Class 12 Chemistry Chapter 9: Amines case study question with answers
One case-based question of the kind the board asks: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| Base | pKb |
|---|---|
| ammonia | 4.75 |
| ethanamine | 3.29 |
| N-ethylethanamine | 3.00 |
| N,N-diethylethanamine | 3.25 |
| benzenamine | 9.38 |
Show answer
Answer: (b) N-ethylethanamine
The smallest pKb, 3.00, belongs to N-ethylethanamine. Benzenamine, with the largest, is the least basic.
Show answer
Answer: (a) 29
Ratio = 10^(4.75 − 3.29) = 101.46 = 2.88 × 10 ≈ 29. 2.9 drops the power of ten; 1.46 is the pKb difference; 0.035 is the inverse ratio.
Answer and marking scheme
In the gas phase only the +I effect acts, so 3° > 2° > 1°. In water the cation is also stabilised by hydrogen bonding with water: (C₂H₅)₃NH⁺ has only one N–H and its bulky ethyl groups hinder the approach of water, so it is less stabilised by solvation than (C₂H₅)₂NH₂⁺. The interplay of inductive effect, solvation and steric hindrance puts the secondary amine first.
- gas phase: only +I effect, so tertiary strongest (1)
- in water the tertiary cation, one N–H and crowded, is less stabilised by solvation (1)
Answer and marking scheme
In benzenamine the –NH₂ is attached directly to the ring; the unshared pair on nitrogen is in conjugation with the ring and less available for protonation. Aniline has five resonance structures against two for the anilinium ion, so it is stabilised more than its cation and accepts a proton less readily: small Kb, large pKb.
- lone pair in conjugation with the ring, less available (1)
- aniline more resonance-stabilised than anilinium, so less basic (1)
← Chapter 8: Aldehydes, Ketones and Carboxylic AcidsChapter 10: Biomolecules →