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Biyani's Think Tank
Concept based notes
Organic Chemistry (B.Sc. Part-II)
Anupama Singh
M.Sc.
Lecturer Deptt. of Science
Biyani Girls College, Jaipur
2
Published by :
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Organic Chemistry 3
Preface
I am glad to present this book, especially designed to serve the needs
of the students. The book has been written keeping in mind the general weakness in understanding the fundamental concepts of the topics. The book is self-explanatory and adopts the “Teach Yourself” style. It is based on question-answer pattern. The language of book is quite easy and understandable based on scientific approach.
Any further improvement in the contents of the book by making corrections, omission and inclusion is keen to be achieved based on suggestions from the readers for which the author shall be obliged.
I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also have been constant source of motivation throughout this Endeavour. They played an active role in coordinating the various stages of this Endeavour and spearheaded the publishing work.
I look forward to receiving valuable suggestions from professors of various educational institutions, other faculty members and students for improvement of the quality of the book. The reader may feel free to send in their comments and suggestions to the under mentioned address.
Author
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Syllabus B.Sc. Part-II
Organic Chemistry
Unit - I
Electromagnetic Spectrum : An Introduction.
Absorption Spectra : Ultraviolet (UV) Absorption Spectroscopy, Absorption Laws (Beer Lamber Law) Molar Absorptivity, Presentation and Analysis of UV Spectra, Types of Electronic Transitions, Effect of Solvent on Transitions, Effect of Conjugation, Concept of Chromophore and Auxochrome, Bathochromic, Hypsochromic, Hyperchromic and Hypochromic Shifts, UV Spectra and Conjugated Enes and Enones.
Infrared (R) Absorption Spectroscopy : Molecular Vibrations, Hook's Law, Selection Rules, Intensity and Position of IR Bands, Measurement of IR Spectrum, Fingerprint Region, Characteristic Absorptions of Various Functional Groups and Interpretation of IR Spectra of Simple Organic compounds.
Alcohols : Classification and Nomenclature.
Monhydric Alcohols : Methods of Formation by Reduction of Aldehydes, Ketones, Carboxylic Acids and Esters, Hydrogen Bonding Acidic Nature, Reactions of Alcohols.
Dihydric Alcohols : Methods of Formation, Chemical Reactions of Vicinal Glycols, Oxidative Cleavage [Pb(OAc)4 and HIO4] and Pinacol Pinacolne Rearrangement.
Trihydric Alcohols : Methods of Formation, Chemicals Reactions of Glycerol.
Unit-II
Phenols : Nomenclature, Structure and Bonding, Preparation of Phenols, Physical Properties and Acidic Character, Comparative Acidic Strength of Alcohols and Phenols, Resonance Stabilization of Phenoxide, Ion Reactions of Phenols, Electrophilic Aromatic Substitution, Acylation and Carboxylation, Mechanisms of Fries Rearrangement, Claisen Rearrangement. Gatterman Synthesis, Hauben-Hoesch Reaction, Lederer Manasse Reaction and Reimer Tiemann Reaction.
Ethers and Epoxides : Nomenclature of Ethers and Methods of their Formation, Physical Properties, Chemical Reactions , Cleavage and Autoxidation, Ziesels' Method.
Synthesis of Epoxides, Acid and Base-Catalyed Ring Opening of Epoxides, Orientation of Epoxide Ring Opening, Reactions of Grignard and Organolithim Reagents with Epoxides.
Unit-III Aldehydes and Ketones : Nomenclature and Structure of the Carbonyl Group. Synthesis of Aldehydes and Ketones with Particular Reference to the Syntheses of
Organic Chemistry 5
Aldehydes from Acid Chlorides, Synthesis of Aldehydes and Ketones using 1,3-Dithianes, Synthesis of Ketones from Nitriles and from Carboxylic Acids, Physical Properties.
Mechanism of Nucleophilic Additions to Carbonyl Group with Particular Emphasis on Benzoin, Aldol, Perking and Knoevenagel Condensations, Condensation with Ammonia and its Derivatives, Writting Reaction, Mannich Reaction.
Use of Acetals as Protecting Group, Oxidation of Aldehydes, Baeyer-Villiger Oxidation of Ketones, Cannizzaro Reaction, MPV, Clemmensen, Wolf-Kishner, LiAl4 and NaBH4 Reductions, Halogenation of Enolizable Ketones.
Unit-IV
Carboxylic Acid : Monemnclature, Structure and Bonding, Physical Properties, Acidity of Carboxylic Acids, Effects of Substituents on Acid Strength, Preparation of Carboxylic Acids, Reactions of Carboxylic Acids, Hell-Volhard-Zelinsky Reaction. Synthesis of Acid Chlorides, Esters and Amides, Reduction of Carboxylic Acids, Mechanism of Decarboyxlation.
Methods of Formation and Chemical Reactions of Halo Acids, Hydroxy acid, Malic, Tartaric and Citric Acids.
Methods of Formation and Chemical Reactions of Unsaturated Monocarboxylic Acid.
Dicarboxylic Acid : Methods of Formation and Effect of Heat and Dehydrating Agents (Succinic, Glutaric and Adipic Acids).
Carboxylic Acid Derivatives : Structure and nomenclature of acid chlorides, esters, amides (urea) and acid anhydrides, Relative stability of acyl derivatives. Physical properties, interconversion of acid derivatives by nucleophilic acyl substitution.
Preparation F Carboxylic Acid Derivatives, Chemical Reactions, Mechanisms of Esterification and Hydrolysis (Acidic and Basic).
Unit-V Organic Compounds of Nitrogen : Preparation of Nitroalkanes and Nitroarenes. Chemical Reactions of Nitroalkenes, Mechanisms of Nucleophilic Substitution in Nitroarenes and their Reductions in Acidic, Neutral and Alkaline Media, Picric Acid.
Halonitroarenes : Reactivity, Structure and Nomenclature of Amines, Physical Properties, Stereochemistry of Amines. Separation of a Mixture of Primary, Secondary and Tertiary Amines. Structural Features Effecting Basicity of Amines. Amine Salts as Phase-Transfer Catalysts.
Preparation of Alkyl, and Aryl Amines (Reduction of Nitro Compounds, Nitriles), Reductive Amination of Aldehydeic and Ketonic Compounds. Gabriefl-Phthalimide Reaction, Hofmann Bromide Reaction.
Reaction of Amines, Electrophilic Aromatic Substitution in Aryl Amines, Reactions of Amines with Nitrous Acid. Diezotisation, Mechanism. Synthetic Transformation of Aryl Diazomnium Salts, Azo Coupling.
□ □ □
6
Content
S. No. Name of Topic
1. Electromagnetic Spectrum
2. Alcohols
3. Phenols
4. Ethers & Epoxides
5. Aldehydes and Ketones
6. Carboxylic Acids
7. Carboxylic Acid Derivatives
8. Organic Compounds of Nitrogen
9. Unsolved Papers 2011 - 2008
□ □ □
Organic Chemistry 7
Chapter-1
ELECTROMAGNETIC SPECTRUM
(ABSORPTION SPECTRA)
Q.1 Define and explain Beer – Lambert Law (Absorption Laws).
Ans.: Lambert Law : It states that the fraction of the incident light absorbed is
independent of the intensity of the source.
Beer Law : It states that the absorption is proportional to the number of
absorbing molecules.
Both these laws are called as Absorption Laws.
According to these laws
log10 (I0 / I) = ε . l . c
Where
I0 Intensity of the incident light
I Intensity of the transmitted light
ε Molar extinction coefficient
t Path length
c Concentration in moles / litre of the solute.
8
So,
A = log10 (I0 / I) Optical density
A = log10 (I0 / I) = ε . l . c
Q.2 Explain the following −
Ans.: (i) Bathochromic or Red Shift : The shift of absorption to a longer
wavelength due to substitution or solvent effect is called as bathochromic
or red shift.
Example – Conjugation of double bonds causes bathochromic shift.
(ii) Hypsochromic or Blue Shift : The shift of absorption to a shorter
wavelength due to substitution or solvent effect is called hypsochromic
shift or blue shift.
Example – In neutral solvent aniline absorbs at 230nm while in acidic
medium it absorbs at 203nm.
(ii) Hyperchromic Shift : If because of a substituent group, intensity of
the band increases, the effect is called as hyperchromic shift.
Example – For benzene Emax is 7,400,
styrene Emax is 14,000
Vinyl group (CH = CH2) in benzene causes hyperchromic shift.
(ii) Hypochromic Shift : If because of a substituent group, the intensity
of absorption band decreases, then the effect is called as hypochromic
shift.
Organic Chemistry 9
Q.3 Explain the Principle of I.R. Spectroscopy.
Ans.: I.R. spectroscopy is basically based upon the study of energy required for
transition between different vibrational levels in a molecule. When we
pass I.R. light from the sample, some frequencies are absorbed while rest
are transmitted. When we plot them against frequency in a graph, then we
obtain I.R. spectrum. Different bonds have different vibrational
frequencies and we can detect them through the absorption bands in I.R.
spectrum.
□ □ □
10
Chapter-2
ALCOHOLS
Q.1 What are Alcohols? How are they classified? How would you
distinguish between them?
Ans.: Alcohols are the alkyl derivatives of water or hydroxy derivatives of
hydrocarbons having general formula R – OH
They can be classified in two different ways −
(i) Classification based upon the number of hydroxyl groups –
(a) Monohydric Alcohols –
Example − CH3 − OH
(b) Dihydric Alcohols –
Example − HO – CH2 – CH2 − OH
(c) Trihydric Alcohols –
Example − HO – CH2 – CH – OH − CH2 – OH
(d) Polyhydric Alcohols –
Example − HO – CH2 – (CH – OH) 4 − CH2 – OH
(ii) Classification based upon the nature of carbon atom bearing hydroxy group –
(a) Primary Alcohols (1°) –
Example − R – CH2 − OH
Organic Chemistry 11
(b) Secondary Alcohols (2°) –
Example − R – CH – OH
R
(c) Tertiary Alcohols (3°) –
Example − R
R – C – OH
R
We can distinguish between primary, secondary and tertiary alcohols either by Lucas test* or Victor Meyer test*.
Q.2 How will you prepare 2−Hexanol by Acid Catalyzed Hydration of the appropriate Alkene?
Ans.: CH3 – CH2 − CH2 – CH – CH2 = CH2 + H2SO4
CH3 – CH2 − CH2 – CH2 – CH – CH3
H2O ∆ OSO3H
CH3 – CH2 − CH2 – CH2 – CH − CH2 + H2SO4
OH
2−Hexanol
12
Q.3 Give reasons –
Ans.: (i) Water has a high B.pt than Ethanol.
Ethanol contains −OH group which is highly polar. Hydrogen is
also having tendency of attracting electro–ve oxygen atom of other
molecule leading to H−bonding. But these bonds are weaker in
camparison to the bonds present in H2O molecule. So it becomes
easy to break these bonds. Hence B.pt of H2O is higher than
ethanol.
(ii) Ethylene Glycol has higher B.pt than that of Propyl Alcohol.
Ethylene glycol having two –OH groups whereas propyl alcohol
having one −OH group. B.Pt. is effected by the extent of hydrogen
bonding.
Q.4 What are Glycols? How is Ethylene Glycol prepared from Alkene?
Explain it by giving its mechanism.
Ans.: The compounds containing two hydroxyl groups are termed as dihydric
alcohols or glycols or diols.
Method of Preparation from Alkene :
H H KMnO4 / OH‾ H H
C = C C C
H H Os O4 / H2 O H H
OH OH
Organic Chemistry 13
Mechanism :
4 /OSO NMO
CH2 = CH2 CH2 CH2
Aq. Tert. Butyl Alcohol
O O
Os
O O
H2O
CH2 CH2
+ H2Os O4
OH OH
(Cisdiol) (Osmic Acid)
Q.5 Write down atleast one method of preparation of Glycerol. How will you prepare Formic Acid, Glycerol Dischlorohydrin from Glycerol?
Ans.: Preparation of Glycerol :
CH2 – OCOR CH2 – OH
CH – OCOR + 3 NaOH 3 RCOO‾Na+ + CH – OH
CH2 – OCOR CH2 – OH
(Oils or Fats) (Salts of Higher (Glycerol)
Fatty Acids)
14
Preparation of Formic Acid :
O
CH2 – OH CH2 – O – C − COOH
HOOC 120°
CH – OH + −H2O CH – OH
HOOC
CH2 – OH CH2 – OH
∆ − CO2 (Glycerol Monooxalate)
CH2 – O – O – H CH2 – OH
CH – OH H2O CH – OH + HCOOH
CH2 – OH CH2 – OH
(Glycerol Monoformate) (Formic Acid)
Preparation of Glycerol Dichlorohydrin :
CH2 – OH CH2 – Cl CH2 – OH CH2 – Cl CH2 – Cl
110°C HCl, 110°C
CH – OH CH – OH + CH – Cl CH – Cl + CH – OH HCl (excess)
CH2 – OH CH2 – OH CH2 – OH CH2 – OH CH2 – Cl
(Glycerol α & β Chlorohydrin) (Glycerol Dichlrohydrin)
Organic Chemistry 15
Q.6 Explain Pinacol –Pinacolone Arrangement with its mechanism.
Ans.: CH3 CH3 CH3
H+
CH3 – C – C − CH3 Acid CH3 – C – C − CH3
OH OH CH3 O
(2, 3 dimethylbutane – 2, 3 diol) (3, 3 dimethylbutanone) (Pinacol) (Pinacolone)
Mechanism : CH3 CH3 CH3 CH3 CH3 CH3 CH3
H+ − H2O
H3C – C – C − CH3 Acid C H3 – C – C − CH3 H3C – C – C − CH3 +
OH :OH OH OH2 OH . . - (Pinacol) (1, 2 – Methylshift)
CH3 CH3 − H+ + Rearrangement
H3C – C – C − CH3 H3C – C – C − CH3
O CH3 OH CH3
(Intra–Molecular (1, 2 − Methylshift) Rearrangement)
Q.7 Complete the following reactions −
Ans.: (i) CH3 Ni CH3
C = O + H2 CH − OH
H H
Invertase
(ii) C12H22 O11 + H2O C6H12 O6 + C6H12 O6 (from yeast) (Glucose) (Fructose)
□ □ □
16
Chapter-3
PHENOLS
Q.1 Write one commercial method of preparation of Phenol? Compare its
physical properties with Cyclohexanol.
Ans.: Commercial Method of Preparation of Phenol (Also called as Dow Process) :
Step−I : FeCl3 + Cl2 Cl + HCl
− HCl
Step−II : 360°C Cl + 2 NaOH O‾Na+ + NaCl + H2O
Pressure
Step−III : O‾Na+ + HCl OH + NaCl
Comparison of Physical Properties :
Phenol −
(i) High B.pt & M.pt.
(ii) Moderately soluble in water
(iii) More polar due to greater acidity
Cyclohexanol −
(i) Low B.pt & M.pt.
(ii) Sparingly soluble in water
(iii) Less polar b’coz these are weakly acidic.
Organic Chemistry 17
Q.2 Explain giving reasons –
Phenols are stronger acids than Alcohols but are weaker N ucleophiles.
Ans.: Phenols are stronger acids than alcohols because these are more polar due
to greater acidity as compared to alcohols. Phenols and alcohols forms
phenoxide and alkoxide ions. Phenoxide ions are more stabilized because
of more resonating structure. Therefore phenols are more acidic than
alcohols but are weaker nuclophiles. Negative charge is spread over
benzene ring in phenoxide ion and hence it lies on carbon atom but in
alcohols it resides on oxygen.
Q.3 Write down methods of preparation of Phenol starting from Cumene
and Coal−Tar.
Ans.: Starting form Cumene : CH3 – CH – CH3
H2SO4 + H3C – CH = CH2 (Cumene) isopropyle benzene
O2, OH‾ 130°C
OH (H3C) 2 – C − O – OH H2O / H+ + C H3COCH3 100°C (Phenol) (Acetone) (Cume−Hydroperoxide)
18
From Coal−Tar :
Middle oil & heavy oil fractions
of coal−tar
Cooled Naphthalene crystallises
Mother liquor
aq. NaOH
Aq. Layer is removed & washed with H2O
Cool CO2 is passed Forms sodium phenoxide
Undergo fractional distillation
Different phenols separated out
Q.4 Explain the following reactions with their mechanism.
Ans.: (i) Fries Rearrangement :− When phenyl acetate is heated with anhyd.
AlCl3, the acyl group migrates from phenolic oxygen to an ortho or para
position of the ring yielding a mixture of hydroxyketones, called as fries
rearrangement.
OH OCOCH3 OH OH COCH3 CH3COCl AlCl3 + −HCl CS2 , ∆ (Phenol) (Phenyl (O−hydroxy COCH3
Acetate) Acetophenone) (P−hydroxy Acetophenone)
Note : At low temp. pera isomers is obtained but at high temp. ortho isomer is formed.
Mechanism : Two types of mechanism are proposed.
Organic Chemistry 19
Intermolecular :
O O . . + + _ + _ R − C − O: R – C – O – AlCl3 : O – AlCl3
+ AlCl3 + [ R – C ≡ O ] − HCl + _ OH O – AlCl2 O – AlCl3 COR COR H H2O −HCl COR
Intramolecular Mechanism :
_ Cl3Al O
+ : O − C − R + _ O – AlCl3 O – AlCl2 OH H COR COR COR −HCl H2O
20
Claisen Re−arrangement : − When allyl aryl ethers are heated up to 200°C, than they get arranged to allyl phenols. This rearrangement is called as claisen’s re−arrangement.
OCH2 – CH = CH2 OH
CH2 – CH = CH2 200°C
Note :− If the ortho positions are occupied then rearrangement of allyl groups taken place at para position.
O − CH2 – CH = *CH2 OH
β γ *CH2 – CH = CH2 H2 H2 C C O CH O CH O H CH2 CH2 *
CH2–CH=CH2 (Cyclic Transition State)
(It is also called as Sigmatropic Rearrangement)
Reimer – Tiemann Reaction : OH OH
CHO 70˚C + CHCl3 + 3 KOH + 3 KCl + 2 H2O
Organic Chemistry 21
Mechanism : It involves intermediate formation of dichlorocarbene which act as nucleophile.
..
3 3 2: ( )OH
ClCCl C Cl CCl Carbene
OH O‾ O H OH‾ : CCl2 C‾Cl2 −H2O H2O − OH‾ O‾ O‾ O CHO CH Cl2 H OH‾ −H+ CH Cl2 − H2O HCl OH CHO
Huben – Hoesch Rx :− It is used where fries rearrangement or friedel – crafts acylation not successful.
22
HCL.NH CH3 C OH OH OH OH ZnCl2, HCl + CH3CN Ether, −10° to 0°
3CH 3CH
H2O ∆ O CH3 C OH OH
3CH
(2, 4, 6 – Trihydroxy Acitophenone)
Note : It is widely used for di and polyhydric phenols.
Organic Chemistry 23
Mechanism :
Step : I . . CH3 – C ≡ N + H+ [ H3C – C ≡ NH . . H3C – C = NH ] Cl‾ H3C – C = NH Cl AlCl3 CH3 − AlCl3 HN = C Cl AlCl3 H3C – C = NH
+ Cl
HN = C HN ≡ C
CH3 CH3 + AlCl4
24
Step : II
CH3
H C = NH OH OH CH3 OH OH + +C = NH OH OH −H+
COCH3 NH = C − CH3 OH OH OH OH H2O / H+ OH OH
Q.5 Write structures of the following − Ans.: OH (i) Resorcinol OH OH (ii) Euqenol O
+
Organic Chemistry 25
CH3 (iii) Bis – phenol A OH −OH
CH3
OH CH (CH3)2 (iv) Thymol CH3
OH (v) Picric acid O2N NO2 NO2
OH (vi) Hydroquinone OH
□ □ □
26
Chapter-4
ETHERS & EPOXIDES
Q.1 What are Ethers? Write down their classification and atleast two
methods of their preparation.
Ans.: These are the dialkyl derivatives of water represented by general formula CnH2n + 2 O and are isomeric with monohydric alcohols.
Classification : Can be classified into two groups –
Symmetrical ethers Unsymmetrical ethers
(These type of ethers posess identical alkyl or aryl group attached to oxygen atom.)
Example –
H5C6 – O – C6H5
(Diphenyl Ether)
H5C6 – H2C − O – CH2 – C6H5
(Dibenzyl Ether)
(They posess two different alkyl or aryl groups attached to oxygen atom.)
Example –
H3C – O – C2H5
(Ethyl Methyl Ether)
H5C6 – O – CH3
(Methyl Phenyl Ether)
Q.2 Explain the following –
Ans.: (i) Williamson Synthesis : It is a laboratory method to prepare both symmetrical and unsymmetrical ethers. It consist of heating alkyl halides with sodium, potassium or alkoxides.
C2H5Br + C2H5ONa C2H5 – O – C2H5 + NaBr
Organic Chemistry 27
Mechanism :
C2H5ONa C2H5O + Na
+ −
C2H5O‾ + C2H5+Br‾ slow [ C2H5O − C2H5 − − − Br ]
fast
C2H5 − O − C2H5 + Br‾
(ii) Alkoxy Mercuration – Demercuration Method : This method is used
to convert alkenes to ethers. Alkenes are treated with mercuric
triflouroacetate in the presence of an alcohol. The resulting product is
reduced with sodium borohydride to give ethers.
R – CH = CH2 + R1 – O – H + Hg(OOCCF3)2 R− CH – CH2
NaBH4 OR1 HgOOCCF3
OR1
R− CH – CH3 + CF3COOH
(iii) Orientation of Cleavage of Epoxides : In symmetrical epoxides, the
two carbons are symmetrical, therefore, nucleophile attack at either of the
C−atom but in unsymmetrical epoxide, nucleophile have preferential
attack depending on the nature of medium.
In Acidic Medium : Nucleophile attacks the more substituted carbon. It
is a SN2 type Rx. Cleavage of carbon – oxygen bond and attack of
28
nuclophile occurs in one step. In transition state, bond breaking exceed
bond making.
Z Z
Z : + − C – C – − +C – C – − C – C −
O+ O + OH
H H
In Basic Medium : Here bond making and bond breaking are nearly
balanced & reactivity is controlled by steric factors attack occurs at less
hindered carbon.
Z Z
Z + − C – C – − C – C – − C – C −
O O − O‾
(iv) Ziesel’s Method : In this process, a known amount of ether is heated
with an excess of conc. hydroiodic acid. The methyl iodide vapours so
produced are passed in cool solution of silver nitrate. Then the percentage
of methoxy group can be calculated.
O – CH3 + HI 130˚ OH + CH3I
Δ
CH3 – I + AgNO3 AgI + CH3NO3
Organic Chemistry 29
Q.3 Explain giving reasons –
Ans.: (i) Diethyl Ethers sometimes explodes on distillation.
Reason : They are highly combustible, volatile & inflammable. They burn in air to produce CO2 and H2O. So during excessive heating, they explode.
(ii) Alkanes are insoluble in water but ethers are sparingly soluble in water.
Reason :Alkanes are not able to form hydrogen bonds with water but ethers are sparingly soluble due to intermolecular hydrogen bonding with water.
R R R
O O O O
R H H H R
□ □ □
30
Chapter-5
ALDEHYDES AND KETONES
Q.1 Why are Aldehydes found to be more reactive than Ketones?
Ans.: Aldehydes have one hydrogen atom and one alkyl or aryl group linked to the carbonyl group while ketones have two alkyl or aryl group but no hydrogen.
R R
C = O C = O
H R
(Aldehyde) (Ketone)
Due to the presence of free hydrogen atom in their molecules aldehydes are more reactive than ketones.
Q.2 How do you account for the following –
Oxidation of Primary Alcohols generally gives Poorer Yield of
Aldehydes than the Oxidation of Secondary Alcohols gives of Ketones.
Ans.: Oxidation of primary alcohols generally gives poorer yield of aldehydes
because in their case there is the risk of further oxidation to give
carboxylic acid.
In case of ketones, there is no risk of further oxidation of ketones as
ketones are not easily oxidized.
Organic Chemistry 31
Q.3 Explain how you will distinguish Aldehydes from Ketones.
Ans.: (i) Aldehydes form a silver mirror on warming with Tollens reagent but ketones do not.
(ii) Aldehydes give red ppt. on warming with Fehling solution or Benedicts solution but ketones do not.
Q.4 Explain the following −
Ans.: (i) Wittig Reaction : It is an important reaction to convert C = O bond into C = C double bond.
Wittig Reaction
C = O C = C
RCHO + PH3P = CHR1 RCH = CHR1 + PH3P = O
(Alkylidentri Phenyl Phosphorane) (Triphenyl Phosphine)
Mechanism : Here in this Rx phosphorous ylide attacks on carbonyl carbon and forms a transition state which undergo elimination to form an alkene.
(a) PH3P = CHR1 PH3P+ − C‾HR1
O‾ P+PH3
(b) R – C – H + PH3P+ − C‾HR1 R – CH − CH R1
O (Transition State)
R – CH = CH − R1 + PH3P = O
32
(ii) Reductive Amination : Some aldehydes and ketones change into
amines on treatment with hydrogen and ammonia in the presence
of a catalyst like nickel. This process is called as reductive
amination. It takes place through the intermediate formation of an
imine − C = NH
NH3, H2, Ni
CHO CH2 – NH2
(Benzaldehyde) (Benzylamine)
O NH CH3
CH3NH2, H2, Ni
C – CH3 CH – CH3
(Acetophenone) (Methyl α–Phenyl Ethyl Amine)
(iii) Bimolecular Reduction of Ketones : In the presence of magnesium,
aldehyes and ketones undergo bimolecular reduction to give
symmetrical glycols called as pinacols.
Organic Chemistry 33
CH3
CH3
CH3 C O
C = O Mg Mg
(Benzene) CH3 C O
CH3
(Acetone) CH3
(Magnesium Salt of Pinacol)
H2O
CH3
CH3 C OH
CH3 C OH
CH3
(Pinacol)
Q.5 Predict the product in the following −
Ans.: (i) CHO COO‾K+
50% KOH
CHO CH2OH
34
(ii) CHO COOH
Ag2O
(CHOH)4 (CHOH)4
CH2OH CH2OH
(Gluconic Acid)
SnCl2
(iii) C6H5C ≡ N (C6H5CHO)
HCl
H+
(iv) C6H5 – CH – CH − CH3 C6H5 – C – CH2 − CH3
OH OH O
Q.6 How will you convert Aldehyde into Ketone by Dithiane Technology?
Ans.: For converting aldehydes into ketones, first dithiane is converted into
lithium cabanion salt by treatment with n−butyl lithium at low
temperature. The anion is then reacted with alkyl halids to yield alkylated
ditheanes. By desulphurization with HgCl2, ketones are obtained.
1, 3 dithiane
H – C – H (CH3)2 CH – C – CH (CH3)2
O O
(Di – Isopropyl Ketone)
Organic Chemistry 35
Mechanism :
H
Dry HCl S H
C = O + CH2 – CH2 – CH2
gas S H
H SH SH
n BuLi / THF
−butane, −30˚
(i) n−Buli/THF/−30˚ S H (CH3)2CH−I S H
(ii) (CH3)2CH−I
S CH(CH3)2 S ‾Li+
S CH(CH3)2 HgCl2 CH(CH3)2
Deprotection O = C
S CH(CH3)2 CH(CH3)2
□ □ □
36
Chapter-6
CARBOXYLIC ACIDS
Q.1 What are Carboxylic Acids? Describe their structure.
Ans.: Carboxylic acids are acidic compounds having one or more carboxylic groups in their molecules.
Structure : Although the carboxyl group is generally represented by the structure
O
− C
OH
But it is a resonance hybrid of the following contributing structures.
. . . . O : : O :‾
− C . . − C O – H O+ − H · ·
(I) (II) (Less stable)
Q.2 Convert Carboxylic Acids into Acid Chloride, Ester and Amide.
Ans.: (i) Carboxylic acid Acid Chloride
RCOOH + PCl5 RCOCl + POCl3 + HCl
Organic Chemistry 37
(ii) Carboxylic acid Ester
H+
RCOOH + R1OH RCOOR1 + H2O
(Carboxylic (Alcohol) (Ester)
Acid)
(iii) Carboxylic acid Amide
RCOOH + NH3 RCOO NH4 −H2O RCONH2
(Amm. Salt) Δ (Acid Amide)
Q.3 Explain why?
(i) The lower members of Carboxylic Acids are soluble in water but higher members are insoluble in water.
Ans.: Lower members of carboxylic acids are soluble in water because they are able to make hydrogen bonds with water as carboxylic group participates in hydrogen bonding both as doner and accepter.
. . . . . . . . − − − H – C = O : − − − H – O : − − − H – O – C = O : − − − H – O : − − −
R H R H
But with the increase in molecular weight solubility decreases because they are not able to make stronger hydrogen bonds.
(ii) The Carboxylic Acids have abnormally higher boiling points.
Ans.: The carboxylic acids have abnormally higher boiling points because in
O
– C – OH group, the –OH part is strongly polarised due to the electron withdrawing carbonyl groups attached to it.
38
Hence it is able to form hydrogen bonds with negative oxygen of carbonyl dipole. Therefore, they usually remain in dimeric state, the molecular hydrogen bonds between hydroxyl hydrogen of one molecule and carbonyl oxygen of other molecules.
O − − − H O
R C C R
O H − − −O
(iii) Most of the Crboxylic Acids exist as dimers.
Ans.: Carboxylic acids are able to form stronger hydrogen bonds. The
negatively polarised oxygen of the carbonyl group is able to form
hydrogen bond with the positively polarised hydrogen of the other
molecule. Therefore, majority of the carboxylic acids exist as cyclic dimers
held together by two hydrogen bond.
O − − − H O
R C C R
O H − − −O
Q.4 How will you Synthesise Citric Acid from Glycerol? what happens when Citric Acid is heated?
Ans.: Citric acid can be synthesized from glycerol in the following ways –
Organic Chemistry 39
CH2 – OH CH2 – Cl CH2 – Cl
HCl [O]
CH – OH CH – OH C = O
Controlled Conditions HNO3
CH2 – OH CH2 – Cl CH2 – Cl
(Glycerol)
C2H5OH KCN
CH2 – COOH CH2 – CN
H3O+ OH
OH − CH – COOH C
CN
CH2 – COOH CH2 – CN
(Citric Acid)
Citric acid act both as α–hydroxy and β–hydroxy acid when it is heated upto 150˚C, it looses water to form aconitic acid and at higher temperature it forms number of products like itaconic acid, acetone, citraconic acid etc.
□ □ □
40
Chapter-7
CARBOXYLIC ACID DERIVATIVES
Q.1 What is Esterification? Explain its mechanism.
Ans.: The reaction of carboxylic acid with an alcohol in the presence of mineral
acid gives esters and this process is called as esterification.
H O
H+
R – C – OH + R – O – H R – C – OR1 + H2O
Mechanism : It occurs in 4 steps –
Step−I :
O O+H‾
R – C + H+ R – C
OH OH
(Protonation of carbonyl oxygen of acid takes place)
Organic Chemistry 41
Step−II :
O+H OH
. . R – C + R − : OH R – C − O+ − R1
O OH H
(Intermediate)
(Attack of alcohol takes place on carbonyl carbon to form intermediate)
Step−III :
OH OH
R – C − O+ − R1 R – C − O+ − R1
OH H +O − H
H
(Transfer of hydrogen from one oxygen atom to second oxygen)
Step−IV :
O − H O
− H+
R – C − O − R1 R – C − OR1
− H2O
+O − H
H
(In this step water molecule is removed to form an ester)
42
Q.2 What happens when basic solution of Bromine reacts with Amide? Also
give its mechanism.
Ans.: When basic solution of bromine reacts with primary amide, the amide is
converted into amine and carbonyl group is lost as CO2.
O
Br2 + NaOH
R – C – NH2 R – NH2 + CO2
H2O
Mechanism : This reaction proceeds in steps –
Step−I : Abstraction of N−H Proton by a Base
O O
. . − H2O + . . R – C – N – H + OH‾ R – C – : N – H
H Br − Br
Step−II : Bromination of Anion O
R – C – : N – H + Br‾
Br
(N – bromoamide)
OH‾
Organic Chemistry 43
Step−III : Abstraction of N−H Proton by a Base
O
. .
R – C – : N‾ + H2O
Br
Step−IV : Elimination of Bromide ion
give Acyl Nitrene O
. .
R – C – : N + Br‾
(acyl nitrene)
Step−V : Intermolecular arrangement of
Acyl nitrene to form isocyanate
O = C = N − R + H − OH
(Isocyanate)
R – N – C = O
R – NH2 – CO2
(Amine) H O − H
(Carbamic acid)
* This Rx. is also called as Hofmann rearrangement or Hofmann degradation.
44
Q.3 Explain the following –
Ans.: (i) Aminolysis : Aminolysis means conversion into amides.
O O
R – C – Cl + NH3 R – C − NH2 + HCl
Note :− Two molecules of ammonia are used because one molecule of
ammonia reacts with acid chloride while second molecule reacts with HCl
to produce ammonium chloride.
Mechanism :
Step−I : Rx. of Acid Chloride with Ammonia :
. .
O : O :
R – C – Cl + : NH3 R – C − Cl
H – N+ – H
H
− Cl
O O H
R – C − NH2 R – C – +N – H
− H+
H
Organic Chemistry 45
Step−II : Rx. of Acid Chloride with Amines to give Amides :
O O
R – C – Cl + RNH2 R – C − NHR + HCl
+ _ HCl + RNH2 RNH3Cl
Step−III : In case of Tertiary Amine, Pyridine can be used instead of excess of Ammonia or Ammines.
O O
Pyridine
R – C – Cl + RNH2 R – C − NHR
+
+
N Cl
H
(ii) Trans−Esterification : Trans−esterification is a process in which an ester react with alcohol to produce a new ester. This reaction takes place either in the presence of an acid or a base.
O
CH3 – C – OC2H5 + CH3 – CH2 – CH2 – OH
(excess)
H+ O
C2H5OH + CH3 – C − OC2H5 – CH2 – CH3
46
(iii) Alcoholysis : Alcoholysis means reaction with an alcohol. The
reaction of acid anhydrides with alcohols gives esters.
O O O O
H+
R – C – O – C – R + R1 – OH R – C – OR1 + R – C – OH
(Ester)
Note :− This reaction proceeds in the presence of an acid.
Mechanism :
+OH O OH O
slow
R – C – O – C – R + R1 – : OH : R – C – O – C – R
(Protonated Anhydride) +O − R1
−H+
H
OH O O − H O – H O O
H+ (fast)
R – C – O – C – R R – C – O – C – R −H+ R – C–OR1 + R–C–OH
(Ester) (Acid)
R’- O O – R’
Organic Chemistry 47
(iv) Friedal Craft Acylation : In F.C. acylation, the acyl group is
introduced into the nucleus of benzenoid compound in the presence of
anhydrous AlCl3 as catalyst.
O
C
O O (i) AlCl3 R O
(anhyd.)
+ R – C – O – C – R + R – C – OH
(ii) HCl
Q.4 Write down at least two methods of the preparation of Acid – Anhydrides.
Ans.: Methods of preparation of Acid – Anhydride
(i) From Acid Chloride :
O O O O
+
R – C – Cl + NaO – C – R R – C – O – C – R + NaCl
Mechanism : . .
O O : O :
. . _ R – C – Cl + R – C – O : R – C – Cl . .
O – C – R
O O − Cl O
R – C – O – C – R
48
(ii) By Dehydration of Carboxylic Acids : When monocarboxylic acid is
heated in presence of acetic anhydride, then acid get dehydrated to
produce acid – anhydride.
O O O O O O
2 R – C – OH + CH3 – C – O – C − CH3 2 CH3 – C – OH + R – C – O – C – R
Mechanism : This reaction proceeds in two steps –
Step−I :
O O O O O
R – C + CH3 – C – O – C − CH3 CH3 – C – O – C–R + CH3COOH
O – H
(First Molecule)
Step−II :
O O O O O
CH3 – C – O – C – R + R – C – O – H CH3COOH + R–C–O–C–R
(Second Molecule)
Q.5 Give Mechanism of the reaction of Acid Chloride with Water to form Carboxylic Acid.
Ans.: O O
base
R – C – Cl + H2O R – C – OH + HCl
Organic Chemistry 49
Since HCl is produced during the reaction therefore this reaction carried out in the presence of a base (either NaOH or pyridine) to remove HCl.
. . − . . − O : O : : O :
. .
R – C – Cl + H2O : R – C – Cl −H+ R – C – Cl
O + O − H OH
(Protonated) (Deprotonated)
R – C – OH H (I) (II)
− Cl
□ □ □
50
Chapter-8
ORGANIC COMPOUNDS OF NITROGEN
Q.1 Explain the structure of Nitroalkanes by giving special emphasis on its
acidic character. How does Nitroalkanes differ from Alkyl Nitriles?
Ans.: In nitroalkanes, the alkyl group is directly attached to nitrogen atom. O
R − N
O In nitroalkanes nitrogen is sp2 hybridised and all the
three sp3hybridised orbitals lie in one plane and unhybridised sp2 orbital
is perpendicular to the plane of sp2 hybridised orbitals.
. . . .
O O O : O : . . +
R – N R – N R – N R − N
O O : O : : O : . . (Classical Structure) (Polar Structure)
Nitroalkanes are acidic in nature because they form salts when dissolved in aq. solution of KOH or NaOH.
Organic Chemistry 51
. .
Example : R – CH2 – NO2 + NaOH R – CH − NO2 + H2O
Na+
Reason : NO2 group is a strong electron withdrawing group and it forms
resonance stabilised carbanion with the abstraction of −
hydrogen by base.
Nitroalkanes differ from alkyl nitriles in various ways –
(i) In nitroalkanes, the alkyl group is
directly attached to nitrogen atom.
In alkyl nitriles, the alkyl group is
not attached to nitrogen but attached
to oxygen atom.
(ii) They are derived from the
following tautomeric structure of
nitrous acid.
They are also called as esters of
nitrons acid because they are
derived from the following
tautomeric structure of nitrous acid.
+ O O
H – N +R,−H R − N
O O
H – O – N = O +R, −H R – O – N = O
Q.2 Explain the following with mechanism –
(i) Hofmann Ammonolysis of Alkyl Halide :
Ans.: In this method SN2 – alkylation of ammonia with an alkyl halide takes place and the aminium salt produced in this process is treated with a base to obtain free amines.
− Primary amine is obtained by using ammonia.
. . SN2 Rx. NaOH
NH3 + R − X R NH3X H2O+NaX+RNH2
− Secondary amine is obtained by using primary amine
52
. . SN2 Rx. NaOH
RNH2 + R − X R2 NH2X H2O+NaX + R2NH
− Tertiary amine is obtained by using secondary amine.
(ii) Gabriel Phthalimide Synthesis :
Ans.: This process is used to prepare primary amines. O O C _ C OH NH N : (KOH / C2H5OH) C (i) deprotonation) C
O O
R − X
O (ii) alkylation
COOH C H2O / H+ N − R (iii) acid catalysed hyrolysis COOH C
+ O
R+NH3
_
: OH (iv) Neutralisation H2O _ COO + R − NH2 _ COO
Organic Chemistry 53
Q.3 Explain Hofmann Test to determine whether a given Amine is 1°, 2° or 3°.
Ans.:
Mixture of 1°, 2°, 3°, 4° amine salt
Neutralized with KOH
Mixture obtained is subjected to distillation
Primary, secondary and tertiary amines get distilled
Quartenary salt is left behind
(non−volatile in nature)
Specific Tests :
(i) Primary amine + diethyl oxalate Oxamide
(crystalline solid)
(ii) Secondary amine + diethyl oxalate Oxamic ester
(oily liquid)
(iii) Tertiary amine + diethyl oxalate No reaction
54
Q.4 Write short notes on the following –
(i) Sandmeyer Reaction :
Ans.: The reaction or aryl diazonium salts with cuprous chloride, cuprous
bromide and cuprous cyanide gives aryl chloride by replacing diazonium
group by nucleophiles like Cl −, Br − and CN −.
N ≡ N Cl
_ CuCl Cl + N2 15 – 60°C
N ≡ N Br
_ CuBr Br + N2 100°C
N ≡ N CN
_ CuCN Cl + N2
(ii) Diazoaminobenzene Rearrangement :
Ans.: Primary and secondary amines reacts with diazonium salts in weakly
Organic Chemistry 55
acidic medium to form N – azo compounds. These compounds can
undergo re−arrangement to form coloured C – azo compounds. + . . N ≡ N + H2N
_ CH3COONa
X H (Diazonium Salt) +
N = N − N − H+ H
N = N − NH ∆ HCl, rearrangement (Diazo−amino benzene)
N = N NH2
(p−amino benzen)
□ □ □
CHAPTER-I