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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 94 of 150

Marks: +1, -0

Identify correct decreasing order of basic strength of amines

$\mathrm{CH}_{3} \mathrm{NH}_{2} > \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} > C_{6} \mathrm{H}_{5} \mathrm{NH}_{2} > \mathrm{NH}_{3}$
$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} > \mathrm{CH}_{3} \mathrm{NH}_{2} > \mathrm{NH}_{3} > C_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
$\mathrm{NH}_{3} > \mathrm{CH}_{3} \mathrm{NH}_{2} > \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} > C_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$
$C_{6} \mathrm{H}_{5} \mathrm{NH}_{2} > \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} > \mathrm{CH}_{3} \mathrm{NH}_{2} > \mathrm{NH}_{3}$

Solution:

Concept:

The basicity of amines depends on the availability of the nitrogen lone pair for donation to a proton.

Explanation:

Methyl groups show the

+I

effect and push electron density towards nitrogen.

This makes the lone pair on nitrogen more readily available.

Therefore, methylamine is more basic than ammonia:

\mathrm{CH}_{3}\mathrm{NH}_{2} > \mathrm{NH}_{3}

Dimethylamine has two methyl groups, so its basicity becomes greater than that of methylamine in aqueous solution:

\left(\mathrm{CH}_{3}\right)_{2}\mathrm{NH} > \mathrm{CH}_{3}\mathrm{NH}_{2}

Aniline is least basic because the lone pair on nitrogen is delocalised by resonance with the benzene ring.

This reduces the availability of the lone pair for protonation.

Hence aniline is less basic than ammonia:

\mathrm{NH}_{3} > \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}

Combining all these comparisons gives the correct decreasing order:

\left(\mathrm{CH}_{3}\right)_{2}\mathrm{NH} > \mathrm{CH}_{3}\mathrm{NH}_{2} > \mathrm{NH}_{3} > \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2}

Answer:

Option B.

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