Question:
Identify the major product obtained when ethyl amine is reacted with excess of methyl iodide ?
Identify the major product obtained when ethyl amine is reacted with excess of methyl iodide ?
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Excess CH\(_3\)I alkylates the amine fully, giving a quaternary ammonium salt.
Updated On: Oct 1, 2026
- Tetramethylammonium iodide
- Ethyltrimethylammonium iodide
- Ethyldimethylamine
- Ethylmethylamine
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Amines react with alkyl halides by nucleophilic substitution. With excess alkyl halide the reaction goes on until the nitrogen carries four groups (exhaustive alkylation).
Step 2: Detailed Explanation
\(\text{C}_2\text{H}_5\text{NH}_2 \to \text{C}_2\text{H}_5\text{NHCH}_3 \to \text{C}_2\text{H}_5\text{N(CH}_3)_2 \to [\text{C}_2\text{H}_5\text{N(CH}_3)_3]^+\text{I}^-\).
Each step adds one methyl group. After three methylations the nitrogen has the ethyl group and three methyl groups.
The final product is ethyltrimethylammonium iodide. The ethyl group stays, so tetramethylammonium iodide cannot form.
Final Answer:
The major product is ethyltrimethylammonium iodide, option (B). \[ \boxed{\text{Ethyltrimethylammonium iodide (B)}} \]
Amines react with alkyl halides by nucleophilic substitution. With excess alkyl halide the reaction goes on until the nitrogen carries four groups (exhaustive alkylation).
Step 2: Detailed Explanation
\(\text{C}_2\text{H}_5\text{NH}_2 \to \text{C}_2\text{H}_5\text{NHCH}_3 \to \text{C}_2\text{H}_5\text{N(CH}_3)_2 \to [\text{C}_2\text{H}_5\text{N(CH}_3)_3]^+\text{I}^-\).
Each step adds one methyl group. After three methylations the nitrogen has the ethyl group and three methyl groups.
The final product is ethyltrimethylammonium iodide. The ethyl group stays, so tetramethylammonium iodide cannot form.
Final Answer:
The major product is ethyltrimethylammonium iodide, option (B). \[ \boxed{\text{Ethyltrimethylammonium iodide (B)}} \]
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