See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Friedel-Crafts acylation (AlCl3). Benzene reacts with succinic anhydride under AlCl3 to give 4-oxo-4-phenylbutanoic acid (a keto-acid: Ph-CO-CH2-CH2-COOH). This is compound (A). Step 2: SOCl2 converts the carboxylic acid of (A) to an acid chloride: Ph-CO-CH2-CH2-COCl. Step 3: NaN3 reacts with the acid chloride to give the acyl azide: Ph-CO-CH2-CH2-CON3. This is compound (B). Step 4: MeOH with the acyl azide triggers a Curtius rearrangement upon heating (or the azide rearranges): the acyl azide loses N2 to form an isocyanate, which reacts with MeOH to give a methyl carbamate. However, the ketone is still present. Product after step 4 would be Ph-CO-CH2-CH2-NH-COOCH3 (methyl carbamate of the amine). Step 5: LiAlH4 reduces both the ketone (to alcohol: Ph-CHOH-) and the carbamate (N-COOCH3 to N-CH3), giving compound (C): Ph-CH(OH)-CH2-CH2-NH-CH3. Step 6 & 7: NaH deprotonates the alcohol of (C) to give an alkoxide, which then reacts with 4-chlorobenzotrifluoride (an aryl chloride activated by the para-CF3 group via SNAr) to displace chloride and form the aryl ether. Step 8: Heating (Delta) drives the SNAr reaction to completion. Product (D): Ph-CH(O-C6H4-CF3(para))-CH2-CH2-NH-CH3, which matches option (a). Why other options fail: - (b) has no nitrogen, inconsistent with the NaN3/Curtius/LiAlH4 steps that introduce N-methyl. - (c) has OH retained and shows N-arylation rather than O-arylation; NaH would preferentially deprotonate the alcohol (pKa ~16) not the amine (pKa ~35), so O-alkylation occurs, not N-arylation. Also the CF3 group is meta in (c), but the reagent shown is para-CF3. - (d) has a para-tolyl group on nitrogen and no CF3, inconsistent with the reagent used. Therefore, the correct answer is A.