Match List I with List II
Choose the correct answer from the options given below:
| LIST I (Amines) | LIST II (pK) |
|---|---|
| A. Aniline | I. 3.25 |
| B. Ethanamine | II. 3.00 |
| C. N-Ethylethanamine | III. 9.38 |
| D. N, N-Diethylethanamine | IV. 3.29 |
Choose the correct answer from the options given below:
- A.A-III, B-IV, C-II, D-I
- B.A-III, B-II, C-I, D-IV
- C.A-III, B-II, C-IV, D-I
- D.A-I, B-IV, C-II, D-III
Show answer and solution
Correct option: A
Solution
Aniline is a very weak base due to lone pair delocalization into the benzene ring, giving it a very high pKb of 9.38 (III).
For the aliphatic amines, basicity order in aqueous solution follows: secondary (diethylamine) > primary (ethanamine) > tertiary (triethylamine), because while alkyl groups increase electron density, the tertiary amine's steric hindrance and poor solvation reduce its basicity below the secondary amine. Lower pKb = stronger base, so: N-Ethylethanamine (secondary) has pKb 3.00 (II, strongest), Ethanamine (primary) has pKb 3.29 (IV), and N,N-Diethylethanamine (tertiary) has pKb 3.25 (I, slightly weaker than secondary due to solvation effects).
Thus A-III, B-IV, C-II, D-I → Option 1.
For the aliphatic amines, basicity order in aqueous solution follows: secondary (diethylamine) > primary (ethanamine) > tertiary (triethylamine), because while alkyl groups increase electron density, the tertiary amine's steric hindrance and poor solvation reduce its basicity below the secondary amine. Lower pKb = stronger base, so: N-Ethylethanamine (secondary) has pKb 3.00 (II, strongest), Ethanamine (primary) has pKb 3.29 (IV), and N,N-Diethylethanamine (tertiary) has pKb 3.25 (I, slightly weaker than secondary due to solvation effects).
Thus A-III, B-IV, C-II, D-I → Option 1.
[JEE Main 2023 · 25 Jan · Shift 2]