See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the compound and conditions: 1-phenylethanol is C6H5-CH(OH)-CH3, a secondary benzylic alcohol. It is placed in acidified aqueous medium (H3O+/H+). Step 2 - Mechanism in acid: Under acidic conditions, the hydroxyl group of 1-phenylethanol gets protonated to form -OH2+, which is a good leaving group. Loss of water then generates a carbocation at the benzylic carbon: C6H5-CH(+)-CH3, a secondary benzylic carbocation (carbonium ion). Step 3 - Why the carbocation causes racemization: The benzylic carbocation is sp2-hybridized and planar. Water can attack from either face (re or si face) with equal probability, regenerating the alcohol as a 50:50 mixture of R and S enantiomers — i.e., a racemate. This explains the loss of optical activity. Step 4 - Why other options fail: - (a) Enolization requires a carbonyl group alpha to a stereocenter; 1-phenylethanol is an alcohol, not a carbonyl compound, so enolization is not applicable. - (c) Carbanion formation would require base, not acid, and secondary benzylic carbanions are not favored under acidic aqueous conditions. - (d) Reversible oxidation-reduction has no basis here; there is no oxidizing or reducing agent present, and simple acid-catalyzed racemization does not involve redox chemistry. Step 5 - Conclusion: The acidic medium protonates the OH, facilitating departure as water to form a planar benzylic carbonium ion, which upon re-attack by water from both faces gives racemic 1-phenylethanol. Therefore, the correct answer is B.