JEE Main 2026 · 23 January · morning · Single correct · +4 / −1
The correct sequence of reagents for the above conversion of X to Y is :
- (a) (i) NaOH(aq)
(ii) Jones reagent
(iii) H₃O⁺ - (b) (i) NaOEt
(ii) B₂H₆/H₂O₂
(iii) Jones reagent - (c) (i) Jones reagent
(ii) NaOEt
(iii) Hot KMnO₄/KOH - (d) (i) B₂H₆/H₂O₂
(ii) NaOEt
(iii) Jones reagent
Answer and solution
Answer (b) (i) NaOEt (ii) B₂H₆/H₂O₂ (iii) Jones reagent
Solution
Chemistry ma’am · 4 steps
The oxygen must move from the inner carbon to the end one, so make an alkene first.
- Spot the shift. In X, C₆H₅CH(Br)CH₃, the Br is on the benzylic carbon. In Y, C₆H₅CH₂COOH, the oxygen sits on the other, terminal carbon. Both have 8 carbons, so nothing is lost or added and the functional group only has to move to the end carbon.
- Eliminate, then hydroborate. NaOEt is a strong base, so it removes HBr and gives styrene, C₆H₅CH=CH₂. Hydroboration-oxidation (B₂H₆, then H₂O₂/OH⁻) adds water anti-Markovnikov, so the OH goes to the less substituted end carbon: C₆H₅CH₂CH₂OH.
- Oxidise. Jones reagent (CrO₃ in aqueous H₂SO₄) oxidises a primary alcohol all the way to a carboxylic acid, giving phenylacetic acid.
C₆H₅CH(Br)CH₃ →(NaOEt) C₆H₅CH=CH₂ →(B₂H₆, H₂O₂) C₆H₅CH₂CH₂OH →(Jones) C₆H₅CH₂COOH
- So. The order is NaOEt, then B₂H₆/H₂O₂, then Jones reagent, which is option (b).
The idea to keep. When the functional group has to move from the inner carbon to the end carbon, eliminate to an alkene first. Then hydroboration-oxidation puts the OH on the less substituted carbon.
Answer from the NTA final answer key. Solution by PrepWiser.