See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: LiAlH4 alone reduces alcohols only in special cases (e.g., allylic/benzylic via SN2 or elimination-addition), but LiAlH4 combined with AlCl3 generates a powerful hydride reagent (alane, AlH3, or related Al-H species) that can perform reductive cleavage of benzylic C–O bonds via an SN1-type or hydride-delivery mechanism. Step 1 – Identify the substrate: The starting material is benzhydrol (diphenylmethanol), Ph2CH–OH. The carbon bearing –OH is benzylic and also secondary with two phenyl groups, making it capable of forming a very stable diphenylmethyl (benzhydryl) carbocation. Step 2 – Role of AlCl3: AlCl3 is a Lewis acid that coordinates to the hydroxyl oxygen, converting –OH into a good leaving group. This facilitates ionization to the highly stable Ph2CH+ (benzhydryl) carbocation. Step 3 – Hydride delivery: LiAlH4 provides hydride (H–). The carbocation Ph2CH+ is captured by H– to give Ph2CH2 (diphenylmethane). Step 4 – Overall transformation: Ph2CH–OH + LiAlH4/AlCl3 → Ph2CH2 (diphenylmethane). This is a reductive deoxygenation (C–O bond cleavage with replacement by C–H). Why other options fail: (a) 'No reaction' is wrong because LiAlH4/AlCl3 is specifically known to cleave benzylic alcohols. (b) Biphenyl would require C–C bond cleavage, which does not occur here. (d) Benzophenone (Ph2C=O) would be an oxidation product, not a reduction product. Therefore, the correct answer is C.