Question 4 of 20intermediate🔧 ApplyShort Answer2 marks

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

Correct Answer

Aniline undergoes exhaustive alkylation with excess methyl iodide in presence of sodium carbonate to form N,N,N-trimethylanilinium iodide as the final product.

The reaction proceeds as follows:

Step 1: C₆H₅NH₂ + CH₃I → C₆H₅NHCH₃ + HI (N-methylaniline)

Step 2: C₆H₅NHCH₃ + CH₃I → C₆H₅N(CH₃)₂ + HI (N,N-dimethylaniline)

Step 3: C₆H₅N(CH₃)₂ + CH₃I → [C₆H₅N(CH₃)₃]⁺I⁻ (N,N,N-trimethylanilinium iodide)

The sodium carbonate neutralizes the HI formed during the reaction, preventing the protonation of amine which would otherwise hinder further alkylation. The final product is a quaternary ammonium salt.

Exercise: Intext Questions | Q: 9.6 | (Page 16)
For More Understanding

Explanation

This question tests the student's understanding of N-alkylation of amines. Aniline, being a primary amine, undergoes successive alkylation with excess alkyl halide (methyl iodide). The reaction proceeds through secondary and tertiary amine intermediates to finally form a quaternary ammonium salt. The sodium carbonate acts as a base to neutralize HI produced, ensuring the reaction proceeds to completion. Note that this is N-alkylation (at nitrogen), not Friedel-Crafts alkylation (at the aromatic ring), which aniline cannot undergo as mentioned in the context.

Solution Steps

  1. Step 1: Identify that aniline (primary amine) reacts with methyl iodide through nucleophilic substitution

  2. Step 2: With excess CH₃I, all three H atoms of -NH₂ group are replaced by methyl groups

  3. Step 3: Write the stepwise reactions showing formation of secondary amine, then tertiary amine, then quaternary ammonium salt

  4. Step 4: Mention the role of sodium carbonate in neutralizing HI formed

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