JEE Chemistry 2026 Question with Solution

Given: Conversion of X to Y is shown.

Find: The correct sequence of reagents.

The solution states that the required strategy is to first form an alkene, then carry out anti-Markovnikov hydration, and finally oxidize the resulting primary alcohol.

1. NaOEt causes dehydrohalogenation of the alkyl halide to give an alkene.
2. $B_2H_6/H_2O_2$ performs hydroboration-oxidation, giving the anti-Markovnikov alcohol.
3. Jones reagent then oxidizes the primary alcohol to the carboxylic acid.

The hint also states that Jones reagent $\left(CrO_3/H_2SO_4\right)$ is a strong oxidant that converts a primary alcohol into a carboxylic acid.

Therefore, the correct option is B.

Given: X is an alkyl halide attached to a benzene side chain, and Y is the corresponding side-chain carboxylic acid.

Find: Which reagent sequence achieves this transformation.

Identify principle: The product requires oxidation at the terminal carbon, so an intermediate primary alcohol is needed before final oxidation.

Apply the sequence shown in the solution:

• Step 1: NaOEt performs elimination to form an alkene.
• Step 2: $B_2H_6/H_2O_2$ gives anti-Markovnikov addition, producing a primary alcohol.
• Step 3: Jones reagent oxidizes that primary alcohol completely to the carboxylic acid.

Hence the sequence is (i) NaOEt (ii) $B_2H_6/H_2O_2$ (iii) Jones reagent.

Therefore, the correct option is B.