See image — Amines Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: HNO2 reacts with primary aliphatic amines to form diazonium ions (R-NH2 + HNO2 → R-N2+), which are highly unstable and immediately lose N2 to generate a carbocation. Step 2 - Identify the carbocation: The NH2 is on the carbon bearing the C(Me)3 group (a neopentyl-type system). Loss of N2 gives a primary carbocation adjacent to a quaternary carbon: the carbocation forms at the carbon that had NH2, i.e., -CH(+)-C(Me)3 with the OH-bearing carbon on the other side. Step 3 - Rearrangement: Primary carbocations are extremely unstable and undergo immediate 1,2-shifts. Here, a methyl group migrates from the adjacent quaternary carbon C(Me)3 to the primary carbocation. This gives a tertiary carbocation at the carbon that was originally C(Me)3, now C(Me)2(+), and the former primary carbocation carbon becomes -CH(Me)-. Step 4 - After methyl migration: The tertiary carbocation is now adjacent to the carbon bearing the OH group. A 1,2-hydride shift then occurs from the carbon bearing OH to the tertiary carbocation, OR alternatively, the OH carbon loses a proton to give a carbonyl. More directly: with the tertiary carbocation now at the carbon next to the C-OH, the OH can migrate or the C-H from C-OH migrates (1,2-hydride shift) to the carbocation, converting C-OH into a carbonyl (C=O). Step 5 - Product: The net result of the methyl migration and hydride shift gives a ketone: the oxygen that was an alcohol becomes a ketone carbonyl, producing propyl-CO-CH2-C(Me)3, which matches option (a). Step 6 - Why other options fail: - (b) and (d) show elimination products (alkenes) which are not the major products in neopentyl-type rearrangements under these conditions. - (c) shows a diol, which would require substitution (SN reaction) at the diazonium carbon; this is not favored over rearrangement with a neopentyl system. - The driving force for rearrangement to give a ketone (via semipinacol-type rearrangement) is strong because it relieves the primary carbocation and forms a stable carbonyl. Therefore, the correct answer is A.