See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: KOBr is a reagent used for the Hofmann bromamide (Hofmann rearrangement) reaction. It converts a primary amide (RCONH2) into a primary amine (RNH2) with loss of one carbon (as CO2). The amide nitrogen migrates to give an isocyanate intermediate which hydrolyzes to an amine. Step 2 - Reaction with KOBr (Hofmann rearrangement): The starting material has a cyclopentane ring with COOH at C1 and CONH2 at C3. KOBr acts on the CONH2 group via Hofmann rearrangement to convert -CONH2 into -NH2 (the carbon of the amide is lost as CO2). So compound (A) is 3-aminocyclopentane-1-carboxylic acid (cyclopentane ring with COOH at C1 and NH2 at C3). Step 3 - Heating (A) with delta: Compound (A) is a gamma-amino acid (the NH2 and COOH are 1,3 on the cyclopentane, making them separated by a suitable number of carbons). On heating, the amino group and the carboxylic acid undergo intramolecular condensation (lactamization) to form a cyclic amide (lactam). The NH2 attacks the COOH, losing water, and forming an amide bond within the ring. This creates a bicyclic lactam: a cyclopentane ring fused with a lactam ring where C(=O)-NH bridges two carbons of the cyclopentane ring. This bicyclic lactam corresponds to option (b), which shows a cyclopentane ring with a C(=O)-NH bridge forming a bicyclic structure. Step 4 - Why other options fail: (a) Shows N-formylcyclobutylamine - there is no cyclobutane in the starting material and no formyl group would form. (c) Shows a bicyclic structure with a CH3 group - no methyl group is present in starting material. (d) Shows an amino ester or amino lactone on a cyclopentane - this would not form under these conditions; heating an amino acid gives a lactam (amide), not a lactone. Therefore, the correct answer is B.