Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Experimental reaction of CH_3 Cl with aniline and anhydrous AlCl_3 does not give o and p -methylaniline. Reason (R): The -NH_2 group of aniline becomes deactivating because of salt formation with anhydrous AlCl_3 and hence yields m -methyl aniline as the product. In the light of the above statements, choose the most appropriate answer from the options given below :

Both A and R are true and (R) is the correct explanation of (A).

Both A and R are true but (R) is not the correct explanation of (A).

Compounds Containing Nitrogen PYQ — JEE Main 2022 · 26 Jul · Shift 1 · Q15 with Solution

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Correct answer: C

Identify the reaction type

The reaction of aniline with $\mathrm{CH_3Cl}$ and anhydrous $\mathrm{AlCl_3}$ is a Friedel–Crafts alkylation condition.

Normally, in Friedel–Crafts alkylation:

$\mathrm{RCl + AlCl_3}$ generates an electrophile.
The aromatic ring undergoes electrophilic substitution.

Behavior of aniline in presence of $\mathrm{AlCl_3}$

Aniline contains a basic amino group $(-\mathrm{NH_2})$ . The lone pair on nitrogen strongly coordinates with the Lewis acid $\mathrm{AlCl_3}$ , forming a complex/salt-like species.

\mathrm{C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2\cdot AlCl_3}

Because the lone pair is tied up in coordination, the amino group loses its strong electron-donating ability. As a result, the benzene ring becomes deactivated toward Friedel–Crafts alkylation.

Hence, aniline does not undergo Friedel–Crafts alkylation under these conditions.

So the assertion:

Experimental reaction of $\mathrm{CH_3Cl}$ with aniline and anhydrous $\mathrm{AlCl_3}$ does not give $o$ - and $p$ -methylaniline.

is true.

Examine the reason

The reason says:

The $-\mathrm{NH_2}$ group of aniline becomes deactivating because of salt formation with anhydrous $\mathrm{AlCl_3}$ and hence yields $m$ -methyl aniline as the product.

Let us split this into two parts:

First part: $-\mathrm{NH_2}$ becomes deactivating due to complex/salt formation with $\mathrm{AlCl_3}$ .
- This is true.
Second part: therefore it gives $m$ -methylaniline.
- This is false.
- In reality, because the ring is strongly deactivated and the amino group complexes with $\mathrm{AlCl_3}$ , Friedel–Crafts alkylation does not proceed satisfactorily. One does not obtain $m$ -methylaniline as the product.

Therefore, the complete Reason statement is false.

Choose the correct option

Assertion (A): True
Reason (R): False

So the correct option is:

\boxed{\text{C: (A) is true, but (R) is false.}}

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View written solutionFree

Correct answer: C

Identify the reaction type

The reaction of aniline with $\mathrm{CH_3Cl}$ and anhydrous $\mathrm{AlCl_3}$ is a Friedel–Crafts alkylation condition.

Normally, in Friedel–Crafts alkylation:

$\mathrm{RCl + AlCl_3}$ generates an electrophile.
The aromatic ring undergoes electrophilic substitution.

Behavior of aniline in presence of $\mathrm{AlCl_3}$

Aniline contains a basic amino group $(-\mathrm{NH_2})$ . The lone pair on nitrogen strongly coordinates with the Lewis acid $\mathrm{AlCl_3}$ , forming a complex/salt-like species.

\mathrm{C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2\cdot AlCl_3}

Because the lone pair is tied up in coordination, the amino group loses its strong electron-donating ability. As a result, the benzene ring becomes deactivated toward Friedel–Crafts alkylation.

Hence, aniline does not undergo Friedel–Crafts alkylation under these conditions.

So the assertion:

Experimental reaction of $\mathrm{CH_3Cl}$ with aniline and anhydrous $\mathrm{AlCl_3}$ does not give $o$ - and $p$ -methylaniline.

is true.

Examine the reason

The reason says:

The $-\mathrm{NH_2}$ group of aniline becomes deactivating because of salt formation with anhydrous $\mathrm{AlCl_3}$ and hence yields $m$ -methyl aniline as the product.

Let us split this into two parts:

First part: $-\mathrm{NH_2}$ becomes deactivating due to complex/salt formation with $\mathrm{AlCl_3}$ .
- This is true.
Second part: therefore it gives $m$ -methylaniline.
- This is false.
- In reality, because the ring is strongly deactivated and the amino group complexes with $\mathrm{AlCl_3}$ , Friedel–Crafts alkylation does not proceed satisfactorily. One does not obtain $m$ -methylaniline as the product.

Therefore, the complete Reason statement is false.

Choose the correct option

Assertion (A): True
Reason (R): False

So the correct option is:

\boxed{\text{C: (A) is true, but (R) is false.}}

Compounds Containing Nitrogen question

Compounds Containing Nitrogen question

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Compounds Containing Nitrogen question

Compounds Containing Nitrogen question

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