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OXFORD INTERNATIONAL PUBLIC SCHOOL

CLASS-XII ( SCIENCE)

HOLIDAY HOME WORK

Sl.no SUBJECT DETAILS

1. English Prepare a fileon the ideal of your lkife with creative writing and

beautiful drawings.

2 Biology Instruction—

1. Select one project for holiday homework.

2. Use filepages.(don’t use rough pages)

3. You can use this below url to get more instruction or answer

https://studyres.com/doc/1861347/investigatory-project-for-class-xii-

biology

4.important project number-5,6,7,12,13,14,15

5.after selecting the project ,send me message(your name,project name)before

starting the project.

INVESTIGATORY PROJECT FOR CLASS XII BIOLOGY

1. Project Report on Malnutrition

2. Biology Project Report on Components of Food

3. Biology Project Report on DNA Fingerprinting

4. Project Report on Pollution

5. Biology Project Report on ABO blood grouping in human beings

6. Biology Project Report on the dispersal of seeds by various agencies

7. Biology Project Report on mosquito species -major diseases caused by

it

8. Biology Project Report on Human diseases

9. Project Report on sleeping Habits in human beings

10. Biology Project Report on Manures and Chemical Fertilizers

11. Project Report on Useful Plants and Animals

12. Biology Project Report on Cancer

13. Biology Project Report on aids

14. Biology Project Report on malaria

15. BIOLOGY PROJECT REPORT ON IMMUNE SYSTEM

3

Physical

Education

Work revise all course done till now

https://studyres.com/doc/1861347/investigatory-project-for-class-xii-biology

https://studyres.com/doc/1861347/investigatory-project-for-class-xii-biology

ANUP

Text Box

Read any novel and write book review of that writing piece and make one drawing describing your daily routine of lockdown in covid 19 pandemic

4. Information

Practice

Do all practical question related to data frame and series in practical

5. Physics. Do the all question of N C E R T chapter 1 and 2. Also solve the question

given in page 2-9.

6. Chemistry Revise all prev. work and sir gives a separate PDF fill of problem Sheet.

7. Maths Refer to page 10-15

PHYSICS WORK SHEET

CHAPTRE -1

CHAPTRE – 2

MATHS PROB. SHEET

TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777

CARBONYL COMPOUNDS 1

CARBONYL COMPOUNDS

EXERCISE # 1

Questions

based on Preparation of Aldehyde and Ketone

Q.1 The hydrolysis of benzal chloride gives–

(A) Benzyl alcohol

(B) Benzoic acid

(C) Benzaldehyde

(D) Benzophenone

Sol.[C]

CH

Cl

Cl

Benzal Chloride

Hydrolysis

HOH

CH OH

OH

–H2O

CHO

Benzaldehyde Therefore option (C) is correct.

Q.2 Ketones are first oxidation products of–

(A) Primary alcohols

(B) Secondary alcohols

(C) Both

(D) None

Sol.[B] Factual

Q.3 Calcium acetate when dry distilled gives–

(A) Formaldehyde

(B) Acetaldehyde

(C) Acetone

(D) Acetic anhydride

Sol.[C] Factual

Q.4 What is the function of BaSO4 in Rosenmund

reactionn–

(A) To stop further oxidation of aldehyde

copletely

(B) To stop further reduction of aldehyde

copletely

(C) Act as a poisonous catalyst

(D) It checks the reactivity of Pd.

Sol.[D] It checks the reactivity of Pd.

Q.5 Toluene on treatment with CrO2Cl2 gives –

(A) Chlorotoluene

(B) Benzyl chloride

(C) Benzaldehyde

(D) Benzoic acid

Sol.[C]

CH3

2CrO2Cl2

(CCl4)

CH3

CrCl2OH

OCrCl2OH

H2O

CHO

Toluene Brown Complex Benzaldehyde

This reaction is called "Etard's reaction".

Chemical Properties of Aldehyde & Ketones (Aldol Condensation, Cannizzaro reaction, Iodoform test)

Questions

based on

Q.6 HCHO with conc. alkali forms two

compounds. The change in oxidation number

would be –

(A) (0 to –2) in both the compounds

(B) (0 to +2) in both the compounds

(C) (0 to +2) in one compound and (0 to –2)

in the second compound

(D) All are correct

Sol.[C]

2.HCHO

[O]

alkali (NaOH)

22–O||

NaOCHOHCH3

HCHO

O||

NaOHC chang inoxidation

number = (0 to +2)

HCHO = CH3–OH – Changin oxidation number

= (0 to – 2)

Therefore option (C) is correct.

Q.7 C3H8O ]O[

SOH/OCrK 42722

C3H6O

.)aq(NaOHI2 CHI3,

In this reaction the first compound is –

(A) CH3CH2CH2OH



(B)

OH|

CHCHCH 33

(C) CH3OCH2CH3

(D) CH3CH2CHO

Sol.[B] CH3–

OH|

–CH CH3 ]O[

H/OCrK 722

O||

CHCCH 33

I2 + NaOH

CHI3 + CH3 –

O| |C –Na

Yellow ppct

Therefore option (B) is correct.

Q.8 CH3–CH2–CHOalkali

.Dil Product. The

product in the above reaction is–

(A) CH3–CH2COOH

(B) CH3–CH2–CH2OH

(C) CH3–CH2–

OH|CH –CH2–CHO

(D) CH3–CH2–

OH|CH – CH

|CH3

CHO

Sol.[D] CH3–CH2–

O| |C –H+CH3–

H|CH –CHO

dil.alkali

CH3–CH2–

OH|CH – CH

|CH3

–

O| |C –H

(Aldol)

–

O| |C –H

This reaction is called Aldol Reaction.

Therefore option (D) is correct.

Q.9 In the Cannizzaro reaction given below –

2Ph–CHO OH

Ph–CH2OH + PhC

the slowest step is –

(A) The attack of OH¯ at the carbonyl group

(B) The transfer of hydride to the carbonyl

group

(C) The abstraction of proton from the

carboxylic group

(D) The deprotonation of Ph - CH2OH

Sol.[B] 2Ph–CHO OH

Ph–CH2OH + PhCO2–

Mechanism

Step I :

Ph – C| |O

–H + OH Ph –

OH|

H–C|O

–H

Conjugate base of hydrate

of aldehyde.

Step II :

Intermolecular hydride ioni transfer

Ph –

OH|

–C|O

H + H– C| |O

–Ph Slow

Ph – C| |O

–OH + H–

H|

–C|O

Ph

Step III :

Ph – C| |O

–OH+NaOH Ph– C| |O

–O– +Na

+ + H2O

Ph–

H|

–C|O

H + H2O Ph –

H|

–C|OH

H

In this reaction mechanism H-transfer process

is slow and difficult :


Q.10 Which of the following statements is wrong –

(A) All methyl ketones give a positive

iodoform test.

(B) Acetaldehyde is the only aldehyde that

gives iodoform test.

(C) All secondary alcohols give positive

iodoform test.

(D) Any alcohol that can be oxidised to an

acetyl group gives a positive iodoform test.

Sol.[C] These compound which contain CH3–

O| |C



or CH2–

OH|CH – group give a positive iodoform

test. So all secondary Alcohol can't be give

positive iodoform test therefore option (C) is

correct.

Q.11 The number of aldols formed by CH3CHO

and CH3–CH2–CHO is –

(A) 2 (B) 3

(C) 4 (D) 1

Sol.[C]

CH3

O| |C – H

+

CH3–CH2–

O| |C –H

CH3–

OH|CH CH2–

O| |C –H

CH3–CH2–

OH|CH –CH

|CH3

O| |C –H

CH3–

OH|CH –CH

|CH3

O| |C –CH3 (Cross Aldol)

CH3– CH2

OH|CH CH2 –

O| |C –H (Cross Aldol)

Four Aldols formed by CH3CHO and

CH3CH2CHO.


Q.12 2D– C|D

= O + OH¯ Cannizzaro

X and Y

(Y is alcohol, D is deuterium)

X and Y will have structure –

(A) D –

H|C|D

– O , D – C| |O

– OH

(B) D –

D|C|D

– O , D – C| |O

– OH

(C) H – C| |O

– O , D –

D|C|D

– OH

(D) None is correct

Sol.[D] Step I :

D – C| |O

– D + OH

D –

OH|

–C| |O

D

Step II :

D –

OH|

–C|O

D + D – C| |O

– D Slow

D –

OH|C| |O

D –

D|

DC|O

Step III :

D– C| |O

–OH + NaOH D – C| |O

–O + Na

+ H2O

D–

D|

–C|O –

+ H2O D –

D|

–C|OH

D

D|

DCDOCD

|| |OHO


Q.13

3

23

O

S)CH( A

HO B; ‘B’ is –

(A)

(CH2)4

CHO

CHO (B)

OH

OH

(C)

OH

CHO

(D)

CHO

Sol.[B]



S)CH( 22 +O3

H|C = O

H|C = O

[A]

Intermolecular

Aldol Condensation

CHO

OH

Aldol

CHO

+ H2O

OH


Q.14 Compound (A) C7H10, on ozonolysis gives

only one compound (B)

C7H10 O3

(ii) Zn/H2O

O

H–C C

O

H

the structure (A) is

(A)

CH2=CH CH=CH2

(B)

(C)

(D)

Sol.[D]

Q.15 In Which of the following crossed aldol

condensations, only one kind of cross aldol is

formed –

(A) CH3CHO & CH3CH2CHO

(B) CH3CHO & (CH3)2CO

(C) (CH3)2CO & (C2H5)2CO

(D) C6H5CHO & CH3CHO

Sol.[D] C6H5–

O| |C –H+

H| |CH2 –

O| |C –HC6H5–

OH|CH –CH2–

O| |C –H.

LH. absent Aldol

In ths reaction only one kind of cross is formed,

because Benzaldehyde can't be have H.


Q.16 In the given reaction final product(s) will be -

Na/liq. NH3

(excess) (A) O3, Zn

H2O (B)

CH3

(A)

CH3–C–CH2–CHO, CH2 CHO

CHO

O

(B)

CH3–C–C–CH3,

CHO

CHO

O O

(C)

O

,

C=O

CH3

CH3

(D) (CH2)3

CHO

CHO

, O

Sol.[B]

CH3–C–C–CH3,

CHO

CHO

O O

Q.17 Alkaline hydrolysis of C4H8Cl2 gives a

compound (A) which on heating with NaOH

and I2 produces a yellow precipitate of

CHI3.The compound (A) should be.

(A) CH3CH2CH2CHO

(B) CH3CH2–C–CH3

O

(C) CH3 – CH2 – CH – CH2

OH OH

(D) CH3 – CH2 – CH – CH2

OH Sol.[B] CH3CH2–C–CH3

O

Q.18 The following statement is true for

Cannizzaro reaction–

(A) The aldehyde is oxidised as well as

reduced

(B) The aldehydes not containing -

Hydrogen atoms give the reaction



(C) The reaction is not given by aldehydes

containing -Hydrogen atoms

(D) All of these

Sol.[D] All of statement are correct

Q.19 Which of the following compounds would

under go the Cannizzaro’s reaction–

(A) Acetaldehyde (B) Benzaldehyde

(C) Propionaldehyde (D) Anisole

Sol.[B] Benzaldehyde would under go the

Cannizzaro’s reaction

Q.20 Product obtained on the addition of an

aqueous alkali to benzaldehyde followed by

acid hydrolysis is–

(A) Benzoic acid (B) Benzyl alcohol

(C) Benzyl benzoate (D) All of the above

Sol.[D] All

Questions

based on Test of Aldehyde acid Ketone

Q.21 Which of the following compounds would

not form a silver mirror with Tollen’s

reagent–

(A) R.CHO (B) Ar.CHO

(C) CH3.CO.R (D) R– C– C–H

O O

Sol.[C] CH3.CO.R would not form a silver mirror

with Tollen’s reagent

Q.22 Acetaldehyde on warming with Fehling

solution gives a red precipitate of –

(A) Elemental copper

(B) Cuprous oxide

(C) Cupric oxide

(D) Mixture of all of the above

Sol.[B] Aldehyde on warming with fehling solution

give a red precipitate of cuprous oxide

RCHO+2Cu(OH)2+NaOH

RCOONa+CH2O+ 3H2O

red ppt.


Q.23 If acetaldehyde is treated with Benedict’s or

Fehling’s solution, the following change

occurs in the system –

(A) Ag+ Agº (B) Cu

+2 Cuº

(C) Cu+2

Cu+ (D) Na

+ Naº

Sol.[C] Fehling solution

R–CHO+2Cu(OH)2 + NaOH

Cu+2

RCOONa + Cu2O + 3H2O

Cu+

Benedict's solution

RCHO + Cu+2

(complex) CH2O + Oxidaton

Red ppt product

In both of there reaction Cu2+

convert in Cu+1


Q.24 For distinction between CH3CHO and

C6H5CHO the reagent used is –

(A) KCN (B) HCN

(C) NH2OH (D) PCl5

Sol.[A] Acetaldehyde react with KCN to form

acetaldehyde cynohydrine.

CH3 – CHO HCl

KCN CH3 –

OH|

–C CN.

While Benzaldehyde react with KCN to form

Benzoin.

2

O| |C –H

Benzaldehyde

KCN C| |O

– –C|OH

Benzoin Therefore option (A) is correct.

Q.25 CH3–CH=CH.CHO is oxidised to

CH3CH=CH.COOH by using–

(A) Alkaline KMnO4

(B) Selenium dioxide

(C) Osmium tetraoxide

(D) Ammonical AgNO3

Sol.[D] Factual

Q.26 Acetaldehyde and acetone can be

distinguished by all the following except–

(A) Iodine + alkali (B) Tollen’s reagent

(C) Fehling solution (D) Schiff reagent

Sol.[A] Acetaldehyde and acetone can be



distinguished by Tollen’s reagent, Fehling

solution, Schiff reagent

Q.27 Schiff’s reagent is–

(A) Magenta solution decolourised with

sulphur dioxide

(B) Magenta solution decolourised with chlorine

(C) Ammonical cobalt chloride solution

(D) Ammonical manganese sulphate solution

Sol.[A] Factual

Q.28 Which of the following does not turn schiff's

reagent to pink –

(A) Formaldehyde

(B) Benzaldehyde

(C) Acetone

(D) Acetaldehyde

Sol.[C] Aceton can't be turn schiff's reagent to pink.

Those carbonyl compound which contain

Aldehyde group give the pink colour to react

with schiff's reagent.

Therefore option. (C) is correct.

Questions

based on Other Chemical Reactions

Q.29 An aldehyde isomeric with allyl alcohol

gives phenyl hydrazone. Pick out a

ketone that too gives a phenyl

hydrazone containing the same percentage

of nitrogen –

(A) Methyl ethyl ketone

(B) Dimethyl ketone

(C) 2– Butanone

(D) 2– Methyl propanone

Sol.[B]

CH2 = CH – CH2 – OH CH3CH2 – CHO

Allyl alcohol (Propanol)

Propanole is isomer of Allyl alcohol.

CH3 – CH2

H

C = O + H2–N–NH–

CH3–CH2

H C = N–NH–

Phenyl hydrozene

Phenyl hydrazone of propane

Both of hydrazone containe same percentage of

Nitrogen. Therefore option (B) is correct.

Q.30 C2H2

4

42

HgSO

SOH Dil P (O)

QPyridine

2SOCl

R Cd)H(C 252 S

The end product in the above sequence of

reactions is –

(A) Ethylethanal

(B) 2-butanone

(C) Propanal

(D) Propanone

Sol.[B] CH2=CH2 H/HgSO4

)P(3CHOCH

]O[

)Q(O| |

OHCCH3

Pyridine

SoCl 2

)R(O| |

OCCH3

(C2H5)2Cd

CH3CH2–

O| |C –CH2–CH3

(S)

2-butanone Therefore option (B) is correct.

Q.31 ? CN,

O2H,EtOH Benzoin.

The reactant is obtained by dry distillation

of the calcium salts of the following

pairs –

(A) C6H5CH2COOH, HCOOH

(B) C6H5COOH, HCOOH

(C) C6H4 (OH)COOH, HCOOH

(D) C6H4 (NH2)COOH, HCOOH

Sol.[B]

C6H5COOH + H

O| |C OH

Benzoic acid Formic acid

C6H5CHO

C2H5OH

CN–

(Benzoin) C6H5–

O| |C –

OH|CH –C6H5

This reaction is called. "benzoin condensation".




Q.32 The product of the reaction between acetone

and ethylene glycol is –

(A)

CH3

CH3

C

O–CH2

O–CH2

(B)

OHOH||CHCHCH 23

(C) CH3–CH–CH–CH2

O (D) None of these

Sol.[A]

CH3

CH3

C = 0 + 2

2

CH—HO|

CHOH

Ethylene glycol Acetone

CH3

CH3

C

Cyclic Ketals

O

O

CH2

CH2

Therefore option (A) is correct.

Q.33 Main product obtained by the oxidation of

C2H5COC3H7 by acidified dichromate –

(A) C3H7 COOH (B) C2H5COOH

(C) C3H8 (D) CH3COOH

Sol.[B] When oxidation of unsymmetrical Ketone occur

product is formed according to popaff's rule.

According to this rule the -CH group belongs

to the alkyl group which has high number of

carbons

CH3–CH2–CH2–

C| |O –CH2–CH3

4KMnO

OH

CH3–CH2COOH + CH3CH2COOH.


Q.34 Following is chemoselective for the reduction

of carbonyl compounds to corresponding

alcohol –

(A) LiAlH4

(B) {(CH3)2CHO} Al / (CH3)2CHOH

(C) B2H6

(D) H2/Pt

Sol.[A] LiAlH4 is more powerful reducing agent it

reduces corbonyl compound to corresponding

alcohol

CH3–

O| |C –H 4LiAlH

CH3–CH2–OH

Aldehyde 1º Alcohol

CH3–

O| |C –CH3 4LiAlH

CH3–

OH|CH –CH3

This reagent has no effect on carbon-carbon

multiple bond

CH3 – CH = CH –

O| |C – H 4LiAlH

CH3 – CH

= CH – CH2 – OH


Q.35

COCH3

22 NHNH)i(

KOHalcoholic)ii(product X

X is –

(A)

C=N NH2

|

CH3

(B)

CH2CH3

(C) Both correct

(D) None is correct

Sol.[B]

O| |C –CH3

22

2

NHNH)i(

OH

3

2

CH||

NH–NC

KOH.Alco

]H[4

3

2

CH|CH + NH2 – NH2

[X] Therefore option (B) is correct.

Q.36 CH3CHO 352 )HAl(OCA

ONaHC 52

B ; ‘A’ and ‘B’ are –

(A) CH3COOC2H5, CH3COCH2COOC2H5

(Tischenko) (Claisen)

(B) CH3COOH, CH3COOC2H5

(C) CH3COOH, C2H5OH

(D) All incorrect

Sol.[A]



CH3CHO 352 )HOC(Al

CH3COOC2H5

(Tischenko)

[A]

aNaHC 52 CH3COCH2COOC2H5

(Claisen)

[B]


Q.37 Which one of the following reactions can be

used to distinguish between benzaldehyde

and benzyl alcohol –

(A) KMnO4/oxidation (B) CrO3 oxidation

(C) Sodium metal (D) Flame test

Sol.[C] Benzylalcohol react with sodium metal and

liberate hydrogen gas. While benzaldehyde can't

be react with 'Na' metal

2C6H5–CH2OH + 2Na C6H5CH2–ONa + H2


True / False type questions

Q.38 Acetone gives iodoform test whereas HCHO

does not.

Sol. CH3

O| |C –CH3 (Acetone) contain, CH3–

O| |C – unit

so it gives iodform test whereas H–

O| |C –H can't

be contain CH3–

O| |C – or CH3–

OH|CH group so it

does not give iodoform Test. Therefore statement is True.

Q.39 wolff-Kishner reduction of acetophenone

gives toluene.

Sol.

CH3

C6H5

C = O OH/NHNH 22 CH3–CH2–C6H5

Acetophenone Ethyl Benzyne Therefore statement is false.

Q.40 Clemmensen reduction is more suitable for

reduction of those carbonyl compounds

which are sensitive to acids.

Sol. R– C| |O

–R ConHCl/HgZn

R–CH2–R This method is not used for acid sensitive

compound for Example : the given conversion can't be possible with reaction

Zn–Hg/HCl

ConHCl/HgZn

C| |O

–CH3

HO HO

CH2–CH3

CH2–CH3

HO Therefore statement is false.

Q.41 Phenylacetaldehyde on treatment with conc.

NaOH undergoes cannizzaro’s reaction.

Sol. Phenylacetaldehyde (C6H5–CH2–

O| |C –H) contain

-Hydrogen So it does not give Cannizzaro's Reaction. Therefore statement is false.

Q.42 Acetaldehyde on treatment with aqueous

NaOH solution gives sodium acetate and

ethyl alcohol.

Sol. Acetaldehyde on treatment with aqueous NaOH

solution gives Aldol compound.

CH3–

O| |C –H+

H|CH2 –

O| |C –H

NaOH

CH3–

OH|CH2 –CH2–

O| |C –H

Aldol

Therefore statement is false.

Fill in the blanks type questions

Q.43 2-Pentanone can be differentiated from

3-pentanone by ...........

Sol. Iodoform test.

Q.44 When 2-butyne is hydrated with dil. H2SO4 in

presence of HgSO4, it gives ...........

Sol. 2-butanone.

Q.45 Cannizzaro reaction is given by aldehydes

which do not have .........



Sol. -hydrogen.

Q.46 Fehling solution ‘A’ consists of an aqueous

solution of copper sulphate while Fehling

solution ‘B’ consists of an alkaline solution

of ..........

Sol. Rochelle Salt.

Q.47 Ethanol vapour is passed over heated copper

and the product is treated with aqueous

NaOH. The product is ...........

Sol. 3-hydroxy butanol.

EXERCISE # 2

Part-A (Only single correct answer type

Questions)

Q.1 In a Cannizzaro reaction the intermediate

which is the best hydride donor is –

(A) C6H5–

OH|

–C|

H

O–

(B) C6H5–

–O

|–C

|H

O–

(C)

O2N

H

O–

O–

(D)

CH3O

H

O–

O–

Sol.[D] – OCH3

is an electron releasing group this group

(– OCH3

) facilitates the release of hydride ion.


Q.2 C6H5COCl2

4

H

BaSOPd

Intermediate

Oxidation

IntermediateonDistillatiDry

SaltCa

A

Compound (A) in above reaction sequence

is–

(A) Benzophenone

(B) Benzaldehyde

(C) Acetophenone

(D) Benzoquinone

Sol.[A] C6H5COCl 2

4

H

BaSOPb

C6H5–

O| |C –H

]O[

C6H5–

O| |C –OH

Ca-salt

Dry Distillation C6H5–

O| |C –C6H5

[A] Benzophenone

Therefore (A) is correct.

Q.3 In the reaction series –

CH3CHO 4

42

KMnO

SOH.dil P 2SOCl

Q

Heat

COONaCH3 R. The product R is –

(A) (CH3CO) 2O

(B) Cl. CH2COOCOCH3

(C) CH3COCH2COOH

(D) Cl2. CHCOOCOCH3

Sol.[A] CH3CHO + 42

4

SOH.dil

KMnO

]P[O| |

OHCCH3 2SOCl

]Q[O| |

ClCCH3

CH3COONa

+NaCl CH3 C

| |O

CH3–

O| |C

O

(R) Therefore optioin (A) is correct.



Q.4

CHO

OH

forms

COOH

OH

on reaction

with –

(A) K2Cr2O7 + H2SO4

(B) Ag2O/OH–

(C) KMnO4

(D) All of these

Sol.[B]

CHO

OH

OAg2

COOH

OH

+2Ag ppt


Q.5 A + B forms

N

N

A and B are –

(A)

NH2

NH2

, CH3– CHO

(B) CH3CHO, CH3 - NH2

(C)

22

22

NHCH|

NHCH

,

CHO|CHO

(D) HCHO, CH3NH2

Sol.[C]

2

2

CH|CH

N–H2 + O = CH

N–H2 + O = O = CH

[A] [B]

–2H2O

2

2

CH|CH

N

N

CH

CH


Q.6 The following preparation of caprolactum

form the oxime of cyclohexanone involves a

rearrangement

called –

NOH

NH

C = O

(A) Pinacol-Pinacolone rearrangement

(B) Claisen rearrengement

(C) Beckmann rearrangement

(D) Curtius rearrangement

Sol.[C]

N–O–H

N

C – O – H

Tautomerisation

N

C = O

This rearrangement called.

Beckmann rearrangement.


Q.7 B 4NaBH CH=CH–CHO

/PtH2 A, then A & B are-

(A)

CH2CH2CHO,

CH=CH–CH2OH

(B)

CH2CH2CH2 OH,

CH=CH–CH2OH

(C)

CH=CH–CH2OH in both cases

(D)

CH2CH2CH2OH in both cases

Sol.[B]

CH=CH–CH2–OH 4NaBH

[B]

CH=CH–CHO Pt/H2

CH2–CH2–CH2–OH

[A]


Q.8 Compound (A) on ozonolysis gives the

following compound

OCH|||

CHO—)(CH—C—CH—CH—CH

3

4223struct



ure of compound (A) is -

(A)

CH2–CH

CH3

CH3

(B)

CH2–CH

CH3

CH3

(C)

CH2–CH

CH3

CH3

(D)

CH2–CH2–CH

CH3

CH3

Sol.[C]

CH2–CH

CH3

CH3

Part-B (One or more than one correct

answer type Questions)

Q.9 Aldehydes and ketones give addition

reactions with –

(A) HCN

(B) NaHSO3

(C) RMgX

(D) ROH

Sol.[A,B,C,D]

Aldehydes and ketones give addition Reaction

with HCN, NaHSO3

RMgx and R–OH

C = O + HCN

C OH

CN

C = O + NaHSO3

C OH

NaSO3

C = O + RMg

C R

OMg

C = 0 + R–OH

C R

OR Therefore option (A), (B), (C) and (D) are

correct.

Q.10 Which of the following reaction are used for

detection presence of carbonyl group?

(A) Reaction with hydroxylamine

(B) Reaction with hydrazine

(C) Reaction with phenylhydrazine

(D) Reaction with semicarbazide hydrochloride

Sol.[A,B,C,D]

Main reaction of carbonyl compounds is

Nucleophilic addition.

Reaction with ammonia derivatives are

nucleophile addition Reaction.

C = O + HO – NH2 C =N – OH

Hydroxylamine Oxime

C = O + NH2 – NH2 C = N – NH2

Hydrazine Hydrozone

C = O + NH2–NH– C = N–NH–

Phenyldrazine Phenylhydrazone

C = O + NH2–NH–

O| |C –NH2 C =N–NH–

O| |C –NH2

Semicarbazide Semicarbazone

Therefore option (A), (B), (C) and (D) are

correct.

Q.11 Which of the following are examples of aldol

condensation ?

(A) 2CH3CHO NaOH.dil

CH3CHOHCH2CHO

(B) 2CH3COCH3 NaOH.dil

CH3COH(CH3)CH2COCH3

(C) 2HCHO NaOH.dil

CH3OH+ HCOOH

(D) C6H5CHO + HCHO NaOH.dil

C6H5CH2OH

Sol.[A,B] Those compound gives Aldol condensation

which contain at least one -H.

Therefore option (A) and (B) are correct.

Q.12 Benzophenone (C6H5COC6H5) does not react

with –

(A) NaHSO3 (B) CH3OH

(C) HCN (D) NH2OH

Sol.[A,B,C] Benzophenon can not be react with NaHSO3,

CH3OH, and HCN, There option (A), (B) and

(C) are correct.

Q.13 Which of the following statements about

benzaldehyde is/are true ?

(A) Reduces Tollen’s reagent

(B) Undergoes aldol condensation

(C) Undergoes Cannizzaro reaction

(D) Does not form an addition compound

with sodium hydrogen sulphite.

Sol.[A,C]



Benzaldehyde Reduces the Tollen's reagent and it's undergoes cannizaro reaction because, it

can't be contain -H atom.

CHO NaOH.dil

CH2OH

+

O| |C Na

+

Therefore option (A) and (C) are correct.

Q.14 Which of the following compounds will give

a yellow precipitate with iodine and alkali ?

(A) 2-Hydroxypropane

(B) Acetophenone

(C) Methyl acetate

(D) Acetamide.

Sol.[A,B]

Those molecules is given yellow prepcipitate

will with I2 and alkali (Haloform Reaction).

Which contain CH3–

O| |C – unit or CH3–

OH|CH – in

the molecule attached to C or H.

CH3–

OH|CH –CH3 and CH3–

O| |C –C6H5 Give

2-hydroxy propane Acetophenone

halogorm Reaction but CH3–

O| |C –OCH3 and

CH3–

O| |C –NH2 can't be give haloform reaction

therefore option (A) mol (B) are correct.

Q.15 Base-catalysed aldol condensation occurs

with –

(A) Propionaldehyde

(B) Benzaldehyde

(C) 2-Methylpropanal

(D) 2, 2-Dimethylpropanal

Sol.[A] Base-catalysed aldol condensation occurs

with Propionaldehyde



Q.16 Under wolff-Kishner reduction condition, the

conversions which may be brought about are–

(A) Benzophenone into diphenylmethane

(B) Benzaldehyde into benzyl alcohol

(C) Cyclohexanone into cyclohexane

(D) Cyclohexanone into cyclohexanol

Sol.[A,C] wolff-Kishner reagent reduce > C = O into > CH2

> C = O OH/NHNH 22 > CH2

neBenzopheno

O| |

HCCHC 5656

OH/NHNH 22

methaneDiphenyl

HCCHHC 56256

OH/NHNH 22

Cyclohexanone Cyclohexane Therefore option (A) and (C) are correct.

Part-C (Assertion & Reasoning type

Questions)

The following questions 17 to 21 consists of

two statements each, printed as Assertion

and Reason. While answering these

questions you are to choose any one of the

following five responses.

(A) If both Assertion and Reason are true and

the Reason is correct explanation of the

Assertion.

(B) If both Assertion and Reason are true but

Reason is not correct explanation of the

Assertion.

(C) If Assertion is true but the Reason is false.

(D) If Assertion is false but Reason is true.

(E) If Assertion & Reason are false.

Q.17 Assertion : Chloral forms a stable hydrate.

Reason : On treatment with an aqueous

solution of NaOH, chloral undergoes

hydrolysis to give chloroform and sodium

formate.

Sol. [B] Hydrate of Chloral is more stable due to Intra-

molecular H-bonding.

CCl3CHO NaOH

CHCl3 + H

O| |C –ONa

–

Chloroform Sodium Formate


Q.18 Assertion : p-Methoxybenzaldehyde is less

reactive than benzaldehyde towards



cyanohydrin towards cyanohydrin formation.

Reason : + R-effect of the methoxy group

increases the electron deficiency of the

carbonyl carbon.

Sol. [C]

CH3 O

O| |C –H

C3

S(+)

+meffe. O| |C –H

C3

S(+)

Due to +m effect on the methoxy group

increases the electron deficiency of the carbonyl

carbon.


Q.19 Assertion : CH3– adds to >C=O group

irrevesibly but CN– ion adds reversibly.

Reason : CH3– ion is much stronger

nucleophile than CN– ion.

Sol.[A] CH3– ion is much stronger nucleophile than CN

–

ion CH3– add to >C = O group irreversible.

Or –

O| |C – + CH3 –

O|

–C|CH3

–

O| |C – + CH3 –

O|

–C|CN

(+)

(–)

–


Q.20 Assertion : Acetaldehyde reduces Fehling’s

solution but benzaldehyde does not.

Reason : Acetaldehyde is a stronger reducing

agent than benzaldehyde.

Sol. [A] Acetaldehyde is a stronger reducing agent than

benzaldehyde so acetaldehyde reduces Fehling's

solution but benzaldehyde does not. Therefore

option (A) is correct.

Q.21 Assertion : Both Grignard reagents and

dialkylcadmium react with acid chlorides to

form tert-alcohols.

Reason : Grignard reagents are more reactive

than dialkyl cadmium.

Sol. [B] Grignard reagents react with acid chloride to

form test-alcohols. but dialkyl cadmium react

with acid chloride to form ketone. Reactivity of

organometallic compounds on the electro

positive character of the metal Mg metal is more

electropositive than so the grignard reagent

(RMgx) are more reactive than dialkyl-

cadmium.

Therefore Assertion is false but reason is true

option (D) is correct.

Part-D Column Matching

Q.22 C8H8O 2INaOH

P + Q

C| |O

–H

NaOH.Conc Q + R

CH3– C| |O

–H 2INaOH

P + S

Column A Column B

(a) P (i) NaOCH||O

(b) Q (ii) CHI3

(c) R (iii) NaOCPh| |O

(d) S (iv) Ph – CH2OH

Sol. (a)-ii, (b)-iii, (c)-iv, (d)-i

Factual

Q.23 Column A Column B

Compound Enol %

(a)

O

O

(i) 99.9

(b) CHCl2CHO (ii) 90

(c) CH3–

O| |C –CH2–

O| |C –CH3 (iii) 2.5

(d) CH3 –

O| |C – H (iv) 1

Sol. (a)-i, (b)-iii, (c)-ii, (d)-iv

Factual



EXERCISE # 3

Part-A Subjective Type Questions

Q.1 Pure HCN fails to react with aldehydes and

ketones.

Sol. HCN is weak acid and have low degree of

dissociation but in presence of a base (even

H2O), the dissociation increases appreciably to

provide appreciable CN¯ to attack >C=O bond.

Q.2 Ketones are less reactive than aldehydes.

Sol. The positive I.E. of two alkyl groups in ketones

makes the carbon atoms less positive and makes

it less reactive in comparison to aldehydes.

R

H C = O

R

H C

+ – O

–

R

R C = O

R

R C

+ – O

–

Less positive charge and thus less reactive

nature.

Q.3 Oximes are more acidic than hydroxyl amine.

Sol. NH2OH NH2O¯

Acid Base

>C=N-OH>C= N..

-

.. :

..O >

C..

– N..

= ....O

Acid Base

In oxime base, delocalization of electron (i.e.

resonance) stabilize it and thus conjugate acid

i.e. oximes are more acidic. No such resonance

exists in hydroxyl amine base (NH2O–)

Q.4 In the laboratory preparation of

acetaldehyde by oxidation of ethanol,

acetaldehyde is distilled out continuously.

Sol. During oxidation of ethanol, acetaldehyde is

formed in I step which should be distilled out,

otherwise further oxidation of acetaldehyde will

lead the formation of acetic acid.

Q.5 Although both >C = O and >C = C<

groupings have double bond they exhibit

different types of addition reactions.

Sol. The >C = C< grouping undergoes electophilic

addition while > C = O undergoes nucleophilic

addition reactions. The difference is because of

the fact that C of >C = O is more electrophilic

than carbon of > C = C < due to more

electronegative nature of O then C. Hence of

> C = O reacts with nucleophiles. The >C = C <

grouping is nucleophilic and hence adds mainly

electrophiles.

Q.6 What happens when 2, 2-Dichloropropane is

treated with aqueous NaOH and the product

treated with amalgamated zinc in presence of

conc. HCl.

Sol.

H3C

H3C C

Cl

Cl

NaOHaq CH3

CH3

C=O HClConc

)Hg(Zn

CH3

CH3

CH2

Q.7 NaHSO3 is used for the purification of

carbonyl compounds.

Sol. Carbonyl compounds form solid additive

products with NaHSO3 which are separated out.

The solid bisulphites of carbonyl compounds on

hydrolysis by dil. acid regenerate original

carbonyl compound and thus this property is

used for the purification of carbonyl compounds

as well as for their separation.

Q.8 Protic acids, Lewis acids and bases all

activate carbonyl group.

Sol. Protic and Lewis acids increase the partial

positive charge on the carbonyl carbon and

hence activate the group.

>C = O + H+ > C

– OH

Bases activate the -methylene component of

the carbonyl compounds by converting them

into carbanions.

R - CH2 – CHO + : B R – HC

– CHO + BH

Note : Proton can be expelled only from -

carbon because such carbanion can stabilize due

to resonance.

Q.9 Differentiate between acetophenone and

benzophenone.

Sol. Acetophenone (C6H5. CO.CH3) gives haloform

test while benzophenone (C6H5. CO. C6H5) not.

Q.10 How will you prepare Diethyl ketone from

propionic acid : (2 steps).

Sol. CH3 .CH2 .COOH 2Ca(OH)



(CH3 .CH2COO)2Caondistillati

dry

CH3.CH2 .COCH2 .CH3

Q.11 An organic compound (A) with molocular

formula C3H6O is not easily oxidised. It gives

on reduction C3H8O (B) which on reaction

with HBr gives a bromide (C). Grignard

reagent obtained from (C) reacts with (A) to

give C6H14O (D). Identify compounds from

(A) to (D).

Sol. CH3COCH3 (Acetone) (A) CH3CHOHCH3 (Isopropyl alcohol); (B) CH3CHBrCH3 (Isopropyl bromide) (C) (CH3)2CH–C–(OH)(CH3)2(2,3-Dimethyl butanol-2)

(D)

Q.12 Two organic compounds ‘A’ and ‘B’ with

molecular formula, C3H6O, react with HCN

in different manner to produce, (C) and (D)

respectively on subsequent hydrolysis of (C)

and (D) gives optically active substances.

Sol. CH3CH2CHO; CH2 = CH CH2OH; (A) (B) CH3CH2CH(OH)CN; CH3 – CH(CN)CH2OH; (C) (D)

CH3CH2 HC*

(OH)COOH;

(E)

CH3 – HC*

(COOH) CH2OH

(F)

Q.13 An organic compound (A) adds Br2 to give

C5H8Br2O. It does not react with Tollen’s

reagent but enters into reaction with phenyl

hydrazine. Ozonolysis of (A) gives an

acetaldehyde and C3H4O2 which readily loses

CO to form acetaldehyde. What is (A) ?

Sol. CH3CH=CHCOCH3

(A)

Q.14 An organic compoud (A), C2Cl3HO, reduces

Tollen’s reagent and on oxidation gives a

monocarboxylic acid (B), C2Cl3HO2. (B) on

distillation with sodalime gave a sweet

smelling liquid (C) containing 89.12% Cl.

What are (A), (B) and (C).

Sol. CCl3CHO, CCl3COOH, CHCl3

(A) (B) (C)

Q.15 An organic compound (A) C9H10O is inert to

Br2 in CCl4. Vigorous oxidation with hot

alkaline permanganate yields benzoic acid.

(A) gives a precipitate with semicarbazide

hydrochloride and with 2,4-dinitrophenyl

hydrazine. Write all possible structures of

(A).

Sol. C6H5COCH2CH3; C6H5CH2COCH3;

(A) (B)

C6H5CH2CH2CHO; C6H5CH(CH3)CHO

(C) (D)

Q.16 An organic compound having molecular

formula C5H10O exists in two chain isomers,

(A) and (B). Isomer (A) undergoes the

Cannizzaro’s reaction to give 2, 2 - dimethyl

propanoic acid and 2, 2-dimethylpropanol-1.

Compound (B) in the presence of dilute alkali

undergoes aldol condensation to from 3-

hydroxy-2-propyl heptanal. Give graphic

representations of (A) and (B).

Sol. (CH3)3CCHO CH3CH2CH2CH2CHO

(A) (B)

Q.17 Two organic compounds (A) and (B) have

same empirical formula CH2O. Vapour

density of (B) is (A) twice the vapour density

of (A). (A) reduces Fehling solution but does

not react with NaHCO3. Compound (B)

neither reacts with NaHCO3 nor reduces

Fehling solution. What are (A) and (B) ? Also

report an isomer of (B) if it reacts with

NaHCO3.

Sol. (A) HCHO (Formaldehyde)

(B) HCOOCH3 (Methyl methanoate)

Q.18 An organic compound (A) with molecular

formula C3H6O is oxidised by Fehling

solution and gives silver mirror with Tollen’s

reagent. (A) gives on reduction C3H8O (B)

with hydrobromic acid followed by heating

with Mg (B) gives (C), which reacts with

ethylene oxide to give C5H12O (D). On

oxidation (D) gives C5H10O2 (E). Idientfy

compounds from (A) to (E).

Sol. CH3CH2CHO; CH3CH2CH2OH;

(A) (B)



CH3CH2CH2MgBr; CH3CH2CH2CH2CH2OH;

(C) (D)

CH3CH2CH2CH2COOH

(E)

Q.19 Write the structure of the product (s) formed

in each case.

(a)

C| |O

CH2CH3

Cl

22

2

ClCH

Cl

(b)

O

CH3

OH;NaOH

SHCHHC

2

256

(c)

O

CH3

+ LiCu (CH3)2 OH.2

etherdiethy l.1

2

(d)

O

+C6H5CHONaOH

,waterethanol

Sol. (a)

Cl|

CCHCH| |O

3Cl

(b)

SCH2C6H5

O

CH3

(c)

CH3

O

CH3

(d)

CHC6H5

O

Q.20 Suggest reasonable structure for compounds

A and B.

O

O

CH3

+ CH2=CH– C| |O

–CH3

Methanol

KOH A

O

O

CH3

CH2 CH2C| |O

CH3

Benzene

])CH(OC[Al 333 B

Sol. (A)

O

O

CH3

CH2CH2 (CO)CH3

(B)

O

O

CH3

Q.21 Identify the reagents appropriate for each step

in the following syntheses -

(a)

OH

OH

HC| |O

CH2CH2CH2CH2C| |O

H

CH| |O

(b)

CH3

CH(CH3)2

CH3C| |O

CH2CH2

23

2

)CH(CH|

CHCHCH| |O

(CH3)2CH 3CCH| |O

Sol. (a)

OH

OH

HIO4 H– C| |O

–CH2–CH2–CH2– C| |O

–H

NaOH/

C| |O

H

(b)



CH3

CH(CH3)2

O3 CH3– C| |O

–CH2–CH2–

23

2

)CH(CH|

CHCH C| |O

–H

NaOH/

C| |O

–CH3

H2O/Zn

Aldol

Condensation

(CH3)2CH

Q.22 Outline reasonable mechansims for each of

the following reactions.

(a)

CH2CH2CH2CH2Br

O

benzene

)CH(KOC 33

O

(76%)

(b) C6H5C| |O

C| |O

C6H5 + C6H5CH2C| |O

CH2C6H5

KOH

O

C6H5

C6H5 C6H5

C6H5

(91 - 96%)

(c) C6H5CH2C| |O

CH2CH3 + CH2=

56

56

HC|

HCCC| |O

OHCH

NaOCH

3

3

O

C6H5 CH3

C6H5

C6H5

(51%)

Sol. (a)

CH2CH2CH2CH2Br

O

benzene

)CH(OK 33

CH2CH2CH2CH2

O

H

O

–h

(b)

C6H5

O| |C

O| |C

C6H5

+

C6H5 CH2

O| |C

CH2 C6H5

KOH

Alodal condensation

C

C6H5

HO

C

C C

C6H5

O| |C

C6H5

H OH

C6H5

Alodal

–2H2O

C

C6H5

C

C6H5

C6H5 C6H5

O| |C

(c)



C6H5

C

CH2 O| |C

C6H5

+

CH2

C6H5 O| |C

CH2 CH3

Na2CO3

CH3OH

CH2

C6H5

O| |C

C|

HC 56

C–OH

C–H

CH3

C6H5

H|

CH

–H2O

C|

HC 56

CH2 C–C6H5

C–CH3 CH

C6H5 O| |C

Part-B Passage based Question

Passage - I (Q. 23 to 27)

Aldehyde without –H atoms on reaction with

concentrated alkali solution undergo an

oxidation-reduction (disproportionation)

reaction. One half of the aldehyde is reduced to

a 1° alcohol and the other is oxidised to a

carboxylic acid. This is known as Cannizzaro

reaction.

2–C–H

O

NaOH%50

– CH2OH +

–C–O–Na

O

(no –H)

Two steps are involved in this reaction.

(i) Attack of OH ion on the carbonyl group of

one aldehyde yields an oxyanion.

H–C–H + OH

O

H–C–H

O

OH

(oxyanion)

(ii) As a result of hydride transfer, the final

product is formed

H–C–H

O

OH

C = O H

H

+

H–C + CH3O

O

OH

Proton

exchange

H–C–ONa + CH3OH

O

– + Na

H–C + CH3OH

O

O

Intramolecular Cannizzaro’s reaction is also

possible

CHO

CHO

NaOH CH2OH

COONa

Q.23 The aldehyde which not shows Cannizzaro’s

reaction is -

(A) HCHO (B) C6H5CHO

(C) CCl3–CHO (D) (CH3)3C. CHO

Sol.[D] The aldehyde which has not -H shows

Cannizzaro’s reaction.

Q.24 The product formed in the following reaction

will be -

CHO

CHO

NaOH product -

(A) CH2OH

CH2OH (B)

CH2OH

COONa

(C)

COONa

COONa (D) All of these

Sol.[B] CHO

CHO

NaOH CH2OH

COONa

Q.25 Mixture of C6H5CHO and HCHO is treated

with NaOH then cannizzaro’s reaction

involves-

(A) Oxidation of HCHO

(B) Reduction of HCHO and Oxidation of

C6H5CHO

(C) Reduction of C6H5OH

(D) Both A & C

Sol.[D] Reduction of HCHO and Oxidation of

C6H5CHO, Reduction of C6H5OH



Q.26 Product (P) in the following reaction

Ph–C–CHO

O

OHHC

HO

52

(P)

(A) Lactic acid (B) Man delic acid

(C) Salicylic acid (D) Malonic acid

Sol.[C] Ph–C–CHO

O

OHHC

HO

52

Ph–C–COOH

O

Q.27 Identify A, B, C in following reaction

C8H6O2OHHC

HO/CH2

52

2 C8H10O2 3PBrC8H8

Br2

(A) (B) (C)

(A) CHO

CHO

CH2.OH

CH2.OH

CH2.Br

CH2.Br

(B) CHO

CHO

OH

OH

OH

Br

Br

Br

(C) CHO

CHO

CH3 OH

OH CH3

Br

Br

(D) All of the above

Sol.[D] All of the above

Passage - II (Q. 28 to 32)

Most common reaction for aldehyde &

ketones are nucleophilic addition reactions.

The reactivity of the carbonyl group arises

from the electronegativity of the oxygen atom

and the resulting polarisation of the C=O

bond. The electrophilic carbon atom is sp2

hybridised and flat, leaving it relatively

undinderd and open to attack from either face

of the double bond. As a nucleophile attacks

the carbonyl group, the carbon atom changes

hybridisation from sp2 to sp

3. The electrons o

the bond are forced out to oxygen atom to

form an alkoxide anion, which protonates to

give the product. A carbonyl group is a weak

base and it can become protonated in an

acidic solution. Protonated carbonyl group is

strongly electropilic, inviting attack by a

weak nucleophilic.

In most cases aldehydes are more reactive

than ketones towards nucleophilic addition.

This reactivity can be explained by electronic

effect and steric effect.

Q.28 In the following reaction the nucleophile that

attacks the ketone

33 CHCCH||O

4NaBH

OHCH–CH 23

33 CHCHCH|OH

(A) B – ··

(B) CH3–CH2– O – ··

(C) H – ··

(D) NaBH4

Sol.[C] Step I

CH3

CH3

C = O + NaBH4

CH3–C–O–BH3Na

|

CH3

|

H

Nucleophile

attack

CH3 –C–OBH3Na +

CH3 |

H

|

Step II

CH3 –

H|

–C|CH3

O BH3Na+B + CH3

CH3 C=O

CH3 –

H|

–C|CH3

O – B Na

4

H2O B(OH)3+NaOH+Cl3– –CH

|CH3

OH In this reaction the Nucleophile is H

–.


Q.29 Which of the following reactions is feasible

favourable.

(A) CH3–CH=O+H2O H

CH3–CH OH

OH



(B) 33 CHCCH||O

+H2O H

OH

|CHCCH

|OH

33

(C)

C| |O

+H2O H

OH|C|OH

(D)

O + H2O H OH

OH

Sol.[D] Steric hindrance and mere alkyl substitutents

make carbonyl compounds less reactive towards

any nucleophile, electron-with drawing groups

and small rings make them more reactive. More

is the angle strain in the cyclic ketones, more is

it's reactivity for nucleophilic addition reactions.

So

O is more reactive than other

carbonyl compound. Which given in option.


Q.30 What will be the product in the following

reactional

CH3– C| |O

–CH2– C| |O

–H COOHCH

BH)COOCH(Na

3

33 P

(A) CH3– C| |O

–CH2 –

H|

–C|

OH

H

(B) CH3– CH|

OH

–CH2– C| |O

–H

(C) CH3– CH|

OH

–CH2–CH2–OH

(D) CH3– CH|

OH

–CH2–COOH

Sol.[A]

CH3COCH2

H

C=O + + –

H|B

(CH3)3

CH3COCH2

H

C=O +

H|B

(CH3COO)3

CH3COCH2

H

C=OB

H

(CH3COO)3

CH3

O| |C –CH2–

H|

H–C|

OH

H–O–H

therefore option (A) is correct.

Q.31 Correct reactivity order towards nucleophilic

addition –

(A) HCHO > CH3–CHO > C6H5–CH2–CHO

(B) HCHO > CH3– C| |O

–CH3 > CH3–CHO

(C) HCHO > CH3– C| |O

–CH3 > C6H5–CHO

(D) HCHO > C6H5–CH2–CHO > CH3CHO

Sol.[D] Reactivity towards Nucleophilic addition a

groupofpowerI

1

– I group

So the reactivity of different carbonyl compounds in decreasing order.

H– O| |C

–H > I

56HC

– CH2–CHO > I3CH

CHO


Q.32 Choose the correct statement about

cannizzarro’s reaction –

(A) Rate determining step involves

nucleophilic attak of _

OH

(B) Rate determining step involves the

nucleophilic attack of carbanion

(C) Rate determining step involves the

nucleophilic attack of hydride ion

(D) All of the above

Sol.[C] According to Reaction mechanism of Cannizzaro reaction

Step I

R–

O| |C –H + OH

R–

OH|

OH–C|O

Step II



R–

OH|

H–C|O–

+H – C| |O

–R Slow R– C| |O

–OH+H–

H|

R–C|

O

attack of hydride Step III

R– C| |O

–OH+NaOH R– C| |O

–O– + Na

+ H

R–

H|

–C|O

H + H2O R–

H|

–C|OH

H

Therefore in two reaction mechanism.Read

determining step (slow step) involves the nucleophilic

attack of hydride. So option (C) is correct.

Passage –III (Q. 33 to 37)

Aldehydes and ketons are specially

susceptible to nucleophilic addition because

carbonyl group C = O

is polar (due to

electronegativity difference between carbon

and oxygen).

C = O

–

Positive charge on carbon makes it reactive

towards the nucleophile. This addition is

catalysed by acid.

Reactivity of carbonyl compound towards

nucleopilic addition increases with increase in

the electron deficiency at carbonyl carbon.

Thus, (– I.E.) groups increase while (+ I.E.)

groups decrease the reactivity of carbonyl

compound.

Q.33 Which among the following is most reactive

to give nucleophilic addition ?

(A) FCH2CHO (B) CICH2CHO

(C) BrCH2CHO (D) ICH2CHO

Sol.[B] CICH2CHO is most reactive to give

nucleophilic addition

Q.34 Select the least reactive carbonyl compound

for nucleophilic addition -

(A)

C6H5–C–C6H5

CH3OH

O

(B)

C6H5–C–CH3

CH3OH

O

(C)

C6H5–C–H

CH3OH

O

(D)

CH3–C–H

CH3OH

O

Sol.[B]

C6H5–C–CH3

CH3OH

O

Q.35 Which one of the carbonyl compound is most

reactive towards NaCN/

H

(A) PhCHO

(B) CH3.O

CHO

(C) CH3

CHO

(D) NC

CHO

Sol.[A] PhCHO carbonyl compound is most

reactive towards NaCN/

H

Q.36 The most reactive compound towards

cyanohydrine formation is

(A) PhCHO

(B) NO2

CHO

(C) HO

CHO

(D) PhCH2CHO

Sol.[D] PhCH2CHO is the most reactive

compound towards cyanohydrine formation

Q.37 The increasing order of the rate of HCN

addition to compounds a – d is -

(a) HCHO (b) CH3COCH3

(d) PhCOCH3 (d) PhCOPh

(A) a < b < c < d (B) d < b < c < a

(C) d < c < b < a (D) d < a < c < b

Sol.[B] d < b < c < a

Passage –IV (Q. 38 to 40)

An organic compound (A) with molecular

formula C5H8 when treated with Na in liq.

NH3 followed by treatment with n-propyl



iodide yields C8H18(B). When treated with

dil. H2SO4 containing Hg2+

, (A) gives the

ketone C5H10O (C). On oxidation with

alkaline KMnO4, (B) gives two isomeric

acids C4H8O2 (D & E).

Q.38 Compound (C) may give following reaction -

(A) Cannizaro (B) Haloform

(C) Both (D) None

Sol.[B] Compound (C) is CH3–CH2–

O| |C –CH3.

and 2-pentanone


Q.39 When (A) is treated with BH3/H2O2, HOH the

compound formed contains following group -

(A) an aldehyde

(B) a ketone

(C) an alcohol

(D) acarboxylic group

Sol.[A] CH3–CH2–CH2–CCH 223 OH/BH

OH|CHCHCHCHCH 2223

CH3–CH2–CH2–CH2–CHO

Pentanaldehyde


Q.40 (D) & (E) on decarboxylation with soda lime

produces -

(A) same alkanes (B) different alkanes

(C) Isomeric alkanes (D) can't be predicted

Sol.[A] CH3–CH2–CH2–

O| |C –OH NOH/CaO

CH3CH2–CH2–CH3

Same alkane is formed therefore option (A) is

correct.

Passage - V (Q. 41 to 45)

An organic compound A(C11H14O) reacts

with 1 mole of Br2 to give C11H14Br2O. A on

treatment with aluminium isopropoxide in

presence of excess of P-benzoquinone to give

B with molecular formula C11H12O. B also

reacts with 1 mole of Br2 to C11H12Br2O. B

reduces Tollen’s reagent and when reacted

H2O2 followed by periodic acids produces

two compounds C(C8H8O) & (C3H4O3)D

reduces ammonical silver nitrate solution

forming malonic acid. C upon permaganate

oxidation gives E (C8H6O4) which is soluble

in aqueous alkali and on heating it give its an

hydride C with molecular formula C8H4O3.

Q.41 What is A in the above passage –

(A)

CH3

CH=CH–CH2–CH2–OH

(B) CH3


(C) CH2–OH

CH=CH–CH2–CH3

(D) none of these

Sol.[B] A CH3



Q.42 Identify B

(A)

OH|

CHOCHCH–CH|OH

2

(B)

OHOH||

CHOCHCHCH 2

CH3

(C)

CH=CH–CH2–CHO

CH3

(D)

O||

CHC 3

CH3

Sol.[C] B

CH=CH–CH2–CHO

CH3


Q.43 What is C in the given passage –

(A)

CHO CH3

(B)

CH2–OH CH3

(C)

OCH3

CH3

(D)

CHO

CHO

TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,

8764333388 CARBOXYLIC ACID & IT'S DERIVATIVES 1

CARBOXYLIC ACIDS & IT'S DERIVATIVES

EXERCISE # 1

Questions

based on Preparation of carboxylic acid and

derivatives

Q.1 Give the order of decarboxylation of the

following acid –

CH3COOH ; CH2 =CH–CH2 – COOH ;

I II

CH2(COOH)2 ;

COOH O2N

NO2

NO2

III IV

(A) I > II > III > IV (B) III > IV > II > I

(C) IV > III > II > I (D) I > III > II > IV

Sol.[C]

CH3 – C – O CH3 + CO2

O unstable (I)

CH2=CH–CH2 –C–O

O Stable due to

conjugation (II)

CH2=CH–CH2 +CO2

H–O–C–CH2 –C – O

O

stabilised due to

resonance

HO–C –CH2

O O

C – O O2N

NO2

NO2 O

+CO2 O2N

NO2

NO2

more stable IV

Therefore stability of carbanion -decarboxy-

lation IV > III > II > I.

Option (C) is correct.

Q.2 The end product Y in the sequence of

reaction:

RX CN

X NaOH

Y is –

(A) An alkene

(B) A carboxylic acid

(C) Sod. salt of carboxylic acid

(D) A ketone

Sol.[C] RX CN

R–CN NaOHD2 R–

O| |C –H+NH3

NaOH

R–

O| |C –O

–Na

+

Sodium Salt of carboxylic acid.

Q.3 The oxidation of toluene with hot KMnO4

gives –

(A) Benzoic acid (B) Benzaldehyde

(C) Benzene (D) Benzyl alcohol

Sol.[A]

CH3

]O[

KmNOHot 4

CH3

+ H2O

Benzoic acid

Q.4 Identity the product Z in the series :

CH3CN OHHC/Na 52 X 2HNO

Y OH/KMnO4 Z

(A) CH3CHO

(B) CH3CH2CONH2

(C) CH3COOH

(D) CH3CH2NHOH

Sol.[C] CH3COOH

Questions

based on Chemical properties of carboxylic acids

and derivatives

Q.5 Which of the following acid does not form

anhydride –

(A) CH3COOH

(B) CH-COOH

CH-COOH

(C)

CH2COOH

CH2COOH

(D) CH-COOH

HOOC-CH Sol.[D] CH-COOH

HOOC-CH

does not form anhydride


CARBOXYLIC ACID & IT'S DERIVATIVES 2

Q.6 Formic acid and formaldehyde can be

distinguished by treating with –

(A) Benedict’s solution (B) Tollen’s reagent

(C) Fehling’s solution (D) NaHCO3

Sol.[D] Formic acid is strong acid than formaldehyde

formic acid react with NaHCO3 and liberated

CO2 but formaldehyde does not react with

NaHCO3.


Q.7 The product formed when adipic acid is

heated–

(A) COOH

(B)

C=O

(C)

O

O

O

(D)

COOH

COOH

Sol.[B]

O

CH2–CH2–C–OH

CH2–CH2–C–OH

O

O

+ CO2 + H2O

Adipic acid Cyclo pentanone

Adipic acid undergo dehydration as well as

decarboxylation to give cyclopentanone when

its heated.

So option (B) is correct.

Q.8 Ethyl acetate react with hydrazine to form -

(A) Acetamide and urea

(B) Acetic acid hydrazide and ethanol

(C) Hydroxamic acid and ethanol

(D) Ethyl isocyanate

Sol.[B]

O

CH3–C–OC2H5 + NH2–NH2

O

CH3–C–NH–NH2 + C2H5OH

Acetic acid

hydrazide

Ethanol


Q.9 X and Y in the following reaction sequence

are–

X 52OP

R– C||O

–NH2OH

Br2 Y

(A) R–CN, RCH2NH2

(B) RNH2, RCN

(C) RCN, RNH2

(D) RCN, RCH2NH2

Sol.[C] H2O+R–CN 52OP

R– C| |O

–NH2 OH

Br2 R–NH2

[X] [Y]


Q.10 Benzoic acid gives benzene on being heated

with X and phenol gives benzene on being

heated with Y. Therefore X and Y are

respectively –

(A) Soda lime and copper

(B) Zinc dust and sodium hydroxide

(C) Zinc dust and soda lime

(D) Soda lime and zinc dust

Sol.[D]

COOH

)elimsoda(

CaO/NaOH

+ CO2

OH

dustZn

+ ZnO

Therefore X and Y are respectively sodalime

Zn-dust.

Option (D) is correct.

Q.11 COOH

is converted into

CHO

by–

(A) (i) LiAlH4 (ii) Cu/

(B) (i) Cu/ (ii) LiAlH4

(C) (i) Ag2O (ii) Cu/

(D) (i)SOCl2(ii)LiAlH4

Sol.[A] COOH

4LiAlH

OH



Cu /

CHO

Therefore option (A) is correct

Q.12

O

.O OHCH3 Y, Y is –

(A)

OH COOCH3

CH2 – CH2 (B)

OCH3 COOH

CH2 – CH2

(C) Both are correct (D) None is correct

Sol.[A]

O

.O

3OCHH

O

.OH C – OCH3

[Y]


Q.13 End product of this conversion

CH3– C| |O

–CH2CH2CH2CO2H H/OH.2

NaBH.1

2

4 is –

(A)

O

O

CH3

(B)

O

O

CH3

(C)

O

O

(D) CH3–

OH

CH –CH2 – CH2CO2H

Sol.[A] CH3– C| |O

–CH2CH2CH2COOH 4NaBH

CH3–C–CH2–CH2–CH2–C=O

O H H–O

H+/H3O+

O

O

CH3


Q.14

COOH + NaHC*O3

—

COONa + CO2

C* is with in the product –

(A) CO2

(B)

COONa

(C) Both

(D) None

Sol.(A)

C–O–H + Na O–C–OH

O

*

O

Sod. Bicarbonate

C–O–Na + H–O–C–O–H

O

*

O

CO2 + H2O


Q.15 Identify the product A in the following

reaction

COOH CH2

3 COOH

CH3COOH + A

(A) CO2 (B) CH3CHO

(C) CH3OH (D) None of these

Sol.[A] COOH CH2

3 COOH

CH3COOH + CO2


Q.16 A carboxylic acid (x) & (y) on heating with

P2O5 (x) give corresponding anhydride but

(y) remain unaffected. Carboxylic acid (y)

is-

(A) CH3COOH (B) HCOOH

(C) both (A & B) (D) CH3CH2COOH

Sol.[B] Carboxylic acid (y) is HCOOH

Q.17 Malonic acid and succinic acid are

distinguished by -

(A) Heating (B) NaHCO3

(C) Both A & B (D) None of these

Sol.[A] By Heating

Q.18 Which of the following statement is correct -

(A) Nucleophilic additions to the carbon-

oxygen double bond is a characteristic

reaction of aldehyde and ketones.



(B) Carboxylic acid and their derivative are

characterized by nucleophilic substitution

reaction.

(C) In both case initial step involves a

nucleophilic attack on the carbonyl

carbon but in second step aldehyde and

ketone accept a proton to yield additional

product whereas in the case of acyl

compound leaving group is ejected to

form substitutional product.

(D) All are correct

Sol.[D] Factual

Q.19 The relative order of reactivity of acyl

derivatives is -

(A) R – C||O

–Cl > R – C||O

– O – C||O

– R >

R – C||O

– NH2 > R – C||O

– OR'

(B) R – C||O

–Cl > R – C||O

– OR' >

R– C||O

– O – C||O

– R > R – C||O

– NH2RBT

(C) R – C||O

– Cl > R – C||O

– O – C||O

– R >

R – C||O

– OR' > R – C||O

– NH2R

(D) None of the above

Sol.[C] R – C||O

– Cl > R – C||O

– O – C||O

– R >

R – C||O

– OR' > R – C||O

– NH2R

Q.20 What is the correct order of alkaline

hydrolysis of different ester CH3COOCH3

[rate = r1] , CH3COOC2H5

[rate = r2] CH3COOC3H7 [rate = r3] -

(A) r1 >r2 > r3 (B) r1 < r2 < r3

(C) r1 < r2 > r3 (D) r1 > r2 < r3

Sol.[A] r1 >r2 > r3

Q.21 Which of the following esters cannot undergo

claisen self condensation -

(A) CH3CH2CH2CH2COOC2H5

(B) C6H5COOC2H5

(C) C6H11CH2COOC2H5

(D) C6H5CH2COOC2H5

Sol.[ B] C6H5COOC2H5 esters cannot undergo claisen

self condensation

Q.22

2O

)A(

C

O

C

O

O

elimsoda/OH2

COOH

Oxidizing agent (A) used is

(A) K2Cr2O7 / H+ (B) AlK . KMnO4

(C) Chromic Acid (D) V2O5

Sol.[D] Oxidizing agent (A) used is V2O5

Q.23 In the reaction

C6H5NH2 Cº50

HCl/NaNO2

(A) KCN

CuCN (B)

OH/H 2 (C)

the product (C) is -

(A) C6H5CH2NH2 (B) C6H5COOH

(C) C6H5OH (D) None of these

Sol.[B] Product (C) is C6H5COOH

Q.24 Benzoic acid on treatment with hydrazoic

acid (N3H) in the presence of concentrated

sulphuric acid gives

(A) Benzamide

(B) Sodium benzoate

(C) Aniline

(D) C6H5CON3

Sol.[C] Factual

Q.25 Number of cross products in the given

reaction: (Without considering stereoisomers)

CH3COOC2H5 + C6H5–CH2–COOC2H5

OHHC

ONaHC

52

52

(A) One (B) Three

(C) Two (D) Four

Sol.[C] Two

Q.26 Arrange following compounds in decreasing

order of reactivity for hydrolysis reactions -

(I) C6H5COCl

(II)

COCl NO2

(III)

COCl CH3



(IV)

C–Cl OHC

O

(A) II > IV > I > III (B) II > IV > III > I

(C) I > II > III > IV (D) IV > III > II > I

Sol.[A] II > IV > I > III

Q.27 In the given reaction sequence :

CH3–CH2–OH

H)ii(

/HO/KMnO)i( 4 (A)

/NH)ii(

SOCl)i(

3

2 (B) KOH/Br2 (C) will be -

(A) Methylamine

(B) Eltylamine

(C) Propylamine

(D) Acetamide

Sol.[A] (C) will be Methylamine

Q.28 Which optically active compound on

reduction with LiAlH4 will give optically

inactive compound ?

(A) CH3–CH–COOH

OCH3

(B) CH3–CH2–CH–COOH

OH

(C) CH3–CH2–CH–COOH

CH2OH

(D) CH3–CH–CH2–COOH

OH Sol.[C] CH3–CH2–CH–COOH

CH2OH

Q.29

CNH2

O

52OP

W

OH

MgBrCH

3

3 X

22 I,)OH(Ca

.)yellowppt( Y

Z will be

(A)

C–CH3

O

(B)

COOH

(C)

C

O

(D)

Sol.[C] Z will be

C

O

True / False type questions

Q.30 Acetonitrile on hydrolysis with a dilute

mineral acid gives acetone.

Sol.

leAcetonitri

NCCH3 H/HOOH

acidAcetic

NHCOOHCH 33

Acetonitrile on hydrolysis with a dilute mineral

acid give acetic acid, not an acetone.

Therefore statement is false.

Q.31 Esterfication involves the heating of a

carboxylic acid with an alcohol in presence of

a protonic acid (H2SO4 or HCl gas) as

catalyst.

Sol. In esterfication reaction :

O

CH3–C–OH + C2H5OH

acid

alcohol

H2SO4

or HCl(gas)

O

CH3C–OC2H5 + H2O

ester

Therefore statement is true.

Fill in the blanks type questions

Q.32 Kolbe’s electrolysis of potassium succinate

gives CO2 and ........

Sol. Ethylene

Q.33 In Hoffmann-bromamide reaction, the

migration of an alkyl or aryl group occurs

from ....... to ..... atom.

Sol. Carbon, nitrogen

Q.34 Ethanenitrile on hydrolysis gives ...........

Sol. Ethanoic acid

Q.35 Hell-Volhard Zelinsky reaction involves the

replacement of an ....... atom from the alkyl

group of a monocarboxylic acid by a .....

atom.

Sol. -hydrogen, halogen



EXERCISE # 2

Part-A (Only single correct answer type

Questions)

Q.1 X HI

Pred CH3COOH 4LiAlH

Y. What

does not true for X and Y –

(A) X is hydrocarbon of general formula

CnH2n+2 while Y belong to alkanol

(B) X can be obtained by reducing

CH3CH2Cl while Y by its hydrolysis

(C) X gives positive litmus test but Y does

not

(D) X and Y both belong to different

homologeous series

Sol.[C] CH3–

CH3HI

Pred CH3COOH 4LiAlH

CH3CH2OH

[X] [Y]

Y give positive litmus test but 'X' does not.


Q.2 The product of which of the following

reaction is capable of changing orange colour

of 272OCr to green colour of Cr

+3 –

(A) CH3(COOH)2

(B) CH3CN OH3

(C) HCN OH3

(D) CH3CONH2 OH2

Sol.[C] HCN OH3 H–COOH +

4NH

Formic acid is strong acid, that its capable of

changing orange coloure of 272OCr to green

colour Cr+3

.

Q.3 In the reaction sequence –

CH3CO

CH3CO O

X

CH3CONH2

Y

CH3C N Z

CH3COOC2H5

(A) NaOH, PCl5, Na + alcohol

(B) NH3, P2O5, aqueous ethanol/H+

(C) NH3, NaOH, Zn + NaOH

(D) NH3, Conc. H2SO4, aqueous methanol

Sol.[B] CH3CO

CH3CO

O

NH3

CH3–C–NH2 + CH3COOH

X

O

P2O5 (Y)

CH3C N + H2O

Aqueous ethanol

CH3COOH + C2H5OH

CH3–C–OC2H5

O


Q.4 OHC–CH2–CH2 –CH2–CH2–OH

is converted into

O

O by –

(A) (i) KMnO4 (ii) H+, H

(B) (i) Na2Cr2O7 (ii) H+,

(C) (i) Ag2O (ii) H+,

(D) All of these Sol.[C] OHC–CH2–CH2 –CH2–CH2–OH

Ag2O

HOOC–CH2–CH2–CH2–CH2–OH

O

O

H+/


Q.5

.O

O .O

O

, on saponification of the given

ester is formed –

(A) OH OH

CH2 – CH2

and OHC–CHO

(B) OH–CH2–CH2–OH and OHC–COOH

(C)

OH OH

CH2 – CH2

and HOOC–COOH

(D) HO–CH2–CH2–COOH and HCOOH



Sol.[C]

CH2

CH2 +

O O H H–O

O H H–O

C

C

O

—

.O

O .O

O


Q.6 A sweet smelling ester, with molar mass 116,

on hydrolysis produces a carboxylic acid and

an alcohol. Alcohol give positive iodoform

reaction which of the following formula could

correspond to above statements –

(A)

CH3–C–O–CH2–CH2–CH2–CH3

O

(B) CH3CH2–O–COC3H7

(C) CH3–O–COC4H9

(D)

CH3–CH2–C–O–CH2–CH2–CH3

O

Sol.[B]

O

C3H7 C–OCH2CH3 OH3

O

C3H7 C–OH + CH3CH2OH

Alcohol give positive

Iodoform reaction


Q.7 YOHCH

NaBH

3

4

O

O

O

4LiAlHX

X and Y are –

(A)

O

O

OH

in both case

(B) OH

OH

in both case

(C) OH

OH

OH

,

O

O

OH

(D) Formation of A and B is not possible

Sol.[C] LiAlH4 is powerful, non selective reducing

agent. It reduces a wide range of functional

group –COOH, COOR, –CN, –

O| |C –NH2

But NaBH4 is more specific and selective for

the reduction of aldehydes and ketones. It has

no effect on ester nitro group, C = C etc.

O

O

OH

OHCH

NaBH

3

4

O

O

O

4LiAlH OH

OH

OH

[Y] [X]


Q.8 CH3–CH2–COOH

423 SOH.ConcNaN

X

by reaction R1

CH3–CH2–COOH

P/Br2 Y

by reaction R2

Which is correct alternate –

(A) X Y

CH3CH2NH2

Br

CH3 – CHCOOH

R1 R2

Schmidt HVZ

(B) X Y

CH3CH2CONH2 CH3CH2COBr

R1 R2

HVZ Schmidt

(C) X Y

CH3CH2NH2 CH3CH2COBr

R1 R2

HVZ Schmidt

(D) None of these

Sol.[A] CH3CH2COO

423 SOH.ConcNaN

CH3CH2NH2 + N2 + CO2

[x]

This reaction is known as Schmidt reaction

[R1]

CH3CH2COOH

P/Br2 CH3

Br

|CHCOOH + HBr

[Y]

This reaction is known as Hell-Volhard-

zelinsky (HVZ) reaction.




Q.9 In the given reaction :

COOH

OH

3PCl

[X]

[X] will be -

(A)

COOH

Cl

(B)

COCl

OH

(C) O

C O

O

(D)

COCl

Cl

Sol.[B]

COCl

OH

Q.10

O

COOH

CH2COOH

X, X is –

(A)

O

C

CH2C

O

O

O (B)

O

COOH

CH3

(C)

O

CH2COOH

(D) None

Sol.[C] X is

O

CH2COOH

Q.11

Br

H C

CH3

O O

is reacted with

OH)iii(

CO)ii(etherMg)i(

3

2 Z.

The final product Z is –

(A)

COOH

H C

CH3

O O

(B)

COOMgBr

H C

CH3

O O

(C)

COOH

H C

CH3

O

(D)

COOH

COOH

CH3

Sol.[C]

Br

H C

CH3

O O

etherMg)i(

MgBr

OMgr C

CH3

O

H

OH

CO

3

2

COOH

H C

CH3

O


Q.12 Y

2

3

Br,KOD)ii(

,HN)i( COOH

2

3

Br,KOH)ii(

,ND)i(,

X, what are X and Y –

(A) X is

NH2 ; Y is

ND2

(B) X is

ND2 ; Y is

NH2

(C) both

ND2

(D) both

NH2

Sol.[A]

COOH 3ND)i(

C–NH2

O

+ H2O

C–N

O

H

H

• •

+

OD

C–N–H

O

Br–Br

• •

C–N

O

Br

H

• •

OD

+



C – N – Br

O

• •

tarrangemenRe

N–C–O + Br

• •

N = C = O + OD • •

N–C=O • •

OD

DO + CO2

ND2

ND – C – O

O • •

D–OD

[Y]

C–OH

O

/ND3

C–ND3

O

C–ND3

O

+ 4KOH + Br2

NH2 + 2Br + K2CO3 + 2D2O


Q.13 An ester (I) with molecular formula C9H10O2

was treated with excess of CH3MgBr and the

complex so formed was treated with H2SO4 to

give an olefin (II). Ozonolysis of (II) gave a

ketone with molecular formula C8H8O which

show positive iodoform test. The structure of (I)

is –

(A) C6H5COOC2H5 (B) C6H5COOC6H5

(C) H3COCH2COC6H5

(D) p– CH3O – C6H4– COCH3

Sol.[A]

O

C6H5COC2H5 + CH3MgBr C6H5–C–OC2H5

CH3

OMgBr

CH3MgBr H3O

+

Con. H2SO4 C6H5–C–CH3

CH3

OH

C6H5–C–CH2

CH3

O3

C6H5–C–CH2O

CH3

O

[II]

Rotophenon


Q.14

NH2 NH2

CH2 – CH2

on reaction with diethyl oxalate

form –

(A) NH2CH2NH– C| |O

COOC2H5

(B)

O N

H

H

N O

(C) H2N–CH2–CH2–N–COOH

H

(D) H2N – CH2 – CH2 – N =

OC2H5

C–COOC2H5

Sol.[B]

CH2

CH2 +

O NH–H C2H5–O

NH–H C2H5–O

C

C

O

—

O N

H

H

N O

+ 2C2H5OH


Q.15 Give the structure of the expected product of

the following reaction

CH3

O

O

+ CH3NH2 — (X)

(A)

CH3

O

NCH3

(B)

CH3

O

NCH3 HO

(C) CH3–NH– C||O

– CH2CH2–

OH

CH –CH3

(D) None



Sol.[C]

CH3

O + CH3–NH–H

O

CH3–NH–C–CH2–CH2–CH–O–

CH3

H+

CH3–NH–C–CH2–CH2–CH–OH

CH3

O


Q.16 Which carboxylic acid (X) of equivalent mass

of 52g / eq loses CO2 when heated to give an

acid (Y) of equivalent mass of 60g/eq.

(A) CH3COOH

(B) CH2(COOH)2

(C) HOOCCH2CH2COOH

(D) HOOC(CH2)3COOH

Sol.[B] The carboxylic acid (X) is CH2(COOH)2

Q.17 59 g of amide obtained from the carboxylic

acid RCOOH, on heating with alkali gave 17g

of ammonia. The acid is –

(A) HCOOH (B) CH3COOH

(C) CH3CH2COOH (D)C6H5COOH

Sol.[B] The acid is CH3COOH

Q.18

COOH

ROH/NH/Na 3 ? Product is –

(A)

COOH

(B)

COOH

(C)

COOH

(D)

COOH

Sol.[B]

COOH

ROH/NH/Na 3

COOH



Q.19 Which one of the following compounds will

give HVZ reaction ?

(A) COOH

(B)

COOH

(C) COOH (D)

COOH

Sol.[B,C]

COOH

COOH , of the compounds

will give HVZ reaction

Q.20 Phenol and benzoic acid may be distinguished

by their reaction with –

(A) Aqueous NaOH

(B) Aqueous NaHCO3

(C) Neutral FeCl3

(D) Aqueous NH3

Sol.(B), (C) Phenols are weaker acids than carboxylic

acid. It is because of this reason that phenols do

not decompose carbonates and bicarbonates

evolving CO2. This test distinguishes phenols

and carboxylic acid.

C6H5OH 3FeClNeutralBlue or violet colour

Benzoic acid does not react with Neutral FeCl3.

Therefore option (B) and (C) are correct.

Q.21 In the given reaction :

O

R– C – OH ]X[

O

R – C – O – CH3

[X] will be -

(A) CH2N2 (B) CH3OH/H

(C) MeCOOH (D) Me2SO4

Sol.[A,B,D] X will be CH2N2, CH3OH/H

, Me2SO4

Q.22 Which of the following statements are true

about HCOOH ?

(A) It is a stronger acid than CH3COOH

(B) It forms formyl chloride with PCl5

(C) It gives CO and H2O on heating with

conc. H2SO4

(D) It reduces Tollen’s reagent

Sol.[A,C,D]

(A) HCOOH is a strong acid than CH3COOH

(B) HCOOH + PCl5 HCOCl + PoCl3 + HCl

Unstable



So the formyl chloride does not form.

(C) HCOOH

42SOH.ConcCO + H2O

(D) Formic acid reduces to tollen's reagent

Therefore option (A), (C) and (D) are correct.

Q.23 The best methods to reduce cyclohex-2-en-1-

one to cyclohex-2-en-1-ol is are –

(A) PCC, CH2Cl2

(B) MPV reduction

(C) NaBH4

(D) Ni/H2

Sol.[B]

O

4NaBH

OH

Cyclohex-2- ene-1-one Cyclohex-2-ene-1-ol

NaBH4 is more specific and selective for the

reduction of aldehydes and ketones. It has no

effect on C = C, C = C nitro group ester.


Q.24 Which of the following on oxidation with

alkaline KMnO4 followed by acidification

with HCl would give benzoic acid ?

(A) Toluene (B) Ethylbenzene

(C) o-Xylene (D) p-Xylene

Sol.[A, B]

CH3

H/KMnO4

COOH

Tolune Benzoic acid

CH2–CH3

H/KMnO4

COOH

Ethyl benzene Benzoic acid

Therefore option (A) and (B) are correct.

Q.25 Nitrogen gas is evolved when HNO2 reacts

with –

(A) C6H5COCH3 (B) C6H5CONH2

(C) CH3CH2NH2 (D) NH2CONH2

Sol.[B,C,D]

C6H5

O| |C –NH2 2HNO

C6H5COOH + N2 + H2O

Benzoic acid

CH3CH2NH2 + 2HNOCH3CH2OH + N2 H2O

NH2CONH2 + HNO2 2N2 3H2O + CO2

Therefore option (B), (C) and (D) are correct.

Q.26 Which of the following on reduction with

LiAIH4 will give ethyl alcohol ?

(A) (CH3CO)2O

(B) CH3COCl

(C) CH3CONH2

(D) CH3COOC2H5

Sol.[A,B, D]

(A)

CH3C

O

O CH3C

O

4LiAlH 2CH3CH2OH

Ethyl alcohol

(B) CH3–

O| |C –Cl 4LiAlH

C2H5OH+LiCl+AlCl3

(C) CH3CH2–

O| |C –NH2 4LiAlH

CH3CH2CH2 NH2+H2O

(D) CH3COOC2H5 4LiAlHCH3CH2OH + CH3OH

So option (A), (B) and (D) are give the ethyl

alcohol by reduction with LiAlH4.

Q.27 Ethyl acetate can be prepared by –

(A) Reaction of ethyl alcohol with acetic acid

in presence of dry HCl gas

(B) Treatment of acetaldehyde with

Al(OC2H5)3

(C) Reaction of acetyl chloride with ethyl

alcohol in presence of acidic medium

(D) Reaction of acetic anhydride with ethyl

alcohol in presence of acidic medium

Sol.[A,B,C,D]

(A) C2H5OH+CH3–

O| |C –OH2

HCH3

O| |C –OC2H5

(B)

O

CH3C + O = C – CH3

H

H

352 )HOC(Al

C

Acetaldehyde O

CH3C–O–CH2–CH3

Ethyl acetate

This reaction is know as



Tischenko reaction.

(C) CH3

O| |C –Cl + C2H5OH CH3

O| |C –OC2H5 + HCl

Ethyl acetate

(D) (CH3CO)2O+C2H5OHCH3

O| |C –OC2H5+CH3COOH

Ethyl acetate

Therefore option (A),(B),(C) and (D) are correct.

Q.28 The formula C4H4O4 can represent –

(A) A cyclic ester of dibasic acid

(B) A cis-dibasic acid

(C) A trans-dibasic acid

(D) An , -unsaturated dibasic acid which

on heating forms a monobasic acid

Sol.[A,B,C,D]

Formula C4H4O4 represent.

(A) A cyclic ester of dibasic acid

CH2–O–C

O

CH3–O–C

O

(B) A cis-dibasic acid

H–C–COOH

H–C–COOH

Maleic acid

(C) A trans-dibasic acid

H–C–COOH

COOH–C–H

Pumaric acid

(D)

HC CH

COOH COOH

CH2=CH–COOH+CO2

Monobasic acid

Therefore option (A),(B),(C) and (D) are correct.

Part-C (Assertion & Reasoning type

Questions)

The following questions 29 to 30 consists of

two statements each, printed as Assertion

and Reason. While answering these

questions you are to choose any one of the

following five responses.

(A) If both Assertion and Reason are true and

the Reason is correct explanation of the

Assertion.

(B) If both Assertion and Reason are true but

Reason is not correct explanation of the

Assertion.

(C) If Assertion is true but the Reason is false.

(D) If Assertion is false but Reason is true.

(E) If Assertion & Reason are false.

Q.29 Assertion : Nitration of benzoic acid gives

m-nitrobenzoic acid.

Reason : carboxylic group increases the

relative electron-density at meta-position.

Sol.[C] –C| |O

–OH, is an electron attracting group thus

cause deactivation of benzene ring particularly

at o, p that m-position, thus electrophilic reagent

can attack at m-position.

Therefore assertion is true but the reason is

false.

So option (C) is correct.

Q.30 Assertion : CH3COCH2COOC2H5 will give

iodoform test.

Reason : It contains CH3CO–group linked to

a carbon atom. Sol.[D] Ethyl acetoacetate (CH3

O| |C –CH2COOC2H5) does

not give iodogorm test, although it contains

CH3–

O| |C

group attached to carbon (methylene

group). This is due to active nature of

methylene group at which iodination occurs and

not on methyl group of CH3CO unit.

Therefore assertion is false but reason is true so

option (D) is correct.

Part-D Column Matching

Q.31 Column-A Column-B (a) H.V.Z. reaction (i) potassium salt of

Monocarboxylic acid

(b) Oxidation of Formic (ii) replacement of

acid -hydrogen by

halogen

(c) –COOH group (iii) H2O + CO2 (d) Salt in Kolbe’s (iv) deactivating

electrolytic reaction



Sol. (A)ii, (B)iii, (C)iv, (D)i ,

Factual

Q.32 Column-A Column-B

(a) Clemmenson (i) ,-unsaturated

reduction compound

(b) Cannizaro’s reaction (ii) red brown

precipitate

(c) Aldol condensation (iii) aldehyde

+ Zn–Hg/HCl

(d) Fehling solution (iv) aldehydes

containing no

-hydrogen

(e) Claisen-Schmidt (v) carbonyl

reaction compounds having

-hydrogen

Sol. (A) iii, (B)iv, (C)v, (D)ii, (E) i

Factual


(Reagents reacting with (Product formed)

PhCH2COOH) (a) CH3MgBr (i) PhCH2COCl

(b) PCl5 (ii) PhCH2COOCH3

(c) NH3, followed by (iii) CH4

heating

(d) CH3OH in the presence(iv) PhCH2CONH2

of conc. H2SO4

Sol. (A) iii, (B)i, (C)iv, (D)ii

Factual


(a) PhCONH2 (i) B2H6/AcOH/H2O

PhCH2OH

(b)

C6H5

O

O

(ii) LiAlH4

CH2OH–(CH2)2–CHOH–C6H5

(c) C6H5CH=CH–COOH (iii) H2/Pd + BaSO4

C6H5–CH=CH2OH

(d) CH3COClCH3–CHO (iv) None

Sol. (A) iv, (B)ii, (C)iv, (D)iii,

Factual


(a) RCN reduction

(i) 1° Amine

(b) RCN OH)ii(

MgBrCH)i(

2

3 (ii) Alcohol

(c) RNC hydrolysis

(iii) Ketone

(d) RNH2 2HNO (iv) Acid

Sol. (A) i, (B)iii, (C)i, iv, (D)ii Factual



EXERCISE # 3

Part-A Subjective Type Questions

Q.1 Fluoro acetic acid is stronger than chloro

acetic acid.

Sol. F–CH2–C–OH F–CH2 –C–O

CH2–N=N

O O

(strong –I power)

–I

Cl–CH2–C–O–H Cl–CH2 –C–O

O O

Anion stabilised by

–I group.

Therefore fluoro acetic acid is stronger than

chloroacetic acid.

Q.2 C – O bond length in formic acid are 1.23 Aº

and 1.36 Aº but in sodium formate both

carbon and oxygen bonds have same value

i.e. 1.27 Aº.

Sol. In sodium formate both carbon and oxygen

bonds have same value because in formate ion

equal resonating structure is formed.

H–C

O–

O

H–C

O

O–

H–C

O

O

–

–

But resonance does not present in formic acid.

Therefore C–O bond length in formic acid are

different.

Q.3 The C – G bond in acid derivative is expected

to be shorter and stronger than alkyl

derivative.

Sol. In acid derivative

O

R–C–G

xÅ

Resonance present

O

R–CH2–G

yÅ no Resonance

O

R – C – G R–C=G

O

Double bond

characters are

present

Therefore C–G bond in acid derivative is

expected to be shorter and stronger than alkyl

derivative.

Q.4 Ester (A) 5PCl B + C

KOH.aq

(CH3)2CHCH2OH

B + C4

2

BaSO

Pd/H (D) 4LiAlH

(CH3)2CHCH2OH

Sol. (A) (CH3)2CHCOOCH2CH(CH3)2 ;

(B) (CH3)2CHCOCl ;

(C) (CH3)2CHCH2Cl

Q.5

COOH

COOH

COOH

A

B

Sol. (A)

COOH

COOH

; (B)

C

C

O

O

O

Q.6

COOH

COOH

4LiAlH

(A) HCl

(B)

)C(

CH2COOH

CH2COOH

BaO

(D) + CO2 + H2O

Sol. (A)

CH2OH

CH2OH

;

(B)

CH2Cl

CH2Cl

;

(C) CN–/H3O

+;

(D)

C=O

Q.7 A OAgMoist 2 B

]O[C

D

(HCOO)2Ca

G2

52

SOCl,2

OHHC/Na H

I

CH3CN OH3 X

A KCN

J OH3

X

Sol. (A) CH3Cl ; (B) CH3OH ;

(C) HCOOH ; (D) Ca(OH)2 ;

(G) HCHO ; (H) CH3Cl ;

(I) KCN ; (J) CH3CN ;

(X) CH3COOH



Q.8 HC CHexcess

COOHCH3 (A)

B + C

HOH

CH3COOH

Sol. (A) CH3CH(OOCCH3)2 ;

(B) CH3CHO ;

(C) (CH3CO)2O

Q.9 On standing in dilute aqueous acid,

compound A is smoothly converted to

mevaionolactone.

Me

CH2COOH

Me

O O

)2(

OH)1( 3

OH

CH3

O

O

Suggest a reasonable mechanism for this

reaction. What other organic product is also

formed ?

Sol.

Me

CH2COOH

Me

O O H2O+

H2C

HO

CH2

OH

CH3 O

CH3

Intermol ester

OH

CH3

O

O

Q.10 Outline reasonable mechanisms for each of

the following reactions.

(a)

O O

+ BrMgCH2CH2CH2CH2MgBr

OH.2

THF.1

2

HO CH2CH2CH2OH

(b)

O S H2NCH2CH2

eoustanspon

O

HS

N

H

Sol. (a)

O O

Br–Mg–CH2 CH2–Mg–Br

+

CH2 – CH2

O

OH . CH2 CH2

CH2 – CH2

OH

OH

CH2 CH2

(b)

O S HNCH2–CH2

• •

• •

H

• •

HS

O–

N–H HS

O

N–H

Q.11 An organic compound A (C4H9NO) on

treatment with bromine and alkali form

another compound B (C3H9N) on treatment

with sodium nitrite and dilute hydrochloric

acid. B yield C (C3H8O). C can be oxidized to

D (C3H6O). Which can also synthesised from

methyl

Sol. (A) (CH3)2CHCONH2; (B) (CH3)2CHNH2;

(C) (CH3)2CHOH; (D) CH3COCH3

Q.12 An organic compound A (C8H8O) an

treatment with NH2OH. HCl give B and C.

Compound B and C can be converted into D

and E respectively by treatment of H2SO4.

Compound B, C, D and E are all isomer of

molecular formula C8H9NO. When D is

boiled with KOH, an oil F (C6H7N) separate

out. F react rapidly with CH3COCl to give

back (D). On other hand. E on boiling with

alkali followed by acidification give a white

solid G (C7H6O2). Identify A to G.

Sol. (A)

–C–CH3

O

; (B)

–C=N–OH

CH3

;



(C)

–C = N

CH3 OH

;(D)

–NH–C–CH3

O

;

(E)

–C–NH–CH3

O

; (F) –NH2

;

(G)

–C–OH

O

Q.13 An organic compound (A) of formula C3H6O

from a monoxime with hydroxylamine and

form iodoform on heating with I2 and NaOH

and sodium acetate. Compound A on reaction

with NaCN and dil H2SO4 gave product B.

Which on hydrolysis produced (C).

Compound (C) on heating gave (D), which on

decarboxylation gave (E) of formula C3H6.

Compound (E) on ozonolysis gave one

molecule of acetaldehyde and one of

methanal. What are A to E ?

Sol. (A) CH3–CO–CH3;

(B)(CH3)2C(OH)CN ;

(C) (CH3)2C(OH)COOH ;

(D) CH2=C–(CH3)–COOH ;

(E) CH3–CH=CH2

Q.14 Compound (A) with an empirical formula,

C7H9N on diazotization give a product which

undergoes reaction with Cu2Cl2 & HCl to

give a compound (B). B on oxidation give a

compound (C). Compound (A) on treatment

with Br2/H2O form C7H6NBr3 (D). Give the

structural formula of A, B, C & D.

Sol.

CH3

NH2

CH3

Cl

COOH

Cl

CH3

NH2

Br

Br Br

Part-B Passage based Question

Passage - I (Q. 15 to 17)

-keto acids are highly unstable acids and

undergo decarboxylation on heating at 90º–

100ºC Via formation of cyclic transition state.

Decarboxylation of -keto acids involves

both free acids and the carboxylate ion. Loss

of CO2 from carboxylate ion yields the

carbanion which is stabilised by resonance.

O

• •

R–C–CH2–C–O

O

R – C – CH2 + CO2

O • •

O • •

R–C=CH2

This carbonion is formed faster than the

simple carbanion i.e. alkyl carbanion that

would be form a simple carboxylate ion

(RCOC–) because it is more stable. It is more

stable of course due to accommodation of the

negative charge by the keto group.

Q.15 Which of the following on heating gives

acetone –

(A)

CH3–C–CH2–C–OH

O O

(B)

CH3–CH2–C–CH2–C–OH

O O

(C)

CH3–C–CH–C–OH

O O

CH

(D) All

Sol.(A)


O O

CH3–C–CH2+CO2 • •

O

CH3–C=CH2

O–

H2O/H+

CH3–C=CH2

OH

CH3–C–CH3

OH




Q.16 Arrange the following in the increasing order

of ease of decarboxylation.

I.


O O

II.

CH3–C–CH2–CH2–C–OH

O O

III.

COOH

NO2 O2N

NO2

IV.

CH2–C–OH

O

(A) I < II < III < IV (B) II < I < IV < III

(C) II < IV < I < III (D) IV < III < II < I

Q.17 What will be the final product in the

following reaction :

CH3–C–H

O

5PClA

KCNB

NaOH

OH2 C HCl

D

E

(A)

CH3–C–NH2

O

(B)

CH3–CH2–C–OC2H5

O

(C) CH3–CH2–CH3

(D)

CH3–CH2–C–OH

O

Sol.(D)

CH3–C–H

O

5PCl Cl CH3CH

Cl [A] KCN

CN CH3CH

CN [B]

H2O/NaCl

COONa CH3CH

COONa [C]

NaCl COOH CH3CH

COOH [C]

CH3–CH2–COOH+CO2


Passage - II (Q. 18 to 20)

Claisen condensation takes place between

two molecules of ester in presence of strong

base like R–ONa in alcohol or NaH in ether.

Two ester molecules may be same or

different. The basic need of this reaction is

the presence at least one -Hydrogen, the

reaction is termed as crossed claissen

condensation.

This reaction is an example of nucleophilic

substitution reaction. In this reaction one

molecules of ester behaves as substrate and

other molecule having -Hydrogen atom

behaves as nucleophile.

The product of the reaction is -Ketoester.

Step (i)

H

CH2–C–OC2H5

• •

O

C2H5O—

CH2–C–OC2H5

O

–

Step (ii)

CH2–C–OC2H5

O

+ CH2–C–OC2H5

O • •

CH3–C–OC2H5

O

CH2–C–OC2H5

• •

O

CH3–C–OC2H5

O

CH2–C–OC2H5

• •

O

CH3–C–CH2–C–OC2H5

O O

This reaction also takes place between ester

and ketone having atleast one -hydrogen.

Ketone behaves as nucleophile. In diesters

intramolecular claissen condensation gives a

cyclic -ketoester.

Q.18 What is X in the following reaction.

C–OC2H5

O

C–OC2H5

O

+

O

H)ii(

OHHC/ONaHC)i( 5252X

(A)

CH2–C–OC2H5

O

O



(B)

C–OC2H5

O

O

(C)

C–C–OC2H5

O

O

O

(D)

C–C–OC2H5

O

O

O

Sol.(C)

O

H + C2H5O

O

+C2H5OH

O

C–OC2H5

O

C–OC2H5

O

+

O

C–OC2H5

O–

C–OC2H5

O

C–C–OC2H5 + C2H5O

O

O

O


Q.19


O


O

OHHC

ONaHC

52

52 A.

What is A–

(A)

CH2–C–OC2H5

O

O

(B)

C–OC2H5

O

O

(C)

C–OC2H5

O

O

(D)

C–CH3

O

O

Sol.(C)

CH2–CH–C–OC2H5

O


O

H

C2H5O


O


O

C–OC2H5

O

O

–C2H5O

CH–C–OC2H5

O–

O

C

CH2

CH2 CH2

OC2H5


Q.20 What will be the product in the following

reactions –


O


O

CH3–CH2–N

OHHC

ONaHC

52

52 Product

(A) CH3–CH2–N O

(B) CH3–CH2–N O

COOC2H5

(C) CH3–CH2–N

COOC2H5

(D)

CH3–CH2–N

O

C–OC2H5

O

Sol.(B)


O


O

CH3–CH2–N

H

C2H5O–




O


O

C2H5–N


O

CH2–CH2

C2H5–N

C–O–

OC2H5

C2H5–N O

C–OC2H5

O


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