QBManager

GENERAL ORGANIC CHEMISTRY

231711 Identify the correct statement.

1 Reaction mechanisms are studied using isotopic labelling

2 Isolation of reactive intermediates is a method of establish reaction mechanism

3 Both (a) and (b) are correct

4 Neither (a) nor (b) are correct

Explanation:

Both statement are correct.
Reaction mechanisms are studied using isotopic labelling.
One of the atoms of one reactant/ reagent replaced with one isotope then the product which this labelled atom is determined.

In the above reaction, $O$ atom of acetic acid in which $OH$ is labeled. It is then observed that $OH$ labeled atom is found in water.
-OH group of acetic acid combined with $H$ atom and form methanol.
Isolation reactive intermediates is a method to established reaction mechanism.
It can proceed through following mechanism
$A \to B \to D$
$A \to C \to D$
Suppose we cannot isolate mechanism A to B and A to $C$ intermediate so we can say that mechanism take place with $A \to B \to D$ .

JCECE - 2017

GENERAL ORGANIC CHEMISTRY

231714 Relative stabilities of the following carbocations will be in the order
$\begin{aligned} \overset{\oplus}{C} H_{3} C H_{3} \overset{\oplus}{C} H_{2} \overset{\oplus}{C} H_{2} OC H_{3} \\ (A) (B) (C) \end{aligned}$

1 C

>

B

>

A

2 C

<

B

<

A

3 B

>

C

>

A

4 C

>

A

>

B

Explanation:

For stability of carbocation depend upon magnitude of charge means $+ I$ group increases stability increases.
original image
It has more power of $+ I$ group, therefore the correct decreasing order of carbocation is $C > B > A$ .

JIPMER-2018

GENERAL ORGANIC CHEMISTRY

231719 Which one of the following reactions proceeds through free radical chain mechanism?

1 Addition of

HBr

on ethane

\to

ionic addition reaction

{CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br

2 Halogenation of benzene in presence of

{FeBr}_{3} \to

Electrophilic substitution reaction

3 Photochemical chlorination of methane

\to

free radical substitution reaction.

4 Hydrolysis with tert-butyl chloride

\to

Nucleophic with aqueous

KOH

substitution reaction

5 Addition of

{NaHSO}_{3}

on acetone

\to

Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Explanation:

(C) :
(a.) Addition of $HBr$ on ethane $\to$ ionic addition reaction
${CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br$
(b.) Halogenation of benzene in presence of ${FeBr}_{3} \to$ Electrophilic substitution reaction
(c.) Photochemical chlorination of methane $\to$ free radical substitution reaction.
(d.) Hydrolysis with tert-butyl chloride $\to$ Nucleophic with aqueous $KOH$ substitution reaction
(e.) Addition of ${NaHSO}_{3}$ on acetone $\to$ Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231721 The stablest radical among the following is

1

C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶

It has

1^{\circ}

free radical

2

{CH}_{3} - {CH}_{2} ⟶

It also has

1^{\circ}

free radical only

3 α

Hydrogen

3

C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶

It has resonance as well as

3 α

- Hydrogen

{CH}_{3} - CH - {CH}_{3} ⟶ 6 α

- Hydrogen

2^{\circ}

free radical

{CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α

- Hydrogen

1^{\circ}

So, on the basis above explanation (c) is more stable

4

{CH}_{3} - \dot{C} H - {CH}_{3}

5

{CH}_{3} - \dot{C} H_{2} - {CH}_{2}

Explanation:

Stability of free radical is similar to that of carbacation i.e. $3^{\circ} > 2^{\circ} > 1^{\circ}$ and electron releasing group increases stability while electron withdrawing group decreases stability and resonance also increases stability.
It can be also determine with the help of hyper conjugation.
It is no. of alpha hydrogen stability directly proportional to no. of alpha hydrogen.
(a.) $C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶$ It has $1^{\circ}$ free radical
(b.) ${CH}_{3} - {CH}_{2} ⟶$ It also has $1^{\circ}$ free radical only $3 α$ Hydrogen
(c.) $C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶$ It has resonance as well as $3 α$ - Hydrogen
${CH}_{3} - CH - {CH}_{3} ⟶ 6 α$ - Hydrogen $2^{\circ}$ free radical
${CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α$ - Hydrogen $1^{\circ}$
So, on the basis above explanation (c) is more stable

Kerala CEE-2011

GENERAL ORGANIC CHEMISTRY

231722 The least stable free radical is

1

{CH}_{3} \dot{C} H_{2}

2

{CH}_{3} {CH}_{2} \dot{C} H_{2}

3

{({CH}_{3})}_{2} \dot{C} H

4

{({CH}_{3})}_{3} \dot{C}

5

\dot{C} H_{3}

Explanation:

Stability of free radical is similar to that of stability of carbocation.
$3^{\circ} > 2^{\circ} > 1^{\circ}$
It can also be explained on the basis of hyper conjugation. Stability of free radical with hyper conjugation which can be determined with the help of $α$ -Hydrogen.
The more number of $α$ -Hydrogen, the more will be stability.
The carbon directly attached to identity species called $α$ -carbon and hydrogen is called $α$ -Hydrogen.
original image
So, the least stable free radical is option (e).

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231711 Identify the correct statement.

1 Reaction mechanisms are studied using isotopic labelling

2 Isolation of reactive intermediates is a method of establish reaction mechanism

3 Both (a) and (b) are correct

4 Neither (a) nor (b) are correct

Explanation:

Both statement are correct.
Reaction mechanisms are studied using isotopic labelling.
One of the atoms of one reactant/ reagent replaced with one isotope then the product which this labelled atom is determined.

In the above reaction, $O$ atom of acetic acid in which $OH$ is labeled. It is then observed that $OH$ labeled atom is found in water.
-OH group of acetic acid combined with $H$ atom and form methanol.
Isolation reactive intermediates is a method to established reaction mechanism.
It can proceed through following mechanism
$A \to B \to D$
$A \to C \to D$
Suppose we cannot isolate mechanism A to B and A to $C$ intermediate so we can say that mechanism take place with $A \to B \to D$ .

JCECE - 2017

GENERAL ORGANIC CHEMISTRY

231714 Relative stabilities of the following carbocations will be in the order
$\begin{aligned} \overset{\oplus}{C} H_{3} C H_{3} \overset{\oplus}{C} H_{2} \overset{\oplus}{C} H_{2} OC H_{3} \\ (A) (B) (C) \end{aligned}$

1 C

>

B

>

A

2 C

<

B

<

A

3 B

>

C

>

A

4 C

>

A

>

B

Explanation:

For stability of carbocation depend upon magnitude of charge means $+ I$ group increases stability increases.
original image
It has more power of $+ I$ group, therefore the correct decreasing order of carbocation is $C > B > A$ .

JIPMER-2018

GENERAL ORGANIC CHEMISTRY

231719 Which one of the following reactions proceeds through free radical chain mechanism?

1 Addition of

HBr

on ethane

\to

ionic addition reaction

{CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br

2 Halogenation of benzene in presence of

{FeBr}_{3} \to

Electrophilic substitution reaction

3 Photochemical chlorination of methane

\to

free radical substitution reaction.

4 Hydrolysis with tert-butyl chloride

\to

Nucleophic with aqueous

KOH

substitution reaction

5 Addition of

{NaHSO}_{3}

on acetone

\to

Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Explanation:

(C) :
(a.) Addition of $HBr$ on ethane $\to$ ionic addition reaction
${CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br$
(b.) Halogenation of benzene in presence of ${FeBr}_{3} \to$ Electrophilic substitution reaction
(c.) Photochemical chlorination of methane $\to$ free radical substitution reaction.
(d.) Hydrolysis with tert-butyl chloride $\to$ Nucleophic with aqueous $KOH$ substitution reaction
(e.) Addition of ${NaHSO}_{3}$ on acetone $\to$ Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231721 The stablest radical among the following is

1

C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶

It has

1^{\circ}

free radical

2

{CH}_{3} - {CH}_{2} ⟶

It also has

1^{\circ}

free radical only

3 α

Hydrogen

3

C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶

It has resonance as well as

3 α

- Hydrogen

{CH}_{3} - CH - {CH}_{3} ⟶ 6 α

- Hydrogen

2^{\circ}

free radical

{CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α

- Hydrogen

1^{\circ}

So, on the basis above explanation (c) is more stable

4

{CH}_{3} - \dot{C} H - {CH}_{3}

5

{CH}_{3} - \dot{C} H_{2} - {CH}_{2}

Explanation:

Stability of free radical is similar to that of carbacation i.e. $3^{\circ} > 2^{\circ} > 1^{\circ}$ and electron releasing group increases stability while electron withdrawing group decreases stability and resonance also increases stability.
It can be also determine with the help of hyper conjugation.
It is no. of alpha hydrogen stability directly proportional to no. of alpha hydrogen.
(a.) $C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶$ It has $1^{\circ}$ free radical
(b.) ${CH}_{3} - {CH}_{2} ⟶$ It also has $1^{\circ}$ free radical only $3 α$ Hydrogen
(c.) $C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶$ It has resonance as well as $3 α$ - Hydrogen
${CH}_{3} - CH - {CH}_{3} ⟶ 6 α$ - Hydrogen $2^{\circ}$ free radical
${CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α$ - Hydrogen $1^{\circ}$
So, on the basis above explanation (c) is more stable

Kerala CEE-2011

GENERAL ORGANIC CHEMISTRY

231722 The least stable free radical is

1

{CH}_{3} \dot{C} H_{2}

2

{CH}_{3} {CH}_{2} \dot{C} H_{2}

3

{({CH}_{3})}_{2} \dot{C} H

4

{({CH}_{3})}_{3} \dot{C}

5

\dot{C} H_{3}

Explanation:

Stability of free radical is similar to that of stability of carbocation.
$3^{\circ} > 2^{\circ} > 1^{\circ}$
It can also be explained on the basis of hyper conjugation. Stability of free radical with hyper conjugation which can be determined with the help of $α$ -Hydrogen.
The more number of $α$ -Hydrogen, the more will be stability.
The carbon directly attached to identity species called $α$ -carbon and hydrogen is called $α$ -Hydrogen.
original image
So, the least stable free radical is option (e).

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231711 Identify the correct statement.

1 Reaction mechanisms are studied using isotopic labelling

2 Isolation of reactive intermediates is a method of establish reaction mechanism

3 Both (a) and (b) are correct

4 Neither (a) nor (b) are correct

Explanation:

Both statement are correct.
Reaction mechanisms are studied using isotopic labelling.
One of the atoms of one reactant/ reagent replaced with one isotope then the product which this labelled atom is determined.

In the above reaction, $O$ atom of acetic acid in which $OH$ is labeled. It is then observed that $OH$ labeled atom is found in water.
-OH group of acetic acid combined with $H$ atom and form methanol.
Isolation reactive intermediates is a method to established reaction mechanism.
It can proceed through following mechanism
$A \to B \to D$
$A \to C \to D$
Suppose we cannot isolate mechanism A to B and A to $C$ intermediate so we can say that mechanism take place with $A \to B \to D$ .

JCECE - 2017

GENERAL ORGANIC CHEMISTRY

231714 Relative stabilities of the following carbocations will be in the order
$\begin{aligned} \overset{\oplus}{C} H_{3} C H_{3} \overset{\oplus}{C} H_{2} \overset{\oplus}{C} H_{2} OC H_{3} \\ (A) (B) (C) \end{aligned}$

1 C

>

B

>

A

2 C

<

B

<

A

3 B

>

C

>

A

4 C

>

A

>

B

Explanation:

For stability of carbocation depend upon magnitude of charge means $+ I$ group increases stability increases.
original image
It has more power of $+ I$ group, therefore the correct decreasing order of carbocation is $C > B > A$ .

JIPMER-2018

GENERAL ORGANIC CHEMISTRY

231719 Which one of the following reactions proceeds through free radical chain mechanism?

1 Addition of

HBr

on ethane

\to

ionic addition reaction

{CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br

2 Halogenation of benzene in presence of

{FeBr}_{3} \to

Electrophilic substitution reaction

3 Photochemical chlorination of methane

\to

free radical substitution reaction.

4 Hydrolysis with tert-butyl chloride

\to

Nucleophic with aqueous

KOH

substitution reaction

5 Addition of

{NaHSO}_{3}

on acetone

\to

Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Explanation:

(C) :
(a.) Addition of $HBr$ on ethane $\to$ ionic addition reaction
${CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br$
(b.) Halogenation of benzene in presence of ${FeBr}_{3} \to$ Electrophilic substitution reaction
(c.) Photochemical chlorination of methane $\to$ free radical substitution reaction.
(d.) Hydrolysis with tert-butyl chloride $\to$ Nucleophic with aqueous $KOH$ substitution reaction
(e.) Addition of ${NaHSO}_{3}$ on acetone $\to$ Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231721 The stablest radical among the following is

1

C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶

It has

1^{\circ}

free radical

2

{CH}_{3} - {CH}_{2} ⟶

It also has

1^{\circ}

free radical only

3 α

Hydrogen

3

C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶

It has resonance as well as

3 α

- Hydrogen

{CH}_{3} - CH - {CH}_{3} ⟶ 6 α

- Hydrogen

2^{\circ}

free radical

{CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α

- Hydrogen

1^{\circ}

So, on the basis above explanation (c) is more stable

4

{CH}_{3} - \dot{C} H - {CH}_{3}

5

{CH}_{3} - \dot{C} H_{2} - {CH}_{2}

Explanation:

Stability of free radical is similar to that of carbacation i.e. $3^{\circ} > 2^{\circ} > 1^{\circ}$ and electron releasing group increases stability while electron withdrawing group decreases stability and resonance also increases stability.
It can be also determine with the help of hyper conjugation.
It is no. of alpha hydrogen stability directly proportional to no. of alpha hydrogen.
(a.) $C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶$ It has $1^{\circ}$ free radical
(b.) ${CH}_{3} - {CH}_{2} ⟶$ It also has $1^{\circ}$ free radical only $3 α$ Hydrogen
(c.) $C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶$ It has resonance as well as $3 α$ - Hydrogen
${CH}_{3} - CH - {CH}_{3} ⟶ 6 α$ - Hydrogen $2^{\circ}$ free radical
${CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α$ - Hydrogen $1^{\circ}$
So, on the basis above explanation (c) is more stable

Kerala CEE-2011

GENERAL ORGANIC CHEMISTRY

231722 The least stable free radical is

1

{CH}_{3} \dot{C} H_{2}

2

{CH}_{3} {CH}_{2} \dot{C} H_{2}

3

{({CH}_{3})}_{2} \dot{C} H

4

{({CH}_{3})}_{3} \dot{C}

5

\dot{C} H_{3}

Explanation:

Stability of free radical is similar to that of stability of carbocation.
$3^{\circ} > 2^{\circ} > 1^{\circ}$
It can also be explained on the basis of hyper conjugation. Stability of free radical with hyper conjugation which can be determined with the help of $α$ -Hydrogen.
The more number of $α$ -Hydrogen, the more will be stability.
The carbon directly attached to identity species called $α$ -carbon and hydrogen is called $α$ -Hydrogen.
original image
So, the least stable free radical is option (e).

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231711 Identify the correct statement.

1 Reaction mechanisms are studied using isotopic labelling

2 Isolation of reactive intermediates is a method of establish reaction mechanism

3 Both (a) and (b) are correct

4 Neither (a) nor (b) are correct

Explanation:

Both statement are correct.
Reaction mechanisms are studied using isotopic labelling.
One of the atoms of one reactant/ reagent replaced with one isotope then the product which this labelled atom is determined.

In the above reaction, $O$ atom of acetic acid in which $OH$ is labeled. It is then observed that $OH$ labeled atom is found in water.
-OH group of acetic acid combined with $H$ atom and form methanol.
Isolation reactive intermediates is a method to established reaction mechanism.
It can proceed through following mechanism
$A \to B \to D$
$A \to C \to D$
Suppose we cannot isolate mechanism A to B and A to $C$ intermediate so we can say that mechanism take place with $A \to B \to D$ .

JCECE - 2017

GENERAL ORGANIC CHEMISTRY

231714 Relative stabilities of the following carbocations will be in the order
$\begin{aligned} \overset{\oplus}{C} H_{3} C H_{3} \overset{\oplus}{C} H_{2} \overset{\oplus}{C} H_{2} OC H_{3} \\ (A) (B) (C) \end{aligned}$

1 C

>

B

>

A

2 C

<

B

<

A

3 B

>

C

>

A

4 C

>

A

>

B

Explanation:

For stability of carbocation depend upon magnitude of charge means $+ I$ group increases stability increases.
original image
It has more power of $+ I$ group, therefore the correct decreasing order of carbocation is $C > B > A$ .

JIPMER-2018

GENERAL ORGANIC CHEMISTRY

231719 Which one of the following reactions proceeds through free radical chain mechanism?

1 Addition of

HBr

on ethane

\to

ionic addition reaction

{CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br

2 Halogenation of benzene in presence of

{FeBr}_{3} \to

Electrophilic substitution reaction

3 Photochemical chlorination of methane

\to

free radical substitution reaction.

4 Hydrolysis with tert-butyl chloride

\to

Nucleophic with aqueous

KOH

substitution reaction

5 Addition of

{NaHSO}_{3}

on acetone

\to

Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Explanation:

(C) :
(a.) Addition of $HBr$ on ethane $\to$ ionic addition reaction
${CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br$
(b.) Halogenation of benzene in presence of ${FeBr}_{3} \to$ Electrophilic substitution reaction
(c.) Photochemical chlorination of methane $\to$ free radical substitution reaction.
(d.) Hydrolysis with tert-butyl chloride $\to$ Nucleophic with aqueous $KOH$ substitution reaction
(e.) Addition of ${NaHSO}_{3}$ on acetone $\to$ Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231721 The stablest radical among the following is

1

C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶

It has

1^{\circ}

free radical

2

{CH}_{3} - {CH}_{2} ⟶

It also has

1^{\circ}

free radical only

3 α

Hydrogen

3

C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶

It has resonance as well as

3 α

- Hydrogen

{CH}_{3} - CH - {CH}_{3} ⟶ 6 α

- Hydrogen

2^{\circ}

free radical

{CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α

- Hydrogen

1^{\circ}

So, on the basis above explanation (c) is more stable

4

{CH}_{3} - \dot{C} H - {CH}_{3}

5

{CH}_{3} - \dot{C} H_{2} - {CH}_{2}

Explanation:

Stability of free radical is similar to that of carbacation i.e. $3^{\circ} > 2^{\circ} > 1^{\circ}$ and electron releasing group increases stability while electron withdrawing group decreases stability and resonance also increases stability.
It can be also determine with the help of hyper conjugation.
It is no. of alpha hydrogen stability directly proportional to no. of alpha hydrogen.
(a.) $C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶$ It has $1^{\circ}$ free radical
(b.) ${CH}_{3} - {CH}_{2} ⟶$ It also has $1^{\circ}$ free radical only $3 α$ Hydrogen
(c.) $C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶$ It has resonance as well as $3 α$ - Hydrogen
${CH}_{3} - CH - {CH}_{3} ⟶ 6 α$ - Hydrogen $2^{\circ}$ free radical
${CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α$ - Hydrogen $1^{\circ}$
So, on the basis above explanation (c) is more stable

Kerala CEE-2011

GENERAL ORGANIC CHEMISTRY

231722 The least stable free radical is

1

{CH}_{3} \dot{C} H_{2}

2

{CH}_{3} {CH}_{2} \dot{C} H_{2}

3

{({CH}_{3})}_{2} \dot{C} H

4

{({CH}_{3})}_{3} \dot{C}

5

\dot{C} H_{3}

Explanation:

Stability of free radical is similar to that of stability of carbocation.
$3^{\circ} > 2^{\circ} > 1^{\circ}$
It can also be explained on the basis of hyper conjugation. Stability of free radical with hyper conjugation which can be determined with the help of $α$ -Hydrogen.
The more number of $α$ -Hydrogen, the more will be stability.
The carbon directly attached to identity species called $α$ -carbon and hydrogen is called $α$ -Hydrogen.
original image
So, the least stable free radical is option (e).

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231711 Identify the correct statement.

1 Reaction mechanisms are studied using isotopic labelling

2 Isolation of reactive intermediates is a method of establish reaction mechanism

3 Both (a) and (b) are correct

4 Neither (a) nor (b) are correct

Explanation:

Both statement are correct.
Reaction mechanisms are studied using isotopic labelling.
One of the atoms of one reactant/ reagent replaced with one isotope then the product which this labelled atom is determined.

In the above reaction, $O$ atom of acetic acid in which $OH$ is labeled. It is then observed that $OH$ labeled atom is found in water.
-OH group of acetic acid combined with $H$ atom and form methanol.
Isolation reactive intermediates is a method to established reaction mechanism.
It can proceed through following mechanism
$A \to B \to D$
$A \to C \to D$
Suppose we cannot isolate mechanism A to B and A to $C$ intermediate so we can say that mechanism take place with $A \to B \to D$ .

JCECE - 2017

GENERAL ORGANIC CHEMISTRY

231714 Relative stabilities of the following carbocations will be in the order
$\begin{aligned} \overset{\oplus}{C} H_{3} C H_{3} \overset{\oplus}{C} H_{2} \overset{\oplus}{C} H_{2} OC H_{3} \\ (A) (B) (C) \end{aligned}$

1 C

>

B

>

A

2 C

<

B

<

A

3 B

>

C

>

A

4 C

>

A

>

B

Explanation:

For stability of carbocation depend upon magnitude of charge means $+ I$ group increases stability increases.
original image
It has more power of $+ I$ group, therefore the correct decreasing order of carbocation is $C > B > A$ .

JIPMER-2018

GENERAL ORGANIC CHEMISTRY

231719 Which one of the following reactions proceeds through free radical chain mechanism?

1 Addition of

HBr

on ethane

\to

ionic addition reaction

{CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br

2 Halogenation of benzene in presence of

{FeBr}_{3} \to

Electrophilic substitution reaction

3 Photochemical chlorination of methane

\to

free radical substitution reaction.

4 Hydrolysis with tert-butyl chloride

\to

Nucleophic with aqueous

KOH

substitution reaction

5 Addition of

{NaHSO}_{3}

on acetone

\to

Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Explanation:

(C) :
(a.) Addition of $HBr$ on ethane $\to$ ionic addition reaction
${CH}_{2} = {CH}_{2} + HBr \to {CH}_{3} - {CH}_{2} - Br$
(b.) Halogenation of benzene in presence of ${FeBr}_{3} \to$ Electrophilic substitution reaction
(c.) Photochemical chlorination of methane $\to$ free radical substitution reaction.
(d.) Hydrolysis with tert-butyl chloride $\to$ Nucleophic with aqueous $KOH$ substitution reaction
(e.) Addition of ${NaHSO}_{3}$ on acetone $\to$ Nucleophic addition reaction
Hence, on the basis of above explanation option (c) is correct
original image

Kerala CEE-2012

GENERAL ORGANIC CHEMISTRY

231721 The stablest radical among the following is

1

C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶

It has

1^{\circ}

free radical

2

{CH}_{3} - {CH}_{2} ⟶

It also has

1^{\circ}

free radical only

3 α

Hydrogen

3

C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶

It has resonance as well as

3 α

- Hydrogen

{CH}_{3} - CH - {CH}_{3} ⟶ 6 α

- Hydrogen

2^{\circ}

free radical

{CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α

- Hydrogen

1^{\circ}

So, on the basis above explanation (c) is more stable

4

{CH}_{3} - \dot{C} H - {CH}_{3}

5

{CH}_{3} - \dot{C} H_{2} - {CH}_{2}

Explanation:

Stability of free radical is similar to that of carbacation i.e. $3^{\circ} > 2^{\circ} > 1^{\circ}$ and electron releasing group increases stability while electron withdrawing group decreases stability and resonance also increases stability.
It can be also determine with the help of hyper conjugation.
It is no. of alpha hydrogen stability directly proportional to no. of alpha hydrogen.
(a.) $C_{6} H_{5} - {CH}_{2} - {CH}_{2} ⟶$ It has $1^{\circ}$ free radical
(b.) ${CH}_{3} - {CH}_{2} ⟶$ It also has $1^{\circ}$ free radical only $3 α$ Hydrogen
(c.) $C_{6} H_{5} - {CH}_{2} - {CH}_{3} ⟶$ It has resonance as well as $3 α$ - Hydrogen
${CH}_{3} - CH - {CH}_{3} ⟶ 6 α$ - Hydrogen $2^{\circ}$ free radical
${CH}_{2} - {CH}_{2} - {CH}_{2} ⟶ 2 α$ - Hydrogen $1^{\circ}$
So, on the basis above explanation (c) is more stable

Kerala CEE-2011

GENERAL ORGANIC CHEMISTRY

231722 The least stable free radical is

1

{CH}_{3} \dot{C} H_{2}

2

{CH}_{3} {CH}_{2} \dot{C} H_{2}

3

{({CH}_{3})}_{2} \dot{C} H

4

{({CH}_{3})}_{3} \dot{C}

5

\dot{C} H_{3}

Explanation:

Stability of free radical is similar to that of stability of carbocation.
$3^{\circ} > 2^{\circ} > 1^{\circ}$
It can also be explained on the basis of hyper conjugation. Stability of free radical with hyper conjugation which can be determined with the help of $α$ -Hydrogen.
The more number of $α$ -Hydrogen, the more will be stability.
The carbon directly attached to identity species called $α$ -carbon and hydrogen is called $α$ -Hydrogen.
original image
So, the least stable free radical is option (e).

Kerala CEE-2012

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