Question
The correct sequence of reagents for the above conversion of X to Y is :
JEE Main
Chemistry
Alcohols Phenols and Ethers
2026
JEE Main 2026 (23 January Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. (i) \( NaOH ( aq )\); \ \ \ (ii) Jones reagent; (iii) \ \ \ \( H _3 O ^ + \)
- B. (i) \( B _2 H _6 / H _2 O _2\); \ \ \ (ii) NaOEt; \ \ \ (iii) Jones reagent
- C. (i) Jones reagent; \ \ \ (ii) NaOEt; \ \ \ (iii) \( Hot KMnO _4 / KOH \)
- D. (i) NaOEt; \ \ \ (ii) \( B _2 H _6 / H _2 O _2\); \ \ \ (iii) Jones reagent
Answer
D. (i) NaOEt; \ \ \ (ii) \( B _2 H _6 / H _2 O _2\); \ \ \ (iii) Jones reagent
Step-by-step solution
The transformation of (X) benzene with Br and CH₃ substituents to (Y) benzene with CH₂COOH requires: First, NaOEt causes elimination of HBr to form an alkene intermediate. Second, B₂H₆/H₂O₂ performs hydroboration-oxidation, adding hydroxyl group in anti-Markovnikov fashion to yield a primary alcohol. Third, Jones reagent oxidizes the primary alcohol to carboxylic acid. The sequence NaOEt → B₂H₆/H₂O₂ → Jones reagent produces the desired carboxylic acid product.
Related: Chemistry — Alcohols Phenols and Ethers · All PYQ Banks