Q.Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

The key idea is that aniline undergoes exhaustive methylation — a nucleophilic substitution where the lone pair on nitrogen attacks methyl iodide repeatedly, and sodium carbonate acts as a base to neutralise the HI formed.

Step 1: Aniline ($\ce{C6H5NH2}$) reacts with one equivalent of $\ce{CH3I}$ to give N-methylaniline.

Step 2: Further methylation yields N,N-dimethylaniline. …

Concept First: Why This Reaction Goes All the Way to the Quaternary Salt

Aniline's nitrogen has a lone pair, making it a good nucleophile. Methyl iodide is an excellent electrophile — iodine is a great leaving group and the methyl carbon is unhindered, so each step is a straightforward $\ce{S_N2}$ reaction at the methyl carbon. Sodium carbonate is a mild base; its job is to neutralise the $\ce{HI}$ that forms at every step, which would otherwise protonate the amine (or the newly alkylated amine) and shut down its nucleophilicity. Because $\ce{Na2CO3}$ keeps regenerating the free base and the question specifies excess $\ce{CH3I}$, the alkylation does not stop at the mono- or di-methylated stage — it proceeds all the way to the quaternary ammonium salt.

Step-by-Step Reactions

Step 1 — Mono-methylation

Aniline's nitrogen attacks $\ce{CH3I}$, displacing iodide. The $\ce{HI}$ formed is mopped up by $\ce{Na2CO3}$, regenerating the free secondary amine, $N$-methylaniline.

Step 2 — Di-methylation

$N$-Methylaniline still has a lone pair, so it reacts again to give $N,N$-dimethylaniline.

Step 3 — Tri-methylation (quaternisation)

$\ce{C6H5N(CH3)2 + CH3I -> C6H5N+(CH3)3 \, I-}$ …

Method: Exhaustive Alkylation (Hofmann Exhaustive Methylation)

This method converts a primary amine into a quaternary ammonium salt by repeated nucleophilic substitution using excess methyl iodide in the presence of a mild base.

Step-by-Step Reasoning

Step 1 — First alkylation

Aniline ($\ce{C6H5NH2}$) acts as a nucleophile. It attacks methyl iodide ($\ce{CH3I}$) in an $\mathrm{S_N2}$ reaction. Sodium carbonate ($\ce{Na2CO3}$) neutralises the $\ce{HI}$ formed, driving the reaction forward.

Step 2 — Second alkylation

The secondary amine ($\ce{C6H5NHCH3}$) is more nucleophilic than aniline, so it reacts again with $\ce{CH3I}$.

Step 3 — Third alkylation

The tertiary amine ($\ce{C6H5N(CH3)2}$) undergoes a third $\mathrm{S_N2}$ attack.

$\ce{C6H5N(CH3)2 + CH3I ->[Na2CO3] C6H5N+(CH3)3 I^-}$ …

Here are the common mistakes students make when tackling the alkylation of aniline with excess methyl iodide, along with how to avoid each.

The Core Concept (The "Why")

Aniline ($C_6H_5NH_2$) is a nucleophile. Methyl iodide ($CH_3I$) is an excellent electrophile. In the presence of a weak base like sodium carbonate ($Na_2CO_3$), the reaction proceeds via N-alkylation (nucleophilic substitution at the nitrogen atom).

Because methyl iodide is in excess, the reaction doesn't stop at mono-alkylation. The product is the quaternary ammonium salt: Trimethylphenylammonium iodide.

Common Mistake #1: Stopping at the Mono- or Di-alkylated Product

• The Mistake: Students write the product as $C_6H_5NHCH_3$ (N-methylaniline) or $C_6H_5N(CH_3)_2$ (N,N-dimethylaniline).
• Why it's wrong: The problem explicitly states "excess of methyl iodide". Each alkylation makes the nitrogen more nucleophilic (due to the +I effect of methyl groups), so the reaction keeps going until all three hydrogens on the nitrogen are replaced.
• How to Avoid: Read the word "excess" as a command. If the reagent is in excess, assume the reaction goes to completion. For aniline + excess alkyl halide, the final product is always the quaternary ammonium salt.

Common Mistake #2: Forgetting the Role of Sodium Carbonate

• The Mistake: Students write the reaction without $Na_2CO_3$ or think it's a catalyst for a different step.
• Why it's wrong: The reaction produces hydrogen iodide (HI) as a byproduct. HI is a strong acid that would protonate the aniline (making it $C_6H_5NH_3^+$), which is not nucleophilic and would stop the reaction. $Na_2CO_3$ is a weak base that neutralizes the HI.

• How to Avoid: Always ask: "What happens to the acid byproduct?" If a base is present, it's there to mop up the acid and keep the amine in its free, nucleophilic form. Write the neutralization as a separate step or note it in the mechanism.

Common Mistake #3: Writing the Wrong Product (Salt vs. Free Base)

• The Mistake: Students write the final product as $C_6H_5N(CH_3)_2$ (a neutral tertiary amine) instead of the salt $C_6H_5N^+(CH_3)_3 \ I^-$.
• Why it's wrong: The nitrogen in the tertiary amine still has a lone pair. With excess $CH_3I$, it will undergo one more $S_N2$ reaction to form the quaternary ammonium salt. The nitrogen becomes positively charged and has no lone pair — it cannot react further. …