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CHXI13:HYDROCARBONS

317887 Which of the following compounds liberate methane when treated with excess of methyl magnesium iodide in dry ether?
(A) ${CH}_{3} - {CH}_{2} - {CH}_{2} OH$
(B) $H_{3} C - {CH}_{2} - C \equiv CH$
(C) $H_{3} C - {CH}_{2} - {CO}_{2} H$
(D) $H_{3} C - {CH}_{2} - CHO$

1 A, B

2 B, C

3 A, B, C

4 All above

Explanation:

The compounds which have active hydrogen atom liberate methane when treated with ${CH}_{3} MgI$ .
$C H_{3} - C H_{2} - C H_{2} - OH + C H_{3} - MgI$
supporting img
$C H_{3} - C H_{2} - C \equiv CH + C H_{3} MgI$

$C H_{3} - C H_{2} - COOH + C H_{3} MgI$

CHXI13:HYDROCARBONS

317888 ${({CH}_{3})}_{3} CMgCl$ on reaction with $D_{2} O$ produces

1

{({CH}_{3})}_{3} CD

2

{({CH}_{3})}_{3} COD

3

(CD)_{3} CD

4

(CD)_{3} COD

Explanation:

${(C H_{3})}_{3} CMgCl + DOD \to {(C H_{3})}_{3} CD + MgCl(OD) .$

CHXI13:HYDROCARBONS

317889 When 2-methylbutane is chlorinated, the percentage of ${({CH}_{3})}_{2} - CH - {CH}_{2} - {CH}_{2} Cl$ is ____
Assume reactivity ratio of
$3^{\circ} H : 2^{\circ} H : 1^{\circ} H = 5 : 3.8 : 1$

1

23 %

2

28 %

3

35 %

4

14 %

Explanation:

Chlorination of 2-methylbutane gives 4 monochlorinated products :
supporting img
The reactivity ratio of $3^{\circ} H : 2^{\circ} H : 1^{\circ} H$ towards chlorination is $5 : 3.8 : 1$
Compound reactivity $\times$ Probability $=$ Number of parts(1) $1 \times 6 = 6$ (2) $5 \times 1 = 5$ (3) $3.8 \times 2 = 7.6$ (4) $1 \times 3 = 3$
Total number of parts $= 21.6$
Percentage of $D = \frac{3}{21.6} \times 100 = 14 %$ .

CHXI13:HYDROCARBONS

317890 Match the following reaction with their mechanism involved
Column I
Column II
A
Cracking
P
Free radical substitution
B
Catalytic reforming
Q
Free radical elimination
C
Halogenation of alkene
R
Free radical addition
D
Wurtz reaction
S
Dehydrogenation

1

A - Q, B - P, C - S, D - R

2

A - Q, B - S, C - P, D - R

3

A - P, B - S, C - R, D - Q

4

A - S, B - Q, C - P, D - R

CHXI13:HYDROCARBONS

317887 Which of the following compounds liberate methane when treated with excess of methyl magnesium iodide in dry ether?
(A) ${CH}_{3} - {CH}_{2} - {CH}_{2} OH$
(B) $H_{3} C - {CH}_{2} - C \equiv CH$
(C) $H_{3} C - {CH}_{2} - {CO}_{2} H$
(D) $H_{3} C - {CH}_{2} - CHO$

1 A, B

2 B, C

3 A, B, C

4 All above

Explanation:

The compounds which have active hydrogen atom liberate methane when treated with ${CH}_{3} MgI$ .
$C H_{3} - C H_{2} - C H_{2} - OH + C H_{3} - MgI$
supporting img
$C H_{3} - C H_{2} - C \equiv CH + C H_{3} MgI$

$C H_{3} - C H_{2} - COOH + C H_{3} MgI$

CHXI13:HYDROCARBONS

317888 ${({CH}_{3})}_{3} CMgCl$ on reaction with $D_{2} O$ produces

1

{({CH}_{3})}_{3} CD

2

{({CH}_{3})}_{3} COD

3

(CD)_{3} CD

4

(CD)_{3} COD

Explanation:

${(C H_{3})}_{3} CMgCl + DOD \to {(C H_{3})}_{3} CD + MgCl(OD) .$

CHXI13:HYDROCARBONS

317889 When 2-methylbutane is chlorinated, the percentage of ${({CH}_{3})}_{2} - CH - {CH}_{2} - {CH}_{2} Cl$ is ____
Assume reactivity ratio of
$3^{\circ} H : 2^{\circ} H : 1^{\circ} H = 5 : 3.8 : 1$

1

23 %

2

28 %

3

35 %

4

14 %

Explanation:

Chlorination of 2-methylbutane gives 4 monochlorinated products :
supporting img
The reactivity ratio of $3^{\circ} H : 2^{\circ} H : 1^{\circ} H$ towards chlorination is $5 : 3.8 : 1$
Compound reactivity $\times$ Probability $=$ Number of parts(1) $1 \times 6 = 6$ (2) $5 \times 1 = 5$ (3) $3.8 \times 2 = 7.6$ (4) $1 \times 3 = 3$
Total number of parts $= 21.6$
Percentage of $D = \frac{3}{21.6} \times 100 = 14 %$ .

CHXI13:HYDROCARBONS

317890 Match the following reaction with their mechanism involved
Column I
Column II
A
Cracking
P
Free radical substitution
B
Catalytic reforming
Q
Free radical elimination
C
Halogenation of alkene
R
Free radical addition
D
Wurtz reaction
S
Dehydrogenation

1

A - Q, B - P, C - S, D - R

2

A - Q, B - S, C - P, D - R

3

A - P, B - S, C - R, D - Q

4

A - S, B - Q, C - P, D - R

CHXI13:HYDROCARBONS

317887 Which of the following compounds liberate methane when treated with excess of methyl magnesium iodide in dry ether?
(A) ${CH}_{3} - {CH}_{2} - {CH}_{2} OH$
(B) $H_{3} C - {CH}_{2} - C \equiv CH$
(C) $H_{3} C - {CH}_{2} - {CO}_{2} H$
(D) $H_{3} C - {CH}_{2} - CHO$

1 A, B

2 B, C

3 A, B, C

4 All above

Explanation:

The compounds which have active hydrogen atom liberate methane when treated with ${CH}_{3} MgI$ .
$C H_{3} - C H_{2} - C H_{2} - OH + C H_{3} - MgI$
supporting img
$C H_{3} - C H_{2} - C \equiv CH + C H_{3} MgI$

$C H_{3} - C H_{2} - COOH + C H_{3} MgI$

CHXI13:HYDROCARBONS

317888 ${({CH}_{3})}_{3} CMgCl$ on reaction with $D_{2} O$ produces

1

{({CH}_{3})}_{3} CD

2

{({CH}_{3})}_{3} COD

3

(CD)_{3} CD

4

(CD)_{3} COD

Explanation:

${(C H_{3})}_{3} CMgCl + DOD \to {(C H_{3})}_{3} CD + MgCl(OD) .$

CHXI13:HYDROCARBONS

317889 When 2-methylbutane is chlorinated, the percentage of ${({CH}_{3})}_{2} - CH - {CH}_{2} - {CH}_{2} Cl$ is ____
Assume reactivity ratio of
$3^{\circ} H : 2^{\circ} H : 1^{\circ} H = 5 : 3.8 : 1$

1

23 %

2

28 %

3

35 %

4

14 %

Explanation:

Chlorination of 2-methylbutane gives 4 monochlorinated products :
supporting img
The reactivity ratio of $3^{\circ} H : 2^{\circ} H : 1^{\circ} H$ towards chlorination is $5 : 3.8 : 1$
Compound reactivity $\times$ Probability $=$ Number of parts(1) $1 \times 6 = 6$ (2) $5 \times 1 = 5$ (3) $3.8 \times 2 = 7.6$ (4) $1 \times 3 = 3$
Total number of parts $= 21.6$
Percentage of $D = \frac{3}{21.6} \times 100 = 14 %$ .

CHXI13:HYDROCARBONS

317890 Match the following reaction with their mechanism involved
Column I
Column II
A
Cracking
P
Free radical substitution
B
Catalytic reforming
Q
Free radical elimination
C
Halogenation of alkene
R
Free radical addition
D
Wurtz reaction
S
Dehydrogenation

1

A - Q, B - P, C - S, D - R

2

A - Q, B - S, C - P, D - R

3

A - P, B - S, C - R, D - Q

4

A - S, B - Q, C - P, D - R

CHXI13:HYDROCARBONS

317887 Which of the following compounds liberate methane when treated with excess of methyl magnesium iodide in dry ether?
(A) ${CH}_{3} - {CH}_{2} - {CH}_{2} OH$
(B) $H_{3} C - {CH}_{2} - C \equiv CH$
(C) $H_{3} C - {CH}_{2} - {CO}_{2} H$
(D) $H_{3} C - {CH}_{2} - CHO$

1 A, B

2 B, C

3 A, B, C

4 All above

Explanation:

The compounds which have active hydrogen atom liberate methane when treated with ${CH}_{3} MgI$ .
$C H_{3} - C H_{2} - C H_{2} - OH + C H_{3} - MgI$
supporting img
$C H_{3} - C H_{2} - C \equiv CH + C H_{3} MgI$

$C H_{3} - C H_{2} - COOH + C H_{3} MgI$

CHXI13:HYDROCARBONS

317888 ${({CH}_{3})}_{3} CMgCl$ on reaction with $D_{2} O$ produces

1

{({CH}_{3})}_{3} CD

2

{({CH}_{3})}_{3} COD

3

(CD)_{3} CD

4

(CD)_{3} COD

Explanation:

${(C H_{3})}_{3} CMgCl + DOD \to {(C H_{3})}_{3} CD + MgCl(OD) .$

CHXI13:HYDROCARBONS

317889 When 2-methylbutane is chlorinated, the percentage of ${({CH}_{3})}_{2} - CH - {CH}_{2} - {CH}_{2} Cl$ is ____
Assume reactivity ratio of
$3^{\circ} H : 2^{\circ} H : 1^{\circ} H = 5 : 3.8 : 1$

1

23 %

2

28 %

3

35 %

4

14 %

Explanation:

Chlorination of 2-methylbutane gives 4 monochlorinated products :
supporting img
The reactivity ratio of $3^{\circ} H : 2^{\circ} H : 1^{\circ} H$ towards chlorination is $5 : 3.8 : 1$
Compound reactivity $\times$ Probability $=$ Number of parts(1) $1 \times 6 = 6$ (2) $5 \times 1 = 5$ (3) $3.8 \times 2 = 7.6$ (4) $1 \times 3 = 3$
Total number of parts $= 21.6$
Percentage of $D = \frac{3}{21.6} \times 100 = 14 %$ .

CHXI13:HYDROCARBONS

317890 Match the following reaction with their mechanism involved
Column I
Column II
A
Cracking
P
Free radical substitution
B
Catalytic reforming
Q
Free radical elimination
C
Halogenation of alkene
R
Free radical addition
D
Wurtz reaction
S
Dehydrogenation

1

A - Q, B - P, C - S, D - R

2

A - Q, B - S, C - P, D - R

3

A - P, B - S, C - R, D - Q

4

A - S, B - Q, C - P, D - R