Q.Write chemical equations for the following reactions:

The key idea is ammonolysis with successive alkylation — ammonia is a nucleophile, and every amine it produces is a still better nucleophile, so alkylation does not stop cleanly at one stage.

(i) Ethanolic $NH_3$ attacks chloroethane ($S_N2$) to give ethanamine — but ethanamine reacts further with more $C_2H_5Cl$, and the substitution continues through the secondary and tertiary amines right up to the quaternary ammonium salt:

(chloroethane → ethanamine → $N$-ethylethanamine → $N,N$-diethylethanamine → quaternary ammonium salt)

(ii) Ammonolysis of benzyl chloride gives benzylamine, which the two moles of $CH_3Cl$ then methylate twice, stopping at the tertiary amine:

(benzyl chloride → benzylamine → $N,N$-dimethylbenzylamine, i.e. $N,N$-dimethylphenylmethanamine) …

The Concept: Nucleophilic Substitution by Ammonia — and Why It Doesn't Stop

Ammonia ($NH_3$) has a lone pair of electrons on nitrogen, making it a good nucleophile. In an alkyl halide like chloroethane ($C_2H_5Cl$) or benzyl chloride ($C_6H_5CH_2Cl$), the carbon attached to chlorine is electrophilic, so ammonia attacks it and displaces chloride in an $S_N2$ reaction. This process is called ammonolysis.

The crucial point: the amine produced still carries a lone pair, and the alkyl group's $+I$ effect makes it an even better nucleophile than ammonia. So the primary amine attacks another molecule of the alkyl halide, giving a secondary amine; the secondary amine attacks again, giving a tertiary amine; and the tertiary amine can attack once more to give a quaternary ammonium salt, which has four carbon groups on nitrogen and can react no further. In each substitution step, the $HCl$ produced is taken up by the excess ammonia/amine present in the mixture (as an ammonium salt).

Step-by-Step Solution

(i) Reaction of ethanolic $NH_3$ with $C_2H_5Cl$

Step 1 — Ammonolysis. The lone pair on nitrogen attacks the carbon bearing chlorine; chloride leaves.

Product: ethanamine (ethylamine, a $1^\circ$ amine). The $HCl$ is taken up by excess $NH_3$ as $NH_4Cl$.

Step 2 — Second alkylation. Ethanamine, a better nucleophile than $NH_3$, attacks another molecule of chloroethane.

Product: $N$-ethylethanamine (diethylamine, a $2^\circ$ amine).

Step 3 — Third alkylation.

Product: $N,N$-diethylethanamine (triethylamine, a $3^\circ$ amine).

Step 4 — Quaternisation. The tertiary amine attacks one last molecule of chloroethane. There is no N–H left to lose, so the product is the salt itself.

Product: the quaternary ammonium salt (tetraethylammonium chloride).

The whole cascade in one line:

Method: Nucleophilic Substitution ($S_N2$) with Successive Alkylation

The governing idea for both parts: a nitrogen nucleophile (first $NH_3$, then each amine it produces) attacks the electrophilic carbon bearing the halide, displacing $Cl^-$. Because every amine formed is itself a nucleophile, alkylation continues stage by stage until either (a) the nitrogen has no capacity left — the quaternary salt — or (b) the question limits the moles of alkyl halide.

Steps for solving any ammonolysis/alkylation problem:

1. Identify the nucleophile ($NH_3$ or an amine — the species with the lone pair).
2. Identify the substrate (the alkyl halide with the leaving group).
3. Write the substitution — one alkyl group transfers to nitrogen per step, releasing $HCl$ (taken up by the excess amine/ammonia present).
4. Ask: does the chain continue? If the alkyl halide is not limited, carry the cascade to the quaternary ammonium salt. If the question fixes the moles, stop at the corresponding stage.

(i) Reaction of ethanolic $NH_3$ with $C_2H_5Cl$

Nothing limits the chloroethane here, so write the full successive-alkylation chain:

Stage by stage (each substitution also releases $HCl$, neutralised by the excess base in the mixture):

1. $C_2H_5Cl + NH_3 \rightarrow C_2H_5NH_2 + HCl$ — ethanamine ($1^\circ$)
2. $C_2H_5NH_2 + C_2H_5Cl \rightarrow (C_2H_5)_2NH + HCl$ — $N$-ethylethanamine ($2^\circ$)
3. $(C_2H_5)_2NH + C_2H_5Cl \rightarrow (C_2H_5)_3N + HCl$ — $N,N$-diethylethanamine ($3^\circ$)
4. $(C_2H_5)_3N + C_2H_5Cl \rightarrow (C_2H_5)_4N^+Cl^-$ — quaternary ammonium salt

(ii) Ammonolysis of benzyl chloride, then two moles of $CH_3Cl$

The question fixes the amount of methyl chloride at two moles, so the chain stops at the tertiary amine: …

Here are the common mistakes students make with this specific question, along with the conceptual fixes to avoid them.

Mistake 1: Stopping part (i) at ethylamine

The Mistake:

Writing only the first step —

— and presenting ethanamine as the answer.

Why it happens:

Students memorise "ammonolysis gives a primary amine" and forget that the primary amine is itself a better nucleophile than ammonia. Nothing in this question limits the amount of chloroethane, so the alkylation keeps going. (A primary amine is the major product only when a large excess of ammonia is deliberately used — a condition this question does not state.)

How to avoid it:

Write the full successive-alkylation cascade:

ending at the quaternary ammonium salt.

Mistake 2: Calling the intermediates in (ii) "quaternary ammonium salts"

The Mistake:

Referring to the protonated intermediates formed during (ii)'s methylation (e.g. $C_6H_5CH_2NH_3^+Cl^-$ or $C_6H_5CH_2NH_2CH_3^+Cl^-$) as "quaternary ammonium salt intermediates".

Why it happens:

Any positively charged nitrogen looks "ammonium-like", so the words get mixed up.

How to avoid it:

• Quaternary means four carbon groups bonded to nitrogen ($R_4N^+$) — no N–H at all. Such a salt cannot be deprotonated back to an amine.
• The intermediates in (ii) are simple alkylammonium salts (protonated amines): they still carry N–H bonds and are freed to the amine by the excess base present.
• In this question, a genuine quaternary salt appears only at the end of part (i): $(C_2H_5)_4N^+Cl^-$.

Mistake 3: Running part (ii) past the tertiary amine

The Mistake:

Adding a third methylation and answering $C_6H_5CH_2N(CH_3)_3^+Cl^-$.

Why it happens:

Students who have just learned the "cascade runs to the quaternary salt" idea from part (i) apply it blindly to part (ii).

How to avoid it:

Read the stoichiometry the question fixes: two moles of $CH_3Cl$ means exactly two methylations —

— stopping at $N,N$-dimethylbenzylamine, a tertiary amine.

Mistake 4: Worrying about $S_N1$ vs $S_N2$ instead of the products

The Mistake:

Spending time debating the mechanism for benzyl chloride and writing no clear product sequence.

How to avoid it: …