JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
The correct order for acidity of the following hydroxyl compound is :
A. CH₃ OH
B. (CH₃)₃ COH
C.
D.
E.
Choose the correct answer from the options given below:
- (a) E > C > D > A > B
- (b) D > E > C > A > B
- (c) E > D > C > B > A
- (d) C > E > D > B > A
Answer and solution
Answer (a) E > C > D > A > B
Solution
Chemistry ma’am · 4 steps
Judge each one by its anion: the better the O⁻ charge is spread or pulled away, the stronger the acid.
- Phenols beat alcohols. In phenoxide the negative charge on oxygen spreads into the ring by resonance. An alkoxide has no ring, so its charge stays stuck on oxygen. That is why C, D and E are all more acidic than A and B.
- Para groups on phenol. -NO₂ pulls electron density out of the ring (−R, −I) and carries the charge onto its own oxygens, so E is the most acidic. -OCH₃ at para pushes electrons in by +R, which crowds the phenoxide charge, so D is a bit weaker than plain phenol C.
- Between the alcohols. Three methyl groups in (CH₃)₃COH push electrons towards oxygen (+I), and the bulky anion is also poorly solvated. So tert-butoxide is less stable than methoxide, and B is weaker than A.
Compound Effect on the anion Approx. pKₐ E: p-nitrophenol −R, −I of NO₂ stabilise it most 7.1 C: phenol ring resonance only 10.0 D: p-methoxyphenol +R of OCH₃ destabilises slightly 10.2 A: CH₃OH no resonance, little +I 15.5 B: (CH₃)₃COH strong +I, poor solvation about 18 - So. The acidity order is E > C > D > A > B, which is option (a).
The idea to keep. For acidity, always look at the conjugate base. Electron-withdrawing groups stabilise the anion and raise acidity; electron-donating groups (+R or +I) destabilise it and lower acidity.
Answer from the NTA final answer key. Solution by PrepWiser.