JEE Main 2023 · 25 January · evening · Single correct · +4 / −1
Match List I with List II
Choose the correct answer from the options given below :
| List I (Amines) | List II (pK_b) | ||
|---|---|---|---|
| A. | Aniline | I. | 3.25 |
| B. | Ethanamine | II. | 3.00 |
| C. | N-Ethylethanamine | III. | 9.38 |
| D. | N, N-Diethylethanamine | IV. | 3.29 |
- (a) A-III, B-II, C-I, D-IV
- (b) A-III, B-II, C-IV, D-I
- (c) A-III, B-IV, C-II, D-I
- (d) A-I, B-IV, C-II, D-III
Answer and solution
Answer (c) A-III, B-IV, C-II, D-I
Solution
Chemistry ma’am · 4 steps
Lower pK_b means a stronger base, so rank these amines by basicity in water, not by alkyl count.
- Read the scale. pK_b = -log K_b, so the strongest base has the smallest pK_b. Sorted, the four values are 3.00 < 3.25 < 3.29 < 9.38.
- Aniline first. In aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is much less available for a proton. Aniline is by far the weakest base here and takes the large value, 9.38.
- Ethylamines in water. In water, three effects decide basicity: alkyl groups push electron density onto N (+I effect), the ammonium ion is stabilised by hydrogen bonding with water, and bulky groups crowd the N. For ethyl groups the balance gives the order 2° > 3° > 1°. The secondary amine has two +I groups and still enough N–H for hydrogen bonding. The tertiary amine has the most +I push, but its ion has only one N–H for hydrogen bonding and three crowding ethyl groups.
Amine Type Basicity rank pK_b Aniline aryl 1°, resonance weakest 9.38 (III) Ethanamine 1° 3rd 3.29 (IV) N-Ethylethanamine 2° strongest 3.00 (II) N,N-Diethylethanamine 3° 2nd 3.25 (I) - So. The match is A-III, B-IV, C-II, D-I, which is option (c).
The idea to keep. In water, ethylamines follow 2° > 3° > 1° and methylamines follow 2° > 1° > 3°. Aryl amines are far weaker because of resonance. Smaller pK_b means stronger base.
Answer from the NTA final answer key. Solution by PrepWiser.