Halogen Derivative

IITIIT-J EE ChE ChEm mis istt r y by N. N.J . sir

ORGA RGA NIC Chem he mIst Ry

R I S . J . N

HALOGEN ALOGEN DERIVAT ERI VATII VES

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1

SUBSTITUTION & ELIMINATION Solvents are most important for reactions since they provide medium for chemical reactions. Molecular collis ions are possible only in gaseous phase or in solvent phase. Solvent

Cl NonPolar

R I S . J . N C — Cl ; Cl Cl

Solvent

; CS2

Polar protic O O H H/ R H/ NH3

Polar

Polar

O

aprotic O

S

C —N

CH3 CH3 DMSO

H

DMF DM F

O

CH3

H O

H

H

O

H

H

H

H

O

H

H

O

O

H

H

(a) (b)

O

O—H

Solvation of cation

H

H

(c) (d)

O—H

Solvation of anions

Polar aprotic solvent Can solvate only cations & not anions CH3

O

CH3

Ans.

O=S

CH3

O S

CH3

Rule 2:

Nucleophiles:- They are e — pair donors . Particles

which donate e — pair to generate covalent or coordinate bonds are nucleophiles.

Strength of Nucleophile Rule-1 (—) charge Conjugate bases are strong nucleophiles compared to acids:-

..  CH3 — .N NH .H ..  CH3 — .O. :

CH— C—O 3

CH3NH2 CH3OH

1. 2.

—

—

R—O

S

H

H

(f)

CH3 S

DMA DM A

O

O

(e)

CH3

CH3

CH3

Polar protic Solvent:– Can solvate both cations & anions H

CH— C—N 3

CH3

CH— C — OH 3

R — OH

 H—S ..  R — .S. :

R — SH + (g) H3O H2O (a) 1>2 (b) 1>2 (d) 1>2 (e) 1<2 (g) 1<2

(c) 1>2 (f) 1<2

For atoms of comparable size, better e — donor is better nucleophile (lower electronegativity better e — donor) .. CH3 — N NH H2 > CH3 — O — H .. R — NH2 > R — OH

.. NH2 <

3.

.. NH2 O

4.

R—O—H

>

R—C—O—H


2

SUBSTITUTION & ELIMINATION Solvents are most important for reactions since they provide medium for chemical reactions. Molecular collis ions are possible only in gaseous phase or in solvent phase. Solvent

Cl NonPolar

R I S . J . N C — Cl ; Cl Cl

Solvent

; CS2

Polar protic O O H H/ R H/ NH3

Polar

Polar

O

aprotic O

S

C —N

CH3 CH3 DMSO

H

DMF DM F

O

CH3

H O

H

H

O

H

H

H

H

O

H

H

O

O

H

H

(a) (b)

O

O—H

Solvation of cation

H

H

(c) (d)

O—H

Solvation of anions

Polar aprotic solvent Can solvate only cations & not anions CH3

O

CH3

Ans.

O=S

CH3

O S

CH3

Rule 2:

Nucleophiles:- They are e — pair donors . Particles

which donate e — pair to generate covalent or coordinate bonds are nucleophiles.

Strength of Nucleophile Rule-1 (—) charge Conjugate bases are strong nucleophiles compared to acids:-

..  CH3 — .N NH .H ..  CH3 — .O. :

CH— C—O 3

CH3NH2 CH3OH

1. 2.

—

—

R—O

S

H

H

(f)

CH3 S

DMA DM A

O

O

(e)

CH3

CH3

CH3

Polar protic Solvent:– Can solvate both cations & anions H

CH— C—N 3

CH3

CH— C — OH 3

R — OH

 H—S ..  R — .S. :

R — SH + (g) H3O H2O (a) 1>2 (b) 1>2 (d) 1>2 (e) 1<2 (g) 1<2

(c) 1>2 (f) 1<2

For atoms of comparable size, better e — donor is better nucleophile (lower electronegativity better e — donor) .. CH3 — N NH H2 > CH3 — O — H .. R — NH2 > R — OH

.. NH2 <

3.

.. NH2 O

4.

R—O—H

>

R—C—O—H


2

O R — O —

5.

—

R—C—O

>

Rule 3:If e — donating atoms are of different size in polar protic solvent Nucleophilicity  size of atom (a) R — OH < R — SH (b) (C6H5)3N < (C6H5)3P     (c) F  Cl  Br  I

(2) (3)

Types of nucleophilic substitution. 1. SN1 2. SN2 3. 4. SNNGP 5. SNi 6. 7. Benzyne.

by stronger nucleophiles or depart from a molecule to generate carbocation are leaving groups. 1  Leaving group ability  Base Base streng strength th  Weak bases are good leaving groups. . .   L R L  R I

—

2.

H

3.

R

—

> Br > Cl

O

O

—

>

O

O

O

Methylsulphonate Trifluoro methylsulphonate (Mesylate) (Triflate) O O O 6. – < – O — S — S — — N O — S — S — CH3 O O

R1

2.

R—L

.. + Nu

C

R2

Θ



R3

R1

R1

R2

Nu

Nu

R2

R3

Characteristics (1) rate = k [R — L]1; rate is independent on concentration of nuclophile. (2) rate is directly proportional to stability of carbocation (3) Rearrangement possible (4) Planar carbocation can be attacked from both sides (5) Supported by polar protic solvent. (6) 2 step reaction (7)

Normally 3° – carbocation & resonance stabilized carbocation support this reaction mechanism if attacking Nuclophile is neutral polar protic solvent.

(8)

Solvolysis (SN1)

Nucleoph ilic su bstitu tion (SN – Reactions) Reactions)

. . . . Nu      R — Nu + L

R3

R3

R3

R — O —

. .

+ L

C

R2

R2

F

– O — S — CH3 < O — S — CF3

5.

Slow    C—L rds.

1.

—

— H > R— O

R—C—O O

4.

>>

R1

R1

— H > OH

O

SN AR SN1’

SN1 — Nucleophilic substitution first order:. .  1. R — L  R  L ..   2. R  Nu  R — Nu

Leaving group (L):- Particles which are substituted

1.

Nucleophile.

R I S . J . N

If e — donating atoms are of different size in polar aprotic solvent Nucleophilicity  basicity



L  L  Leaving Leaving group Solvent (4)

Ph — CH2 – Cl

H

O

H

Ph – CH2 — OH + HCl

4 Compon Compon ents

(1)

R  Substrate


3

Q.1

Compare the rate of SN1 reaction:-

(a)

Ph — CH2 — Cl

(3)

EtOH Ph — CH2 — Br 

(4)

O || H—C—OH Ph — CH2 — Br 

(5)

PhOH Ph — CH2 — Br 

Ph — CH — Cl Ph

(i)

(ii)

Ph

Ans:

Ph — C — Cl Ph

Ph — CH2 — OH Ph — CH2 — OCH3 Ph — CH2 — OEt O || Ph — CH2 — O — C — H Ph — CH2 — OPh

R I S . J . N (iii)

I

(1) (2) (3)

I

I

(4) (5)

(b) (i) I

(ii) I

(iii) I

Q.3

Which SN1 reaction is expected to occur rapidly.

1. (a) (CH3)3 CCl + H

(c)

O (d)

(b) (CH3)3 CBr + H

(i) (ii) CH3 — O — CH 2 — Cl (i) CH3 — O — CH 2 — CH2 — Cl (ii)

(e)

(f) (g)

2. (a)

(b)

Q.2

(1) (2)

Cl + tO — 

3. (a)

(a) (iii) > (ii) > (i) (b) (ii) > (iii) > (i) (c) (i) > (ii) > (iii) (d) (i) > (ii) (e) (i) > (ii) (f) (i) > (ii) (g) (i) > (ii) 1

SN reactions are also known as solvolysis as .. Θ solvent molecules behave as Nu . Write the product in each case. H2O Ph — CH2 — Br   Ph — CH2 — Br

CH3OH  

1.0 M

Cl + EtO — 

(b)

80%H2O   20%EtOH

CH3

O — Et + Cl —

O — Et + Cl —

1.0 M 1.0 M

CH3

Ans:

O — Et + Cl —

1.0 M 1.0 M

CH3 — O — CH = CH — CH 2 — Cl & CH3 — O — CH 2 — CH = CH — CH 2 — Cl 100%  (i)  Ph — CH — Cl H2O

Ph — CH — Cl

H  (CH3)3 C OH + HBr

Cl + EtO — 

2.0 M

(ii)

O

H (CH3)3 C OH+ HCl

(iii)

CH—Cl 2

CH—Cl 2

O

Cl + EtO — 

O — Et + Cl —

Ans.

1.0 M 1.0 M 1.0 M 2.0 M (1) b > a (2) a < b (3) a = b

Q.4

Compare rate of SN1 reaction

1.

2.

(a)

Cl + H

(b)

Br + H

(a)

Cl + H

O

H 

O

H  O H  O

(b)

Cl + CH3

H 


4

3.

Cl + H

H  O Cl + H H 

(a) (b)

Ans.

Br (better leaving group)

(1) (b)

(3) (a)

Me

Ans.

O

CH—Br 2

Cl

Q.6

(double bond character)

(3) a > c > b > d (6) b > a

2. CH2 = CH — Cl

Me

OTS

5. CH3 — CH2 — F

CH3 CH3

Me

6. CH— C — CH— Br 7. 32

(b)

Cl

CH3

t–Bu

t–Bu

Cl

CH3

4. CH— C — F 3

(a)

3.

Cl

Me

OTS Me

2.

(b) (2) b > a (5) a < b

These substrates cannot give SN1. Explain

1.

Compare rate of SN1

Me

(a) (1) a > b (4) a < b

R I S . J . N H better solvation

Me

1.

6.

Br

(2) (b)H

Q.5

O

(O, F & CH3 – O cannot be substituted strong bars & weak leaving groups)

SN2 reaction nucleophilic substitution

Cl (a) I

Cl (b)

Cl

O

3.

SN2 — Reaction Nucleophilic Substitution 2nd order . . Nu + R — L  Nu – – R – – L

Cl

Br

O

O

O

(a)

(b)

(c)

OTS H

H

(d)

Characteristics:-

x

x

1.

Nu

y

OTS

Nu - - - C - - - L 

+ C—L x

y

z

Nu — C

4.

. .

Nu — R + L

z

+L

y

z

(a)

(b)

2.

Br

5. (a)

Br

(b)

3. 4. 5.

1 . Bulkiness in x,y & z . . rate  strength of Nu . . rate  Leaving group ability of L Inversion of configuration takes place. ‘

rate of reaction 


5

Br + CH3O — 

(ii) Characteristics Compare rate of SN2 reaction:– Q.1 1. CH3 — CH2 — CH2 — Br & CH—CH—Br 3

CH3 (i) 2.

Ans. Q.3

(ii)

CH3 — CH2 — CH2 — CH2 — Cl (i) CH3 — CH2 — CH2 — CH2 — I (ii)

3.

5.

Q.2

(A)(i)

(1) 1 > 2 (4) 1 < 2

Q.4

1.

(2) 2 > 1 (5) 1 > 2

Write the products of SN2 reaction:– KSH  H H  Br

Br

Cl

2.

D

H D

H

(ii) (3) 1 > 2

2

Compare rate of SN reaction:– — Cl + CH3 — CH2 — O 

or Cl+ CH3 — CH2 — OH 

(ii)

SN2 X (Walden unreason not possible)

Cl

(ii)

Cl

Ans.

Br

Cl

(i)

(D) ii > i

1– Bromobicyclo [2.2.1] heptane is extremely unreactive in either SN1 or SN2 reaction. Explain: SN1 X

R I S . J . N Br

(i)

2.0 M (B) I < ii (C) ii > i

(Bridge Head position)

CH3 — CH — CH 2 — Cl (i) CH3 — CH2 — CH2 — CH2 — Cl | CH3 (ii)

4.

1.0 M (A) i > ii

ACO —  

CH3 CH3

3.

KSH  Cl 

H

Et

4.

CH3

Cl — CH— C — CH— CH 2— Cl 2 2

KSH  

CH3

Cl+ CH3 — CH2 — O — 

(B)(i)

NaCN (1 eq)

Cl

Br + (C6H5)3N 

(C)(i)

Br + CH3O — 

(D) (i) 1.0 M

NaCN  6. 

5.

Br + (C6H5)3P 

(ii)

I

CH2

Cl + CH3 — CH2 — S — 

(ii)

Br

7.

KSH(1eq)  

I

KSH  Br — CH2 — CH = CH — Br  (1eq)

1.0 M


6

Q.5

SN1

Reactant (1) (2)

Cl CH2 = CH — Cl

(3) (4) (5)

SN2

CH2 — Cl

R I S . J . N CH2 = CH — CH 2 — Cl

Cl

(6) (7)

CH3 — O — CH 2 — Cl

CH—N — CH— Cl 2 3 CH3

(8) (9)

CH3 — F CH3

CH— C — Cl 3 CH3

(10)

(11)

Br

CH 3— C — Et Cl

CH—Cl 2

(12)

Cl

(13)

CH3

(14)

CH3 —C — CH 2 — Cl CH3

(15)

— CH— Cl 2


7

Q.6

Explain – OH + CH3 — O — CH3  No. Reaction H  CH — O — CH  2CH OH 3

3

Q.14 Explain:–

(a)

TSCl

NaCN

 A  B OH 

H

(b)

14 CH— CH 2 2

R I S . J . N

KSH + CH3 — O

CH 3 

S

14 CH— CH—OAC 2 2

O

Compare rate of hydrolysis H

– ACO / ACOH    

I

O

Q.9

H

CH3 — S — CH 2 — CD2 — OH

D

Q.8

O

3

CH3

Q.7

H CH3 — S — CH2 — CD2 — I

H

Ph S

H

+

14 CH— CH—OAC 2 2

H

CH3

SPh

Cl

(a)

Q.10

H

Cl

(b)

Cl

Q.15

(c)

HO

Explain mustard gas hydrolysis at a very high rate:– Cl — CH2 — CH2 — S — CH2 — CH2 —Cl

Q.11 Explain:–

CH3

CH3

Q.16

OTS

– ACO / ACOH  

OTS

OTS

+

HBr  Br  CCl4 H

H

– Ph ACO / ACOH   H

H TSO

CH3

Q.17

(1)

H — O — CH2 — CH2 — CH2 — Br

O

 

CH3

Q.12

H H

(2)

HBr  Br  CCl 4 OH

(3)

Compare rate of hydrolysis:– SH

SH

I

I

(a)

H — O — CH2 — CH2 — Br   NH2 —CH2 — CH2 — CH2 — CH2 — Br

OH/H2O   

CH3

Q.13

+  + Br + H

(b)

SN1’

H2O CH3 —CH=CH—CH2 —Cl 

CH3 —CH=CH—CH2 —OH+ CH3 — CH —CH=CH2 | OH


8

Cl

HO

2  

Q.18

Q.19

EtOH  

Cl

Q.20

R — X + Reagent  Product

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

. . Nu

R I S . J . N Reagent

Product

aq. KOH

OH

aq. NaOH

OH

Na2CO3 (aq.)

OH

Moist Ag2O

OH

RO — /R—OH

R — NH2

R — O — .. NH3 .. R— NH2

NaSH

SH

NH3

dry Ag2O

Ag

O

Ag

KCN

 CN

AgCN

Ag — CN:


9

IIT-J EE ChEmist r y by N.J . Sir

ORGANIC chemist r y

DPP NO-01 1.

Which alkyl halide would you expect to react more rapidly by an S N2 mechanism? Explain your answer (A) CH3CH2CH2Br or (CH3)2 CHBr (B) CH3CH2CH2CH2Cl or CH3 CH2 CH2 CH2I (C) (D) (E)

2.

Time: 30 minutes

 CH3 2 CHCH2 Cl or CH3 CH2 CH2 CH2 CI CH3 2 CHCH2CH2Cl or CH3CH2CH CH3 CH2CI

R I S . J . N C6H5Br or CH3CH2CH2 CH2 CH2CI

Which S N2 reaction of each pair would you expect to take place more rapidly in a protic solvent? (A)

(1)

CH3CH2CH2Cl  CH3CH2O   CH3 CH2CH2OCH2CH3  Cl OR

(B)

(2)

CH3CH2CH2 Cl  CH3 CH2 OH  CH3 CH2 CH2OCH2 CH3  HCl

(1)

CH3CH2CH2Cl  CH3CH2O   CH3 CH2CH2OCH2CH3  Cl OR

(C)



(2)

CH3CH2CH2Cl  CH3CH2S  CH3CH2CH2SCH2CH3  Cl

(1)

CH3CH2CH2Br   C6H5 3 N  CH3 CH2CH2N  C6H5 3  Br 



OR

(D)



(2)

CH3CH2CH2Br   C6H5 3 P  CH3CH2CH2P  C6H5 3  Br 

(1)

CH3CH2CH2Br 1.0M3  CH3O  1.0M  CH3CH2CH2OCH3  Br  OR

(2)

3.

CH3CH2CH2Br 1.0M  CH3O   2.0M  CH3CH2CH2OCH3  Br 

Which S N1 reaction of each pair would you expect to take place more rapidly? Explain your answer. (A)

(1)

 CH3 3 CCl  H2O   CH3 3 COH  HCl OR

(2)

(B)

(1)

 CH3 3 CBr  H2O   CH3 3 COH  HBr  CH3 3 CCl  H2O   CH3 3 COH  HCl OR

(2)

(C)

(1)

 CH3 3 CCl  CH3OH   CH3 3 COCH3  HCl EtOH  CH3 3 COCH2CH3  Cl   CH3 3 CCl  1.0M   CH3CH2O  1.0M   OR

(2)

(D)

(1) (2)

(E)

(1) (2)

EtOH   CH3 3 COCH2 CH3  Cl  CH3 3 CCl   2.0M  CH3CH2O  1.0M  EtOH  CH3 3 COCH2CH3  Cl   CH3 3 CCl  1.0M   CH3CH2O  1.0M   EtOH   CH3 3 COCH2 CH3  Cl  CH3 3 CCl  1.0M   CH3CH2O   2.0M   CH3 3 CCl  H2O   CH3 3 COH  HCl

C6H5Cl  H2O  C6H5OH  HCl


10

4.

Write conformational structures for the substitution products of the following deuteri compound:

(D)

4,Chloro-1-butanol + NaH  H2  heat   C4H8CIONa   C4H8O  NaCl Et 2O Et 2O

(E)

Propyne  NaNH2

Cl H

(A)

l CH3OH

D

?

NH

CH3I  3   C4H6  Nal C3H3Na  liq.NH3

H

R I S . J . N Cl

H

(B)

l

CH3OH

H

8.

When the alkyl bromides (listed here) were subjected to hydrolises in a mixture of ethanol and water (80% C2H5OH/20% H2O) at 55oC, the rates of the reaction showed the following order: (CH3)3CBr > CH3Br > CH3CH2Br > (CH3)2CHBr Provide an explanation for this order of reactivity

9.

What would be the effect of increasing solvent polarity on the rate of each of the following nucleophilic substitutions reactions?

?

H

Cl

(C)

l

H

H

CH3OH

D

?

Cl

CH3

(D)

H2 O

CH3OH

H

?

(a)

Nu: + R  L  R  Nu+ + :L –

(b)

R  L+  R+ + :L

D

10.

5.

1-Bromobicyclo [2.2.1] heptane is extremely unreactive in either SN2 or SN1 reaction explanations for this behaviour.

6.

When ethyl bromide reacts with potassium cyanide in methanol, the major product is some CH3CH2NC is formed as well, however. Write Lewis structures for the cyanic both products, land provide a mechanistic explanation of the course of the reaction

Competition experiments are those in which two reactants at the same concentration (or one reactant with two reactive sites) compete for a regent. Predict the major product resulting from each of the following competition experiments: CH3

(a)

(A)

Br

F

+ NaI (1 mol) H

acetone

C5 H8FI

NaBr

(B) 1, 4–Dichlorohexane (1 mole) + NaI (1 mole)   C 6H10 ICl  NaCl acetone (C)

CH3

Cl l



CH3

Give structures for the products of each of the following reactions: H

CH2

CH3

(b)

7.

Cl CH2 C

Cl

C

CH 2

CH 2

Cl H 2O

CH 3

11.

Predict the structure of the product of this reaction: HS H

H Cl

NaOH in aqueous EtOH

C6 H10S

1, 2–Dibromoethane (1 mole) + NaSCH2CH2SNa   C 4H8S 2  2NaBr


11

R I S . J . N

IIT-J EE ChEmist r y by N.J . sir

ORGANIC chemist r y

Comparison of SN1, SN2, E1 & E2

Reactant

Neutral Nucleophiles or Bases

R

O

HH

O

HR

S

.. NH3

HH

S

H

Weak Nucleophil es

Strong Bases /Nucleophiles

O || — I , SH , CN , CH3 – C –O

RO / EtO / CH3 O / O /Et3N1

Ph — CH2 — X

Ph— CH2 —CH2 —X



..   ..  : , Br , SCN , R– S , .N. , N , .N.  

(X  F) CH3 —X







SN2

SN2

SN2

SN1(Solvolysis)

SN2

SN2

SN2

SN2

E2 (formation of conjugated double. Bond) SN2 {Exception in sterically hirdered base

SN2

R—CH2 —X

O /  OH or Et3N,

SN2

E2 is major product}

SN1(low Temp.)

R— CH —R | Br

R | R— C —Br | R

E1 ( high Temp.

SN2

E2

SN1 (low Temp.)

SN1 (low Temp.)

E2

E1(high Temp.)

E1 (high Temp.)

12



ORGANIC chemist r y

DPP-02

Time: 30 Minutes

Comparison of SN1, SN2, E1 & E2 (1)

Ph — CH2 — Br

  H2O

KSH   KCN    EtONa/EtOH 

(2)

(9)

CH3 | CH3 — C — CH — CH3 | | CH3 Br

(10)

(a)

(3)

(4)

R I

(b)

Cl CH3 — CH2 — Br

CH3 —O / CH3OH  

H2O  CH3OH  

EtOH   

CH3 — O — +

Br

O — + CH3 — Br

 

CH3

(11)

To prepare CH3—O—CH

, which of

CH3 the following will give better yield:– (1) CH3 — ONa + CH3 CH — CH3 


ORGANIC chemist r y

DPP-02

Time: 30 Minutes

Comparison of SN1, SN2, E1 & E2 (1)

KSH   KCN    EtONa/EtOH 

(2)

Cl (3)

  H2O

Ph — CH2 — Br

CH3 —O / CH3OH  

H2O  CH3OH    EtONa/EtOH  

Cl

(11)

(6)

CH3 (7)

NaI /Acetone   NaN3     CH3 —C—O || O  EtO/EtOH

NaCN   H  EtONa/EtOH    NaI /Acetone  OTS 

(12)

(13)

, which of

Styrene (C6H5 — CH = CH 2) is to be prepared by dehydrohalogenation using alc.KOH. Which of the following will give better yield. Ph alc.KOH  (1)  CH3—CH Br Ph | CH2 — CH2 — Br

(b)

NaH

 O — H  (14)

alc.KOH  

Write mechanism which accounts for the following products. (a) HO — CH2 — CH2 — Br

O

NH2 —CH2 —CH2 —CH2 —CH2 — Br

OH/H2O 

CH3 —CH2 —CH2 —Br   (B)



O — + CH3 — Br

OH/H2O 

Br

(A)



Br

To prepare CH3—O—CH

(2)

EtONa/EtOH CH2=CH— CH —CH3   | NaI /Acetone  

(8)

CH3 — O — +

CH3 the following will give better yield:– (1) CH3 — ONa + CH3 — CH — CH3  | CH3  (2) CH3 — CH — ONa + CH3 — I | CH3

 /

R-2-chloropentane

H

(a)

CH3

O OH    Et3N

(5)

EtOH   

R I S . J . N (10)

(b)

CH3 — CH2 — Br

(4)

(9)

CH3 | CH3 — C — CH — CH3 | | CH3 Br

N H

Major Product of reactions (E1, E2, SN1 & SN2)


13

(a)

Cl

CH3 —CH2 — CH2 — CH2 — CH2 —Br + CH3 — CH2 — O —

H

(b)

CH3CH2OH  50C

H

 I   CH3OH

D Cl

(b)

CH3 —CH2 — CH2 — CH2 — CH2 —Br +



  OH/50C

O —

 I   CH3OH

H

(c) D

R I S . J . N H

(c) (d)

 CH3OH /50C

—

Br + CH3O

Cl

O —  

OH / 50C

Br +

H2O /CH3OH CH 3  

(d)

H

D

Cl

(e)

  50C I /Acetone

(16)

Cl

(f)

CH3OH/ CH3  

Give major product (E1, E2, SN1 or SN2) (a) CH3 — CH2 — CH2 — Br CH3 —OH + CH3 — O —   50C (b)

25C

CH3 — CH2 — CH2 — Br O —

+

(g)

(h)

(i)

(15)

CH3OH 3-Chloropentane + CH3O —   

CH3

50C

(c)

O || 3-Chloropentane + CH3 — C — O — O || CH3 —C—OH   50C  25C  OH + (R) -2-bromobutane 

(j)

 (S)-3-bromo-3-methylhexane  

(k)

 (S)-2-bromooctane  CH3OH /50C

Et

(d)

Et

   CH3OH

(a) D H

 CH3OH  + OH  50C

Br

CH3OH   25C

Et

(e)

Et

Et

(16)

Cl

I

Br

Et

I

H

50C   CH3OH

H Et

25 C CH3OH

Major product (E1,E2,SN1 or SN2)

 + SH C — Br



OH   50C

(a) (b)

 Consider the reaction of I with CH3 —CH2 — Cl. Would you expect the reaction to be SN1 or SN2 The rate constant of the reaction at 60°C is 5 × 10 –5 L mol –1 sec –1. What is the reaction rate  if [ I ] = 0.1 mol/L and [CH3 — CH2 — Cl ] = 0.1 mol/L.


14


ORGANIC chemist r y

DPP- 03

Time: 30 Minutes —

E1CB - Mechanism Hoffmann V/S Saytzeff alkene (elimination)

Q.5

1.

EtONa   EtOH

Ph — CH2 — CH — CH3 | F OH/   

R I S . J . N Q.6

+ N

I

+

+

Saytzeff

L.G.

CH3

Hoffmann

% Hoffmann

 – N (Me)3  – S (Et)2 – F — Br

CH3

—

OH   

Q.7

+ N

L.G. Tendency

—

90%

Q.8

74% 70% 19%

OH   

+ N

—

Q.9

EtO/EtOH   N(Me)3 +

OH   

+

95%

5%

OH   

+ N

Q.10

OH/   

+ N

T.S. has more carbanion character.

+ N(Me)3

Q.1

Q.2

Hoffmann exhaustive methylation Q.1



2. Ag2O(moist) 3. 

EtO/  

F

Q.3

Q.2

NH2

N

Q.3

Me | Me—CH2 — C — CH2 — Me |  N(Me)3

EtO/  

EtO/  

1. CH3 I   2. Ag2O(moist) 3.  1. CH3 I   2. Ag2O 3. 

1. CH3 I   2. Ag2O 3. 

Q.4 N H

N+ Q.4

1. CH3I CH3 — CH2 — CH2 — NH2   

Q.5 N H

1. CH3I   2. Ag2O(moist) 3. 


15

Ph

1. CH3 I   2. Ag2O 3. 

Q.6 N H

Q.4

N —

O Q.7

N

Me

1. CH3 I 1. CH3 I   (A)   (B) 2. Ag2O 2. Ag2O 3.  3. 

CHCl2 — CF3

Q.9

O — || OH  CH3 — C —CH2 —CH2 —F  

—

Q.11

E

1

Q.1

Q.2

 

Me

1.H2O2    2. 

Q.5

R I S . J . N alc.KOH   

Q.8

Q.10

N

O Me

Me

EtO  CH3 — C — CH2 — CH2 — F  || O — EtO CH3 — C — CH2 — CH2 — F    || O

CH3 |  Ph — CH2 — CH2 — N — Me  | O — H2O2    (A) 

Me—N—Me

Q.6

CH3 — N — CH2 — CH2 — CH3 | Ph MCPBA  (A)    MCPBA   

Q.7

N

Q.8

O ||  Ph — C — O — CH2 — CH3 

Q.9

S ||  CH3 —CH2 — C —S—CH2 —CH3 

Q.10

Q.11

O ||  CH3 — CH2 — C — O — CH2 —CH3  O O—C—CH3   O—C—CH 3 O

Ph

Q.3

N —

O

Me

 

Me


16

H ALOGEN DERIVATIVES Classification

Aliphatic

R I S . J . N Unsaturated

Aromatic

CH2 =CH–Cl (Vinyl) CH2 =CH–CH2–Cl (Allyl)

Saturated

(Benzyne)

CH3–CH2–CH2–Cl (1°)

Monohalogen

CH3–CH–CH3 (2°)

CH3–CH

dihalogen

Cl

SN1/SN2

(SN AR)

CH3–C–CH3 (3°) Cl Cl

CH2–Cl

Cl NO2

Cl CH3

Side chain

Cl

(gem) (geminal)

CH3–CH–CH (Vic) (Vicinal) Cl

Cl

CH 2–CH2–CH2 (Isolated) Cl

Cl

CHCl3 (Chloroform)

Polyhalogen

CCl4 (Carbon tetrachloride pyrene)

General Method of Preparation

(1)

Q.1

Form alkanes XX  R—X + H—X R—H h X2 = Cl2 or Br 2 Cl2 = Non – Selective ; Br 2 = Selective X2  F2 and X2  I2 F2 = uncontrollable ; I2 = reversible Cl2  % =_ _ _ + % = _ _ _ h



R1

: R2

: R3

 1: 3.8 : 4.5

Br 2  _ _ _ _ _ _ h

Q.2

CH3

Cl2

hv

---- % of 3° mono-chloroproduct

hv

Q.3

Br 2

CH3—CH—CH2—CH 3

Cl2/ h

CH3

Br2/ h

(2)

Monochloro

Monobromo

From alkenes

R—CH = CH2

HX CCl4

R— CH —CH3 | X : Markownikoff addition; Classical carbocation HBr R — CH = CH 2 R—CH2 —CH2 —Br R–O–OR


17

Anti – markownikoff addition peroxide effect or Kharash effect Q.1

HCl(R–O–OR)

HBr( R–O–OR)

A

(1)

C

h  N–Br 

C

HBr/CBr 4

 + Br N

C

O

B

D

O

O C

HCl CCl4

C H 3 — CH =C H 2

Mechanism:–

O

O

C Q.2

Write mechanism:– CCl4 (a)  Peroxide

R I S . J . N CCl3

Cl



I

CCl3

BrCCl3  Peroxide

(c)

Br

(3)

CH2 = CH – CH 2 – H + N

(2)

CBr 3

ICBr 3  Peroxide

(b)

SO2Cl2   h

R—Cl + SO2  HCl  

C O

(3)

C C O



NBS  CH3 

Q.1

O

 CH2 = HC – CH2 + H–N

  CH2 = CH – CH2 + Br CH2 = CH – CH 2 – Br

Reed’s Reacti on

R—H

C

CH2Br

SO2Cl2  Sulphuryl Chloride Mechanism.

(free radical substitution)

O



 Cl

O

Imp.

C

NBS

NBS  

N–Br (N-Bromo Succinimide)

Cl2 CH3 – CH  CH2  CH2 – CH  CH2 500C | Cl

at high temperature free radical substitution and not Non-classical carbocation addition

O

(4)

NBS  Q.4 

Q.3

Br

O

   Cl R—H + Cl  H—Cl + R +  R— Cl. (Reaction rate is slow; selective and better method then direct halogenation)

2.

Br +

O

h  Cl + Cl—S—Cl  Cl—S  SO2 +

1.

NBS  

Q.2

(5)

From alcohols

(a)

C

O

NBS  CH3 — CH = CH 2  CH2 — CH = CH2 | Br

Darzen’s process

CH3

H

OH D

SOCl2 

CH3

H

Cl D

+ SO2 + HCl (Retention of configuration)


18

CH3

CH3 H

SOCl2 .. N

OH

Cl

H

(2)

D

D

+ SO2 + HCl (inversion of configuration) SOCl2

OH Q.1

Cl Cl

A

Cl P

Cl

Cl

CH3

CH3

D

H

OH

H Ph

SOCl2

(3)

B

Cl

C

Cl

SOCl2 .. N

D

R—OH + Lucas Reagent (HCl/ZnCl2) HCl/ZnCl2 R—OH   R—Cl (Turbidity) Test todistinguish1 /2 /3  alcohol.

(6)

Cl

P

HI   KBr 

R—OH (c)

R—OH (SN2)

R—OH

Cl

Cl

O

Cl

P

Cl

Cl

Cl P—Cl  Cl Cl

HI  (excess)

No. Reaction

R— I + R’– I

PCl5  

R—Cl

i 2 Cl (SN (or SN )

Q.1

Q.2

O

O CH3

HI  

OH

+

I

(1)

1

R—O—R Cl Cl P—Cl Cl Cl



’ R—Cl + R —Cl + POCl3

Q.3

O CH3 – I +

CH3 – I

HI(Conc.)  

(high temperature SN2 dominates) PCl5

Cl

R—OH + R’– I (SN1) R—O—R’

R—Cl + H3PO3

+ POCl3

HI   (anhydrous) R—OH + R’– I (SN2) HI(Conc.) R—O—R’  

R— I

PCl3  

Cl

From ethers

O +

+

R—O—R’

3  alcohol   within sec onds  2  alcohol   within minutes   1  alcohol   w ithin hrs 

R—OH



R—C—H || O +

(4)

(b)

Cl | R — C — H + POCl3 | Cl

R I S . J . N SOCl2 .. N

Q.2

R — C — R1  || O +

Cl | R — C — R1 + POCl3 | Cl

OH

+

HI(anhydrous)  

OH


19

(7)

Halide exch ange Reacti on

(a)

(2)

Finkelstein Reaction NaI   acetone

Cl

I

(b)

+ NaCl.

O

Br + NaCl.

NaBr acetone

Halo for m Reaction

I2 /NaOH CH2—C—H   O

Swartz Reaction NaF/DMF  R—F + NaCl. R—Cl  or AgF

CHI3

+

+ H—C—O — Na

(yellow)

R I S . J . N O

(8)

H.V.Z. Reactio n (Hell Volhar d Zelinesky )

O

O

NaOH/ I

2  C—CH3 

Q.1

RedP  CH —C—OH CH3—C—OH  2 Cl

O

2

Q.2

Cl

O Imp

H—C—OH

× (a)

O

O

Ph—C—OH

×

C—OH

×

R—C—OH

Nucleophilic Substitution

K

1.AgOH  R — Br 2.Br2 / CCl4 / 

J

Q.1 Q.2

ONa

+

I

R–C C–Ag

KNO2

AgNO2

G

 

(4)

RSNa

ONa

+

CH3 — Cl



alc.AgNO3

AgCN

RONa

ONa

Q.5

ONa +

+

CH2–Br 

Br

A B C D E F

NH3 

CH3 CH —N +—CH 3

Q.4

L

Hoffmann exhaustive alkylation

CH3 —Cl

Q.3

R’Na

R—Cl

RCuLi

H

CH3 — Cl

dry. Ag2O

, 2

 (SN2)

ONa + CH3 — Cl

KCN

R’MgX

Properties of Halogen Derivatives (1) Williamson ether synthesis

R—ONa + R1 — Cl R—O—R1 + NaCl

NaOH/ I2  

C—CH3

Q.3

Hunds diecker Reaction

O

NaOH/ I2  

CH3—C O

3

Cl —

CH3




20

(5)

With AgNO3(aq) & AgNO3(alc.) Gives ppt. with Subst rate

aq. AgNO3

1.

C—C—C—Cl

2.

C—C—C

alc. AgNO3

Cl C

R I S . J . N

C—C—C

3.

Cl

Cl

4.

Cl

5.

Cl 6. 7.

C = C — Cl

8.

C = C – CH2 – Cl

9.

Cl

10.

CH2–Cl


21

Q.7

EXERCISE – I Q.1

Which one of the following compounds will be most reactive for S N1 reactions : Br Cl Cl I O

O (B)

(A)

(C)

Me Q.8

O (D)

Consider the following groups : (I) — OAc (II) — OMe (III) — O — SO 2 – Me (IV) — O — SO2 — CF3 The order of leaving group nature is : (A) I > II > III > IV (B) IV > III > I > II (C) III > II > I > IV (D) II > III > IV > I

HO

OH     H II

H

Et B Steps I and II are (A) I can’t be S N1 (C) I SN1 & II S N2

O Q.2

Which of the following compounds is most rapidly hydrolysed by SN1 mechanism. (A) C6H5Cl (B) Cl – CH2 – CH = CH 2 (C) (C6H5)3CCl (D) C6H5CH2Cl Me Cl

OH     I

H

Et

R I S . J . N Q.9

Me OH Et A

(B) II can’t be S N1 (D) I SN2 & II SN1

What are A & B in the following reaction ? Cl ( i) CH3CHO Mg / THF      A       B (ii ) aq. NH4 Cl

Q.3

(I)

CH2 – Br

(II)CH3 – CH2 (IV)

(II) H3C

CH2 – Br

(A)

CH3

CH

CHOHCH3

Br

&

Br

Cl

CH2 – Br

CH3 (A) II > III > IV > I (C) III > IV > II > I Q.4

Br MgCl

Arrange the following compounds in order of decreasing rate of hydrolysis for S N1 reaction :

(B)

CH2 – Br

Br (D) None of these

Q.10

In the given reaction the product [P] is : (A) CH3 – CH = CH – CH 2 – Br Br (B) CH3 – CH – CH = CH 2 (C) CH2 = CH – CH = CH 2 Br

CHOHCH 3

MgCl

(C)

Consider the given reaction :

SN 1'

&

MgBr

MgCl

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

HBr    [P] CH3 – CH = CH – CH2 – OH 

Cl

&

CHOHCH 3

Which will give white ppt. with AgNO 3? (A)

Cl

(B)

(C)

CH2Cl

(D) Both A & C

Cl

Q.11

When ethyl bromide is treated with moist Ag 2O, main product is : (A) Ethyl ether (B) Ethanol (C) Ethoxy ethane (D) All of the above

(D) CH3 – CH – CH 2 – CH2 – OH Q.5

The given compound CH 3 – O – CH2 – Br gives which one of the following reactions : (A) Only SN1 (B) Only SN2 (C) SN1 as well as SN2 (D) E1

Q.12

When ethyl bromide is treated with dry Ag 2O,main product is : (A) Ethyl ether (B) Ethanol (C) Ethoxy ethane (D) All of the above

Q.6

Among the bromides I-III given below, the order of reactivity in SN1 reaction is : O

Q.13

Consider the S N1 solvolysis of the following halides in aqueous formic acid : Br CH3 Br CH3 (I) CH – CH – CH 3 (II) CH3 Br

(I)

(II)

Br (A) III > I > II (C) II > III > I

(III) Br

Br (B) III > II > I (D) II > I > III

(III) C6H5 – CH – C6H5 (IV) Br


22

Which one of the following is correct sequence of the halides given above in the decreasing order of their reactivity? (A) III > IV > II > I (B) II > IV > I > III (C) I > II > III > IV (D) III > I > II > IV Q.14

D

(B)

Cl

OH

HI SN1

   ?

Q.18

H H Major product is : D D

In the given pair in which pair the first compound is more reactive than second for S N1 reaction (A)

D

D I

(A)

I

(B)

CH2Cl

D

Cl

I

R I S . J . N D

(C)

Cl

(C)

D

Cl

Cl

Cl

(D)

Q.15

Cl

CH2 – Cl

(C)

CH3 CH2 – Cl

Br (A) I > II > III > IV (C) III > IV > II > I

For the given reaction R1 R1 HOH    R – C – OH R – C – X 

(A) C6H5 – C – Br

(B) CH2 = CH – C – Br

C2H5

Br

(C)

C6H5 – C

OCH3

C2H5 Br

C6H5 – C

NO2

(D)

(B)

CH3

CH3

(D)

CH3

CH3

R2 R2 Which substrate will give maximum racemisation ? CH3 CH3

NO2

Q.21

Br

Arrange the following compounds in decreasing order of their reactivity for hydrolysis reaction (I) C6H5 – CH2 – Br (II) C6H5 – CH – C2H5

(III)

(D) C2H5 – C – Br

D

(D)

Br CH3 Br

C2H5 C2H5

(C) C6H5 – C – Br

Which compound undergoes hydrolysis by the S N1 mechanism at the fastest rate ? CH3 Br CH3

(C) Q.17

C2H5 H

Q.20

(B)

OCH3

(A)

Which one of the following compounds will give enantiomeric pair on treatment with HOH ? C2H5 CH3 (A) C6H5 – C – I (B) CH3 – C – Br

CH2 – Cl

H CH2 – Cl

Q.16

Q.19

Which of the following is most reactive toward SN1.

(A)

(D) None of these

Br Br

(IV)

CH

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

NH3

In the given reaction : Cl Cl

CH3OH      [X]

O Cl (excess) Cl

OCH3 Cl

Cl

(A)

(B) O

OCH3

O

OCH3


23

Cl

OCH3 Cl

OCH3

(C) O

(A)

O

20%

Cl

Cl

NaOH      A. excess

(A)

(C)

R I S . J . N OH

100%

OH

OH

(D)

(D)

Cl

OH

Compare the two methods shown for preparation of carboxylic acids :

100%

the

Mg Method 1 : RBr     RMgBr

Q.26

Which of the following nucleophile will show minimum reactivity towards S N2 reaction : (A) Me3CO (B) MeO H (C) (D) Me2CHO O

Q.27

In the given reaction OH H NaOH      [X] H 25C Cl OH OH (A) (B) H H

diethyl ether

(I) CO2      RCO2H  (ii) H3O

NaCN     RCN Method 2 : RBr  H2O, HCl      RCO2H heat

Which of the following statements correctly describes this conversion ? 

(D) Q.24

Br CO2H Both method 1 and method 2 are appropriate for carrying out this conversion Neither method 1 nor method 2 is appropriate for carrying out this conversion. Method 1 will work well, but method 2 is not appropriate Method 2 will work well, but method 1 is not appropriate

The major product in the given reaction CH2Br + NH3  Br CH2NH2 CH2NH2 (A)

(B)

Br CH2Br

(C)

NH2

(D) All of these

OH H H OH

(C) Mixture of (A) and (B) (D)

H

Q.29

In the given reaction : O (i ) NaN3      [X] ( ii ) HOH

[X] will be OH

(A)

OH

(C) Mixture of (A) and (B)

N3

(B)

Et 2 O / 

the products are :

H

Which of the following can not give S N1 reaction easily ? Br Br Br Br (A) (B) (C) (D)

N3

In the given reaction : Br CH 3  C  CNa       

O

Q.28

NH2 Q.25

80% C  C – CH 3

Cl

(B)

(C)

+ 20%

OH

(C)

80%

C  C – CH3

is

O

(B)

Br

(B)

Q.22

(A)

and

(D)

Cl

Cl

Q.23

C  C – CH3

(D)

OH OH


24

Q.30

In the given reaction : CH3 – CH – CH2 – CH2 – CH – CH3 OTs

(C)

OTs

(i) SH (one equivalent)

Q.36

(ii) KOH [x] [X] will be : OTs

Q.31

Q.32

Q.33

Q.34

R I S . J . N S

CH3

CH3

(D)

OCH3

Q.37

Q.38

H D

Me D

H

(B) Me Et

Which reaction proceeds faster with NaI is DMSO. CH2Cl Cl CH3 (A) (B)

is

Cl

Correct order of rate of hydrolysis for following compounds is Br Br (I)

(II)

Br

Br

(III)

(IV)

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

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

Which reaction conditions (reagents) is suitable for the following reaction : ?  

H D

SH SH Me (D) H Et D H

–

(D)

Cl

H

H SH

(C) HS Et

NO2

In which of the following, replacement of Cl most difficult? (A) (B) Cl Cl (C)

Q.39

SH

Q.35

S

H KSH D    

(A) Me Et

(D)

CH3

The reactivity of 2-bromo-2-methylbutane (I), 1bromopentane (II) and 2-bromopentane(III) towards SN2 displacement is such that : (A) I > II > III (B) I > III > II (C) II > III > I (D) II > I > III

H

CH3 CH2 – Cl

(C)

S

Non-occurrence of the following reaction Br – + CH3OH  BrCH3 + OH – , is due to (A) Attacking nucleophile is stronger one (B) Leaving group is a strong base (C) Alcohols are not good substrate (D) Hydroxide ions are weak bases

Me Et

(B) H CH2 – Cl

For CH3Br + OH –  CH3OH + Br – the rate of reaction is given by the expression : (A) rate = k [CH 3Br] (B) rate = k [OH – ] – (C) rate = k [CH3Br] [OH ] (D) rate = k[CH 3Br]° [OH – ]°

I

Which of the following is most reactive toward S N2. CH2 – Cl CH2 – Cl

S

S

(C)

(D)

(A)

(A) CH3 – CH – CH 2 – CH2 – CH – CH 3 (B) CH3 – CH – CH 2 – CH2 – CH – CH 3

CH3

CH2Cl

Cl

Br – C – H

H – C – OH

D

D

(A) Br 2 / CCl 4 (B) SOBr 2 (C) HBr / conc. H2SO4 (D) PBr 3

Q.40

Arrange these compounds in order of increasing SN2 reaction rate : O Cl (I)

Br (II)

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

Cl

Br

(III)

(IV)

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


25

Q.41

The reaction of SOCl 2 on alkanols to form alkyl chlorides gives good yields because (A) Alkyl chlorides are immiscible with SOCl 2 (B) The other products of the reaction are gaseous and escape out (C) Alcohol and SOCl 2 are soluble in water (D) The reaction does not occurs via intermediate formation of an alkyl chloro sulphite

Q.46

The major product formed in the following reaction is : OH H SO 4 CH3  2    heat CH3

H3C CH3

CH3

(A)

CH3 Q.42

SOCl

2     (A) OH  Pyridine

H

H3C

(B) Cl

Cl

(C)

(D) H2C = C

H

Q.47

D

H2C

CH3

(A)

Cl <

Cl

SOCl2

    (A) OH 

(B)

(C)

CH3

CH3

(B) Cl

Cl

(D) H2C = C

Br (D) CH3 – CH2 – Cl < CD 3 – CD2 – Cl

H

D

Q.48

In the given reaction : ..

.

HOH CH3 – CH2 – S    [X] .. – CH2 – CH2 – Br  [X] will be :

* (A) CH3 – CH2 – S – CH 2 – CH2 – OH * (B) CH3 – CH2 – S – CH2 – CH 2 – OH (C) 1 : 1 mixture of (A) and (B) (D) 2 : 1 mixture of (A) and (B)

Q.45

H2 O

   (X) Me2C = CH – CH2 – CH2 – Cl  Major product of above reaction is OH (A) Me – C – CH2 – CH2 – CH2 Me (B) Me2C = CH – CH 2 – CH2 – OH (C) Me2C = CH – CH – CH2 – OH OH

OH

(D)

CMe2

<

I

H

D

D

(C) H2C = CH2

Cl <

Cl

D Major product (A) is :

Q.44

CH3

In the given pairs, which pair represent correct order of rate dehydrohalogenation reaction

CH3

(A) H

CH3

(D)

H3C

H

D

(C) H2C = CH2

H

H3C

CH3

CH3

D

Q.43

CH3

R I S . J . N

D Major product (A) is : CH3

(A) H

CH3 (B)

Q.49

CH3 C 2H5 O  H D      ? Major product is : C 2H5OH Br H CH3 H3C H3C H (A) (B) C=C C=C CH3 H H H3C H3C CH3 (C) (D) C=C C=C D H D

C6H5 CH3 H Br H C6H5 CH3 C (A) C H H C (C) C C6H5

CH3 H H CH3

alcoholic, KOH        A E 2

C6H5

CH3

C6H5

C

(B) C6H5 CH3

C C6H5

H

(D) None is correct C6H5


26

Q.50

The rate of elimination (using EtONa) of :. Br Br (I)

(II)

H  / MeOH

     OH 

(D) H Et

(III) Br

t-Bu (A) I > II > III (C) II > I > III

Me

t-Bu

Q.5

(B) I > III > II (D) III > I > II

SN2 reaction will be negligible in Br Br Br (A)

Q.2

SN1 & SN2 is not favourable in (A) H2C = CH – Cl (B) Ph – CH 2 – Cl (C) Ph – Cl (D) H2C = CH – CH2 – Cl

Rate of SN2 depends on (A) Conc. of Nucleophile (B) Conc. of substrate (C) Nature of leaving group (D) Nature of solvent

Q.7

Correct statement(s) for the product(s) of following reaction. Cl2 / 500C      CH2 = CH – CH2 – Ph  (A) Four different products are formed (B) Two optically active products are formed (C) The optically active compound formed here can also be made by the reaction of HCl (D) The reaction path is free radical substitution.

Q.8

In the given pair in which pair the first compound is more reactive than second to S N2 reaction.

(B)

Cl

(C)

(D) Ph – CH – CH – CH 3 CH3 Cl

Cl Q.3

Q.4

Which of the following statements is/are true ? (A) CH3 – CH2 – CH2 – I will react more readily than (CH3)2CHI for S N2 reactions. (B) CH3 – CH 2 – CH2 – Cl will react more readily than CH3 – CH2 – CH2 – Br for SN2 reactions. (C) CH3 – CH2 – CH2 – CH2 – Br will react more readily than (CH 3)3C – CH2 – Br for S N2 reactions. (D) CH3 – O – C6H4 – CH2Br will react more readily than NO 2 – C6H5 – CH2Br for S N2 reactions. In which of the following case configuration about chiral C* is retained : Me (A) H

(D)

Q.6

SN1 & SN2 product are same in (excluding stereoisomer) Cl (A)

(C)

R I S . J . N EXERCISE – II

Q.1

(B)

Br

CH3Br Na        OH 

D

Me

SOCl2

CH3ONa

         OH 

(B) H D Me

(A)

Cl

Cl

(B)

Cl

(C)

Cl

Cl

(D)

Cl

Cl

Q.9

A gem dichloride is formed in the reaction : (A) CH3CHO and PCl 5 (B) CH3COCH3 and PCl 5 (C)CH2 = CH2 and Cl 2 (D) CH2 = CHCl and HCl

Q.10

Match List – I with List – II for given S N2 reaction & select the correct answer from the codes given below Z – CH2Br + CH3O  Z – CH2 – OCH3 + Br Li st – I (A) H –

Li st – II (relative reactivi ty)

(B) CH3 – (C) C2H5 – (D)

CH3 CH3

PCl3

CH2Cl

CH –

(P) 0.1 (Q) 3 (R) 1

(S) 100

CH3ONa

        OH 

(C) H D


27

Q.11

In which product formation takes place according to Hoffmann’s rule (A) CH3CH2 – CH – CH3

EXERCISE – III Q.1

t–Bu O K

Identify A, B, C, D, E and F in the following series of reaction.



Br 2 h

Br

(B) CH3CH2 – CH – CH 3 CH3CH2OK 

Q.2

OH 

NBS

E

C

F

What are the products of the following reactions? CH3 (a) CH3 – C – Cl + OCH 3

OH

(D) CH3CH2CH – CH3

Na

+C

R I S . J . N CH3

CH3



CH3 CH3



S(CH3)2

Q.12

B

alc.KOH D

Br CH3 (C) CH3CH2 – CH – N —– CH 3

aq.KOH

A

(b) CH3 – C – O – + CH3 – X 

Which of following are correct for given reaction

CH3



N OH

Q.3

Complete the following by providing the structure of (A),(B), (C) and (D). PBr 3 Alc.KOH    (A)      (B) (i) CH3CH2CH2OH 

(A) Major product of reaction is

NH3 HBr     (C)     (D)

N



Alc.KOH H / H2 O     (A)  (ii) CH3CH2CH2I      (B)

CH3

SOCl2 H      (C)   (D) LiAlH4

(B) Major product is

N

Alc.KOH NBS    (A)      (iii) CH3CH2CH = CH2  Light

(C) The reaction is thermal elimination reaction (E1CB) (D) The reaction is E 2 reaction Q.13

Match the List I (reaction) with List II (reaction intermediate) and select the correct answer using the codes given below the Lists. List I

HBr    (C) (B) 

CH3CHO / H2O HBr       (A)     (iv) CH3CH2MgBr  Alc.KOH     (C) (B) 

Q.4

CH3 – CH2I reacts more rapidly with strong base in comparison 2 I Li to CD3CHList II

Q.5

Propose a mechanism for the following reactions. OH CH3 CH3 H2O C – Br CH3 CH3

Q.6

Each of the following alcohols has been subjected to acid catalyzed dehydration and yields a mixture of two isomeric alkenes. Identify the two alkenes in each case, and predict which one is the major product on the basis of the Zaitsev rule. H3C OH (a) (CH3)2CCH(CH3)2 (b)

List II

alc.KOH / 

     (A) CF3 – CHCl 2  CF2 = CCl 2 (1) Transition state CH3 

H (B) CH3 – C – OH    

CH3 CH3 – C = CH2

(2) Carbocation

CH3

alc.KOH (C) CH3 – CH2 – Br      

CH2 = CH2

(3) Carbanion

Br

OH OH

alc.KOH / 

     (D) CH3 – C – CH 3  CH3 CH3 – C = CH2 CH3

(4) Free radical

(c) H


28

Q.7

Give the major product (with proper explanation) when following halogen compounds are treated with sodium ethoxide. CH2Br (a) CH3 – CH – CHCH3 (b) CH3 Br CH3

Q.2

Arrange the following compounds in order of increasing dipole moment [IIT 1996] Toluene

m-dichlorobenzene

I

II

o-dichlorobenzene

p-dichlorobenzene

III CH3

(c) Cl Q.8

Q.9

KCN reacts with R – I to give alkyl cyanide, while AgCN results in isocyanide as major product.

Q.10

Predict the product(s) and write the mechanism of each of the following reactions. HI (1mole) 

O Q.11

Q.3

Q.4

excess HI

(ii)

O

CH2 – CH3 

(CH3)3CMgCl reaction with D 2O produces : [IIT 1997]

(A) (CH3)3CD

(B) (CH3)3OD

(C) (CD3)3CD

(D) (CH3)3OD

The order of reactivity of the following alkyl halides for a SN2 reaction is : [IIT 2000]

(D) R – I > RBr > R – Cl > R – F

CH = CH2

Q.5

A hydrocarbon C8H10 (A) on ozonolysis gives compound C4H6O2 (B) only. The compound (B) can also be obtained from the alkyl bromide C3H5Br (C) upon treatment with magnesium in dry ether followed by CO 2 and acidification. Identify (A), (B) and (C) and also give equations for the reactions.

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides [IIT 1990] due to (A) The formation of less stable carbonium ion (B) Resonance stabilization (C) The inductive effect (D) sp2 hybridised carbon attached to the halogen

Identify the set of reagents / reaction conditions ‘X’ and ‘Y’ in the following set of transformation : CH3

–

CH2

–

CH2Br

X   Product

Y   CH3 – CH –CH 3

[IIT 2002]

Br (A) X = dilute aqueous NaOH, 20°C; Y = HBr / acetic acid, 20°C

Treatment of 2-bromobutane with hot alcoholic KOH gives a mixture of three isomeric butenes (A), (B) and (C). Ozonolysis of the minor product (A), gives formaldehyde and another aldehyde in equimolar amounts. What are the structural formulae of (A),(B) and (C)?

EXERCISE – IV(A) Q.1

(D) IV < II < I < III

(C) R – Cl > R – Br > RF > R I

(iii) CH3CH2CH = CH2  CH2 = CH – CH = CH 2 (iv) OH – CH2 – CH2CH = CH2  O

Q.13

(C) IV < I < III < II

(B) R – F > R – Br > R – Cl > R – I

(ii) CH3CH2CH = CH2  CH3CH2CH2CH2NH2

Q.12

(B) IV < I < II < III

(A) RF > RC > R – Br > R – I



Convert (i)

(A) I < IV < II < III

R I S . J . N

2-chloro-3-methylbutane on treatment with alcoholic potash gives 2-methylbutene-2 as major product.

(i)

IV

(B) X = concentrated alcoholic NaOH, 80°C; Y = HBr / acetic acid 20°C (C) X = dilute aqueous NaOH, 20°C; Y = Br 2/CHCl3, 0°C (D) X = concentrated alcoholic NaOH, 80°C; Y = Br 2 / CHCl 3, 0°C

Q.6

CH3MgBr + Ethyl ester  which can be formed as [IIT 2003] product (excess)

CH2CH3

(A) HO

CH2CH3

(B)

CH3

HO

CH2CH2CH3

CH2CH3

CH2CH3

CH2CH3

CH3

(C) HO

CH2CH3 CH3

(D) HO

CH3 CH3


29

Q.7

Match the following

[IIT 2006] Column II

Column I

Q.8

Q.12

O C NH C

(A) CH3 – CHBr – CD3 on (P) E1 reaction treatment with alc. KOH gives CH2 = CH – CD3 as a major product (B) Ph – CHBr – CH 3 reacts (Q) E2 reaction faster than Ph – CHBr – CD3 (C) Ph – CD 2 –CH2Br (R) E1 cb reaction on treatment with C2H5OD/C2H5O – gives Ph – CD = CH 2 as the major product. (D) PhCH2CH2Br and (S) First order reaction PhCD2CH2Br react with same rate The reagent(s) for the following conversion, ? H Br 

Me

(A)

(C)

Q.1

SPh

Identify the major reactions : Cl

product in [IIthe following [IIT 1993]

Predict the structure of the intermediates/products [IIT 1996] in the following reaction sequenceBr Ph H Na I     C H Acetone MeO Ph C6H5

Q.4

(a) C6H5CH2CHCl

SPh

Alcoholic KOH   A + B Write Heat

structures of (A) and (B).

SPh

(D)

HI( excess ) (b) (CH3)2CHOCH3       A + B Write Heat

NO2

In the reaction

C H2Cl

Q.3

SPh

NO2

Me

O

Br

An alkyl halide X of formula C 6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C 6H12). Both alkenes on hydrogenation give 2, 4-dimethylbutane. Predict the structures of X, Y and Z. [IIT 1996]

F

Br

(D)

Q.2

(B)

NO2

C N C

HBr     ? (ii) C6H5COOH + CH3Mg I  ? + ?

F

O

KOH

NO2

Me

C N

CH2Cl

EXERCISE – IV(B)

Br

SPh

O

O–CH2

dimethylformamide

Q.10

C N C

Br (B)

O

 

(C)

C N–CH2 C O

P hS Na F  

Me

O

alcoholic (i) C6H5 – CH2 – CH – CH 3      ? 

[IIT 2008]

(A)

CH2Cl

O

H

The major product of the following reaction is

Me

(ii) Br

O

(A) alcoholic KOH (B) alcoholic KOH followed by NaNH 2 (C) aqueous KOH followed by NaNH 2 (D) Zn/CH3OH Q.9

[IIT-JEE 2011

(i) KOH

R I S . J . N [IIT 2007]

Br

The major product of the following reaction is

Q.5

NO2

 the products are OCH 3  HBr

structures of A and B. [IIT 1998] Complete the following reaction with appropriate structures of products / reagents. [IIT 1998] CH =CH2 ( i)NaNH (3equi .)

Br 2 2     (A)          (B)

(A) Br (C)

OCH3 and H2 (B) Br and CH3OH (D)

OH and CH3Br [IIT-JEE 2010]

Q.11

(II) CH3I

Br and CH3Br

The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane [IIT-JEE 2011] using alcoholic KOH is

Q.6

What would be major product ? [IIT-JEE 2000] CH3 C H OH

5 2    ? CH3 – C – CH 2Br  

CH3


30

ANSWER KEY EXERCISE - I Ques. 1 An s. A Ques. 21 An s. A Ques. 44 An s. B

2 B 22 B 42 B

3 A 23 C 43 A

5 C 25 B 45 D

6 A 26 A 46 C

7 C 27 D 47 A

8 C 28 C 48 C

9 B 29 C 49 A

10 11 12 13 14 15 16 17 18 19 20 D B C A C C B B B C C 30 31 32 33 34 35 36 37 38 39 40 C C B C D B D D A D C 50 A

R I S . J . N

EXERCISE – II Ques. 1 An s. A,C Ques. 9 An s. A,B,D

2 B,C

3 A,C

(A) S ; (B) Q ; (C) R ; (D) P

Q.10

Q.11

4 A,C

5 A,B,C

A, C, D

6 A,B,C,D

Q.12

A, C

7 A,B,C,D

Q.13

8 B,D

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

EXERCISE – III Q.2

Q.1 Q.3

(i) A, CH 3CH2CH2Br ; B, CH3CH = CH2 ; C, CH3CHBrCH3; D, CH3CHNH2CH3 (ii) A, CH 3CH = CH2; B, CH3CHOHCH3; C, CH3CHClCH3; D, CH3CH2CH2 (iii) A, CH3CHBrCH = CH2; B, CH2 = CH – CH = CH2; C, CH3CHBrCH = CH2 & CH3CH = CH – CH 2Br (iv) A, CH3CH2CHOHCH3; B, CH3CH2CHBrCH3; C, CH3CH = CH – CH3 The elimination of HI (or DI) in presence of strong base shows E 2 elimination. The rate determining step involves breaking up of C – H (or C – D) bond. The C – D bond being stronger than C– H and thus elimination is faster in case of CH3 – CH2I.

Q.4

Q.5 Q.8

4 B 24 A 44 C

Q.6

Stability of alkene by -hydrogen

Q.7

KOH(alc )

H3C – CH – CH – CH 3       CH3 – C = CH – CH 3 + CH3 – CH – CH = CH 2

CH3 CH3 major minor Elimination occurs according to saytzeff rule. The major product is one which involves elimination of H from less hydrogenated carbon. CH3 Cl

Q.9

KCN is an ionic compound [K +(:C  ) – ] in which both C and N carry a lone pair electron. Carbon carrying lone pair of electrons is more reactive and thus alkyl attacks carbon to give alkyl cyanide AgCN being covalent has Ag – C  N : structure with lone pair on N thus R attacks on N atom and R – N  C is formed. 

Q.10

Q.11

Br

Q.13

Cyclopropyl bromide

Q.12

Mg ether

(i) CO2

MgBr (ii) H O 3

C – OH

ozonolysis

O Cyclopropane (B) carboxylic acid

(C)

CC

Dicyclopropyl (A) acetylene

EXERCISE – IV(A)

Ques. An s.

1 B,D

2 B

3 A

Q.7 Q.8

A – Q ;B – Q; C – R, S; D – P, S B

Q.9

A

Q.10

D

4 D

5 B

6 D


31

Halogen Derivative

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