See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: NBS (N-Bromosuccinimide) reacts with KOBr (potassium hypobromite) under Hofmann-like degradation conditions. The reagent combination NBS/KOBr effectively generates hypobromite (OBr-) in situ, which is used for the Hofmann bromamide degradation (Hofmann rearrangement) of amides to amines. Step 1: Identify the starting material. The question states NBS reacts with KOBr to give product A. The starting material (implied from context and the answer choices) is succinimide or a related four-membered ring amide. Given the answer choices involve four-membered rings, the starting material is likely beta-lactam or a cyclic imide. However, looking at the context more carefully: NBS itself is N-bromosuccinimide. When NBS reacts with KOBr (a base/oxidant system), the Hofmann rearrangement of the amide nitrogen occurs. Step 2: NBS (N-bromosuccinimide) under basic conditions (KOBr provides OBr- and KOH/base) undergoes Hofmann rearrangement. In NBS, the nitrogen bears a bromine. Under basic aqueous conditions, NBS can hydrolyze and rearrange. The N-Br bond in NBS undergoes Hofmann-type rearrangement: the carbonyl adjacent to N-Br loses CO2 (decarboxylation via isocyanate intermediate), converting the amide nitrogen into an amine. Step 3: NBS is the succinimide N-bromo derivative. Hofmann degradation of NBS: The nitrogen with Br in the succinimide ring undergoes rearrangement. One of the C=O groups adjacent to N undergoes Hofmann rearrangement, generating an isocyanate intermediate. The ring opens partially. The product retains a four-membered ring character with -NH2 and -COO- groups. Step 4: The Hofmann rearrangement of NBS (a cyclic imide) with KOBr (base) gives a product where one carbonyl is converted: the nitrogen migrates, losing one carbon as CO2, giving a beta-amino acid derivative. The product is the carboxylate salt form (due to basic KOBr conditions) of 3-aminocyclobutane... actually for a five-membered succinimide ring undergoing Hofmann on one carbonyl: succinimide -> opens to give 3-aminopropionic acid (beta-alanine) carboxylate. But given four-membered ring answers, the starting cyclic imide must be a four-membered one (glutarimide analog smaller). Step 5: The answer is B - the four-membered ring with COO- (carboxylate anion) and NH2. This makes sense because under the basic conditions of KOBr, the carboxylic acid product would be deprotonated to give the carboxylate anion (-COO-), and the Hofmann rearrangement converts one amide C=O to NH2 with loss of CO2, leaving the other carboxylate intact. The basic medium (KOBr is basic) ensures the product exists as the carboxylate salt rather than free acid. Why other options fail: - (a) Shows COOH (protonated acid) - incorrect because KOBr is basic, product would be deprotonated - (c) Shows a cyclic anhydride - Hofmann rearrangement does not produce anhydrides - (d) Shows a beta-lactam (cyclic amide) - this is a starting material type, not the Hofmann product Therefore, the correct answer is B.