QBManager

GENERAL ORGANIC CHEMISTRY

231325 Which of the following carbocations is most stable?

1 I

2 II

3 III

4 IV

Explanation:

We know that, the order of stability of carbocation is $3^{\circ} > 2^{\circ} > 1^{\circ} >$ methyl carbocation.
Hence, the (ii) carbocation is $3^{\circ}$ carbocation and it is most stable all of them.

AMU-2018

GENERAL ORGANIC CHEMISTRY

231326 The correct order of acidity of the following is

1 (I)

>

(IV)

>

(II)

>

(III)

2 (IV)

>

(II)

>

(I)

>

(III)

3 (IV)

>

(I)

>

(II)

>

(III)

4 (IV)

>

(II)

>

(III)

>

(I)

Explanation:

Electron-donation groups decrease while electron withdrawing groups increase the acidity of phenols.
Decreasing order of acidity in nitro phenol -
P-nitrophenol $>$ O-nitrophenol $>$ M-nitrophenol $>$ phenol

TS EAMCET 10.08.2021

GENERAL ORGANIC CHEMISTRY

231328 Arrange the following carbanions in order of their decreasing stability:
(i) $H_{3} C - C \equiv C^{-}$
(ii) $H - C \equiv C^{-}$
(iii) $H_{3} C - {CH}_{2}^{-}$

1 i

>

ii

>

iii

2 ii

>

i

>

iii

3 iii

>

ii

>

i

4 iii

>

i

>

ii

Explanation:

The order of decreasing stability of carbons is:
$H - \underset{II}{C} \equiv C^{-} > {CH}_{3} - C \equiv C^{-} > {CH}_{3} - {CH}_{2}^{-}$
sp-hybridised carbon atom is more electronegative than ${sp}^{3}$ - hybridized carbon atom and hence, can accommodate the negative charge more effectively. So, I and II are more stable than III.
$- {CH}_{3}$ group has $+ I$ effect, therefore, it intensifies the negative charge and hence, destabilizes the carbanion.
${CH}_{3} \to C \equiv C^{-}$

AMU-2017

GENERAL ORGANIC CHEMISTRY

231331 Order of the base strength of the compounds
original image is

1 iv

>

ii

>

i

>

ii

2 iii

>

ii

>

iv

>

i

3 ii

>

iii

>

iv

>

i

4 ii

>

iii

>

i

>

iv

Explanation:

(D) :

II $>$ III $>$ I $>$ IV . The phenoxide ion is resource stabilized so it is least basic.

AMU - 2010

GENERAL ORGANIC CHEMISTRY

231338 The basicity of compounds I, II, III and IV
$\begin{array}{cccc} {CH}_{3} {NH}_{2} & {({CH}_{3})}_{2} {NH}_{I} & {({CH}_{3})}_{3} N & C_{6} H_{5} {CH}_{2} {NH}_{2} \\ II & III & IV \end{array}$ varies in the order:

1 I

>

II

>

III

>

IV

2 II

>

I

>

III

>

IV

3 III

>

I

>

II

>

IV

4 IV

>

I

>

II

>

III

Explanation:

In aquous medium alkyl amines and ammonia react with water and form ammonium respectively due to this reason the basicity is change due to steric hindrance.
Hence the order of basicity is-
${({CH}_{3})}_{2} NH > {CH}_{3} {NH}_{2} > {({CH}_{3})}_{3} N > C_{6} H_{5} {CH}_{2} {NH}_{2}$ Option (b) is correct.
This phenomena only occures in polar solvent.
In any non-polar solvent, the normal basicity order is forward as we expected.

AMU-2005

GENERAL ORGANIC CHEMISTRY

231325 Which of the following carbocations is most stable?

1 I

2 II

3 III

4 IV

Explanation:

We know that, the order of stability of carbocation is $3^{\circ} > 2^{\circ} > 1^{\circ} >$ methyl carbocation.
Hence, the (ii) carbocation is $3^{\circ}$ carbocation and it is most stable all of them.

AMU-2018

GENERAL ORGANIC CHEMISTRY

231326 The correct order of acidity of the following is

1 (I)

>

(IV)

>

(II)

>

(III)

2 (IV)

>

(II)

>

(I)

>

(III)

3 (IV)

>

(I)

>

(II)

>

(III)

4 (IV)

>

(II)

>

(III)

>

(I)

Explanation:

Electron-donation groups decrease while electron withdrawing groups increase the acidity of phenols.
Decreasing order of acidity in nitro phenol -
P-nitrophenol $>$ O-nitrophenol $>$ M-nitrophenol $>$ phenol

TS EAMCET 10.08.2021

GENERAL ORGANIC CHEMISTRY

231328 Arrange the following carbanions in order of their decreasing stability:
(i) $H_{3} C - C \equiv C^{-}$
(ii) $H - C \equiv C^{-}$
(iii) $H_{3} C - {CH}_{2}^{-}$

1 i

>

ii

>

iii

2 ii

>

i

>

iii

3 iii

>

ii

>

i

4 iii

>

i

>

ii

Explanation:

The order of decreasing stability of carbons is:
$H - \underset{II}{C} \equiv C^{-} > {CH}_{3} - C \equiv C^{-} > {CH}_{3} - {CH}_{2}^{-}$
sp-hybridised carbon atom is more electronegative than ${sp}^{3}$ - hybridized carbon atom and hence, can accommodate the negative charge more effectively. So, I and II are more stable than III.
$- {CH}_{3}$ group has $+ I$ effect, therefore, it intensifies the negative charge and hence, destabilizes the carbanion.
${CH}_{3} \to C \equiv C^{-}$

AMU-2017

GENERAL ORGANIC CHEMISTRY

231331 Order of the base strength of the compounds
original image is

1 iv

>

ii

>

i

>

ii

2 iii

>

ii

>

iv

>

i

3 ii

>

iii

>

iv

>

i

4 ii

>

iii

>

i

>

iv

Explanation:

(D) :

II $>$ III $>$ I $>$ IV . The phenoxide ion is resource stabilized so it is least basic.

AMU - 2010

GENERAL ORGANIC CHEMISTRY

231338 The basicity of compounds I, II, III and IV
$\begin{array}{cccc} {CH}_{3} {NH}_{2} & {({CH}_{3})}_{2} {NH}_{I} & {({CH}_{3})}_{3} N & C_{6} H_{5} {CH}_{2} {NH}_{2} \\ II & III & IV \end{array}$ varies in the order:

1 I

>

II

>

III

>

IV

2 II

>

I

>

III

>

IV

3 III

>

I

>

II

>

IV

4 IV

>

I

>

II

>

III

Explanation:

In aquous medium alkyl amines and ammonia react with water and form ammonium respectively due to this reason the basicity is change due to steric hindrance.
Hence the order of basicity is-
${({CH}_{3})}_{2} NH > {CH}_{3} {NH}_{2} > {({CH}_{3})}_{3} N > C_{6} H_{5} {CH}_{2} {NH}_{2}$ Option (b) is correct.
This phenomena only occures in polar solvent.
In any non-polar solvent, the normal basicity order is forward as we expected.

AMU-2005

GENERAL ORGANIC CHEMISTRY

231325 Which of the following carbocations is most stable?

1 I

2 II

3 III

4 IV

Explanation:

We know that, the order of stability of carbocation is $3^{\circ} > 2^{\circ} > 1^{\circ} >$ methyl carbocation.
Hence, the (ii) carbocation is $3^{\circ}$ carbocation and it is most stable all of them.

AMU-2018

GENERAL ORGANIC CHEMISTRY

231326 The correct order of acidity of the following is

1 (I)

>

(IV)

>

(II)

>

(III)

2 (IV)

>

(II)

>

(I)

>

(III)

3 (IV)

>

(I)

>

(II)

>

(III)

4 (IV)

>

(II)

>

(III)

>

(I)

Explanation:

Electron-donation groups decrease while electron withdrawing groups increase the acidity of phenols.
Decreasing order of acidity in nitro phenol -
P-nitrophenol $>$ O-nitrophenol $>$ M-nitrophenol $>$ phenol

TS EAMCET 10.08.2021

GENERAL ORGANIC CHEMISTRY

231328 Arrange the following carbanions in order of their decreasing stability:
(i) $H_{3} C - C \equiv C^{-}$
(ii) $H - C \equiv C^{-}$
(iii) $H_{3} C - {CH}_{2}^{-}$

1 i

>

ii

>

iii

2 ii

>

i

>

iii

3 iii

>

ii

>

i

4 iii

>

i

>

ii

Explanation:

The order of decreasing stability of carbons is:
$H - \underset{II}{C} \equiv C^{-} > {CH}_{3} - C \equiv C^{-} > {CH}_{3} - {CH}_{2}^{-}$
sp-hybridised carbon atom is more electronegative than ${sp}^{3}$ - hybridized carbon atom and hence, can accommodate the negative charge more effectively. So, I and II are more stable than III.
$- {CH}_{3}$ group has $+ I$ effect, therefore, it intensifies the negative charge and hence, destabilizes the carbanion.
${CH}_{3} \to C \equiv C^{-}$

AMU-2017

GENERAL ORGANIC CHEMISTRY

231331 Order of the base strength of the compounds
original image is

1 iv

>

ii

>

i

>

ii

2 iii

>

ii

>

iv

>

i

3 ii

>

iii

>

iv

>

i

4 ii

>

iii

>

i

>

iv

Explanation:

(D) :

II $>$ III $>$ I $>$ IV . The phenoxide ion is resource stabilized so it is least basic.

AMU - 2010

GENERAL ORGANIC CHEMISTRY

231338 The basicity of compounds I, II, III and IV
$\begin{array}{cccc} {CH}_{3} {NH}_{2} & {({CH}_{3})}_{2} {NH}_{I} & {({CH}_{3})}_{3} N & C_{6} H_{5} {CH}_{2} {NH}_{2} \\ II & III & IV \end{array}$ varies in the order:

1 I

>

II

>

III

>

IV

2 II

>

I

>

III

>

IV

3 III

>

I

>

II

>

IV

4 IV

>

I

>

II

>

III

Explanation:

In aquous medium alkyl amines and ammonia react with water and form ammonium respectively due to this reason the basicity is change due to steric hindrance.
Hence the order of basicity is-
${({CH}_{3})}_{2} NH > {CH}_{3} {NH}_{2} > {({CH}_{3})}_{3} N > C_{6} H_{5} {CH}_{2} {NH}_{2}$ Option (b) is correct.
This phenomena only occures in polar solvent.
In any non-polar solvent, the normal basicity order is forward as we expected.

AMU-2005

GENERAL ORGANIC CHEMISTRY

231325 Which of the following carbocations is most stable?

1 I

2 II

3 III

4 IV

Explanation:

We know that, the order of stability of carbocation is $3^{\circ} > 2^{\circ} > 1^{\circ} >$ methyl carbocation.
Hence, the (ii) carbocation is $3^{\circ}$ carbocation and it is most stable all of them.

AMU-2018

GENERAL ORGANIC CHEMISTRY

231326 The correct order of acidity of the following is

1 (I)

>

(IV)

>

(II)

>

(III)

2 (IV)

>

(II)

>

(I)

>

(III)

3 (IV)

>

(I)

>

(II)

>

(III)

4 (IV)

>

(II)

>

(III)

>

(I)

Explanation:

Electron-donation groups decrease while electron withdrawing groups increase the acidity of phenols.
Decreasing order of acidity in nitro phenol -
P-nitrophenol $>$ O-nitrophenol $>$ M-nitrophenol $>$ phenol

TS EAMCET 10.08.2021

GENERAL ORGANIC CHEMISTRY

231328 Arrange the following carbanions in order of their decreasing stability:
(i) $H_{3} C - C \equiv C^{-}$
(ii) $H - C \equiv C^{-}$
(iii) $H_{3} C - {CH}_{2}^{-}$

1 i

>

ii

>

iii

2 ii

>

i

>

iii

3 iii

>

ii

>

i

4 iii

>

i

>

ii

Explanation:

The order of decreasing stability of carbons is:
$H - \underset{II}{C} \equiv C^{-} > {CH}_{3} - C \equiv C^{-} > {CH}_{3} - {CH}_{2}^{-}$
sp-hybridised carbon atom is more electronegative than ${sp}^{3}$ - hybridized carbon atom and hence, can accommodate the negative charge more effectively. So, I and II are more stable than III.
$- {CH}_{3}$ group has $+ I$ effect, therefore, it intensifies the negative charge and hence, destabilizes the carbanion.
${CH}_{3} \to C \equiv C^{-}$

AMU-2017

GENERAL ORGANIC CHEMISTRY

231331 Order of the base strength of the compounds
original image is

1 iv

>

ii

>

i

>

ii

2 iii

>

ii

>

iv

>

i

3 ii

>

iii

>

iv

>

i

4 ii

>

iii

>

i

>

iv

Explanation:

(D) :

II $>$ III $>$ I $>$ IV . The phenoxide ion is resource stabilized so it is least basic.

AMU - 2010

GENERAL ORGANIC CHEMISTRY

231338 The basicity of compounds I, II, III and IV
$\begin{array}{cccc} {CH}_{3} {NH}_{2} & {({CH}_{3})}_{2} {NH}_{I} & {({CH}_{3})}_{3} N & C_{6} H_{5} {CH}_{2} {NH}_{2} \\ II & III & IV \end{array}$ varies in the order:

1 I

>

II

>

III

>

IV

2 II

>

I

>

III

>

IV

3 III

>

I

>

II

>

IV

4 IV

>

I

>

II

>

III

Explanation:

In aquous medium alkyl amines and ammonia react with water and form ammonium respectively due to this reason the basicity is change due to steric hindrance.
Hence the order of basicity is-
${({CH}_{3})}_{2} NH > {CH}_{3} {NH}_{2} > {({CH}_{3})}_{3} N > C_{6} H_{5} {CH}_{2} {NH}_{2}$ Option (b) is correct.
This phenomena only occures in polar solvent.
In any non-polar solvent, the normal basicity order is forward as we expected.

AMU-2005

GENERAL ORGANIC CHEMISTRY

231325 Which of the following carbocations is most stable?

1 I

2 II

3 III

4 IV

Explanation:

We know that, the order of stability of carbocation is $3^{\circ} > 2^{\circ} > 1^{\circ} >$ methyl carbocation.
Hence, the (ii) carbocation is $3^{\circ}$ carbocation and it is most stable all of them.

AMU-2018

GENERAL ORGANIC CHEMISTRY

231326 The correct order of acidity of the following is

1 (I)

>

(IV)

>

(II)

>

(III)

2 (IV)

>

(II)

>

(I)

>

(III)

3 (IV)

>

(I)

>

(II)

>

(III)

4 (IV)

>

(II)

>

(III)

>

(I)

Explanation:

Electron-donation groups decrease while electron withdrawing groups increase the acidity of phenols.
Decreasing order of acidity in nitro phenol -
P-nitrophenol $>$ O-nitrophenol $>$ M-nitrophenol $>$ phenol

TS EAMCET 10.08.2021

GENERAL ORGANIC CHEMISTRY

231328 Arrange the following carbanions in order of their decreasing stability:
(i) $H_{3} C - C \equiv C^{-}$
(ii) $H - C \equiv C^{-}$
(iii) $H_{3} C - {CH}_{2}^{-}$

1 i

>

ii

>

iii

2 ii

>

i

>

iii

3 iii

>

ii

>

i

4 iii

>

i

>

ii

Explanation:

The order of decreasing stability of carbons is:
$H - \underset{II}{C} \equiv C^{-} > {CH}_{3} - C \equiv C^{-} > {CH}_{3} - {CH}_{2}^{-}$
sp-hybridised carbon atom is more electronegative than ${sp}^{3}$ - hybridized carbon atom and hence, can accommodate the negative charge more effectively. So, I and II are more stable than III.
$- {CH}_{3}$ group has $+ I$ effect, therefore, it intensifies the negative charge and hence, destabilizes the carbanion.
${CH}_{3} \to C \equiv C^{-}$

AMU-2017

GENERAL ORGANIC CHEMISTRY

231331 Order of the base strength of the compounds
original image is

1 iv

>

ii

>

i

>

ii

2 iii

>

ii

>

iv

>

i

3 ii

>

iii

>

iv

>

i

4 ii

>

iii

>

i

>

iv

Explanation:

(D) :

II $>$ III $>$ I $>$ IV . The phenoxide ion is resource stabilized so it is least basic.

AMU - 2010

GENERAL ORGANIC CHEMISTRY

231338 The basicity of compounds I, II, III and IV
$\begin{array}{cccc} {CH}_{3} {NH}_{2} & {({CH}_{3})}_{2} {NH}_{I} & {({CH}_{3})}_{3} N & C_{6} H_{5} {CH}_{2} {NH}_{2} \\ II & III & IV \end{array}$ varies in the order:

1 I

>

II

>

III

>

IV

2 II

>

I

>

III

>

IV

3 III

>

I

>

II

>

IV

4 IV

>

I

>

II

>

III

Explanation:

In aquous medium alkyl amines and ammonia react with water and form ammonium respectively due to this reason the basicity is change due to steric hindrance.
Hence the order of basicity is-
${({CH}_{3})}_{2} NH > {CH}_{3} {NH}_{2} > {({CH}_{3})}_{3} N > C_{6} H_{5} {CH}_{2} {NH}_{2}$ Option (b) is correct.
This phenomena only occures in polar solvent.
In any non-polar solvent, the normal basicity order is forward as we expected.

AMU-2005