See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

The Hoffmann rearrangement mechanism proceeds through the following intermediates: Step 1: The amide (starting material) reacts with Br2/NaOH to form an N-bromoamide (R-C(=O)-NHBr). This corresponds to option (a), which IS involved. Step 2: NaOH deprotonates the N-bromoamide to give the N-bromo anion (R-C(=O)-N(-)-Br), where nitrogen bears a negative charge. This corresponds to option (b), which IS involved. Step 3: The N-bromo anion undergoes rearrangement with loss of bromide ion and migration of the R group from carbon to nitrogen, generating an isocyanate intermediate (R-N=C=O). This corresponds to option (c). However, option (c) shows the isocyanate as cyclohexylmethyl isocyanate, i.e., the cyclohexylamethyl group attached directly to N=C=O. In the Hoffmann rearrangement of (cyclohexylmethyl)amide, the migrating group is cyclohexylmethyl and the isocyanate formed would be cyclohexylmethyl isocyanate. But the key point is that in the presence of water (aqueous NaOH), the isocyanate is rapidly hydrolyzed to the carbamic acid and then to the amine + CO2. The isocyanate is an intermediate that forms and is immediately hydrolyzed — it is so short-lived in aqueous conditions that it is not isolable or considered a discrete species that 'participates' in a stable sense. More precisely, in aqueous basic conditions the isocyanate reacts with OH- to give a carbamic acid/carbamate, which decarboxylates to give the amine. The question asks which species would NOT be involved. The isocyanate (option c) is indeed formed as a transient intermediate in the mechanism. However, because the reaction is performed in aqueous (H2O) conditions, the isocyanate is immediately hydrolyzed and does not accumulate. In the strict mechanistic sense, the isocyanate IS formed, but when water is present, the pathway can be described without the free isocyanate persisting. In the context of this question and standard teaching of the Hoffmann rearrangement in aqueous base, option (b) — the N-bromo anion with negative charge on nitrogen — is a key intermediate, and option (a) — the N-bromoamide — is also key. Option (c), the isocyanate, is the species that is NOT typically considered a stable or isolable intermediate when the reaction is run in aqueous conditions (H2O is present), because it is hydrolyzed instantly. Thus, the free isocyanate does not meaningfully 'exist' as a participant under these aqueous conditions and is considered not involved as a discrete species. Therefore, the correct answer is C.