCH CH CH

3CH

Br

(I) 3 3 2 2 2 3 2(CH ) CCH CH CH CH(CH )




180

(J) 3 2 3CH CHCH CHCHCH3CH

3CH3CH(K) 3 2 4(CH CH ) C

(L)3 2 2 3CH CHCH CHCH CH

3CH OH(M) 3 2 2 3CH CH CHOCH CH

2 2 2 3CH CH CH CH

(N)

3CH

Br(O)

3CH

3NCH

(P)

OH

2 3CH CH(Q) 3 2 2 2 3CH OCH CH CH OCH

(R) 3 2 2 2 3CH CH CH OCH CH (S) 3 2 2 2CH CHCH CH CH OH

3CH

(T)3 2 3CH CH CHCH

2NH(U)

3 2 3CH CH CHCHCl

(V) 3 2 2 3CH CHCH CH CH3CH

(W) 3CH CBr

2 3CH CH

3CH

(X)OH

(Y)Br

(Z) 3 2CH CHNH3CH

35. Write IUPAC names of the following compounds and classifythem into primary, secondary and tertiary amines.

(A) 3 2 2(CH ) CHNH (B) 3 2 2 2CH (CH ) NH

(C) 3 3 2CH NHCH(CH ) (D) 3 3 2(CH ) CNH


181



(E) 6 5 3C H NHCH (F) 3 2 2 3(CH CH ) NCH

(G) 6 4 2m - BrC H NH

(H) 3 2 2 2 2 2 2CH CH CH CH CH CH NH

(I) 3 2 2 2 2 2 3CH CH CH NHCH CH CH CH

(J) 3 2 2 3CH CHCH NHCHCH CH

3CH 3CH

(K)3 2 2 2 3CH CH CH NCH CH

2 3CH CH

(L)2NH

(M) 3 2 2 2 2 2 2CH CHCH CH CH CH CH NH3CH

(N) 3 2 2 2 2 3CH CH CH NHCH CH CHCH3CH

(O) 33 2(CH CH ) N (P)3CH

3H C 2NH

(Q)

COOH

2NH(R)

COOH

N

H

(S)

COOH

N(T)

COOH

2NH




182

(U)2NH

COOH(V)

2NH

CN

(W)2NH

CHO(X)

2NH

Cl

(Y)2NH

OH(Z)

2NH

O O

36. Write the structures of the following organic compounds.(A) 2-Chloro-3-methylpentane(B) p-Bromochlorobenzene(C) 1-Chloro-4-ethylcyclohexane(D) 2-(2-Chlorophenyl)-1-iodooctane(E) 2-Bromobutane(F) 4-tert-Butyl-3-iodoheptane(G) 1-Bromo-4-sec-butyl-2-methylbenzene(H) 1,4-Dibromobut-2-ene(I) 2,3-Dimethylhexane(J) 4-isopropyl-2,4,5-trimethylheptane(K) 4,4-Diethyldecane(L) 2,2-Dimethyl-4-propyloctane(M) 4-isobutyl-2,5-dimethyloctane(N) 4-(1,1-dimethylethyl)octane(O) 2-Methylbutan-2-ol(P) 1-Phenylpropan-2-ol


183



(Q) 3,5-Dimethylhexane-1,3,5-triol(R) 2,3-Diethylphenol(S) 1-Ethoxypropane(T) 2-Ethoxy-3-methylpentane(U) Cyclohexylmethanol(V) 3-Cyclohexylpentan-3-ol(W) Cyclopent-3-en-1-ol(X) para-toluidine(Y) meta-cresol(Z) para-xylene

37. Draw the structures of the following compounds.(A) 3-Methylbutanal(B) p-Nitropropiophenone(C) p-Methylbenzaldehyde(D) 4-Methylpent-3-en-2-one(E) 4-Chloropentan-2-one(F) 3-Bromo-4-phenylpentanoic acid(G) p,p’-Dihydroxybenzophenone(H) Hex-2-en-4-ynoic acid(I) 2-Methyl-2-bromo-1-propanamine(J) N-Ethylethanamine(K) 5-Methylhexan-1-amine(L) Methyldipropylamine(M) N,N-dimethylpentan-3-amine(N) Cyclohexylethylmethylamine(O) 5-Ethyl-2-methyloctane(P) 1,3-Dimethylcyclohexane(Q) 2,3,3,4-Tetramethylheptane




184

(R) Propylcyclopentane(S) 2-Methyl-4-(1-methylethyl)octane(T) 2,6-Dimethyl-4-(2-methylpropyl) decane(U) 3,3-Dimethylcyclopentene(V) Ethyl vinyl ether(W) 6-Bromo-2,3-dimethyl-2-hexene(X) Allyl alcohol(Y) Cyclooctyne(Z) isopropylacetylene

38. Write a structure for each of the following compounds:(A) isopropyl alcohol (B) isopentyl fluoride(C) sec-butyl iodide (D) neopentyl chloride(E) tert-butylamine (F) isooctyl bromide(G) 2-Hexyne (H) 1-Bromo-1-pentyne(I) 5-Ethyl-3-octyne (J) Propargyl bromide(K) Methylacetylene (L) Diethylacetylene(M) Vinylacetylene (N) di-tert-butylacetylene(O) Methoxyethyne (P) Cyclopentylacetylene(Q) sec-butyl-tert-butylacetylene (R) 5,6-Dimethyl-2-heptyne(S) m-Ethylphenol (T) 2-Chloroanthracene(U) p-Nitrobenzenesulfonic acid(V) m-Chlorostyrene(W) (E)-2-Phenyl-2-pentene (X) o-Nitroanisole(Y) o-Bromoaniline (Z) 2,4-Dichlorotoluene

39. Write a structural formula for each of the following compounds.(A) sec-butyl tert-butyl ether (B) isoheptyl alcohol(C) sec-butylamine (D) neopentyl bromide(E) 1,1-Dimethylcyclohexane (F) 4,5-Diisopropylnonane(G) Triethylamine (H) Cyclopentylcyclohexane


185



(I) 4-tert-butylheptane (J) 5,5-Dibromo-2-methyloctane(K) 1-Methylcyclopentanol (L) 3-Ethoxy-2-methylhexane(M) 5-(1,2-Dimethylpropy)nonane (N) 3,4-Dimethyloctane(O) m-Dichlorobenzene (P) 3-Benzylpentane(Q) p-Bromophenol (R) m-Chlorotoluene(S) o-Nitroaniline(T) 2,5-Dinitrobenzaldehyde(U) 2-Bromo-4-iodo-1-nitrobenzene(V) o-xylene (W) 2-phenylhexane(X) 2-Ethanoyloxybenzoic acid(Y) 2-Methoxycarbonylbenzoic acid(Z) 2-Oxocyclohexane carboxylic acid

40. Give the systematic name for each of the following compounds.

(A)

3CH3 3CH CHCH = CHCH (B) 2 2 3BrCH CH CH = CCH

2 3CH CH

(C) 3 2 3CH CH C = CCHCH3CH

3CH3CH(D)

3CH3H C

(E)Br

(F) 3 2 2 2 3CH CH = CHOCH CH CH CH

(G) 3 2 2 2 3CH CH CHCH = CHCH CH CHCH

Br Br

(H)

2 3CH CH

3 2CH CHC = C

2 2 3CH CH CHCH3H C

3CH

(I)3CH

3CH




186

(J)2 3CH CH

C = C2 2 3CH CH CHCH

3H C

3H C

(K) 2 2 3BrCH CH C CCH (L) 3 2 2 3CH OCH C CCH CH

(M) 3 2 2 3CH CH CHC CCH CHCHClBr

(N)3 2CH CH CHC CH

2 2 3CH CH CH

(O) 3 2 3CH C CCH CHCHBr

(P)3 2 3CH C CCH CHCH

2 2 3CH CH CH

(Q) 3 2 3CH C CCH CCH

3CH

3CH

(R)3 2 3CH CHCH C CCHCH

3CHCl

(S)

(T)3CH

3CH

(U) 2 2 2 3CH = CHCH C CCH CH

(V) 2 2HOCH CH C CH


187



(W)3 2 2CH CH = CCH CH = CH

3CH

(X) 3 2CH CH = CHCH = CHCH = CH

(Y) 33 2 2CH CH = CCH CHCH CH3CH3CH

(Z) 3 2 2 2CH CH CH = CCH CH C CH

2CH = CH

41. Give the systematic name for each of the following compounds:

(A) 3 2 2 2 2CH C CCH CH CH CH = CH

(B) C=C2 2HOCH CH

2 3CH CH

H H

(C) 3 2 2 2CH CH C CCH CH C CH

(D)Cl

(E)3CH

(F)

2CH Br

(G)

2CHBr

(H) 3H COH

3CH (I)Br

COOH

(J)Br

3CH(K)

3OCH

2 3CH CH(L)

3SO H

Cl Cl




188

(M)

2 3CH CH

Cl(N)

2CH = CH

2NO(O

BrBr

Br

(P) 3CH (Q)

CN

Cl (R) 3 2 2CH CH CH C N

(S)O

3 2 2 2 3CH CH CH COCH CHCH3CH

(T) O

3 2 3CH CH COCCHO

(U)O

3 2 2 2 2 3 2CH CH CH CH CH CN(CH ) (V) O

33 2 2CH CH CH COCH

(W)O

NH2(X)

O

3 2 2 2CH CH CH CH CCl

(Y)COOH

(Z)

COOH

O CHO42. Name the following compounds:

(A)3 2 2 2 2CH CH CHCH CH CH COH

O2 3CH CH (B) 3 2 2 2CH CH CH CH CCl

O

(C) 3 2 2 2 3CH CH COCH CH CHO

(D)3COCCH

O O

(E) 3 2 2 2CH CH CH CH C N (F) 2 2 3CH = CHCH CNHCHO


189



(G) 3 2 2 3CH CH COCCH CHO O

(H)C

2 3CH CH

3CH H

2CH COOH

(I) 3 2 2 3 2CH CH CH CN(CH )O

(J)C

2CH C N

3CH H

2 2 3CH CH CH

(K) 3 2 2CH CH CHCH CH

3CH

O(L)

2 2 2CH CH CH CHO

(M) 3 2 2 2 2 3CH CH CH CCH CH CHO

(N) 3 2 2 2CH CH CHCH CH CH2 3CH CH O

(O)3 2 2 2 3CH CHCH CCH CH CH

3CH

O

(P) 2 2 2 2 3CH = CHCCH CH CH CHO

(Q) 3 2 2 2 3CH CHCH CH CCH CHO

OH(R)

C N

O

(S) 3 2 2 2CH CH CHCH CNHO

HC=O

(T) Br Cl

OH




190

(U)

OH

ClOH

Br2H C(V) Br

OH

ClCH

(W)CH

3H C (X)

Br

3H CCl

OH

OH

3CH

(Y) Cl OH2CH

3H C(Z) 2CH

43. Write the IUPAC names of following compounds

(A) 3H C 2CH(B)

CH

3H C

(C)CH

3H C

2CH

2CH (D)O

O

(E) 3H CO

2H C O 2CH (F)

O

2H C2CH 2CH

(G)2H C

O 2CH (H)3H C

Br O

N3CH

H

(I)

Br O

NCl

H

Br(J)

Br O

NCl

H

ClI


191



(K)

Br O

N3CH

Cl

3CHI (L) 3CHO

O

(M)3CHO

O

3CH(N)

3H CCl O

OH

(O) OH

O

CHCH

CHCH

3H C(P)

O

OHBr

3CHNH

O

(Q)3CH

HC OH

Cl O

(R)2CH

(S)

CH

(T)

3SO H

CHOCOOH

(U) CN

OH

2NH(V) 2CH

COOH

O

(W)OO3H C

O OH

(X)




192

(Y)Br OH

(Z)

OO3H C

O 3CH

44. Write the IUPAC names of following compounds

(A)OO3H C

O H

(B)OO3H C

O Cl

(C)

OO3H C

O 2NH

(D)

O

OH

Cl

3CHO

HC

3CH

(E)

O

OH

O

3CHOHC (F)

3H C 3CH

3CH3H C

(G)

3H C 3CH

3CH3H C3CH

3CH

(H)

ClCl

Cl

Cl

Cl

(I)

ClCl

Cl3CH3H C

3H C

Cl3CH

Br3CH

(J)O

3CHO

O


193



(K)OO

OH (L)O

2NH

O

(M)OO

OH (M)

O

O

(O)O

O (P) 3H–C–N–CH

O

2 3CH CH

(Q) H–C–ClO

(R)2H–C–NH

O

(S) H–C–HO

(T)O

NH

(U)O

2NH (V)O

OH

(W)O

OH (X)O

OH

(Y)O

(Z) H–C–O

O




194

AnswersSubjective Approach

0201. Five basic components are mainly- Secondary prefix, Primary

prefix, Word root, Primary suffix and Secondary suffix. Theirrole are Secondary prefix-It tells about the nature and position

of substituent if any. Primary prefix-It is used to specify acyclic, cyclic, bicyclic

or spiro nature of main chain. Word root-It tells about the longest continuous carbon

chain which includes functional group, carbon carbonmultiple bond (double or triple bond) and substituent atminimum position.

Primary suffix-It tells about saturated or unsaturatednature of carbon-carbon bonds in the parent carbonchain.

Secondary suffix-It tells about the main functionalgroups present in the compound

02. Prefix used for them are tabulated belowSubstituent Prefix Substituent Prefix

F Fluoro OR Alkoxy

Cl Chloro NO Nitroso

Br Bromo NO2 Nitro


195



I Iodo C6H5 Phenyl

OCH3 Methoxy CH3 MethylOC2H5 Ethoxy C2H5 Ethyl

03. Suffix used for following functional groups areFunctional group Secondary suffixCarboxylic acid (–COOH) oic acidSulphonic acid (–SO3H) Sulphonic acidEster (–COOR) oateAcid halide (–COX ) oyl halideAmide (–CONH2) amideNitrile (–CN) nitrileAldehyde (–CHO) alKetone (–CO–) oneAlcohol (–OH) olAmine (–NH2) amine

04. Benzene is the correct IUPAC name of Cyclohexa-1,3,5-triene.As aromatic compounds are different from alkenes and polyenein chemical reactions so 1° Prefix, Word root and 1° Suffixcombines to give a name ‘Benzene’. If one hydrogen is removedfrom benzene ring then the skeleton that remains is called asphenyl group. In IUPAC nomenclature, if benzene ring is parentchain then benzene is used but if in any compound benzenering is present as substituent then phenyl is used.

05. Correct IUPAC of common groups areGroups IUPAC namen propyl Propyliso propyl 1-Methylethyln butyl butyliso butyl 2-Methylpropylsec butyl 1-Methylpropyl




196

tert butyl 1,1-Dimethylethylneo pentyl 2,2-Dimethylpropyl

06. The correct IUPAC name of given common compounds areCommon compounds IUPAC namen butane Butane

iso butane 2-Methylpropane

neo pentane 2,2-Dimethylpropane

acetone Propanoneiso butene 2-Methylpropene

formaldehyde Methanal

acetic acid Ethanoic acid07. These functional groups are commonly Carboxylic acid, Ester,

Acid halide, Amide, Nitrile and Aldehyde08. Structure of given compounds are

(A)Br

F3CHBr

3CH (B)

3CH

3H C

3CH

(C)3CH

3H C

3CH

Cl

HC (D)

3CH

(E) 3CH

OH

OH

3CH

3H C

09. Main chain is

3CH

3H C3CH

3CH

I

3CH

3CH

3CH3CH3H C

3H C3H C

II10. 2, 4 respectively


197



11. Substituents in I are- Bromo, Nitro & Propyl and substituentsin II are Chloro, methyl & 1-Methoxyethyl

12. Position of substituents areI- 1,2,4; II- 1,1,3,4,5;III- 1,2,3,5; IV- 1,2,3,4,5 ;V- 1,1,2,3,3,4,5,6; VI- 1,1,2,5,5

13 I- 3-Bromo-5-ethyl-4-nitrooctaneII- 2-Chloro-6-(1-methoxyethyl)-3-methylnonaneIII- 5-Bromo-6-(1-methoxyethyl)-3-methylnonaneIV- 1,4-DimethylcyclohexaneV- 1,2,4-TrimethylcyclohexaneVI- 3-Ethyl-2-methylhexaneVII- 3,5-Dimethyl-4-propylheptaneVIII- 5-Ethyl-3-methyl-4-propylnonane

14. I- 1-Methylethyl; II- 2-Methylpropyl;III- 1-Methylpropyl; IV-2,2-Dimethylpropyl;V- 3-Methylbutyl; VI- 1,1-Dimethylethyl

15. I- 7-Ethyl-2,4,5 ,6-tetramethyldeca- 1,8-dieneII- 3-Ethynylhepta-1,5-diene

III- Prop-1-enylcyclobutene

IV- 1,3-DiethylcyclobuteneV- 1-Ethyl-2-methylcyclohexa-1,4-diene

16. Carboxylic acid > Sulphonic acid > Ester > Acid halide >Amide > Nitrile > Aldehyde > Ketone > Alcohol > Amine

17. When CHO is directly attached as a substitution then wordformyl is used but if one or more carbon is present betweenmain chain and CHO then ‘oxo’ is used.

18. Two different prefixes are used for ester on the basis ofattachment with main chain. 2° Prefix alkoxycarbonyl is used




198

when ester group is directly attached to main chain by CO groupwhile alkanoyloxy is used when directly attached atom is O.For example:

O3H C

COOH

3CHO

O

O MethoxycarbonylEthanoyloxy

19. Both multiple bonds denotes unsaturation in carbon carbonbonds. So position of minima for unsaturation will decide thelowest position for ene or yne. For ex.

12H C3CH

3 5642

Hex-1-en-4-yne 1

3CH3 5 642

Hex-4-en-1-yne

HC

20. Carboxylic acid and alcohol combines together and looses awater molecule to form ester as a product. In nomenclature ofester, alkyl part of alcohol is written first followed by alkyl partof carboxylic acid as alkylalkanoate

21. When two same complex substituents is presents in anymolecule then to indicate its repeatition two times, word ‘bis’ isused. For example

1,4-Bis-(1,1-Dimethylethyl) cyclohexane

3CH3CH

3CH3CH

3CH

3H C

22. When two same or different carboxylic acid combines togetherand loose a water molecule then acid anhydride is formed as aproduct. IUPAC name of smallest anhydride and its next higherhomologues are Methanoic anhydride and Ethanoic methanoicanhydride respectively.

23. Word methylene is used to a carbon bounded with 2 hydrogenatoms. In case of substituent, word methylene is used for =CH2group and ethylidene for =CHCH3


199



24. The structure are

2H CCl

Vinyl chloride

2H C ClAllyl chloride

Benzyl chloride

Cl

25. Structures are

Benzene

Benzyne

O

O

PhPh

Benzil

BenzylO

Benzoyl

Benzal

Benzo

26. I) 1-(1,1-Dimethyl ethyl)-4-(1-methyl ethyl) cyclohexane or1-Iso propyl-4-tert butyl cyclohexane

II) 1,4-Bis- (1,1-Dimethyl ethyl) cyclohexane or1,4-Bis( tert. Butyl) cyclohexane

27 I) Pent-4-yn-1-olII) 3-Chloro-5-methylcyclohexanolIII) 2-Methylcyclopent-2-enoneIV) 3-Methylcyclobut-2-enamineV) 4,5,6,8-Tetramethylnona-1,8-dien-3-one

28. I) 1-MethylethylethanoateII) 1-Methylethyl-2-methylpropanoateIII) 1-Methylethyl-3-methylbut-2-enoateIV) 2-Methyl 3-oxo-butyl-4-chloropentanoate

29. I) Ethanoic anhydrideII) Ethanoic-2-methylpropanoic anhydrideIII) Ethanoic-3-methylbut-2-enoic anhydrideIV) Butanoic-3-oxobutanoic anhydride




200

30. I) 1,4-Bis-(1,1-dimethylethyl)cyclohexeneII) 1,1,1-Trichloro-2,2-bis-(4-chlorophenyl)ethaneIII) 3,7-Bis-(dichloromethyl)nona-2,7-diene

31. IUPAC Name Nature of halideA. 2-chloro-3-methylbutane 2° Alkyl halideB. 3-chloro-4-methylhexane 2° Alkyl halideC. 1-Iodo-2,2-dimethylbutane 1° Alkyl halideD. 1-Bromo-3,3-dimethyl-1-phenylbutane 2° benzyl halideE. 2-Bromo-3-methylbutane 2° Alkyl halideF. 1-Bromo-2-ethyl-2-methylbutane 1° Alkyl halideG. 3-chloro-3-methylpentane 3° Alkyl halideH. 3-chloro-5-methylhex-2-ene vinyl halideI. 4-Bromo-4-methylpent-2-ene 3° Allylic halideJ. 1-chloro-4-(2-methylpropyl)benzene Aryl halide

K.1-chloromethyl-2-(2,2-dimethyl

propyl) benzene 1° benzyl halide

L. 1-Bromo-2-(1-methylpropyl)benzene Aryl halideM. 2-Chlorobutane 2° alkyl halideN. 1-chloro-5-methylhexane 2° Alkyl halideO. Bromocyclohexane 2° Alkyl halideP. 2-Fluoropropane 2° Alkyl halideQ. 1-chloropentane 1° Alkyl halideR. 1-chloro-4-methylcyclohexane 2° Alkyl halideS. 1-chloro-4-methylpentan-4-ol 1° Alkyl halideT. 4-chloro-2-methylhexane 2° Alkyl halideU. 4-chloro-2,6-dimethylhexane 2° Alkyl halideV. 4-chloro-3-ethylcyclohexanol 2° Alkyl halideW. 1-Bromo-2-nitrobenzene Aryl halide


201



X. 1-(1-Brmoethyl)-4-nitrobenzene 1° benzyl halideY. 3-Fluorobenzene carbaldehyde Aryl halide

Z.3-(1-Fluoroethyl)benzene

carboxylic acid 2° Alkyl halide

32. A. 2-Bromo-3-chlorobutaneB. 1-Bromo-1-chloro-1,2,2-trifluoroethaneC. 1-Bromo-4-chlorobut-2-yneD. 1,1,1,2,3,3,3-Heptochloro-2-trichloromethylpropaneE. 3-Bromo-2,2-bis-(4-chlorophenyl)butaneF. 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-eneG. MethoxyethaneH. EthoxyethaneI. 4-MethoxyoctaneJ. 1-PropoxybutaneK. 2-(1-methylethoxy)pentaneL. 3-Methyl-1-(1-methylethoxy)butaneM. 1-MethoxypropaneN. Propan-1-olO. 4-Propylnonan-1-olP. 4-Methyl-5-(1-methylethyl)octaneQ. 6-chloro-4-ethyl-3-methyloctaneR. 5-Methyl-3-propylhexan-1-olS. 2,3-Dimethyl-6-(2-methylpropyl)decaneT. N-Methyl-6-methylheptan-3-amineU. 3-Ethyl-2,3,6-trimethyloctaneV. 1,4-Dichloro-5-methylheptaneW. 5-(1,1-Dimethylpropyl)nonaneX. 5-(2-Ethylbutyl)-3,3-dimethyldecane




202

Y. Pent-4-en-1-olZ. 2-Bromo-3-chlorohept-5-en-1-ol

33. A. 2,2,4-Trimethylpentan-3-ol B. 5-Ethylheptan-2,4-diolC. Butane-2,3-diol D. Propane-1,2,3-triolE. 2-Methylphenol F. 4-MethylphenolG. 2,5-Dimethylphenol H. 2,6-DimethylphenolI. 1-Methoxy-2-methylpropane J. EthoxybenzeneK. 1-Phenoxyheptane L. 2-EthoxybutaneM. 2,2,4-Trimethylhexane N. 2,2-DimethylbutaneO. 2,5-Dimethylheptane P. 3,3-DiethylhexaneQ. 3,3-Diethyl-4-methyl-5-propyloctaneR. 3-Methyl-4-propylheptaneS. 5-Ethyl-4,4-dimethyloctaneT. 4-(1-methylethyl)octaneU. 1-Ethyl-2-methylcyclopentaneV. EthylcyclobutaneW. 4-Ethyl-1,2-dimethylcyclohexaneX. 3,6-DimethyldecaneY. 2-CyclopropylpentaneZ. 1-Ethyl-3-(2-methylpropyl)cyclohexane

34. A. 4-MethylpentanalB. 6-chloro-4-ethylhexan-3-oneC. But-2-enalD. Pentan-2,4-dioneE. 3,3,5-Trimethylhexan-2-oneF. 3,3-Dimethylbutanoic acidG. Benzene-1,4-dicarbaldehydeH. 5-Bromo-2-methyloctane


203



I. 2,2,6-Trimethylheptane J. 2,3,5-TrimethylhexaneK. 3,3-Diethylpentane L. 5-Methylhexan-3-olM. 3-EthoxyheptaneN. 1-Bromo-4-methylcyclohexaneO. N,N-DimethylcyclohexanamineP. 3-Ethylcyclohexanol Q. 1,3-DimethoxypropaneR. 1-Ethoxypropane S. 4-Methylpentan-1-olT. Butan-2-amine U. 2-chlorobutaneV. 2-methylpentane W. 2-Bromo-2-methylbutaneX. Cyclohexanol Y. BromocyclopentaneZ. Propan-2-amine

35. A. Propan-2-amine 1° amineB. Propan-1-amine 1° amineC. N-Methylpropan-2-amine 2° amineD. 2-Methylpropan-2-amine 1° amineE. N-Methylaniline 2° amineF. N-Ethyl-N-methylethanamine 3° amineG. 3-Bromoaniline 1° amineH. Hexan-1-amine 1° amineI. N-propylbutan-1-amine 2° amineJ. N-(2-methylpropyl)butan-2-amine 2° amineK. N,N-Diethylpropan-1-amine 3° amineL. Cyclohexanamine 1° amineM. 6-Methylheptan-1-amine 1° amineN. N-Propyl-3-methylbutan-1-amine 2° amineO. N,N-Diethylethanamine 3° amineP. 2,5-Dimethylcyclohexanamine 1° amineQ. 3-Aminomethylbenzenecarboxylic acid 1° amine




204

R. 3-(Methylamino)benzenecarboxylic acid 2° amine

S.3-(N-Ethyl-N-methylamino)

benzenecarboxylic acid 3° amine

T. 3-Aminobenzene carboxylic acid 1° amineU. 3-Amino-2-ethylbutanoic acid 1° amineV. 3-Amino-2-ethylbutane nitrile 1° amineW. 3-Amino-2-ethylbutanal 1° amineX. 3-Chloropentan-2-amine 1° amineY. 2-Aminopentan-3-ol 1° amineZ. Methyl-3-amino-2-ethylbutanoate 2° amine

36. A.Cl

B.

Cl

Br

C.

Cl

D. I Cl

E.Br

F.I

G.

Br

H. BrBr

I. J.

K. L.


205



M. N.

O.OH

P.OH

Ph

Q.OHOHOH R.

OH

S. OEt T.OEt

U. 2CH OHV.

OH

W.OH

X.3CH

2NH

Y.

OH

3CHZ.

3CH

3CH

37. A.H

OB.

2O N

O

C.

CHO

3CHD.

O




206

E.O Cl

F.Br O

OHPh

G.

O

OHHOH. OH

O

I. 2NHBr

J.NH

K.2NH L. N

M. N N. N

O. P.

Q. R.

S. T.

U. V. O

W. Br X. OH

Y. Z.


207



38. A.OH

B. F

C.I

D. Cl

E.2NH F. Br

G. H. Br

I. J. Br

K. L.

M. N.

O. O P.

Q. R.

S.OH

T.Cl

U.

2NO

3SO H

V.Cl

W. Ph X.3OCH

2NO

Y.2NHBr Z.

3CHCl

Cl




208

39. A. O B. OH

C.2NH D. Br

E. F.

G.N

H.

I. J.BrBr

K.OH

L.O

M. N.

O.

Cl

ClP.

Q.

OH

BrR.

3CH

Cl

S. 2NO2NH

T.2NO

CHO

2NO


209



U.

2NOBr

I

V. 3CH3CH

W.Ph

X.COOH

3O–C–CHO

Y.COOH

3C–OCH

O

Z.COOH

O

40. A. 4-Methylpent-2-eneB. 1-Bromo-4-methylhex-3-eneC. 2,3,4-Trimethylhex-3-eneD. 1,5-DimethylcyclohexeneE. BromocyclopentaneF. 1-ButoxypropeneG. 3,8-Dibromonon-4-eneH. 4-Ethyl-3,7-dimethyloct-3-eneI. 1,5-DimethylcyclopenteneJ. 3-Ethyl-2-methylhept-2-eneK. 5-Bromopent-2-yneL. 1-Methoxypent-2-yneM. 6-Bromo-2-chlorooct-4-yneN. 3-Ethylhex-1-yneO. 5-Bromohex-2-yneP. 5-Methyloct-2-yneQ. 5,5-Dimethylhex-2-yneR. 6-Chloro-2-methylhept-3-yneS. Cycloocta-1,5-diene




210

T. 1,6-Dimethylcyclohexa-1,3-dieneU. Hept-1-en-4-yneV. But-3-yne-1-olW. 4-Methylhexa-1,4-dieneX. Hepta-1,3,5-trieneY. 3,5-Dimethylhept-2-eneZ. 5-Ethenyloct-5-en-1-yne

41. A. Oct-1-en-6-yneB. Hex-3-en-1-olC. Oct-1,5-diyneD. ChlorocyclohexaneE. 1-Methylcyclohepta-1,3,5-trieneF. 1-Bromo-1-phenylmethaneG. 1,1-Dibromo-1-phenylmethaneH. 2,6-DimethylphenolI. 3-Bromobenzene carboxylic acidJ. 1-Bromo-2-methylbenzeneK. 1-Ethyl-3-methoxybenzeneL. 3,5-Dichlorobenzene sulphonic acidM. 2-Chloro-4-ethyl-1-phenylbenzeneN. 4-NitrophenyletheneO. 1,2,4-TribromobenzeneP. 1-Cyclohexyl-4-methylbenzeneQ. 3-ChlorobenzenecarbonitrileR. ButanenitrileS. 2-MethylpropylbutanoateT. Ethanoic propanoic anhydrideU. N,N-DimethylhexanamideV. Methylbutanoate


211



W. 2-AminocyclopentanoneX. Pentanoyl chlorideY. Cyclopentane carboxylic acidZ. 3-Formyl-5-oxocyclohexane carboxylic acid

42. A. 5-Ethylheptanoic acidB. Pentanoyl chlorideC. PropylpropanoateD. Benzene carboxylic ethanoic anhydrideE. Pentane nitrileF. N-Methylbut-3-enamideG. Propanoic anhydrideH. 3-Methylpentanoic acidI. N,N-DimethylbutanamideJ. 3-MethylhexanenitrileK. 3-MethylhexanalL. 4-PhenylbutanalM. Heptan-4-oneN. 4-EthylhexanalO. 2-Methylheptan-4-oneP. Hept-1-en-3-oneQ. 6-Hydroxyheptan-3-oneR. 2-oxocyclohexanecarbonitrileS. 3-FormylpentanamideT. 4-Bromo-2-chloro-3-cyclopropylcyclohexanolU. 3-Bromo-2-chloro-5-methylenecyclohexane-1,4-diolV. 2-Bromo-4-chloro-5-ethynylcyclohexanolW. 4-Ethynyl-1-methylcyclohexeneX. 3-Bromo-4-chlorooct-3-en-6-yne-2,5-diol




212

Y. 4-Chloro-5-(1-methylethenyl)cyclopenta-1,3-diene-1-olZ. 5-Prop-2-enylcyclopenta-1,3-diene

43. A. Octa-1,3,7-trieneB. Hept-5-en-1-yneC. 4-Ethenyl-3-ethylhept-6-en-1-yneD. CyclobutylmethylcyclopropanecarboxylateE. Prop-2-enyl-3-ethylpent-4-enoateF. 4-Methylenehexa-1,5-dien-3-oneG. 2-Ethenylbut-3-enalH. N-Methyl-3-bromobutanamideI. N-bromo-3-bromo-4-chlorobut-2-enamideJ. N-2-Chloro-1-iodoethyl-3-bromo-4-chlorobut-2-enamideK. N-Ethyl-N-methyl-3-bromo-4-chloro-2-iodobutanamideL. 2-OxobutanalM. 2-(1-Oxo ethyl)pent-3-enalN. 5-Chloro-6-cyclopropylhept-3-enoic acidO. 3,4,5-Triethynyl-6-methyloct-7-ynoic acidP. 2-Bromo-4-(N-formylamino)-3-methylbutanoic acidQ. 3-Chloro-4-cyclopropyl-2-methylhept-6-ynoic acidR. PhenyletheneS. PhenylethyneT. 2-Formyl-4-sulphobenzenecarboxylic acidU. 2-Amino-5-hydroxybenzenecarbonitrileV. 3-Ethenyl-4-formylbenzenecarboxylic acidW. 4-Methoxycarbonylbenzenecarboxylic acidX. PhenylbenzeneY. 4-(2-Bromophenyl)phenolZ. Methyl-4-(1-oxo ethyl)benzenecarboxylate


213



44. A. Methyl-4-formylbenzenecarboxylateB. Methyl-4-chlorocarbonylbenzenecarboxylateC. Methyl-4-carbamoylbenzenecarboxylateD. 3-Chloro-4-(1-oxo ethyl)-2-methylhept-6-ynoic acidE. 4-Formyl-2-methyl-3-oxohept-6-ynoic acidF. 1,4-Bis-(1-methylethyl)benzeneG. 1,4-Bis-(1,1-dimethylethyl)benzeneH. 1,1,1-Trichloro-2,2-bis-(4-chlorophenyl)ethaneI. 3 - B r o m o- 2 , 2 - d i c h l or o- 1 , 1 - b i s- ( 4 - c h l or o - 3 -

methylethylphenyl)benzeneJ. Methyl-2-oxocyclohexanecarboxylateK. 6-Oxocyclohex-2-enecarboxylic acidL. 6-Oxocyclohex-3-enecarboxamideM. 5,6-Dioxocyclohex-2-enecarbaldehydeN. CyclohexylbenzenecarboxylateO. PhenylcyclohexanecarboxylateP. N-Ethyl-N-methylmethanamideQ. Methanoyl chlorideR. MethanamideS. MethanalT. 2-MethylaminocyclohexanoneU. 2-AminomethylcyclohexanoneV. 2-HydroxymethylcyclohexanoneW. 2-HydroxycyclohexanoneX. 6-Hydroxy-2,2-dimethylcyclohexanoneY. O,O-Dimethylcyclohex-2-enoneZ. Phenylmethanoate




214

02Exercise

Objective Approach

Single Correct Questions (SCQ) :

01. The correct IUPAC name of the alkane is :-

(A) 2-Ethyl-4-methyl hexane (B) 3, 5-Dimethyl heptane(C) 5-Ethyl-3-methyl hexane (D) 3, 5-Dimethyl hexane

02. Which of the following is the correct priority order of functionalgroups?

(A) 3–SO H > –COR > –CHO > –OH

(B) 3 2–COOH > –SO H > –CN > –CONH

(C) –CN > –CHO > –COR > –OH

(D) –COOR > –COX > –CHO > –CN03. The correct IUPAC name of the following compound

Cl – C – C – C – H

2NO Br

Et H I

F is –


215



(A) 2-Bromo-1-chloro-1-ethyl-3-fluoro-3-iodo-1-nitro propane(B) 2-Bromo-3-chloro-1-fluoro -l-iodo-3-nitro pentane(C) 2-Bromo-3-chloro-3-ethyl-1-fluoro-1-iodo-3-nitro propane(D) 4-Bromo-3-chloro-5-fluoro -5-iodo-3-nitro pentane

04. The correct structure of 6-Ethyl-2,3,5-trimethyl nonane.

(A) (B)

(C) (D)

05. The correct IUPAC name of incorrectly named 2,3-Diethyl butaneis(A) 3,4-Dimethyl hexane (B) 2,3-Dimethyl hexane(C) 2-Ethyl-3-methyl pentane (D) 2-Ethyl butane.

06 The common & IUPAC names for the following alkyl grouprespectively is–

3 2 2 2(CH ) CH – CH – CH –

(A) Iso pentyl & 3-Methyl butyl(B) Iso pentyl & 2-Methyl butyl(C) Sec-pentyl & 1,1-Dimethyl propyl(D) Neo-pentyl & 3-Methyl butyl

07. Select the structure with correct numbering in the chain.

(A)2 2H C = CH – CH – C CH

12345

(B)3 2CH – CH = CH – CH – C CH

1 2 3 4 5 6

(C)2 2 2CH = CH – CH = CH – CH – CH = CH

7 6 5 4 3 2 1

(D) 2 2CH = CH – CH = CH – CH – C CH

1 2 3 4 5 6 7




216

08. The correct IUPAC name of the compound

2 2 3H C = CH – CH – CH – CH2 5C H

is:

(A) 4-Methyl hex -1-ene (B) 4-Ethyl pent-1-ene(C) 2-Ethyl pent -4-ene (D) 3-Methyl hex-1-ene

09. The correct structure for 2-Ethyl-3-methyl hexa-1,4-diene is–

(A) (B)

(C) (D)

10. In which of the following compounds the carbon atom chainhas been correctly numbered ?

(A)3 2 2 2 3CH – CH – CH – CH – CH – CH – CH

2 5C H 3CH

1234567

(B) 3 2 3CH – CH – C = CH – CH

3 2CH – CH

1 2 3

4 5

(C) 3 2CH – CH – CH – C CH12345

2 2 3CH – CH – CH

(D) 3CH – C – CH = CH – CHO3CH

123453CH

11. The correct IUPAC name of the following compoundCl I

FBr is–

(A) 1-Chloro-1-iodo-2-fluoro methyl -4-bromo but-1-ene.(B) 1-Bromo -4-chloro -3-fluoro methyl -4-iodo but -1-ene.(C) 4-Bromo -1-fluoro -2-chloro iodo methyl butene(D) 4-Bromo -1-chloro -2-fluoro methyl -1-iodo but -1-ene.


217



12. In which of the following, acyclic chain is main chain.

(A) (B)

(C) (D)

13. In which of the following compound IUPAC numbering iscorrect–

(A)3

265 4

1

(B)3 4

5612

(C) 32

1654 (D) 3

216

5 4

14. The correct IUPAC name of the following compound is–(A) 2-Cyclohexyl butane(B) Iso-butyl cyclohexane(C) (1-Methyl propyl) cyclohexane(D) 1-cyclohexyl-1-methyl propane

15. In which of the following, ring is the main chain.

(A) (B)

(C) (D)

16. The correct IUPAC name of the compound is-

(A) 1-Ethenylcyclopenta-2,4-diene(B) 5-Ethenylcyclopenta-1,3-diene(C) Cyclopenta-2,4-dienylethene(D) Cyclopenta-1,3-dienylethene




218

17. The correct IUPAC name of 3 2 2 2 3CH –CH–CH –CH –CH –C–CHCl

Cl

OHis:

(A) 6,6-Dichloroheptan-2-ol (B) 2,2-Dichloroheptan-6-ol(C) 6,6-Dichloroheptan-2-al (D) 6,6-Dichloroheptan-2-one.

18. The correct IUPAC name of compound 3SO HO

(A) 6-Propyl-1-methyl-4-oxohept-6-ene-1-sulphonic acid.(B) 5-Oxo-7-propyloct-7-ene-2-sulphonic acid.(C) 2-Propyl-7-sulphooct-1-ene-4-one.(D) 7-Methylene-5-oxodecane-2-sulphonic acid.

19. The correct IUPAC name of the compound 2NH2NH is

(A) 4-Ethyl-3-methylnonane-2,8-diamine(B) 6-Ethyl-7-methylnonane-2,8-diamine(C) 5-Ethyl-1,6,7-trimethylheptane-1,7-diamine(D) 4-Ethyl-2-methylnonane-2,7-diamine

20. The correct structure of 6-Chloro-4-hydroxycyclohex-2-ene-1-sulphonic acid is

(A)3SO H

OH

Cl (B)3SO H

ClHO

(C)

3SO H

ClOH

(D)Cl

HO

3SO H


219



21. The correct IUPAC name of following compound COOHBr

is(A) 3-Bromobut-2-enoic acid (B) 3-Bromobut-3-enoic acid(C) 2-Bromobut-1-en-4-oic acid (D) 3-bromo-3-carboxypropene

22. The IUPAC name of the following compound

3 2 2CH – CH = CH – CH – CH – CH – COOH

2NH is

(A) 2-Aminohept-5-enoic acid(B) 6-Aminohex-2-enecarboxylic acid(C) 2-Aminohept-4-enoic acid(D) 6-Aminohept-2-enoic acid.

23. The correct IUPAC name of following compound

COOH

Br2NH

is

(A) 4-Aminomethyl-3-bromocyclohex-5-ene-1-carboxylic acid(B) 2-Aminomethyl-5-carboxycyclohex-3-en-1-bromine(C) 4-Aminomethyl-5-bromocyclohex-2-ene-1-carboxylic acid(D) 3-Bromo-4-aminomethylcyclohex-5-en-1-oic acid

24. The IUPAC name of the compound O

OO is

(A) Butanoic anhydride (B) Dibutanoic anhydride(C) Butanoyl oxybutanoate (D) 1-oxopropyl butanoate

25. The correct IUPAC name of compound O

O

O is

(A) Butanedicarboxylic anhydride(B) Cyclobutanedioic anhydride(C) Cyclobutanedicarboxylic anhydride(D) Butanedioic anhydride




220

26. The IUPAC name of compound is

(A) Bicyclo [2,2,1] hexane (B) Bicyclo [2,2,1] heptane(C) Spiro [2,2,1] hexane (D) Spiro [2,2,1] heptane

27. Aspirin is

(A)3O – C – CH

O

COOH(B)

3O – C – CHO

COOH

(C) 3C – OCHOOH

(D)3O – C – CH

O

OH

28. Which of the following compounds has wrong IUPAC name ?[AIEEE 02]

(A) 3 2 2 2 3CH CH CH COOCH CH Ethyl butanoate

(B) CH – CH – CH – CHO 3-Methyl-butanal3 2

CH3

(C) CH – CH – CH – CH 2-Methyl-3-butanol3 3

OH CH3

(D) CH – CH – C – CH CH 2-Methylpentan-3-one3 2 3

CH3

O

29. The IUPAC name of the compound is HO

[AIEEE 04]

(A) 3, 3-Dimethyl cyclohexanol(B) 1, 1-Dimethyl-3-hydroxy cyclohexane(C) 3, 3-Dimethyl-1-hydroxy cyclohexane(D) 1, 1-Dimethyl-3-cyclohexanol


221



30. The IUPAC name of the compound shown below is :

Cl

Br[AIEEE 06]

(A) 3-Bromo-1-chlorocyclohexene(B) 1-Bromo-3-chlorocyclohexene(C) 2-Bromo-6-chlorocyclohex-1-ene(D) 6-Bromo-2-chlorocyclohexene

31. The IUPAC name of is [AIEEE 07]

(A) 3-Ethyl-4,4-dimethylheptane(B) 1, 1-Diethyl-2, 2-dimethylpentane(C) 4, 4-Dimethyl-5, 5-diethylpentane(D) 5, 5-Diethyl-4, 4-dimethylpentane

32. The correct decreasing order of priority for the functionalgroups of organic compounds in the IUPAC system of nomen-clature is [AIEEE 08]

(A) 3 2-COOH.- SO H,-CONH - CHO

(B) 3 2-SO H, -COOH,-CONH , -CHO

(C) 3 2-CHO,-COOH,-SO H,-CONH

(D) 2 3-CONH - CHO,-SO H,-COOH

33. The IUPAC name of neopentane is : [AIEEE 09](A) 2, 2-Dimethylpropane (B) 2-Methylpropane(C) 2,2-Dimethylbutane (D) 2-Methylbutane

34. The IUPAC name of 6 5C H COCl is [IIT 06]

(A) Benzene chloro ketone (B) Benzoyl chloride(C) Chloro phenyl ketone (D) Benzene carbonyl chloride




222

35. The IUPAC name of the following compound CN

OH

Br is

[IIT 09](A) 4-Bromo-3-cyanophenol(B) 2-Bromo-5-hydroxybenzonitrile(C) 2-Cyano-4-hydroxybromobenzene(D) 6-Bromo-3-hydroxybenzonitrile

36. The IUPAC name of 3 2 3CH – CH = C – CH CH

2 2 3CH CH CHis: [CPMT 02]

(A) 3-Propylhex-2-ene (B) 3-Propylhex-3-ene(C) 4-Ethylhex-4-ene (D) 3-Ethylhex-2-ene

37. The IUPAC name of 3CH

is : [AIIMS 03]

(A) 3-Methyl cyclohexene (B) 1-Methyl cyclohex-2-ene(C) 6-Methyl cyclohexene (D) 1-Methyl cyclohex-5-ene

38. The IUPAC name of the compound 3 2 2CH – CH – CH – CH OH3OCH

is: [BHU 04](A) 2-Methoxy-1-butanol (B) 3-Methoxy-1-butanol(C) 1-Methoxy-1-butanol (D) 1,2-Methoxy butanol

39. The compound which contains all the four 1°, 2°, 3° and 4° carbonatom is: [CET (J & K) 05](A) 2,3-Dimethylpentane(B) 3-Ethyl-2,3-dimethylpentane(C) 2,3,4-Trimethylpentane(D) 3,3-Dimethylpentane


223



40. Which of the following compounds has isopropyl group?[BHU 05]

(A) 2,2,3,3-Tetramethylpentane(B) 2,2-Dimethylpentane(C) 2,2,3-Trimethylpentane (D) 2-Methylpentane

41. The IUPAC name of 3 3 2(CH ) C – CH = CH is: [AMU (Med.) 05](A) Hex-1-ene (B) 2,2-Dimethyl but-3-ene(C) 2,2-Dimethyl pent-4-ene (D) 3,3-Dimethyl but-1-ene

42. The compound Buta-1,2-diene has: [UGET (Med.) 06](A) only sp hybridized carbon atom(B) only sp2 hybridized carbon atom(C) both sp and sp2 hybridized carbn atoms(D) sp, sp2 and sp3 hybridized carbn atoms

43. The IUPAC name of O

Cl is : [AIPMT 06]

(A) 2-Ethyl-3-methyl butanoyl chloride(B) 2,3-Dimethyl pentanoyl chloride(C) 3,4-Dimethyl pentanoyl chloride(D) 1-Chloro-1-oxo-2,3-dimethyl pentane .

44. The IUPAC name of 2 2Cl – CH – CH = C – CH OH2 3CH CH

is: [DPMT 06]

(A) 1-Chloro-2-ethyl-4-hydroxybut-2-ene(B) 4-Hydroxy-1-chloro-2-ethylbut-2-ene(C) 4-Chloro-2-ethylbut-2-en-1-ol(D) 2-Ethyl-4-chlorobut-2-en-1-ol

45. The IUPAC name of the following compound is :

[UGET (Med.) 07](A) Bicyclo [2,2,0] octane (B) Bicyclo [0,2,2] hexane(C) Bicyclo [2,1,1] hexane (D) Bicyclo [2,2,0] hexane




224

46. The IUPAC name of3CH

3 2 2CH – CH – C – CH – CH OHO

is [AIIMS 07]

(A) 1-Hydroxy-4-methylpentan-3-one(B) 5-Hydroxy-2-methylpentan-3-one(C) 4-Methyl-3-oxopentan-1-ol(D) Hexan-1-ol-3-one

47. The correct IUPAC name of the compound CHO

2NO

3OCH

is :

[AFMC 07](A) 2-Formyl-5-methoxynitrobenzene(B) 4-Formyl-3-nitroanisole(C) 4-Methoxy-2-nitrobenzaldehyde(D) 4-Methoxy-6-nitrobenzaldehyde

48. The IUPAC name of the compound

2 3H C = CH – CH – CH – CH – CH

3CH 2 5C H OH is: [DPMT 07]

(A) 3-Ethyl-4-methylhex-5-en-2-ol(B) 3-Methyl-4-ethylhex-1-en-5-ol(C) 3-Ethyl-2-hydroxy-4-methylhex-5-ene(D) None of the above

49. How many sigma and pi bonds are there in tetracyano ethylenemolecule? [BHU (Mains) 08](A) 9 and 9 (B) 9 and 7

(C) 5 and 9 (D) 5 and 7

50. The IUPAC name of 3 2 3(CH ) CHCH is: [BHU (S) 08]

(A) Isopropyl methane (B) 2-Methyl propane(C) Trimethyl methane (D) Dimethyl ethane


225



51. The IUPAC name of 3 2 2CH – CH – CH – C = CH

2CH 3CH

3CH

is :

[Punjab PMET (Med.) 2008; CPMT 08](A) 2-Methyl-3-ethyl-1-pentene(B) 3-Ethyl-4-methyl-4-pentene(C) 3-Ethyl-2-methyl-1-pentene(D) 3-Methyl-2-ethyl-1-pentene

52. The IUPAC name of tertiary butyl iodide is: [CET 08](A) 1-Iodo-3-methyl propane (B) 2-Iodo-2-methyl propane(C) 4-Iodobutane (D) 2-Iodobutane

53. The correct structure of 4-Bromo-3-methyl but-1-ene is:

(A) 3 2BrCH = C(CH ) [AIIMS 08]

(B) 2 3 2 2H C = C(CH )CH CH Br

(C) 2 3 2H C = CH - CH(CH ) - CH Br

(D) 3 3 2CH - C(CH ) = CHCH Br

54. The IUPAC name of the compound C – N 3CHO

3CH is :

[JIPMER (Med.) 08](A) Cyclopropionamide(B) N-Methyl cyclopropanamide.(C) N,N-Dimethyl cyclopropane carboxamide(D) None of these

55. The correct IUPAC name of 2H C = CH - C CH is :[AIPMT 09](A) But-3-en-1-yne (B) But-1-en-3-yne(C) But-1-yn-3-ene (D) But-3-yn-1-ene




226

56. The IUPAC name of the following compound

3CH - CH = CH - C CH is: [AIPMT 10]

(A) Pent-4-yn-2-ene (B) Pent-2-en-4-yne(C) Pent-3-en-1-yne (D) Pent-1-yn-3-ene

57. The IUPAC nomenclature of 3 3 3 2(CH ) C - CH = C(CH ) is:

[AFMC 10](A) 2,4,4-Trimethyl pent-3-ene(B) 2,4,4-Trimethyl pent-2-ene(C) 2,2,4-Trimethyl pent-3-ene(D) 2,2,4-Trimethyl pent-2-ene

58. Which of the following is a correct name according to IUPACrules? [CET (J & K) 11](A) 2,3-Diethyl hexane (B) 3-Ethyl-2-methyl pentane(C) 3,4-Dimethyl pentane (D) 2-Ethyl-2-methyl pentane

59. The IUPAC name of O

OH is [CET (Karnataka) 11]

(A) Prop-2-enoic acid (B) But-1-enoic acid(C) But-3-enoic acid (D) Pent-4-enoic acid

60. The correct IUPAC name for the compound is :


(A) 3-(1-Ethylpropyl) hex-1-ene (B) 3-Ethyl-4-ethenyl heptane(C) 3-Ethyl-4-propyl hex-5-ene (D) 4-Ethyl-3-propyl hex-1-ene

61. The IUPAC name of 3H C

2 3CH CH

I

ClC = C is: [AIPMT 11]

(A) 2-Chloro-3-iodopent-2-ene(B) 2-Chloro-3-iodopent-2-ene

(C) 3-Iodo-4-chloropent-3-ene (D) 3-Iodo-4-chloropent-3-ene


227



62. The IUPAC name of OH O

is : [CET (Karnataka) 12]

(A) 2-Hydroxy-4-pentanone (B) 4-Hydroxy-2-pentanone(C) 2-Oxo-4-pentanol (D) 4-keto-2-pentanol

63. The IUPAC name of the compound

3 2 2 2 3H C – CH – CH – CH – CH – CH CH

3CH2 3CH CH is: [PMT (Kerala) 12]

(A) 3-Ethyl-5-methyl heptane(B) 5-Ethyl-3-methyl heptane(C) 3,5-Diethylhexane(D) 1,1-Diethyl-3-methylpentane

64. Which nomenclature is not according to IUPAC system?

(A) 2 2Br – CH – CH = CH1-Bromoprop-2-ene


(B)3CH

3 2 2 3CH – CH – C – CH – CH – CH

4-Bromo-2,4-dimethylhexaneBr 3CH

(C) 3 2 3CH – CH – CH – CH – CH

2-Methyl-3-phenylpentane3CH

(D) 3 2 2 2CH – C – CH – CH – CH COOH

5-Oxo hexanoic acidO

65. The structure of isobutyl group in compound is: [NEET 13]

(A) 3 2 2 2CH - CH - CH - CH - (B) 3 2 3CH – CH – CH – CH

(C) 2CH – CH –3CH3CH

(D)3CH

3CH – C –

3CH




228

Multiple Correct Questions (MCQ) :66. Correct IUPAC names as per rules are

(A) 1,1,1-Trichloro-2,2-diphenylethane(B) 3-Cyclopentylbut-1-ene(C) Cyclopentane-2,4-diene(D) 1-Bromo-2,3-dichlorocyclopropane

67. In which of the given molecules cyclic part is/are parent chain:

(A)

Br

Br

(B)Br

(C) (D)

68. Which of the following is/are incorrect IUPAC name:–

(A)3 3CH –C–CH–CH

O 3CH 2-Methylbutan-3-one

(B)2HC C–CH–CH=CH

2HC=CH 3-Ethenylpent-1-en-4-yne

(C)

OH3CH

2 2 2CH CH NH

3-(2-Aminoethyl)-2-methylcyclohexan-1-ol

(D)3CH –CH–C–CH–OH

O 3CH3CH 4-Methyl-3-oxopentan-2-ol


229



69. Which is/are the correct IUPAC name of the correspondingstructures

(A)

OH3CH

2 3NHCH CH

3-(N-Ethylamino)-2-methylcyclohexan-1-ol

(B) 3CH –CH–CH–CHO

ClBr3-Bromo-2-chlorobutanal

(C) 2 2CH –CH–CH –COCl

3COOCH

CNMethyl 4-chlorocarbonyl

-3-cyanobutanoate

(D) CHO Cyclohexanal70. Which of the following IUPAC name are correct.

(A) 2C–NH

O2-Methylcyclo

pentanecarboxamide

(B)COCl

Cyclohexanoyl chloride

(C)3CH

C N2-Methylcyclo

butanecarbonitrite

(D)Br

3COOCH Methyl 2-bromocyclohexanecarboxylate

71. Which indicated alkyl group is/are correct.

(A) Iso-propyl (B) Sec-butyl

(C) Iso-butyl (D) neo-pentyl




230

72. Which of the following is/are correct IUPAC name.

(A)O

2NH

Cl

2-Chlorocyclohexanecarboxamide

(B)3CH O

3OCH Methyl-2-methylcyclohexane carboxylate

(C)OCl

Cl 2-chloro cyclohexanecarbonyl chloride

(D) HO O

2-Oxocyclohexane carbaldehyde

73. The correct combination of names for isomeric alcohols withmolecular formula 4 10C H O is/are [IIT Advanced-2014]

(A) tert-butanol and 2-methylpropan-2-ol(B) tert-butanol and 1,1-dimethylethan-1-ol(C) n-butanol and butan-1-ol(D) isobutyl alcohol and 2-methylpropan-1-ol

74. Which of the following compounds are named correctly ?

(A) 3 2 2 2 2(CH ) CHCH CH CH CHO (5-methyl-1-hexanal)

(B) 3 2 2(CH ) CHCH C C–COOH (5-methyl-2-hexynoic acid)

(C) 3 2 2 2 3CH CH CH CH CH(CH )COOH (2-methylhexanoic acid)

(D) 3 2 3CH CH CH=CH–COCH (3-hexen-5-one)


231



Assertion / Reason type Questions (A/R) :Each question has 5 choice (A), (B), (C), (D) & (E) out of whichonly one is correct.(A) Statement-1 is true, Statement-2 is true and statement-2 is


not correct explanation for statement 1.(C) Statement-1 is true and Statement-2 is false.(D) Statement-1 is false, Statement-2 is true(E) Both Statement-1 and Statement-2 is false.

75. Statement-1:The locant (2,6,7) is preferred over the locant (3,4,8).Statement-2:The first term, i.e. 2 in the first set is lower than the

first term, i.e. 3 in the second set.

76. Statement-1:The IUPAC name of 3CH CH=CH–C CH is pent-2-en-4-yne and not pent-3-en-1-yne.

Statement-2:While deciding the locants for double bond andtriple bonds, lowest locant rule for multiple bondsis followed.

77. Statement-1:The correct IUPAC name of the compound,

CH3CH

3CH is (1-methylethyl) cyclopentane.

Statement-2:It is named as a derivative of cyclopentane becausethe number of carbon atoms in the ring is more thanin the side chain.

78. Statement-1:H

CN is called cyclohexane nitrile.

Statement-2:It contains six carbon atoms in the ring and –CN asthe functional group.




232

79. Statement-1:The IUPAC name for the compound

2 5 2C H –C–CH OH

2CH is 2-ethyl prop-2-en-1-ol.

Statement-2:Etlhyl (C2H5-) rather than methylene (=CH2) isconsidered as the substituent group because ‘e’ ofethyl comes first in alphabetical order than ‘m’ ofmethylene.

80. Statement-1:The IUPAC name of citric acid is 2-hydroxypropane 1,2,3,-tricarboxylic acid

COOHHOOCCOOH

OHStatement-2:When an unbranched carbon chain is directly

linked to more then two like functional groups, thenit is named as derivative of parent alkane whichdoes not include the C-atoms of the functionalgroups.

81. Statement-1:The IUPAC name of the compound,

2 2OHC–CH –CH –COOH is butane-3-formyl-1-oicacid.

Statement-2:COOH is considered as substituent group whileCHO is considered as the principal functionalgroup.

82. Statement-1:The IUPAC name for the compound

6 5 2 2C H COOCH CH COOH i s 3 - b e n z o y l o x ypropanoic acid.

Statement-2: 6 5 2C H CH O is called benzoyloxy group.

83. Statement-1: O is commonly called as acetone.

Statement-2:Common name is derived from 5 components i.e.Secondary prefix + Primary prefix + Word root +Primary Suffix + Secondary Suffix.


233



Match the Column type Questions (MTC) :84. Match the compounds written in Column-I with their name in

Column-II :-Column - I Column - II

(A)COOH

CHO (P) 2-(2-Oxoethyl) cyclohexanecarboxylic acid

(B)COOH

CHO (Q) 2-Aminomethyl cyclohexanecarboxylic acid

(C)2NH

COOH(R) 2-N-Methylaminocyclohexane

carboxylic acid

(D) 3CH

COOHNH

(S) 2-Formyl cyclohexanecarboxylic acid

85. Structural formula of carboxylic acid IUPAC nameColumn - I Column - II

(A) Cl COOH

Br

(P) 2 – aminbenzoic acid

(B)

Ph

3CH –CH–COOH (Q) 2 – Methyl - 2- phenyl- 3-buten- 1- oic acid

(C)COOH

NH2

(R)2,2,3-trifluoro

propanoic acid

(D) 2 2CH F - CF - COOH (S)2-Bromo-4-chlorocy

clohexanecarboxylic acid




234

Comprehension type Questions :Comprehension 01 :-A saturated hydrocarbon (A) has five membered ring. Threealkyl groups attached to the ring adjacent to each other withfollowing observations.(A) First group has only two carbon atoms.(B) Second group has four carbon atom with all hydrogens are

chemically same(C) Third group has total five carbon atoms and its main chain

contains three carbon atoms with ethyl as a substituents.86. The IUPAC name of compound (A) is

(A) 2 - ( 1 , 1 - D i me t h y le t h y l) - 1 - ( 1 - e t h y lp r o p yl ) - 3 -ethylcyclopentane

(B) 2- (1 ,1 -Dimeth yleth yl)- 1-e th yl-3- (1 -e th ylp ropyl)cyclopentane

(C) 1-(1 ,1-Dimet hylethyl)-3-ethyl-2-(1-methylpropyl)cyclopentane

(D) 1-(1,1-Dimethylethyl)-2-(1-methylethyl)-3-(1-methylpropyl)cyclopentane

87. How many 1° hydrogens are present in compound (A).(A) 15 (B) 18(C) 12 (D) 9

88. How many 3° carbons are present in compound (A).(A) 3 (B) 2(C) 4 (D) 5Comprehension 02 :-During nomenclature of unsaturated hydrocarbon, lowest locantis given to multiple bond rather than any substituent, as thepriority of multiple bond is more than substituent. When thelocant for multiple bond is same from either side than doublebond, then double bond is given priority over triple bond. For


235



selection of main chain if the size of main chain & number ofmultiple bonds are equal then the priority is given to lowest setof locants for multiple bond.

89. IUPAC name of following compound is

(A) 4-Ethenylhepta-1,3-dien-6-yne(B) 4-(prop-2-ynyl) hexa-1,3,5-triene(C) 3-(prop-2-ynly) hexa-1,3,5-triene(D) 4-Ethenylhexa-4,6-dien-1-yne.

90. IUPAC name of following compound

(A) 3-Ethenylpenta-1-en-4-yne(B) 3-Ethynylpenta-1,4-diene(C) 2,2-Diethenylpropyne (D) 1,1-DiethenylpropyneComprehension 03 :-The priorities for citing principal groups in a carboxylic acidderivative are as follows :

Carboxylic acid > Sulphonic acid > ester > acid halide >amide > cyanideAll of these groups have citation priority over aldehydes andketons as well as the other functional groups.When these functional groups act as a substituent then following2° prefix is used for them.

Group 2° prefix—COOH carboxy

2 5

O– C–OC H ethoxy carbonyl

3

O–O– C–CH

ethanoyl oxy




236

2

O– C–NH carbamoyl

3

O– C–OCH methoxy carbonyl

2

O–CH – C –OH

carboxy methyl

O

– C–Cl chlorocarbonyl

91. Which of the following structure has correct loctans?

(A) 2 3CO CHCl–C

O

CN

6

123

4

5COOH

(B) 2 3CO CH

2CO H

Cl–C

O

CN

12

345

6

(C) 2 3CO CH

2CO H

Cl–C

O

CN

16

543

2

(D) 2 3CO CH

2CO H

Cl–C

O

CN

54

321

6

92. Which of the following is the Correct IUPAC name of givencompound is compound?

3C–NH–CH

OCl

(A) p-acetamido chlorocyclohexane(B) 4-chloro-N-methylcyclohexane carboxamide(C) N-methyl amido chlorocyclohexane(D) N-Methyl-4-chlorocyclohexane carboxamide


237



93. For the given compound which statements is correct about theabove compound ?

3CH –C–NH

O

2CO H

(A)3CH –C–NH–

O group is principal group

(B) 3CH –C–NH–

O

will be considered as substituent group

(C) Locant number 1 will be assigned to —COOH(D) p-acetamidobenzoic acid is its IUPAC nameComprehension 4 :-In the IUPAC nomenclature, when a substituent is complex ,then prefixes like di, tri, tetra etc are replaced by bis, tris, tetrakis- etc. Bused on this information answer the following questions :

94. The structural formula of 1,3 – Bis (bromo methyl) benzene

(A)

2CHBr

3CH(B)

2CHBr

3CH

(C)

2Br–CH

3CH(D)

2CH –Br

2CH –Br

95. The IUPAC name of 2 4C CH OH is(A) 1,2,3,4 – Tetra hydroxy butane(B) tris (hydroxy ethyl) ethane(C) tetrakis (hydroxy methyl) methane

(D) 1,1,1,1 – Tetrahydroxy methyl alcohol




238

96. Structural formula of tris – (2–chloroethyl) amine is

(a) 2 2 3N CH - CH - Cl (b)3 3N(CH–CH )

Cl

(c) 3 2 2CH - CH - N - Cl (d) None of these

Integer type Questions :97. How many of the functional groups have high priority than

aldehyde in seniority order when naming polyfunctional com-pounds.

Carboxylic acid(I)

Sulphonic acid(II)

Ester(III)

Acid halide(IV)

Amide(V)

Nitrile(VI)

Ketone(VII)

Alcohol(VIII)

Amine(IX)

98. How many them are incorrectly named according to IUPACname.

OH

1,1-Dimethylethanol

I

AcetoneII

O

2-EthylbutaneIII

Et

COOH

Pentan-2-oic acid(IV)

CN

Pentan-2-carbonitrile(V)

CHO

Pentan-2-al(VI)

1,6-Dimethylcyclohexene

(VII)

CNPropane-1,2,3-tricarbonitrile

(VIII)

NC CN

Propan-1,2,3-triol(IX)

HO OHOH


239



99. How many of them have correct common name.

Iso-octane(I)

COOH

COOHMaleic acid

(II)

COOH

COOHSuccinic acid

(III)

Carbolic acid(IV)

OH

Aspirin(V)

3OCOCHCOOH

Syrene(VI)

Cumene(VII)

Cinnamic acid(VIII)

CH=CH–COOH

OHCOOH

Salicylic acid(IX)

100. How many groups among the following are correctly named.

n-butyl(I)

Sec-butyl(II)

Iso-butyl(III)

tert-butyl(IV)

Phenyl(V)

n-propyl(VI)

Iso-propyl(VII)

neo-pentyl(VIII)

Iso-hexyl(IX)




240

AnswersObjective Approach

0201. (B) 02. (C) 03. (B) 04. (B) 05. (A)06. (A) 07. (D) 08. (A) 09. (C) 10. (D)11. (A) 12. (C) 13. (B) 14. (C) 15. (A)16. (B) 17. (A) 18. (B) 19. (A) 20. (B)21. (B) 22. (A) 23. (C) 24. (A) 25. (D)26. (B) 27. (B) 28. (C) 29. (A) 30. (A)31. (A) 32. (A) 33. (A) 34. (D) 35. (B)36. (D) 37. (A) 38. (A) 39. (B) 40. (D)41. (D) 42. (D) 43. (B) 44. (C) 45. (D)46. (B) 47. (C) 48. (A) 49. (A) 50. (B)51. (C) 52. (B) 53. (C) 54. (C) 55. (B)56. (C) 57. (B) 58. (B) 59. (C) 60. (D)61. (A) 62. (B) 63. (A) 64. (A) 65. (C)66. (A,B,D) 67. (A,B,D) 68. (A,B,D) 69. (A,B,C,D) 70. (A,C,D)71. (A,B,C,D) 72. (A,B,C,D)73. (A,C,D) 74. (A,B,C) 75. (A)76. (D) 77. (A) 78. (D) 79. (C) 80. (A)81. (E) 82. (B) 83. (C) 84. (A–S ; B–P ; C–Q ;D–R) 85. (A–S ; B–Q ; C–P ; D–R) 86. (B)87. (B) 88. (C) 89. (A) 90. (B) 91. (A)92. (B) 93. (B) 94. (D) 95. (C) 96. (A)97. (6) 98. (A) 99. (4) 100. (6) 101. (12)102. (6)


241



01Test

Basic understanding of organicchemistry & nomenclature

SECTION-ISingle Correct Questions (SCQ) :

1. Functional group which is absent in Penicillin

R – C – N S

OH

O

COOHPenicillin

N is

(A) 2° amide (B) 3° amine

(C) Carboxylic acid (D) 3° amide

2. Which of the following is not the homolog of 3 2CH CH COOH.

(A) 3CH – COOH (B) 3CH – CH – COOH

3CH

(C) 3 2 3CH COOCH CH (D) HCOOH




242

3. Open chain molecule with lowest molecular mass that contains2 sp2 hybridized carbon and 2 sp3 hybridized carbon have totalcarbon(A) 2 (B) 3(C) 4 (D) 5

4. Total number of primary, secondary, tertiary & quaternarycarbon in given molecule respectively is

(A) 6, 0, 3, 1 (B) 7, 0, 3, 1(C) 7, 1, 3, 2 (D) 7, 1, 3, 1

5. Total number of primary, secondary & tertiary hydrogensrespectively present in given compound is

O(A) 9, 6, 1 (B) 9, 4, 1(C) 11, 4, 1 (D) 11, 6, 1

6. Molecular formula of anthracene is

(A) 14 14C H (B) 14 10C H

(C) 14 12C H (D) 12 12C H

7. Total number of sp2–sp2 bond present in given molecule

(A) 2 (B) 3(C) 4 (D) 5

8. In which of the following compound IUPAC numbering iscorrect–

(A)3

265 4

1

(B)3 4

5612


243



(C) 32

1654 (D) 3

216

5 4

9. The correct IUPAC name of compound 3SO HO

(A) 6-Propyl-1-methyl-4-oxohept-6-ene-1-sulphonic acid.

(B) 5-Oxo-7-propyloct-7-ene-2-sulphonic acid.

(C) 2-Propyl-7-sulphooct-1-ene-4-one.

(D) 7-Methylene-5-oxodecane-2-sulphonic acid.

10. The correct structure of 6-Chloro-4-hydroxycyclohex-2-ene-1-sulphonic acid is

(A)3SO H

OH

Cl (B)3SO H

ClHO

(C)

3SO H

ClOH

(D)Cl

HO

3SO H

SECTION-IIMultiple Correct Questions (MCQ) :

11. Functional group is present in vitamin C

O

HO

O

OHVitamin C

HOOH

is:

(A) Alcohol (B) Enol(C) Ester (D) Ether




244

12. Which molecules will exist at room temperature ?

(A)O

(B)O

(C)O

(D)

13. Which molecule have ( ) gamma position ?

(A) N+ (B) N+

(C) N+ (D) S+

14. The pair of compounds having the same general formula.

(A) and (B) and C=C=CH

H

H

H

(C) and HC C–C CH (D) and

15. The correct structure/s of 2,3,4-Trimethylhexane is/are

(A) (B)

(C) (D)

16. Which of the following pair/s have same IUPAC naming

(A)COOH

COOCH3

Br&

OCOCH3

BrCOOH


245



(B)F

BrI

ClI

BrCl

I&

I

F

(C) &

(D)2NH

OH

Cl

F& OHCl

2H NF

17. Which of the following IUPAC names are incorrect(A) 3-Ethyl-5-methylhepta-1,6diene(B) 3-Methyl-5-ethylhepta-1,6-diene(C) 3,4-Dimethylpentane(D) 1-Bromo-3-chloro-4-fluorobut-2-ene

SECTION-IIIAssertion / Reason type Questions (A/R) :This section has 5 questions. Each question has 5 choice (A),(B), (C), (D) & (E) out of which ONLY ONE is correct.(A) Statement-1 is true, Statement-2 is true and statement-2 is


not correct explanation for statement 1.(C) Statement-1 is true and Statement-2 is false.(D) Statement-1 is false, Statement-2 is true.(E) Both Statement-1 and Statement-2 is false.

18. Statement-1: Sec-butyl amine is a secondary amine.

Statement-2: In secondary amine, 2NH group attached withsecondary carbon.




246

19. Statement-1: Oxalic acid & formic acid are homologue.Statement-2: Homologues are those compounds which have

same nature of chemical reaction and differ in2 n(CH ) unit.

20. Statement-1: Loss of one hydrogen from alkane leads to alkylgroup.

Statement-2: Butane have four types of alkyl groups.

21. Statement-1: O is commonly called as acetone.

Statement-2: Common name is derived from 5 components i.e.Secondary prefix + Primary prefix + Word root +Primary Suffix + Secondary Suffix.

22. Statement-1: Cyclohexane is saturated while cyclohexene isunsaturated.

Statement-2: Saturated compound is that which have onlysingle bond while unsaturated compound is thatwhich have multiple bond of any form.

SECTION-IVMatch the Column type Questions (MTC) :

23. Match the compounds written in Column-I with their name inColumn-II :-

Column - I Column - II

(A)COOH

CHO (P) 2-(2-Oxoethyl) cyclohexanecarboxylic acid

(B)COOH

CHO (Q) 2-Aminomethyl cyclohexanecarboxylic acid

(C)2NH

COOH(R) 2-N-Methylaminocyclohexane

carboxylic acid


247



(D) 3CH

COOHNH

(S) 2-Formyl cyclohexanecarboxylic acid

24. Match the compounds written in Column-I with its nature inColumn-II :-


(A) (P) Aliphatic

(B) (Q) Aromatic(C) (R) Alicyclic

(D)O

(S) Heterocyclic

(T) Homocyclic

SECTION-VComprehension type Questions :

Paragraph for Question Nos. 25-26A saturated hydrocarbon (A) has five membered ring. Three alkyl groups attached to the

ring adjacent to each other with following observations.(A) First group has only two carbon atoms.(B) Second group has four carbon atom with all hydrogens are

chemically same(C) Third group has total five carbon atoms and its main chain

contains three carbon atoms with ethyl as a substituents.25. The IUPAC name of compound (A) is

(A) 2 - ( 1 , 1 - D i me t h y le t h y l) - 1 - ( 1 - e t h y lp r o p yl ) - 3 -ethylcyclopentane

(B) 2- (1 ,1 -Dimeth yleth yl)- 1-e th yl-3- (1 -e th ylp ropyl)cyclopentane




248

(C) 1-(1 ,1-Dimet hylethyl)-3-ethyl-2-(1-methylpropyl)cyclopentane

(D) 1-(1,1-Dimethylethyl)-2-(1-methylethyl)-3-(1-methylpropyl)cyclopentane

26. How many 1° hydrogens are present in compound (A).(A) 15 (B) 18(C) 12 (D) 9

Paragraph for Question Nos. 27-28During nomenclature of unsaturated hydrocarbon, lowest locantis given to multiple bond rather than any substituent, as thepriority of multiple bond is more than substituent. When thelocant for multiple bond is same from either side than doublebond, then double bond is given priority over triple bond. Forselection of main chain if the size of main chain & number ofmultiple bonds are equal then the priority is given to lowest setof locants for multiple bond.

27. IUPAC name of following compound is

(A) 4-Ethenylhepta-1,3-dien-6-yne(B) 4-(prop-2-ynyl) hexa-1,3,5-triene(C) 3-(prop-2-ynly) hexa-1,3,5-triene(D) 4-Ethenylhexa-4,6-dien-1-yne.

28. IUPAC name of following compound

(A) 3-Ethenylpenta-1-en-4-yne(B) 3-Ethynylpenta-1,4-diene(C) 2,2-Diethenylpropyne (D) 1,1-Diethenylpropyne


249



SECTION-VIInteger type Questions :

29. Total number of homocyclic compounds among the givenmolecule is

O

NNNNNNNNNNNNNNNNNNNNNNN S O

OH

OHHO

O O

30. Total number of secondary hydrogens present in is




250

02Test



01. Functional group absent in ‘Heroin’ is

(Heroin)

O

3H C–C–O 3O–C–CH

O

O

3CH

N

(A) Ester (B) Ether(C) Tertiary amine (D) Tertiary amide

02. The number of sp sp sigma bonds in ‘Capillin’ is

(Capillin)3CH –C C–C C–C

O

(A) 1 (B) 2(C) 3 (D) 4


251



03. The total number of lone pair of electrons in ‘Dioxane’ is O

O

(A) 1 (B) 2(C) 3 (D) 4

04. Which of the following lactum have highest angle strain.

(A)

O

N–Hα-Lactum

(B)N–H

O

β-Lactum

(C)N–H

O

γ-Lactum(D)

O

N–H

δ-Lactum

05. Total number of secondary carbon present in ‘Decalin’ is

(Decalin)

(A) 6 (B) 8(C) 10 (D) 4

06. Total number of tertiary hydrogen present in ‘Steroid’

R

is

(A) 7 (B) 6(C) 5 (C) 4




252

07. Which among the following is heterocyclic compound.

(A) OH

Menthol(B)

NH

Indole

(C)

OH

Phenol(D)

OH

O-Cresol

3CH

08. Which among the following is aromatic in nature.

(A) (B)

(C) (D)

09. Which among the following is secondary alcohol.

(A) OH (B)OH

(C) OH (D) OH

10. Which among the following will not exist at room temperature.

(A) (B)

(C) (D)


11. Which of the following is/are the homologue of Ethanoic acid

3(CH COOH) .

(A) H–C–OH

O

(Formic acid)(B)

COOH

(Butanoic acid)


253



(C)

COOH

COOH(Succinic acid)

(D)COOH

(Propanoic acid)

12. Which among the following compound have atleast one 2spcarbon present.

(A)HO

NH

Adrenaline

HOOH

3CH (B)

O

HOEstrone

(C)Cubane

(D)Cumene

13. Which among the following have structure & molecular formulacorrectly matched.

(A) 6 6(C H ) (B) 10 8(C H )

(C) 14 10(C H ) (D) 14 10(C H )

14. Which among the following have correct number of substitu-ents present(s).

(A)

OH

O

OH

Cl(3)

(B)

O

(2) Cl

(C)

COOH

(3)(D)

O

Cl

(2)




254

15. Which among the following structure & IUPAC name are cor-rectly matched.

(A)Et

2-Ethylbutane

(B)O

OH

OH 3- Hydroxybutanoic acid

(C)O

6 - Methylcyclohex-2-enone

(D)C–ClO

Benzene carbonyl chloride

16. Which among the following structure & their common nameare correctly matched.

(A)O

(Acetone)(B)

2NO2O N

2NO

OH

(Picric acid)

(C)

3CH

(Toluene)(D) H–C–H

O

(Formaldehyde)17. Which among the following alkyl groups & their name are cor-

rectly matched.

(A) 3 2 2 2CH –CH –CH –CH –(n-butyl) (B)

3 2 3CH –CH–CH –CH

(sec-butyl)

(C) 3 2CH –CH–CH –3CH

(iso-butyl)(D)

3CH –C–3CH

3CH(tert-butyl)


255






18. Statement – 1 : Hydroquinone

OH

OH is alicyclic compound.

Statement – 2 : Alicyclic compounds are simply the aliphaticcyclic compounds.

19. Statement – 1 : Azulene, is aromatic in nature.

Statement – 2 : Aromatic compounds are cyclic compoundswith each carbon 2sp or sp hybridized, mol-ecule as a whole planner with total

electrons are 4n + 2 in number.

20. Statement – 1 : General formula of & aresame.

Statement – 2 : General formula of any molecule is derived byremoving hydrogen from alkane which havegeneral formula n 2n 2C H .

21. Statement – 1: Tert-butyl amine is tertiary amine.Statement – 2 : Tertiary amine is that amine in which 2NH

group is attached to tertiary carbon.




256

22. Statement – 1 : Organic molecule with molecular formula2 7C H will not exist at room temperature.

Statement – 2 : Carbon have valency four, so compound with 2carbon can maximum accumulate 6 hydrogensin it.


23. Match the compound given in Column – I with their IUPACname in Column – II.

Column – I Column – II

A. 2NH

O

P. 2-N-Methylaminocyclohexanone

B.3CH

ONH

Q. 2-Aminomethylcyclohexanone

C.COOH

CHO R. 2-Oxocyclohexanecarboxylic acid

D.COOH

O S. 2- Formylcyclohexanecarboxylic acid

24. Match the compound given in Column – I with their character-istics in column – II.

Column – I Column – II

A. P. Aromatic

B.O

Q. Alicyclic

C. R. Anti-aromatic


257



D. S. Homocyclic

T. Heterocyclic


Paragraph for Question Nos. 25-26

Formaldehyde 2CH O is known to all biologists because of itsusefulness as a tissue preservative. When pure, formaldehydetrimerises to give trioxane, 3 6 3C H O , which surprisingly enough,has no aldehydic or ketonic group. On monobromination reac-tion trioxane give only one monobromo derivative 3 5 3C H BrO .

25. The structure of formaldehyde is

(A) H–O–H (B) H–C–H

O

(C) C=OH

H(D) H–C–O–H

26. The structure of Trioxane is

(A)3 2CH –CH –C–H

O(B)

3 3CH –C–CH

O

(C) OOO

(D) OOO

Paragraph for Question Nos. 27-28Testosterone is one of the most important male steroid hormones.When testosterone is dehydrated by treatment with acid, rear-rangement occurs to yield the product shown.

OH

OTestosterone

O

2H O / H+




258

27. The degree of unsaturation (DU) present in Testosterone is(A) 5 (B) 6 (C) 7 (D) 8

28. Functional group present in product of dehydration ofTestoterone is

(A) Ester (B) Ketone (C) Ether (D)Acid

anhydrideSECTION-VI

Integer type Questions :29. Total number of aromatic compounds present in given molecule

are

I II III

IVOV VI

VII VIII IX30. Total number of heterocyclic compounds present in given mol-

ecules are

OI

NHII

NIII

OH

2NHIVN

V

NH

VI

COOH3O–C–CH

O

VIIHO

NH

OVIII

IXS


259



03Test



1. Incorrect statement for given structures is

I

II

III(A) I, II & III have same empirical formula(B) I, II & III have same general formula(C) I, II & III have same molecule formula(D) I, II & III are homologue

2. Which statement among these is incorrect.(A) CH2=CH2 has sp2 hybridization for each carbon with bond

angle 120°.(B) HC CH has sp hybridization for each carbon with bond

angle 180°.(C) H2C=C=CH2 has central carbon sp hybridized with bond

angle 180°.(D) HC C–C CH has sp hybridization for only two carbons

with bond angle 180°.




260

3. All the members of a homologus series have same(A) Functional groups (B) Molecular mass(C) Weight (D) Physical properties

4. Number of types of carbon & hydrogen atoms present in givencompound is

(A) 1, 2 (B) 2, 2(C) 2, 1 (D) 1, 1

5. Which among the following is not alicyclic compound.

(A)O

(B)

(C) (D)

6. Which of the following alkane have all the four types of Cpresent.

(A) 3 3(CH ) CH (B) 2 5 3(C H ) CH

(C) 3 3 2 3 2(CH ) CCH CH(CH ) (D) 3 4(CH ) C

7. IUPAC name of sec-butyl group is(A) 2-Methylethyl (B) 1,1-Dimethylethyl(C) 2-Methylpropyl (D) 1-Methylpropyl

8. IUPAC name of compound

Cl

Br is

(A) 3-Bromo-7-chloro-7-ethyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-3-methylnonane

(B) 3-Bromo-7-chloro-5-(1,1-dimethylethyl)-7-ethyl-3-methyl-5-(2-methylpropyl)nonane


261



(C) 3-Bromo-7-chloro-7-ethyl-3-methyl-5-(1,1-dimethylethyl)-5-(2-methylpropyl)nonane

(D) 3-Bromo-5-(1,1-dimethylethyl)-5-(2-methylethyl)-7-chloro-7-ethyl-3-methylnonane

9. IUPAC name of the given structure is

(A) 3-Ethynyl-4-ethenyl-7-(2-methylbutyl)-8-methylnona-1,7-diene

(B) 4-Ethenyl-3-ethynyl-5,8-dimethyl-7-propylnona-1,7-diene(C) 3,4-Diethenyl-5,8-dimethyl-7-propylnona-7-ene-1-yne(D) 4-Ethenyl-3-ethynyl-7-(2-methylbutyl)-8-methylnona-1,7-

diene

10. The IUPAC name of the compound 3CHH–C–N

O

2 3CH –CH is

(A) N-Ethyl-N-methylmethanamide(B) N-Methyl-N-ethylmethanamide(C) N-Ethyl-N-methylformamide(D) N-Ethylmethylmethanamide


11. Neohexane contains(A) Four 1° carbon atoms and two 2° hydrogen(B) Twelve 1° hydrogen but no 3° carbon(C) Two 2° carbon & one 4° carbon(D) One 4° carbon and two 3° hydrogen

12. The pair of compounds having the same general formula.

(A) and (B) and C C CH

H

H

H

(C) and HC C C CH (D) and




262

13. The correct structure/s of 2,3,4-Trimethylhexane is/are

(A) (B)

(C) (D)

14. Correct structure of 5-Ethynyl-1,3,6-heptatriene

(A) 2H C2CH

CH(B)

2H C 2CH

2CH

(C) 2H C2CH

CH(D) 3H C

2CHCH

15. Structure of 2-Ethenyl-3-methylcyclohexa-1,3-diene is/are

(A) (B)

(C) (D)

16. Which of the following is correct IUPAC name:

(A) 2 2 2H N–C–CH –C–NHOO

Propanediamide

(B) 3C–NHCH

O

N-Methyl cyclohexanecarboxamide

(C) 3 2 2 3H CO–C–CH –CH –O–C–CHOO

Methyl 3-ethanoyloxy

propanoate

(D) 3C–OCH

OBrMethyl 5-bromocyclohex-2-enecarboxylate


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17. Which of the following is correct structure of 4-(4-Methoxyphenyl)-4-oxo-butanoic acid.

(A) O

O OMe

HO (B) OO

OMeHO

(C)

OMe

COOH

O

(D)O

O

HO

OMe




18. Statement-1: Neohydrocarbons contain a tertiary carbon atom.Statement-2: Whenever a carbon atom is bonded to three carbon

atoms it is tertiary.

19. Statement-1:

O

Br

3CHClC

2NH 3 - Br om o- 2 - c h l or o- 3 -

methylpentanamide is incorrect IUPAC name.Statement-2: In case of chain terminating senior most functional

group numbering start from itself.




264

20. Statement-1: Esters are formed by reaction of carboxylic acid &alcohol.

Statement-2: Amides are formed by reaction of carboxylic acid& amines.

21. Statement-1: –OH group attached to carbon leads to formationof many types of function group such as alcohol,phenol, oxime, enol & carboxylic acid.

Statement-2: –X group attached to carbon leads to formation ofmany types of functional group such as alkylhalide, vinyl halide, Allyl halide, Aryl halide &acid halide.

22. Statement-1: Each carbon atom of cyclooctatetraene (COT) issp2 hybridized but molecule is non-planner.

Statement-2: Cyclooctatetraene (COT) is tub-shaped.


23. Match the compound written in Column-I with its nature writtenin Column-II :-


(A) 3H C

O

OCocaine

MeOOCN

(P) Carbocyclic

(B)3CH

O 3CH


(Q) Heterocyclic

(C)3CHO

COOH

O

Aspirin

(R) Aromatic


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(D)N

2NH

HNN

NAdenine

(S) Saturated

(T) Unsaturated24. Match the compounds written in Column-I with their name in

Column-II :-Column - I Column - II

(A)3C–O–CH–CH

3CH

O

OH

(P)Methyl

cyclopentylethanoate

(B)2 3CH –COOCH

(Q)Isopropyl 3-hydroxy

cyclohexaneocarboxylate

(C)COOH

O

OHC(R)

3-Carbamoylcyclobutanecarboxylic acid

(D) 2NH

HOOC

O

(S)4-Formyl-2-oxocyclohexane

carboxylic acid


Paragraph for Question Nos. 25-26Have you ever thought that when you read anything from “anybook” your eyes use an organic compound (retinal) to convertvisible light into nerve impulses. When you pick up this bookfrom any place, your muscles do chemical reactions on sugarsto give you energy. As you read the words and sentences of thisbook, gaps between your brain cells are being bridged by simpleorganic molecules (neurotransmitter amines) so that nerveimpulse can be passed around your brain and you can




266

understand all matters correctly and you did all that withoutconsciously thinking about it.

3CH3H C

O3CH

11-Cis retinal3H C

H

3CH

(Absorbs light when we see)25. Deegree of unsaturation (DU) present in 11-cis-retinal is

(A) 6 (B) 7(C) 8 (D) 5

26. Which statement is incorrect for cis-retinal(A) cis-retinal is an unsaturated compound.(B) cis-retinal is a homocyclic compound.(C) cis-retinal have aldehydic group.(D) cis-retinal is isolated polyenes.

Paragraph for Question Nos. 27-28A hydrocarbon (A) has six membered ring in which there is nounsaturation. Two alkyl groups are attached to the ring adjacentto each other. One group has 2 carbon atoms and another has 4carbon atoms. The larger alkyl group has main chain of threecarbon atom of which second carbon is substituted.

27. How many primary secondary & tertiary carbons are present inthis compound respectively.(A) 2, 4, 6 (B) 3, 5, 4(C) 3, 6, 3 (D) 4, 6, 2

28. What is the correct IUPAC name of this compound.(A) 1-(2-Methylpropyl)-2-ethylcyclohexane(B) 1-Ethyl-2-butylcyclohexane(C) 1-Ethyl-2-(2-methylpropyl)cyclohexane(D) 2-Ethyl-1-isobutylcyclohexane


267



SECTION-VIInteger type Questions :

29. How many of the compounds among these are unsaturated innature.

3H C 3CH

SH

SHCompound with

worst smellI

3H C3CH

SH

Compound withworst smell

II

O

3CHO

OOlean sex pheromone

of olive fly(III)

Compound fromcakes & biscuits

(IV)

O

O

HO

N

O

3CHH

Paracetamole(V)

N3CHN

Nicotine(VI)

N 3CH

3CHS

O

HNH

R

OCOOH

Penicilin(VII)

ONH

HNO

Indigo dye(VIII)

O

SO

2NH

ClO

NH

OOH

Furosemide (Sulpha drug)(IX)

30. How many of the compound among these have all atoms sp2

hybridized which form the ring.

N(I)

NH

(II)

(III)O

(IV)S

(V)

(VI)

O

(VII)

(VIII)HN

N

(IX)H H




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Test 011. (B) 2. (C) 3. (C) 4. (B) 5. (C)6. (B) 7. (C) 8. (B) 9. (B) 10. (B)11. (A,B,C,) 12. (A,C,D) 13. (B,C,D) 14. (C,D) 15.(A,B,C,D)16. (C,D) 17. (B,C,D) 18. (E) 19. (D) 20. (B)21. (C) 22. (A) 23. (A-S ; B-P ; C-Q ; D-R)24. (A–Q,T ; B–P,R,T ; C–P ; D–Q,S) 25. (B) 26. (B)27. (A) 28. (B) 29. (6) 30. (4)

Test 021. (D) 2. (C) 3. (D) 4. (A) 5. (B)6. (C) 7. (B) 8. (C) 9. (B) 10. (A)11. (A,B,D) 12. (A,B,D) 13. (A,B,C,D) 14. (A,B,C,D) 15. (B,C,D)16. (A,B,C,D) 17. () 18. (D) 19. (A) 20. (A)21. (E) 22. (A) 23. (A–Q ; B–P ; C–S ; D–R)24. (A–P,S ; B–P,T ; C–R,S ; D–Q,S) 25. (B) 26. (C)27. (B) 28. (B) 29. (5) 30. (6)

Test 031. (C) 2. (D) 3. (A) 4. (C) 5. (D)6. (C) 7. (D) 8. (B) 9. (C) 10. (A)11. (A,B) 12. (C,D) 13. (A,B,C,D) 14. (A,B) 15. (B,D)16. (A,B,C,D) 17. (A,C) 18. (D) 19. (D) 20. (B)21. (B) 22. (B) 23. (A–Q,R,T ; B–P,T ; C–P,R,T ;D–Q,R,T) 24. (A–Q ; B–P ; C–S ; D–R) 25. (B) 26. (D)27. (C) 28. (C) 29. (7) 30. (7)

Answers


Samplebook-Tools of organic chemistry

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