JEE Main 202623 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 23 January
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Alcohols Phenols and Ethers
JEE Main 2026 · 23 January · Morning Shift · Chemistry · Alcohols Phenols and Ethers
The correct sequence of reagents for the above conversion of X to Y...
The correct sequence of reagents for the above conversion of X to Y is :
Options
- A(i) (\mathrm{NaOH}(\mathrm{aq})); \ \ \ (ii) Jones reagent; (iii) \ \ \ (\mathrm{H}_3 \mathrm{O}^{+})
- B(i) (\mathrm{B}_2 \mathrm{H}_6 / \mathrm{H}_2 \mathrm{O}_2); \ \ \ (ii) NaOEt; \ \ \ (iii) Jones reagent
- C(i) Jones reagent; \ \ \ (ii) NaOEt; \ \ \ (iii) (\mathrm{Hot} \mathrm{KMnO}_4 / \mathrm{KOH})
- D(i) NaOEt; \ \ \ (ii) (\mathrm{B}_2 \mathrm{H}_6 / \mathrm{H}_2 \mathrm{O}_2); \ \ \ (iii) Jones reagent
Correct answer
D. (i) NaOEt; \ \ \ (ii) (\mathrm{B}_2 \mathrm{H}_6 / \mathrm{H}_2 \mathrm{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.
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.