See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify intermediates A and B. - Ph-CO2H (benzoic acid) reacts with SOCl2 to give A = Ph-COCl (benzoyl chloride, an acid chloride). - Ph-COCl reacts with Me2NH (dimethylamine) to give B = Ph-CON(CH3)2 (N,N-dimethylbenzamide, a tertiary amide). Step 2: Identify what reagent C must do. - Reagent C must convert Ph-CON(CH3)2 (a tertiary amide) into Ph-CHO (benzaldehyde, an aldehyde). - This requires partial reduction of the amide to the aldehyde oxidation state, without over-reducing to the alcohol. Step 3: Evaluate each option. (a) LiAlH4: A powerful, unselective hydride reagent. It reduces amides all the way to amines (Ph-CH2-NMe2), not to aldehydes. Incorrect. (b) NaBH4: A mild reducing agent that does not reduce amides at all under normal conditions. Incorrect. (c) LiAlH(t-BuO)3 (lithium tri-tert-butoxyaluminohydride): A hindered, selective hydride reagent. It reduces acid chlorides to aldehydes and, importantly, reduces tertiary amides selectively to aldehydes via a stable tetrahedral intermediate that does not collapse further under these mild conditions. This is the Bouveault-Blanc-type selective reduction: the bulky reagent delivers one hydride to the amide carbonyl, forming a stable aminol anion intermediate that upon workup gives the aldehyde. This is the classic method for converting a Weinreb amide or tertiary amide to an aldehyde using this reagent. Correct. (d) PCC/CH2Cl2: PCC is an oxidant used to oxidize primary alcohols to aldehydes or secondary alcohols to ketones. It cannot reduce an amide to an aldehyde. Incorrect. Step 4: Conclusion. LiAlH(t-BuO)3 selectively reduces the tertiary amide B (Ph-CONMe2) to benzaldehyde (Ph-CHO) because its bulkiness prevents over-reduction. Therefore, the correct answer is C.