Q.Aniline is treated with excess $\text{CH}_3\text{I}$ (exhaustive alkylation). Name the final quaternary product and explain why alkylation of an amine, unlike acylation, is difficult to stop cleanly at the monoalkylated stage.

Direct nitration of aniline uses a strongly acidic (nitrating) medium, which protonates $-\text{NH}_2$ to the meta-directing anilinium ion and also oxidises aniline, giving substantial meta product and tar. To avoid this, the amino group is first acetylated (with acetic anhydride) to acetanilide, $\text{C}_6\text{H}_5\text{NHCOCH}_3$. The $-\text{NHCOCH}_3$ group is still o,p-directing but less activating, so nit …

Understanding Acetylation and Ring Activation

When we talk about acetylation of aniline, we need to be clear about what is being acetylated. Aniline has a nucleophilic amino group that readily reacts with acetic anhydride or acetyl chloride to form acetanilide. This is indeed a monosubstituted product—one acetyl group attaches to the nitrogen.

The question then asks us to evaluate whether the reason given (about relative activating effects) correctly explains this observation.

Step-by-Step Analysis

1. What happens during acetylation of aniline?

Aniline ($C_6H_5NH_2$) reacts with an acetylating agent like $(CH_3CO)_2O$ to give:

The product is acetanilide, where one acetyl group is attached to nitrogen. This is a monosubstituted product, so Assertion (A) is true.

2. Why does acetylation stop at one acetyl group?

Nitrogen in aniline has one lone pair and two hydrogens. After the first acetylation, we get $-NHCOCH_3$, which still has one hydrogen but the nitrogen is now less nucleophilic (the electron-withdrawing carbonyl reduces the availability of the lone pair). A second acetylation is possible under forcing conditions, but under normal conditions we get predominantly the monoacetyl product.

The real reason for monosubstitution is simply that the first acetylation satisfies the typical reaction conditions and the resulting amide nitrogen is much less nucleophilic than the original amine.

3. Now let's examine the Reason (R): Is $-NHCOCH_3$ more activating than $-NH_2$?

This is where we need to understand activating effects toward electrophilic aromatic substitution.

The $-NH_2$ group is one of the most powerful activating groups for the benzene ring. It donates electron density through resonance (the lone pair on nitrogen delocalizes into the ring), making the ring electron-rich and highly reactive toward electrophiles.

When we convert $-NH_2$ to $-NHCOCH_3$, the lone pair on nitrogen is now partially delocalized into the carbonyl group ($C=O$) through resonance:

This means less electron density is available for donation to the benzene ring. The acetyl group is electron-withdrawing by resonance, competing with the ring for nitrogen's lone pair.