Question:

The major product ‘B’ of the following reaction sequence is

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Hydroboration oxidation gives anti-Markovnikov alcohol. In terminal alkenes, the \(-OH\) group is added to the less substituted carbon.
Updated On: Jun 26, 2026
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The Correct Option is D

Solution and Explanation

Step 1: Reaction of 1-phenylethanol with HBr.
The given compound is \[ C_6H_5CH(OH)CH_3 \] On reaction with HBr, the \(-OH\) group is replaced by bromine.
So, the product formed is \[ C_6H_5CHBrCH_3 \]

Step 2: Reaction with alcoholic KOH and heat.
Alcoholic KOH causes dehydrohalogenation.
Thus, \[ C_6H_5CHBrCH_3 \] loses HBr to form styrene: \[ C_6H_5CH=CH_2 \] Therefore, \[ A=\text{styrene} \]

Step 3: Hydroboration oxidation of styrene.
Styrene undergoes hydroboration oxidation with \[ B_2H_6 \] followed by \[ NaOH/H_2O_2 \] This reaction gives anti-Markovnikov addition of water.
Hence, the \(-OH\) group attaches to the terminal carbon of the double bond.
So, \[ C_6H_5CH=CH_2 \longrightarrow C_6H_5CH_2CH_2OH \]

Step 4: Final conclusion.
The major product \(B\) is \[ \boxed{C_6H_5CH_2CH_2OH} \] which is 2-phenylethanol.
Hence, the correct option is \[ \boxed{(4)} \]
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