See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: The Beckmann rearrangement converts a ketoxime to an amide using acid (H2SO4). The key rule is that the group ANTI (trans) to the OH on nitrogen migrates to nitrogen. The group that migrates ends up bonded to nitrogen, and the other group (the one syn to OH) becomes the carbonyl carbon. Step 1: Identify the geometry of each oxime. For the LEFT oxime (gives A): The structure drawn shows the nitrogen double-bonded to the carbon bearing CH3 and Ph. The OH is on nitrogen. In the drawn structure, the lone pair is shown on the same side as a specific substituent. Based on the structural drawing shown, the OH is positioned such that CH3 is anti to OH (trans to OH). In Beckmann rearrangement, the group anti to OH migrates. Therefore CH3 migrates to N... wait, let me reconsider by looking at the actual structural drawings carefully. Step 2: Re-examine the drawings. Left oxime: The carbon bears CH3 (upper left) and Ph (upper right), with N=C and N-OH. The lone pair on N is shown on the left side. The OH is on the right side of N. The group anti to OH on the C=N would be the group on the same side as the lone pair direction. In the left structure, Ph is drawn on the right side of C and OH is on the right side of N, making Ph syn to OH and CH3 anti to OH. The anti group (CH3) does NOT migrate - actually the anti group DOES migrate in Beckmann. So CH3 migrates to N, giving: Ph-C(=O)-NH-CH3. This is product A = Ph-C(=O)-NH-CH3. Right oxime: The carbon bears Ph (upper left) and CH3 (upper right), with N=C and N-OH. Here Ph is on the left side of C and OH is on the right of N, making Ph anti to OH. So Ph migrates to N, giving: CH3-C(=O)-NH-Ph. This is product B = CH3-C(=O)-NH-Ph. Step 3: Match to options. A = Ph-C(=O)-NH-CH3 and B = CH3-C(=O)-NH-Ph corresponds to option (c). Step 4: Why other options fail. (a) Both products identical as Ph-C(=O)-NH-CH3 - ignores geometry difference between the two oximes. (b) Both products identical as CH3-C(=O)-NH-Ph - ignores geometry difference. (d) Reverses A and B: gives CH3-C(=O)-NH-Ph for A and Ph-C(=O)-NH-CH3 for B, which contradicts the anti migration rule as applied to these specific geometric isomers. Therefore, the correct answer is C.