Which of the following amines has lowest pKb value?

Q: Which of the following amines has lowest pKb value?

**Correct option: (A) C₆H₅–N(CH₃)₂ (N,N-dimethylaniline)** Lowest pKb = **strongest base**. - **Two electron-donating (+I) methyl groups** on nitrogen push electron density onto the N, making the lone pair most available → strongest base → **lowest pKb**. **Basicity order (in aqueous solution):** C₆H₅N(CH₃)₂ > C₆H₅NH(CH₃) > C₆H₅NH₂ > p-O₂N–C₆H₄–NH₂

Q: Why does lowest pKb mean strongest base?

$pK_b = -\log K_b$. A stronger base has a larger Kb (accepts protons more readily), so its pKb is smaller. Therefore the amine with the lowest pKb value is the strongest base among the given set.

Amines

Basicity of aromatic amines (pKb)

Options

(A)C₆H₅ – N(CH₃)₂

(B)C₆H₅ – NH(CH₃)

(C)C₆H₅ – NH₂

(D)O₂N–C₆H₄–NH₂ (para-nitroaniline: benzene ring with NO₂ and NH₂ at para positions)

Official Answer

Correct option: (A) C₆H₅–N(CH₃)₂ (N,N-dimethylaniline)

Lowest pKb = strongest base.

Two electron-donating (+I) methyl groups on nitrogen push electron density onto the N, making the lone pair most available → strongest base → lowest pKb.

Basicity order (in aqueous solution):

C₆H₅N(CH₃)₂ > C₆H₅NH(CH₃) > C₆H₅NH₂ > p-O₂N–C₆H₄–NH₂

pKbbasicity of aminesN,N-dimethylanilineelectron donating groupinductive effectp-nitroaniline weak baselone pair availabilityaromatic amine

Marking Scheme

11 mark: correct option (A) C₆H₅–N(CH₃)₂.
2Award only for the N,N-dimethylaniline choice (strongest base = lowest pKb).

Hint

Lowest pKb = strongest base = most available N lone pair; +I methyls on N help, para –NO₂ hurts most.

Quick Oral Answer

N,N-dimethylaniline has the lowest pKb because its two electron-releasing methyl groups make the nitrogen lone pair most available, so it is the strongest base of the four.

Analysis & Explanation

Concept

Basicity of an amine depends on the availability of the nitrogen lone pair. Electron-donating groups increase it (lower pKb, stronger base); electron-withdrawing groups decrease it (higher pKb, weaker base).

Comparing the choices

(A) N,N-dimethylaniline — two +I methyl groups raise electron density on N the most → strongest base, lowest pKb.
(B) N-methylaniline — one methyl, moderately strong.
(C) Aniline — lone pair delocalised into the ring, weaker base.
(D) p-nitroaniline — the –NO₂ group at para withdraws electrons strongly (−R, −I), pulling the lone pair away → weakest base, highest pKb.

Exam trap

Remember the inverse relationship: lowest pKb means strongest base, not weakest. Students who pick p-nitroaniline reverse the logic.

Note on gas vs solution

In aqueous solution the order can be modulated by solvation, but among these aryl amines the alkyl-substituted N,N-dimethylaniline is the accepted strongest base and p-nitroaniline the weakest.

Common Mistakes

1Confusing low pKb with weak base and selecting p-nitroaniline.
2Ignoring the additive +I effect of two methyl groups in N,N-dimethylaniline.
3Assuming aniline is the strongest base among aryl amines.

Interesting Facts

In the gas phase, without solvent effects, amine basicity follows purely the inductive order 3° > 2° > 1° > NH₃; water's solvation of the protonated ion reshuffles this for simple alkylamines.

The strongly deactivating –NO₂ group makes p-nitroaniline ( $pK_b \approx 13$ ) roughly ten-thousand times weaker a base than aniline.

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Frequently Asked Questions

Why does lowest pKb mean strongest base?

$pK_b = -\log K_b$ . A stronger base has a larger Kb (accepts protons more readily), so its pKb is smaller. Therefore the amine with the lowest pKb value is the strongest base among the given set.

Why is p-nitroaniline the weakest base here?

The para –NO₂ group withdraws electron density through both –I and –R effects, delocalising the nitrogen lone pair into the ring and nitro group. This makes the lone pair least available for protonation, giving p-nitroaniline the highest pKb and weakest basicity.