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1: When isobutane is monochlorinated in the presence of ultraviolet light, the product obtained in

higher yield is

(a) n-butyl chloride (b) iso butyl chloride

(c) sec-butyl chloride (d) tert-butyl chloride

Solution:

=

=

% of isobutyl chloride =

% of tertbutyl chloride=

The answer is (b)

2: The product of reaction CH3CH2CH2MgBr + HCCCH3

is

(a) CH3 CH2 CH3 (b) CH3 CH2CH2 C C CH3

(c) CH3 CH2CH2 OH (d) CH3 CH2 CHO

Solution: Grignard reagent reacts with any compound having active hydrogen producing an alkane.

nC3H7MgBr + HC CCH3 nC3H8 + CH3C CMgBr

The answer is (a)

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3:

The major product (A) in above given reaction is:

(a) (b)

(c) (d)

Solution: The reaction proceeds as follows:

Markovnikov's rule is not applicable because of carbocation rearrangement.

The answer is (c)

4: 1-butyne can be distinguished from 2-butyne by using

(a) bromine water (Br2 in CCl4)

(b) coldalk. KMnO4 (Baeyer's Reagent)

(c) ammoniacal solution of silver (Tollen's Reagent)

(d) diethyl ether

Solution: Terminal alkynes form a precipitate with Tollen's reagent.

RC CH + Ag+

RC C_: Ag

++ H2

ppt. of Silver acetylide

1-butyne, a terminal alkyne gives this reaction but

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2-butyne, a non-terminal alkyne does not give this reaction.

The answer is (c)

5: The ozonolysis of a triple bond produces

(a) a mixture of aldehydes and ketones

(b) a mixture of ketones and carboxylic acids

(c) a mixture of carboxylic acids

(d) none of the above

Solution:

The answer is (c)

6: One of isomers (A) of 2-butene gives a racemic mixture on addition of Br2 whereas another isomer

(B) gives a meso compound on addition of Br2. (A) and (B) are respectively:

(a) cis and trans-2-butene (b) trans and cis-2-butene

(c) both are cis (d) none of the above

Solution: Cis isomer on anti-addition gives a racemic mixture and a meso compound on syn-addition but

a trans isomer gives a meso compound on anti-addition and a racemic mixture on syn addition. Br2 also

adds in anti manner.

The answer is (a)

7: An alkene with the lowest heat of hydrogenation is

(a) CH3 CH = CH CH3 (b)

(c) (d)

Solution: The more substituted an alkene the more stable it is and lower its heat of hydrogenation.

The answer is (b)

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8: Which of the following halides does not form precipitate with alcoholic silver nitrate?

(a) ethyl chloride (b) 3-chloropropene

(c) chlorobenzene (d) isobutyl chloride

Solution: Only aryl and vinyl halides are incapable of forming silver halide precipitates with alc.AgNO3.

The answer is (c)

9:

The major product is

(a) (b)

(c) (d)

Solution: The conjugation of double bond with an aromatic ring yields a highly stable system.

Unusually stable

Hence, (a) being highly stable is the major product.

The answer is (a)

10: Which of the following aryl halides undergoes hydrolysis most readily with aq. NaOH.

(a) (b)

(c) (d)

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Solution: The greater the number of electron withdrawing groups at ortho/para positions to halogen,

the greater the reactivity of aryl halides towards nucleophilic substitution.

11: gives as the major product on hydrolysis. The yield of

the product will be maximum when

(a) water is solvent and X = I (b) dimethyl sulfoxide is solvent and X = I

(c) water is solvent and X = Br (d) water is solvent and X = Cl.

Solution: The mechanism of reaction must be SN2, because if it was SN1 the carbo-cation

rearrangement would yield (CH3)3COH as the major product. SN2reaction is assisted by an aprotic

solvent-like dimethyl sulfoxide and a good leaving group like iodide.

The answer is (b)

12: Hydrolysis by SN2 mechanism will be shown by

(a) C6H5CH2Br (b) CH3Br

(c) CH2 = CH CH2Br (d) (CH3)3 C Br

Solution: (a), (c), and (d) can form highly stable carbo-cations and thus show SN1 mechanism. CH3Br

is a (1) halide and has a greater tendency towards SN2 reaction.

The answer is (b)

13:

(D) is

(a) 1,2,3-tribromo-3-

methyl butane (b) 1,2,3-tribromo-2-methyl

butane

(c) 1,1,3-tribromopentane

(d) 1,3,3-tribromo-3-methyl butane

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Solution:

The answer is (a)

14: What is the order of reactivity of the following compounds towards aq. NaOH?

(a) IV > II > III > I (b) III > II > I > IV

(c) I > II > III > IV (d) IV > I > III > II

Solution: (IV) is an alkyl halide and undergoes substitution fastest through SN2 mechanism.

(I), (II), and (III) are aryl halides and undergo substitution through bimolecular displacement

reaction where reactivity is increased by electron-withdrawing groups ortho/para to halogen and

decreased by electron-releasing groups.

The answer is (a)

15: A hydrocarbon (A) (molecular formula C6H12) decolorises Br2 in CCl4 and is oxidised by hot

acidified KMnO4 to a resolvable carboxylic acid, C4H9COOH.

Determine (A).

Solution: Since the carboxylic acid, C4H9COOH is resolvable (optically active) it must have a chiral

centre and only structure of the acid that is optically acive is

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And, the alkene which on oxidation with hot acidified KMnO4 gives the acid is

Hence, (A) is 3-methyl pentene

16: (a) Outline synthesis of propyne from isopropyl bromide.

(b) One mole of a hydrocarbon (A) reacts with one mole of bromine giving a dibromo compound

C5H10Br2. (A) on treatment with cold alkaline KMnO4 forms compound of molecular formula

C5H12O2. On ozonolysis, (A) gives equimolar quantities of propanone and ethanal. Deduce the

structural formula of (A).

Solution: (a)

(b) Hydrocarbon (A) must be an alkene because

(i) it shows addition reaction

(ii) undergoesozonolysis reaction giving propanone and ethanal

Hence, (A) should be

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17: An alkane (A), C5 H12 on chlorination at 300C gives a mixture of four different monochlorinated

derivatives (B), (C), (D), and (E). Two of these derivatives give the same stable alkene (F) on

dehydrohalogenation. On oxidation with hot alkaline KMnO4 followed by acidification, (F) gives twoproducts (G) and (H). Give structures of (A) to (H).

Solution: The only alkane of molecular formula C5H12 that gives four monochlorinated isomers is

Out of these, (C) and (D) give the same alkene on dehydrohalogenation.

On oxidation with hot alkaline, KMnO4 (F) gives

18: Calculate the percentages of various isomers formed during mono-chlorination of 2,3-dimethyl

pentane at room temperature. The relative reactivity orders of 3, 2, 1 H atoms are 5 : 3.8 : 1.

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Solution :

Relative proportions:

A : B : C : D : E : F

= 6 1 : 1 5 : 1 5 : 3 1 : 2 3.8 : 3 1

= 6 : 5 : 5 : 3 : 7.6 : 3

% of :

A = = 20.27%

C, B = = 16.89% each

F, D = = 10.13% each

E = = 25.67%

19: An alkyl halide C6H13Cl (A) on treatment with tertiary butoxide gives two isomeric alkene B (asmajor product) and C (as minor product). B and C have molecular formula C6H12 and both on

hydrogenation give 2,3-dimethyl butane. What are A, B and C?

Solution: Two alkene which on hydrogenation give 2,3-dimethyl butane are

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For (A) to give these alkenes on dehydrohalogenation, its structure should be

Because of bulky base being used here less substituted alkene is the major product.

Hence, (B) is [2, 3-dimethyl but-1-ene]

And, (C) is [2,3-dimethyl but-2-ene]

20: Two isomeric organic compounds (A) and (B) of molecular formula C4H8Br2 on hydrolysis gave

two compounds (C) and (D) of formula C4H8O. (C) and (D) gave two isomeric acids (E) and (F) of

formula C4H8O2 on oxidation. Both the acids on decarboxylation with

soda-lime gave propane. Identify (A) to (F).

Solution: There are only two carboxylic acids having formula C4H8O2. They are

and (F)

on decarboxylation both give propane.

21: An organic compound (A) having molecular formula C7 H15Cl yields a white precipitate when

treated with AgNO3. On treatment with alcoholic potash, (A) yields a mixture of two isomericproducts (B) and (C) having formula C7H14. The ozonolysis of this mixture yielded four compounds.

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, , HCHO and .

Identify (A), (B), and (C).

Solution: (B) and (C) must be alkenes containing seven carbon atoms each. (Since alkyl halide fromwhich they are formed contains seven C atoms).

Hence, and O = CH2 must be ozonolysis products of one of alkenes (B) and

and must be obtained from (C).

[\ the number of C atoms adds upto 7]

Hence, (B) and (C) are

+ HCHO

(B) and (C) can be obtained by dehydrohalogenation as :

22: An unsaturated hydrocarbon (A) (C6 H10) readily gives (B) on treatment with NaNH2 in liquid

NH3. When (B) reacts with 1-chloropropane the compound (C) is obtained. On partial hydrogenationwith Lindlar's catalyst (C) gives (D) (formula = C9 H18). On ozonolysis, (D) gives 2,2-dimethylpropanal

and 1-butanal. Identify (A), (B), (C) and (D).

Solution: (A) should be an alkyne as indicated by molecular formula C6H10 and since it reacts with

NaNH2 in liquid NH3(A) should be a terminal alkyne.

Hence, (A) should be of type RC CH

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The products of ozonolysis are given as

and

Hence, R is tert-butyl group. R =

Hence, (A) : (CH3)3 CC CH 3, 3-dimethyl but-1-yne

(B) : (CH3)3 CC C_Na

+

(C) : (CH3)3 CC CCH2CH2CH3 2,2-dimethyl hept-3-yne

(D) : Cis-2,2-dimethyl hept-3-ene

23: A dihalogen derivative (A) of a hydrocarbon having two carbon atoms reacts with alcoholic potashand forms another hydrocarbon which gives a red precipitate with ammonical cuprous chloride.

Compound `A' gives an aldehyde when treated with aqueous KOH. Determine (A).

Solution: (A) can be or CH3CH Cl2

Both can react with alcoholic potash forming ethyne.

But only CH3CH Cl2 can form aldehyde with aqueous KOH.

Hence, (A) = CH3CHCl2

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CH3CH2C CH2CH3

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racemic

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G : C C

CH3

Br CH3

Br

trans

H : as F

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C

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(colourless)

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KMnO4C C

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(colourless)

( MnO2)(purple)

brown-black precipitate)

(loss of colour and formation of ppts)

C CH2SO4

(conc.)C C

H

(Solubility in sulphuric acid)

HSO4 Heat

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C C H2 C C

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