JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
The correct order of pKₐ values for the following compounds is:
- (a) a>b>c>d
- (b) c > a > d > b
- (c) b>d>a>c
- (d) b > a > d > c
Answer and solution
Answer (c) b>d>a>c
Solution
Chemistry ma’am · 3 steps
Ask what each para group does to the phenoxide ion: a donor makes it weaker, so pKa goes up.
- Link pKa to stability. A phenol is acidic because it loses H⁺ to give a phenoxide ion. Anything that stabilises the negative charge on that ion makes the phenol a stronger acid with a lower pKa. Anything that pushes more electron density onto it makes the acid weaker with a higher pKa.
- Rank the para groups. -NMe₂ is a strong +R donor, so it destabilises the phenoxide the most. Isopropyl is a mild donor through +I and hyperconjugation. Plain phenol has no substituent at all. -NO₂ is a strong -R and -I group, and it spreads the negative charge onto its oxygens.
Compound Para group and its effect Phenoxide ion pKa b -N(CH₃)₂, strong +R donor least stable highest d -CH(CH₃)₂, weak +I donor slightly destabilised a little above phenol a -H, no effect reference about 10 c -NO₂, strong -R, -I most stable lowest - So. The pKa order is b > d > a > c, which is option (c).
The idea to keep. For substituted phenols, electron-withdrawing groups lower pKa and donating groups raise it. A resonance donor like -NR₂ beats an alkyl group, which only donates weakly.
Answer from the NTA final answer key. Solution by PrepWiser.