See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is HOCH2CH2CH2-C(=O)-OCH2CH3, which is ethyl 4-hydroxybutanoate (an ester with a primary alcohol terminus separated by 3 carbons from the ester carbonyl). Step 2: Reaction with PCC to give (A). PCC selectively oxidizes primary alcohols to aldehydes without affecting esters. The -CH2OH end is oxidized to -CHO. So (A) is OHC-CH2CH2-C(=O)-OCH2CH3, i.e., ethyl 4-oxobutanoate (ethyl succinate semialdehyde). Step 3: Reaction with H2C=CHMgBr (vinyl Grignard, 1 molar equivalent) to give (B). The Grignard reagent H2C=CHMgBr adds to the aldehyde carbonyl (aldehydes are more reactive than esters toward Grignard reagents, and with only 1 molar equivalent, addition occurs selectively at the aldehyde). The vinyl group adds to the aldehyde carbon giving an alkoxide at that carbon. So (B) is H2C=CH-CH(OMgBr)-CH2CH2-C(=O)-OCH2CH3. Step 4: Workup with NH4Cl/H2O to give (C). Aqueous ammonium chloride protonates the alkoxide. (C) is H2C=CH-CH(OH)-CH2CH2-C(=O)-OCH2CH3, a secondary allylic alcohol with an ester group. Step 5: KOH/H2O then H3O+ (saponification and acidification). Base hydrolysis (saponification) of the ester followed by acidification gives the carboxylic acid. (C) becomes H2C=CH-CH(OH)-CH2CH2-C(=O)-OH, i.e., 6-hydroxy-7-octenoic acid (a hydroxy acid with a secondary allylic OH and a terminal carboxylic acid). Step 6: Reaction with acetic anhydride (CH3CO)2O / Pyridine to give (D). Acetic anhydride with pyridine acetylates alcohols selectively (not carboxylic acids under mild conditions, but in this context the OH is acetylated). The secondary allylic -OH is converted to -O-C(=O)-CH3 (acetate ester). The carboxylic acid remains as -C(=O)-OH. So (D) is H2C=CH-CH(OC(=O)CH3)-CH2-CH2-C(=O)-OH. This matches option (a): H2C=CH-CH(O-C(=O)-CH3)-CH2-CH2-C(=O)-OH. Why other options fail: - Option (b) has a branched structure with a tertiary alcohol and carboxylic acid on the same carbon, inconsistent with the reaction sequence. - Option (c) places the OH on a tertiary carbon adjacent to the vinyl group and forms an anhydride-like linkage with the carboxylate end, which would require different chemistry. - Option (d) lacks the acetyl group and has the OH in the wrong position. Therefore, the correct answer is A.