Question:medium

The most reactive amine towards dilute hydrochloric acid is:

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Secondary amines like dimethylamine are more basic than primary and tertiary amines due to stronger inductive effects and lower steric hindrance, making them more reactive towards dilute acids.
Updated On: Mar 20, 2026
  • \( \text{CH}_3\text{NH}_2 \)
  • \( (\text{CH}_3)_2\text{NH} \)
  • \( (\text{CH}_3)_3\text{N} \)
  • \( \text{C}_6\text{H}_5\text{NH}_2 \)
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The Correct Option is B

Solution and Explanation

The basicity of amines, which dictates their reactivity with dilute \( \text{HCl} \), is influenced by electron-donating groups and steric hindrance.Step 1: Evaluate the basicity of various amines. Methylamine (\( \text{CH}_3\text{NH}_2 \)), a primary amine, exhibits less basicity due to a weaker electron-donating inductive effect from the methyl group compared to secondary amines, resulting in lower electron density on nitrogen. Dimethylamine (\( (\text{CH}_3)_2\text{NH} \)), a secondary amine, possesses two methyl groups, leading to a more pronounced electron-donating inductive effect and increased electron density on nitrogen. This makes it the most basic among the options. Its lower steric hindrance compared to tertiary amines also contributes to its high reactivity with dilute \( \text{HCl} \). Trimethylamine (\( (\text{CH}_3)_3\text{N} \)), a tertiary amine, has three methyl groups, resulting in a strong inductive effect. However, significant steric hindrance around the nitrogen atom impedes lone pair availability for protonation, reducing its reactivity relative to dimethylamine. Aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)) shows reduced basicity compared to aliphatic amines because its nitrogen lone pair is delocalized into the aromatic ring through resonance. This delocalization diminishes the lone pair's availability for protonation, rendering aniline the least reactive amine in this comparison.Step 2: Conclusion. Dimethylamine demonstrates the highest reactivity towards dilute \( \text{HCl} \) owing to its substantial electron-donating groups and minimal steric hindrance.\[(\text{CH}_3)_2\text{NH} + \text{HCl} \to (\text{CH}_3)_2\text{NH}_2^+ + \text{Cl}^-.\] Final Answer:\[\boxed{(\text{CH}_3)_2\text{NH}}\]
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