Amines: Tricky Questions Solved (5)

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Question

Arrange the following in decreasing order of basicity in aqueous solution: methylamine (CH3NH2\text{CH}_3\text{NH}_2), dimethylamine ((CH3)2NH(\text{CH}_3)_2\text{NH}), trimethylamine ((CH3)3N(\text{CH}_3)_3\text{N}), and ammonia (NH3\text{NH}_3). Then explain the JEE/NEET trap: why the gas-phase order differs from the aqueous-phase order.

Solution — Step by Step

For amines in aqueous solution, basicity depends on:

  1. +I+I (inductive donation) of alkyl groups — increases basicity by pushing electrons onto N, stabilising the protonated form.
  2. Steric hindrance — bulky alkyl groups make the lone pair harder to access, decreasing basicity.
  3. Solvation of the protonated cationRnNH4n+\text{R}_n\text{NH}_{4-n}^+ with more N-H\text{N-H} bonds is better hydrogen-bonded by water; more H atoms = more solvation = more stable cation = higher basicity.
  • NH3\text{NH}_3: no alkyl groups (no +I+I), small (no steric hindrance), best solvated (44 N-H bonds in NH4+\text{NH}_4^+).
  • CH3NH2\text{CH}_3\text{NH}_2: one alkyl (+I+I helps), small (low steric), 33 N-H bonds (good solvation).
  • (CH3)2NH(\text{CH}_3)_2\text{NH}: two alkyl (+I+I stronger), moderate steric, 22 N-H bonds (decent solvation).
  • (CH3)3N(\text{CH}_3)_3\text{N}: three alkyl (+I+I strongest), high steric, 11 N-H bond (poor solvation).

The three effects compete. In aqueous solution, the dominant interplay produces this order:

(CH3)2NH>CH3NH2>(CH3)3N>NH3(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3

Approximate pKb\text{p}K_b values in water at 25°C25\text{°C}:

  • (CH3)2NH(\text{CH}_3)_2\text{NH}: 3.273.27
  • CH3NH2\text{CH}_3\text{NH}_2: 3.363.36
  • (CH3)3N(\text{CH}_3)_3\text{N}: 4.194.19
  • NH3\text{NH}_3: 4.754.75

Smaller pKb\text{p}K_b = stronger base. The order matches.

In the gas phase (no solvent), there’s no solvation effect — only +I+I and steric (which is small for these). So basicity simply tracks +I+I:

(CH3)3N>(CH3)2NH>CH3NH2>NH3(gas phase)(\text{CH}_3)_3\text{N} > (\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{NH}_3 \quad \text{(gas phase)}

JEE Main has tested this distinction (gas vs aqueous) repeatedly. The question always specifies the medium — read carefully.

Why This Works

Basicity isn’t a single property — it’s a measure of equilibrium RnNH3n+H2ORnNH4n++OH\text{R}_n\text{NH}_{3-n} + \text{H}_2\text{O} \rightleftharpoons \text{R}_n\text{NH}_{4-n}^+ + \text{OH}^-. The position of this equilibrium depends on the medium because solvent stabilisation of the products matters.

In water, R3NH+\text{R}_3\text{NH}^+ has only one N-H bond available for H-bonding, so it’s poorly solvated. In gas phase, no solvation — so +I+I wins and tertiary amine becomes the strongest base.

Alternative Method

Use pKa\text{p}K_a of conjugate acid: a stronger base has a higher pKa\text{p}K_a for its conjugate acid. So compare pKa\text{p}K_a of RnNH4n+\text{R}_n\text{NH}_{4-n}^+ across the four amines. Same ordering.

For exam purposes, memorise the aqueous-phase order: secondary > primary > tertiary > NH₃ for methyl-substituted amines. For ethyl-substituted, the order shifts slightly but the same principle applies. Always assume aqueous unless the problem says otherwise.

Common Mistake

Three traps:

  1. Ignoring solvation. Students who only consider +I+I get the gas-phase order in an aqueous problem. NEET specifically tests the aqueous order — different from inductive prediction.

  2. Confusing amines with anilines. Anilines (aryl amines) are less basic than ammonia because the lone pair delocalises into the ring. Don’t apply the methylamine reasoning to aniline.

  3. Wrong sign of pKb\text{p}K_b comparison. Smaller pKb\text{p}K_b = stronger base. Many students reverse this and rank larger pKb\text{p}K_b as more basic.

Final answer: Aqueous: (CH3)2NH>CH3NH2>(CH3)3N>NH3(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3. Gas phase reverses solvation effect.

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