Question:medium

Write the reason of the following: Ethylamine is soluble in water, while aniline is not soluble in water.

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Compare the size of the carbon part and the availability of the nitrogen lone pair for hydrogen bonding with water.
Updated On: Jul 10, 2026
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Solution and Explanation

Step 1: Balance of two parts. Solubility in water is a tug of war between the polar \(-NH_2\) head, which likes water, and the carbon skeleton, which dislikes water. Whichever part dominates decides the result.
Step 2: Ethylamine. Here the water-hating part is only a two-carbon ethyl group, which is small. The \(-NH_2\) group hydrogen bonds strongly with water, so the water-loving part wins and ethylamine mixes freely with water.
Step 3: Aniline. Here the water-hating part is a bulky benzene ring, a large non-polar surface that resists mixing with water. Moreover, the nitrogen lone pair is drawn into the ring by resonance, weakening the hydrogen bonds it can offer to water.
Step 4: Net effect. With a big hydrophobic ring and a lone pair that is tied up in resonance, the hydrophobic character of aniline dominates, so it stays largely insoluble in water while ethylamine dissolves.
\[\boxed{\text{Small alkyl + free lone pair = soluble; bulky aryl + delocalised lone pair = insoluble}}\]
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