See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Bromolactonization is an electrophilic cyclization where an unsaturated carboxylic acid reacts with Br2 to form a cyclic ester (lactone). The carboxylate oxygen acts as an internal nucleophile attacking the bromonium ion intermediate. Step 1: Identify the substrate. The starting material is pent-4-enoic acid: CH2=CH-CH2-CH2-COOH. It has a terminal alkene at C4-C5 and a carboxylic acid at C1. Step 2: NaHCO3 deprotonates the carboxylic acid to form the carboxylate anion (RCO2-), which is a better nucleophile. Step 3: Br2 reacts with the terminal double bond to form a bromonium ion intermediate at C4-C5. Step 4: The carboxylate oxygen (from C1) attacks the bromonium ion intramolecularly. For a 5-membered lactone (gamma-lactone), the oxygen attacks C4 (the internal carbon of the bromonium, which is more electrophilic due to Markovnikov-like selectivity and ring-size preference). This gives a 5-membered ring: O-C(=O)-CH2-CH2-CH(ring-O)-CH2Br, i.e., the oxygen bridges C1 carbonyl to C4, and Br ends up on C5 (terminal carbon) as -CH2Br. Step 5: The product is a gamma-butyrolactone with a -CH2Br group at the carbon adjacent to the ring oxygen. This corresponds to a 5-membered ring lactone with a bromomethyl (CH2Br) substituent, which is option (b). Why other options fail: - (a) shows an OH group instead of Br — this would require water as nucleophile, not bromolactonization. - (c) is a simple dibromide addition product (open chain), no lactonization occurred. - (d) shows a -CH2CH2OH side chain on a lactone — this would require a different carbon skeleton and does not arise from pent-4-enoic acid under these conditions. The anti addition of bromonium and the intramolecular attack by the carboxylate on the terminal (less substituted) carbon gives the 5-membered lactone with the -CH2Br group at the carbon bearing the ring oxygen. Therefore, the correct answer is B.