Q.Chloromethane on treatment with excess of ammonia yields mainly

The key idea is that the reaction of an alkyl halide with excess ammonia follows SN2 mechanism, and the product distribution is controlled by the concentration of ammonia.

1. Chloromethane reacts with ammonia via SN2 to give methanamine ($\mathrm{CH_3NH_2}$).
2. Methanamine is more nucleophilic than ammonia, so it can further react with chloromethane to form secondary and tertiary amines. …

Why This Question Trips Students Up

The instinct is to think: "Chloromethane reacts with ammonia to give a primary amine, which then reacts further to give secondary, tertiary, and finally quaternary ammonium salts." That's true — but only when ammonia is limiting. Here, ammonia is in excess, and that changes everything.

The key insight: when you have a huge excess of ammonia, the very first substitution product (methanamine) is vastly outnumbered by ammonia molecules. So instead of methanamine finding another chloromethane to react with, it almost always bumps into an ammonia molecule instead. The reaction gets "stuck" at the first step.

Step-by-Step Reasoning

1. The reaction is an $\mathrm{S_N2}$ nucleophilic substitution.

Ammonia ($\mathrm{NH_3}$) acts as the nucleophile, attacking the electrophilic carbon in chloromethane ($\mathrm{CH_3Cl}$). The leaving group is chloride ($\mathrm{Cl^-}$). The immediate product is the methylammonium ion:

This is a salt. To get the free amine, a base (here, another ammonia molecule) deprotonates it:

So one molecule of chloromethane consumes two molecules of ammonia to produce one molecule of methanamine.

2. Now methanamine can react further — but with what?

Methanamine ($\mathrm{CH_3NH_2}$) is itself a nucleophile (it has a lone pair on nitrogen). It can attack another chloromethane molecule:

Then deprotonation by ammonia gives dimethylamine:

This secondary amine can go on to form trimethylamine, and eventually tetramethylammonium chloride.

3. Here's the critical point: competition for chloromethane.

Both ammonia and methanamine are competing for the limited chloromethane. The relative rates depend on two things:

• Nucleophilicity: Methanamine is a better nucleophile than ammonia (alkyl groups donate electron density, making the lone pair more available).
• Concentration: Ammonia is in huge excess — the problem says "excess of ammonia."

The concentration factor dominates. Even though methanamine is more reactive, there are so many ammonia molecules that the probability of a chloromethane molecule encountering methanamine is very low. Almost every chloromethane molecule gets intercepted by ammonia first.

Concept: Nucleophilic Substitution & Ammonolysis of Alkyl Halides

Method: Hofmann's Ammonolysis (Exhaustive Alkylation)

Steps:

1. Identify the reaction type

   Chloromethane ($\mathrm{CH_3Cl}$) reacts with ammonia ($\mathrm{NH_3}$) via nucleophilic substitution ($\mathrm{S_N2}$). Ammonia acts as the nucleophile, attacking the electrophilic carbon.

2. First substitution

   $\mathrm{CH_3Cl + NH_3 \rightarrow CH_3NH_2}$ (methanamine) + $\mathrm{HCl}$

3. Further substitutions (excess ammonia)

   The product methanamine is more nucleophilic than ammonia itself. With excess $\mathrm{CH_3Cl}$, it undergoes further alkylation:

   • $\mathrm{CH_3NH_2 + CH_3Cl \rightarrow (CH_3)_2NH}$ (N-methylmethanamine)
   • $\mathrm{(CH_3)_2NH + CH_3Cl \rightarrow (CH_3)_3N}$ (N,N-dimethylmethanamine)

4. Key insight — excess ammonia

   The question specifies excess of ammonia. This means ammonia is in large excess relative to chloromethane. …

Let's break this down. The question tests your understanding of the ammonolysis of alkyl halides — the nucleophilic substitution of alkyl halides with ammonia.

The Core Concept

When an alkyl halide (like chloromethane, $\mathrm{CH_3Cl}$) reacts with excess ammonia ($\mathrm{NH_3}$), the reaction proceeds stepwise:

1. First step:

   $\mathrm{CH_3Cl + NH_3 \rightarrow CH_3NH_3^+Cl^-}$

   (protonated primary amine)

2. Deprotonation (by excess $\mathrm{NH_3}$):

   $\mathrm{CH_3NH_3^+ + NH_3 \rightarrow CH_3NH_2 + NH_4^+}$

   (free methanamine)

3. Further alkylation:

   Methanamine ($\mathrm{CH_3NH_2}$) is more nucleophilic than ammonia, so it reacts with more $\mathrm{CH_3Cl}$ to give secondary, then tertiary amines, and finally the quaternary ammonium salt.

Key result: With excess alkyl halide, the major product is the quaternary salt.

With excess ammonia, the major product is the primary amine (methanamine).

Common Mistakes & How to Avoid Them

✗ Mistake 1: Choosing the tertiary amine (i) as the main product

• Why it happens: Students think "excess alkyl halide" means more substitution, but the question says excess ammonia.
• How to avoid: Always check which reagent is in excess.
  • Excess $\mathrm{NH_3}$ → primary amine is major.
  • Excess $\mathrm{RX}$ → further alkylation occurs.

✗ Mistake 2: Choosing the secondary amine (ii)

• Why it happens: Students forget that the primary amine is formed first and is more reactive, but with excess ammonia, the primary amine is quickly removed from further reaction by the large pool of $\mathrm{NH_3}$.
• How to avoid: Remember: Excess ammonia favours the first substitution product because $\mathrm{NH_3}$ outcompetes the amine for the alkyl halide.

✗ Mistake 3: Choosing "mixture in equal proportion" (iv)

• Why it happens: Students think all steps occur equally, but reaction rates differ — each step is slower than the previous because the nucleophilicity changes.
• How to avoid: …