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Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc

Isomerism

Alkyl Halide

Method of Preparation of Organic Halide

Nucleophilic Substitution

Dr. Sangeeta Khanna Ph.D


Test Date – 30.7.2016 Topic: Isomerism & Alkyl Halide

READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hour duration. 2. The maximum marks are 311. 3. This test consists of 75 questions. 4. Keep your mobiles switched off during Test in the Halls.

(Single Correct Choice Type) Negative Marking [-1]

This Section contains 45 multiple choice questions. Each question has four choices A), B), C) and D) out

of which ONLY ONE is correct. Marks: 45 × 4 = 180

1. The configuration of the chiral centre and the geometry of the double bond in the following molecule can be described by

a. R and E b. S and E c. R and Z d. S and Z

C

2. XCHCHNaSCH 23

Θ

3

The reaction is fastest when X is:

a. – OH b. – F c. – I d. 3

| |

CH

O

CO

C

3. Given the number of N and M? a. 6, 6 b. 6, 4 c. 4, 4 d. 7, 3 B Sol.

F

C

C

OH HOOC

C

Me Me

H

CH3

CH3

H3C

Cl2, h

CH2Cl

CH3

H3C

(i) and (ii) (+) and (-)-isomer

CH3

CH3

H3C +

Cl

+

CH3

H3C Cl

CH3

+

(iii)

CH3

CH2Cl

H3C

(iv) and (v) (+) and (-) isomer

(vi)

CH3

CH3

H3C

Cl2, h C5H11Cl (N isomeric products)

Fractional

distillation M (isomeric products)



Thus, a total of six isomers will be formed. Since enantiomers have same b.p., they cannot be separated by fractional distillation, i.e., (i) and (ii) will distil together and similarly (iv) and (v) will distil together. Thus, out of six isomers formed, four can be separated by fractional distillation

4. Rate of reaction with aqueous ethanol follows the order: a. P > Q > S > R b. Q > P > R > S c. P > R > Q > S d. R > P > S > Q A 5. The reaction proceed by the mechanism:

a. SN1 b. SN2 c. SE2 d. SNi B

6. Consider the following nucleophiles

IVIIIIII

Θ| |

3

ΘΘ| |

| |

Θ| |

| |3 O

O

CCH ,O-Ph ,O

O

O

SPh ,O

O

O

SCF

when attached to sp3 hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order: a. I > II > III > IV b. I > II > IV > III c. IV > I > II > III d. IV > III > II > I

B 7. Find the product of the following reaction: a. b. c. d. B 8. The rate of SN1 reaction is fastest with :

a. b. c. d. A

O – CH2 – Br

(P)

CH3 Br

(Q) (R)

Br

(S)

Br

OH

+ SOCl2

N H

+ SO2 + HCl

H Cl

Br CH3OH

OCH3

OCH3

CH

Br

CH

Br

CH

Br

NO2 CH2 Br



9. The order of decreasing nucleophilicities of the following species is :

a. Θ

3

Θ

3

Θ

3 OCOCHOCHSCH b. Θ

3

Θ

3

Θ

3 OCHSCHOCOCH

c. Θ

3

Θ

3

Θ

3 OCHOCOCHSCH d. Θ

3

Θ

3

Θ

3 SCHOCOCHOCH

A

10. Arrange

Increasing order of reactivity in SN1 reactions a. I < II < III < IV b. I < II < IV < III c. IV < II < III < I d. I < III < II < IV

A Sol. First compound is a secondary halide and the other three are tertiary halides. Also the reactivity follows the

order: chloride < bromide < iodide. Thus the reactivity order is I < II < III < IV. 11. A dextrorotatory optically active alkyl halide undergoes hydrolysis by SN2 mechanism. The resulting

alcohol is

a. dextrorotatory b. laevorotary c. optically inactive due to racemization d. may be dextro-or laevorotatory D

Sol. SN2 reactions occur with inversion of configuration. Therefore, an optically active reactant gives an optically active product whose sign of rotation cannot be predicted.

12. CH3 – Br + Nu– CH3 – Nu + Br – The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is

[Nu– = (A) PhO–, (B) CH3COO–, (C) HO–, (D) CH3O–]

a. D > C > A > B b. D > C > B > A c. A > B > C > D d. B > D > C > A A

Sol. PhO– is less resonance stabilized than AcO– ion and hence is a stronger nucleophile than AcO–. Therefore, the overall nucleophilicity decreases in the order: CH3O

– (D) > HO– (C) > PhO– (A) > AcO– (B)

13. The SN1 reactivity of the following halides will be in the order (i) (CH3)3CBr (ii) (C6H5)2CHBr (iii) (C6H5)2C(CH3)Br (iv) (CH3)2CHBr (v) C2H5Br a. (v) > (iv) > (i) > (ii) > (iii) b. (ii) > (i) > (iii) > (v) > (iv) c. (i) > (iii) > (v) > (ii) > (iv) d. (iii) > (ii) > (i) > (iv) > (v) D

Sol. More stable the carbocation, more reactive is the alkyl halide in SN1 reactions. Since the stability of the carbocations decreasing in the order:

5223332563256 HCHC)CH(C)CH(HC)HC()CH(C)HC(

, therefore, reactivity of the alkyl

halides decreases in the order: (iii) > (ii) > (i) > (iv) > (v) 14. Of the following alkyl halide, one with the lowest boiling point is

a. Ethyl bromide b. Isopropyl bromide c. n-Butyl bromide d. Methyl bromide D

Sol. For the same halogen, smaller the size of the alkyl group, weaker the van der Waals forces of attraction and hence lower is the b.p. Therefore, option

15. SN1 reactions are favoured by

a. non-polar solvents b. bulky groups on the carbon atom attached to the halogen atom

Cl

CH3

(II),

CH3

(III), Br

Cl

(I),

CH3

(IV) in I



c. small groups on carbon atom attached to the halogen atom d. Electron withdrawing group on substrate B

Sol. Bulky groups o the atom attached to the halogen atom will hinder attack by SN2 mechanism due to steric hindrance and hence would favour SN1 mechanism.

16. Neopentyl alcohol on treatment with HBr gives

a. Neopentyl bromide b. 2-Bromo-2-methybutane c. 2-Methyl-2-butene d. 2-Methyl-1-butene B

Sol.

etanmethy lbu2Bromo2

32

3|

|3

Br32

3|

3shif t Methy l

2,12

3

2

|

|3

O2H

H

alcohol Neopenty l

2

3

3

|

|3 CHCH

CH

Br

CCHCHCH

CH

CCHHC

CH

CH

CCHOHCH

CH

CH

CCH

17. Find the major product of the following reaction: a. b. c. d. D 18. Find the product of the following reaction:

a. b. c. d.

A

Sol. In presence of neutral nucleophile, mechanism is SN1

19. Which of the following is least reactive towards nucleophilic displacement reaction when treated with

aqueous KOH?

a. 2, 4, 6-Trinitrochlorobenzene b. 2, 4-Dinitrochlorobenzene

c. 4-Nitrochlorobenzene d. 3-Nitrochlorobenzene

D

Sol. Cl is activated towards nucleophilic substitution reaction by presence of NO2 groups at o- and p-

positions and not at m-position.

20. A compound has vapour density 29. On warming with a solution PCl5, it gives non terminal gem

dihalides the compound is:

a. CH3CH2CHO b. CH3COCH3 c. CH3CHOHCH3 d. CH2 = CHCH2Br.

B

Sol. If V.D. is 29, the Mol. wt, is 2 × 29 = 58. Thus, the compound in CH3COCH3.

21. 1, 4-Dibromobutane (0.1 mole) is treated with Na2S (0.1 mole) is aqueous ethanol, the product formed

is

a. BrCH2CH2CH2CH2SH b. HSCH2CH2CH2CH2SH

CH2 – Br KSH

Br

CH2 – Br

SH

S

CH2 – SH

SH

CH2 – SH

Br

Cl

H2O

OH

OH

OH



c. BrCH2CH2CH2S

–Na+ d.

D

Sol.

22. In the reaction,

252Ni

2HKCN NHHCBA

a. A is CH3I b. B is CH3NC c. A is C2H5I d. B is C2H5NC A

Sol. 252]H[

)B(3

KCN

)A(3 NHHCCNCHICH

23. In dinitrochlorobenzene, chlorine is readily replaced. This is because

a. NO2 makes the ring electron rich at ortho and para positions b. NO2 withdraws e– from meta-position c. NO2 donates e– at m-position

d. NO2 withdraws e– from ortho/para-positions D

Sol. NO2 group withdraws electrons from o- and p-positions and hence activates the Cl towards nucleophilic substitution reactions

24. The correct increasing order of reactivity of halides for SN1 reaction is

a. CH3CH2X < (CH3)2CHX < CH2 = CHCH2X b. (CH3)2CHX < CH3CH2X < CH2 = CHCH2X c. (CH3)2CHX < CH2 = CHCH2X < CH3CH2X d. CH2 = CHCH2X < (CH3)2CHX < CH3CH2X A

Sol. The correct increasing order of reactivity follows the sequence: 1° < 2° < allyl < benzyl

25. Rate of SN1 attack will be

a. b. c. d.

B 26. Which statement is not correct for nucleophilic substitution reaction?

a. Nucleophilic substitution reaction is mainly of two types: unimolecular or bimolecular b. Nucleophilicity of incoming group should be more than the nucleophilicity of leaving group c. Substitution reaction is carried out in the presence of polar solvent d. Aryl SN2 reaction takes place via formation of carbocation D

S

Br – CH2

CH2

CH2 – Br

CH2

Na2S

-NaBr Br – CH2

CH2 CH2

CH2 -NaBr

Na+ - S

CH2 CH2

CH2 CH2

S

Cl

<

Cl

<

Cl

Cl

<

Cl

<

Cl

Cl

<

Cl

<

Cl

Cl

<

Cl

<

Cl



27. ACH

CH

CCHKCN )2(

HBr )1(2

3|

3 . Product ‘A’ is

a. 2, 2-Dimethyl propane nitrile b. 3-Methylbutane nitrile c. 2-Methyl propane isocyanide d. 2- Methyl propane nitrile A

Sol. 3

3|

|33

3|

|3 CH

CH

CN

CCH ,CH

CH

Br

CCH

28. In the given reaction:

[X] will be: a. b.

c. d. C

Sol. 29. The structure of the major product formed in the following reaction is a. b. c. d. D Sol. Aralkyl halides are more reactive than aryl halides, therefore, only the halogen in the side chain is

displaced.

30. CH3CH2CH2 – I NaCN

CH3CH2CH2CN

Which of the following solvent can be used in this reaction

a. Hexane b. H2O c. alcohol d. 3

3

CH

CHN

O

CH /\||

l CH3 HOH

[X]

OH CH3 CH3

OH

CH3

Cl

I

NaCN

DMF

CN

CN

Cl

I

NC

CN

Cl CN

I

I

CH2Cl NaCN

DMF

I

CH2CN

H3C H3C

OH and CH3 CH3

OH



D Sol. It is SN2 & aprotic polar solvent will favour. 31. Arrange the following in decreasing order of reactivity towards SN1 reaction. a. P > Q > R > S b. S > R > P > Q c. Q > S > P > R d. Q > S > R > P C 32. Which of the SN2 reaction is fastest?

a. CH3Br + HC C– CH3C CH + Br–

b. CH3Br + HC CH CH3C CH + HBr

c. CH3CH2Br + HC C– CH3CH2C CH + Br–

d. CH3CH2Br + HC CH CH3CH2C CH + Br–

A 33. What is the correct increasing order of reactivity of the following in the SN2 reaction?

I. (CH3)2CHCH2Cl II. CH3CH2CH(Cl)CH3

III. C6H5Cl IV. p – O2N – C6H4 – CH2Cl

a. I < II < III < IV b. III < II < I < IV c. II < III < IV < I d. III < I < II < IV D

34. Which of the following are the examples of strong nucleophiles but weak base in protic solvents? I. CH3S

– II. CH3O– III. I– IV. H2O V. F–

a. Both I and IV b. Both IV and V c. Both I and III d. Both III and IV C

35. The specific rotation of optically pure (R)-2-butanol is – 13.52° at 25°C. An optically pure sample of (R)-2—bromobutane was treated with aqueous NaOH in order to form 2-butanol via SN2 reaction. What would be the specific rotation of the product assuming 100% yield?

a. 0° b. – 13.52° c. + 13.52° c. None of these C

36. Which of the following reaction cannot be used to prepare arylhalide;

a. b. c. d.

B 37. Cyanide anion has two atoms that have lone pair of electrons. Either could act as nucleophile in the

reaction. Yet in the vast majority of the cases, carbon acts as nucleophile and forms a bond to the

substrate, why?

a. Nitrogen is strongly solvated in hydroxylic solvents, it will be less reactive

b. Carbon is more reactive in polar aprotic solvents, normally used for these SN2 reactions

c. Carbon is less electronegative than N

d. The lone pair of electrons on C is in sp hybrid orbital

C

Br

CH2 – Br

Br

Br

(P) (Q) (R) (S)

N2Cl KI + I2

+ Cl2 FeCl3

+ ICl



38. When CH3CH2Br is treated with aqueous NaCN, CH3CH2CN is formed while CH3CH2NC is formed

when AgCN is used in the similar reaction. it is due to

a. negative charge on carbon atom in NaCN

b. covalent nature of Ag C bond in AgCN

c. covalent nature of C – N bond in NaCN

d. ionic nature of Ag – C bond in AgCN

A,B

39. Which one of the two iodine will be more reactive in the SN1 and SN2 reactions?

a. A will be faster in SN1 reaction but slower in SN2

b. A will be a faster both in SN1 and SN2 reactions

c. A and B will be equally reactive

d. B will be faster in both SN1 and SN2 reactions

B

40. In SN1 reaction, racemisation occurs if the reaction occurs at a stereogenic centre, however, 50:50

mixture of enantiomers are rarely obtained, why?

a. Usually one enantiomer is more stable than other

b. Retention of configuration is always favoured in reaction

c. Hydroxylic solvent favour retention of configuration

d. There is steric hindrance to the approach of nucleophile from the front side as some of the leaving

group are not departed completely which favour inversion of configuration more than retention

D

41. Pick out the most reactive alkyl halide for an SN1 reaction.

a. b.

c. d.

A

42. Pick out the compound which reacts fastest in the presence of AgNO3.

a. (CH3)3CCl b. (CH3)2CHCH2Cl c. (CH3)2CHCl d. CH3CH2Cl A

43. Which of the following statement(s) is/are true for SN1 reaction?

I. The rate of SN1 reaction depends on concentration of alkyl halide

II. The rate of SN1 reaction depends on concentration of nucleophile

III. SN1 reaction of alkyl halides are favoured by non-polar solvents

a. Only I b. Only II c. Only III d. Both I and III A

I I

A

B

Ph

Br

Ph

Br

Br

Ph

Ph

Br



44. Pick out the strongest substrate(s) for a SN1 reaction. a. Only I b. Only II c. Only III d. Only IV

D 45. Pick out the following factor(s) which promote a SN1 reaction:

I. Temperature II. Concentration of nucleophile III. Concentration of alkyl halide IV. Aprotic solvents a. Both I and II b. Both II and III c. Both III and IV d. Both I and III D

SECTION – B (ASSERTION & REASON)

This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D)

out of which ONLY ONE is correct. (5 × 4 = 20 Marks)

(a) Both Statement I and Statement II are correct and Statement II is the correct explanation of

Statement I (b) Both Statement I and Statement II are correct but Statement II is not the correct explanation of

Statement I (c) Statement I is correct but Statement II is incorrect (d) Statement II is correct but Statement I is incorrect

1. Consider the following two bromide I and II, undergoing solvolysis reaction in boiling ethanol:

Statement I A is less reactive than B in the given solvolysis reaction. Statement II Carbocation of (B) is more stable due to methyl shift. a. (a) b. (b) c. (c) d. (d) A

2. Statement I When 3-bromo propene, which contain a labelled 13C at C-1 position is refluxed with methanol, following products were obtained.

32

13

232

13

2Boil

OHCH2

13

2 OCHHCCHCHOCHCHCHCHBrCHCHCH 3

Statement II Methanol has an acidic proton bonded to oxygen. a. (a) b. (b) c. (c) d. (d) B

3. Statement – I: Optically active 2-iodobutane on treatment with NaI in acetone undergoes racemization Statement – II: Repeated Walden inversions on the reactant and its products eventually gives a racemic mixture.

a. (a) b. (c) c. (c) d. (d)

Br

C2H5

I.

C2H5 Br

II.

Br

C6H5

III.

IV. C6H5

Br

CH3 H3C

H3C

CH3

Br Ethanol

Product A.

B.

Product Ethanol

CH3 H3C

CH3

Br

CH3



A Sol. Statement – II is the correct explanation of Statement – I.

)(

32

|

|3

)(32

|

3 CHCH

H

CCH -

CHCHHCCH

I

II

4. Statement – I: SN2 reaction of an optically active alkyl halide with an aqueous solution of KOH always gives an alcohol with opposite sign of rotation. Statement – II: SN2 reactions always proceed with inversion of configuration.

a. (a) b. (c) c. (c) d. (d) D

Sol. Correct Assertion: SN2 reaction of an optically active halide with aqueous solution of KOH always gives an alcohol which is also optically active but its rotation can be +ve or –ve.

5. Statement – I : A Bulky anion is a good nucleophile Statement – II: Steric hindrance oppose nucleophilic attack. a. (a) b. (c) c. (c) d. (d) D

SECTION – C (Paragraph Type)

This Section contains 3 paragraph. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 10 × 4 = 40 Marks

Paragraph – 1 Optical isomerism arises due to chirality of molecules. Chiral molecules may or may not contain chiral carbons. If a molecule contains only one chiral carbon atom, it exist in two stereoisomers which are always optically active. If, however, the molecule contains two chiral carbon atoms, the number of stereoisomers increases to either 3 or 4 depending up whether the two chiral carbon atoms are similar or dissimilar. The number of stereoisomers which are non-superimposable mirror images of each other are called enantiomers while those which are not mirror images of each other are called diatereomers. Enantiomers are always chiral molecules whereas diastereomers may or may not be.

1. Which of the following compounds can have non-superimposable mirror image?

a. b. c. d.

D

Sol. a, b & c contain plane of symmetry and are optically inactive. So, they form superimposable mirror image.

2. Which of the following are optically inactive?

a. b. c. d.

D Sol.

OH

OH

OH

H

H

H

CH3

CH3

OH

OH

H

H

H

CH3

CH3

HO

OH

OH

H

H

H

CH3

CH3

Br

OH H

H

CH3

CH3

H

HO

HO

Cl

H

C = C = C = C = C

C2H5

H

H3C

H

H

CH3

CH3 – CH – COOH

D

C = C = C = C

H3C

H

CH3

H

Cl

H

C = C = C = C = C

C2H5

H

Optically active

;

H

CH3

H3C

H

; H3C – CH – COOH

D Optically active



‘d’ will show geometrical Isomerism

3. How many optically active stereoisomers are possible for butane-2, 3-diol?

a. 1 b. 2 c. 3 d. 4

B

Sol. Butane-2, 3-diol (CH3 - 3CHHOHCHOHC

) has two similar chiral carbon atoms. Therefore, like

tartaric acid it has three stereisomeric forms (d-, - and meso). Out of these only two (d- and ) are

optically active

4. The number of enantiomers of the compound CH3CHBrCHBrCOOH is

a. 0 b. 1 c. 3 d. 4

D

Sol. The compound HBrCOOHCHBrCCH3

contains two dissimilar chiral carbon atoms and hence 22 = 4

enantiomers are possible. Passage – 2

In a SN2 reaction, back side attack takes place which leads to inversion of configuration at -carbon which is known as Walden’s inversion:

5. What is true regarding the following reaction? a. The product would be a dextro isomer

b. The product would be a laevo isomer c. The product would be a racemic mixture d. The product would be either laevo or dextro isomer D

6. What can be correctly predicted regarding product of the following reaction?

a. The product would be achiral

b. The product would be optically active c. The product would be racemic mixture d. The product would contain unequal amount of enantiomers A

CN–

+ C X NC - - - - C - - - - X

–

–

NC C + X–

Transition state

CH3

H Br

C2H5

(laevo)

+ H C C–Na

+ H C C H + NaBr

CH3

C2H5

H

Cl

CH3

(1.0 mol)

+ NaCN(aq) (1.0 mol)

H



7. What can be predicted correctly about product of the following SN2 reaction? a. Product would be enantiomer of starting compound

b. Product would be a pure enantiomer c. Product would be a racemic mixture d. Product would be achiral

C Passage – 3

Nucleophilic aliphatic substitution reaction proceeds either by SN1 mechanism or by SN2 mechanism. The SN1 is a two step unimolecular reaction in which a carbocation is first formed as an intermediate in a slow rate determining step followed by an attack by a nucleophile in a fast step. Since carbocation is planar, the attack by a nucleophile can take place from either side of the plane. An optical active substrate will give rise to the formation of both (+) and (-) forms of the product. In most of the cases, the product usually coexists of 5-20% inverted product and 80-95% racemic mixture. The more stable the carbocation, the greater is the proportion of racemisation.

8. Which of the following compound will give SN1 reaction?

(I) C6H5 – CH2 – Br (II) CH2 = CH – CH2 – Br (III) CH3 – CH2 – Br (IV) BrCH

CH

CH

CCH 2

3

3

|

|3

Select the correct answer from the codes given below: a. (I), (II) and (III) b. (I), (II) and (IV) c. (II), (III) and (IV) d. (I), (III) and (IV) B

9. Which one of the following substrates will give maximum racemisation in the following reaction?

OH

R

R

CRX

R

R

CR|

|HX

O2H|

|

a. 32

3

3

|

|56 CHCH

CH

CH

CHC b. Br

CH

HC

CCHCH

3

52

|

|2

c. d. C Sol. It will form most stable carbocation (a) X = CH3 No leaving group

CH3

CH3

Br H

H OH + NaOH (aq)

C6H5 C

Br

CH3

OCH3 C6H5 C

Br

NH3

NO2



10. Consider the following reactions:

(I) OHH

CH

CHCClH

CH

CHC)(

3

|56

water%5

Acetone%95

3

|56

(II) OHH

CH

CHCClH

CH

CHC)(

3

|56

water%10

Acetone%90

3

|56

(III) OHH

CH

CHCClH

CH

CHC)(

3

|56

water%20

Acetone%80

3

|56

(IV) OHH

CH

CHCClH

CH

CHC)(

3

|56

water%100

3

|56

The correct decreasing order of proportion of inverted product would be a. (I) > (II) > (III) > (IV) b. (II) > (I) > (III) > (IV) c. (III) > (II) > (I) > (IV) d. (IV) > (III) > (II) > (I) A

Sol. It will form most stable carbocation

SECTION – D (More than One Answer Type) No Negative Marking

This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which One or More than one answer may be correct. 8 × 5 = 40

1. Which of the following compounds will not give SN2 reaction?

a. b. H2C = CH - Cl c. d.

A, B, C, D 2. Which of the following compounds will give SN1 reaction?

a. H3C – CH2 – Br b. c. ClCHOCH 23

d.

C, D 3. Which of the following cannot be prepared by direct halogenation of benzene with X2 in presence of

Lewis Acid

a. Iodobenzene b. Chlorobenzene c. Bromobenzene d. Fluorobenzene A,D

4. Which of the following statements regarding the SN1 reaction shown by alkyl halide is correct?

a. The added nucleophile plays no kinetic role in SN1 reaction. b. The SN1 reaction involves only inversion of configuration of the optically active substrate c. The SN1 reaction on the chiral starting material ends up with racemization of the product. d. The more stable the carbocation intermediate faster is the SN1 reaction A,C,D

Sol.SN1 reaction involve racemisation of configuration. 5. Which reaction results in the formation of a pair of enantiomers?

a. b. c. d. B,D

Cl

Br Br

Br

Ph

I

CH2CH3

H3C

Br H

I

Acetone

Br H

H2O

Br H

CH3OH OH H

HBr



6. The correct statement(s) concerning the structure E, F and G is (are)

a. E, F and G are resonance structures b. E, F and E, G are tautomers c. F and G are geometrical isomers d. F and G are diastereomers

B,C, D 7. In which of the following reaction(s), the major product is a primary alkyl bromide?

a. b.

c. d.

B,C,D Sol. (A)

(B)

(C)

(D)

8. In which of the following pair(s), the first one is stronger nucleophile than the second one?

a. CH3O–, (CH3)3CO– b. (CH3)2N

–, 2NH

c. C6H5O–, (CH3)2CHO– d. HS–, HO–

A,D

CH3

O

H3C

H3C

(E)

CH3

OH

H3C

H3C

(F)

CH3

OH H3C

H3C

(G)

+ Br2 h COOAg Br2/CCl4

CH2

+ HBr (C6H5CO)2O2 OH

PBr3

+ Br2 h

Br

Secondary (major)

COOAg + Br2

Br

Primary + AgBr + CO2

CH2

+ HBr (C6H5CO)2O2

Br

Anti-Markownikoff’s addition (primary)

OH PBr3 Br



SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 × 2 = 16 Marks

1. Match the Column – I with Column – II.

Column – I Column – II

A.

(P) Plane of symmetry

B.

(Q)

Optically inactive

C.

(R) Optically active

D.

(S)

Geometrical isomerism

Sol. A r; B p, q; C p, q, s; D p, q 2. Match the Column – I with Column – II.

Column – I Column – II

A.

dark

HBr2

|

|

3 CH

CH

CH

CH

CCH

3

3

(P)

Carbocation Intermediate

B.

(Q)

Free Radical Reaction

C.

(R)

Rearrangement

D.

H

HOH2

|

|56 CHCH

CH

HC

CHC

3

52

(S)

Markovnikoff’s addition

Sol. A P,S; B q; C p; D p, r, s

C = C = C

HO

CH3

Br

Cl

CH3

CH3

Br

OH

H3C

Cl C = C = C = C

CH3

Cl

H

H

Cl

CH3

Cl

H3C

CH = CH – CH3 HBr

Peroxide

CH3 Cl2

FeCl3



SECTION – F (Integer Type) No Negative Marking

This Section contains 5 questions. The answer to each question is a single digit integer ranging from 0 to 10. 5 × 3 = 15

1. Consider the following solvolysis reaction.

Heat

5232

|

|

|

|56 OHHCCHCH

H

Br

C

CH

CH

CHC

3

3

How many principal products would be formed by SN1 reaction? Sol. 6 Intermediate are

52

|

|HCHC

Me

Me

CPh

and

Isomer 4

52|

|

|

Nu52

|

|

HCH

Me

C

Me

Nu

CPhHCH

Me

C

Me

CPh

2. How many of the following undergo solvolysis reaction faster than benzyl chloride? Sol. 5 [iii, iv, v, vi, vii] 3. If 2, 4-dimethyl pentane is subjected to free radical chlorination reaction, how many different

monochlorinated products would be formed? Sol. 4

CHCl2 CHCl2

Cl

CH2Cl

H3C

i.

ii.

iii.

CH2Cl

MeO

CH2Cl ClCH2

OCOCH3

iv.

v.

vi.

CH3 Cl

CH3 O CH2Cl CH2Cl

O2N

vii.

viii.

ix.

CH2Cl

ClH2C

x.

+ Cl2 hv

Cl

(±) +

Cl

+

Cl



4. If 1,3-butadiene is treated with excess of bromine in CCl4, how many different tetrabromides would be formed?

Sol. 3 5. Consider the following reaction, How many different monobromo derivatives would be produced? Sol. 4

(I) has two optically active enantiomers and (II) has two geometrical isomers.

H3C CH2

H3C CH2

C = CH2 + N Br

O

O

hv

CCl4

CH3 CH2

CH3 CH2

C = CH2 + NBS

CH3 CH

CH3 CH2

C = CH2 +

Br

CH3 CH

CH3 CH2 C CH2

Br

(I) (II)

+ Br2 CCl4 H

H

Br

Br

CH2Br

CH2Br

+ H

H

Br

(±)

Br

CH2Br

CH2Br

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