See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The Ei (pyrolytic syn-elimination) reaction proceeds via a concerted cyclic 5-membered transition state in which the hydrogen and the leaving group (here the amine oxide, -N+Me2O-) must be syn-periplanar (cis) to each other. This is the Cope elimination, which is strictly syn-specific. Step 1 - Identify the stereochemistry of starting materials: - Syn isomer: The NMe2(+)O(-) group is on a wedge at C1 and Ph is on a dash at C2. These two substituents are on opposite faces (one up, one down), meaning they are trans to each other on the ring. - Anti isomer: Both NMe2(+)O(-) at C1 (wedge) and Ph at C2 (wedge) are on the same face, meaning they are cis to each other. Step 2 - Determine which H is removed in each case: For Cope elimination (syn-Ei), the oxygen of the amine oxide abstracts a beta-hydrogen that is syn (cis) to it in a 5-membered cyclic TS. For the SYN substrate (C1-NR2O- up, C2-Ph down): - The amine oxide oxygen at C1 must abstract a H from C2 that is on the same face (syn). In the syn isomer, C2 has Ph on the dash (axial-down) face. The H on the same face as the N-oxide at C1 (up/equatorial face) at C2 would be the equatorial H. The elimination occurs toward C2, removing the H syn to the N-oxide. Since Ph is on the opposite face from N-oxide, the double bond forms between C1 and C2, placing Ph on C2 as an allylic substituent. This gives a product where Ph is on the ring carbon adjacent to the double bond - this corresponds to structure (P): 3-phenylcyclohex-1-ene (Ph on C adjacent to double bond, exocyclic-like arrangement, matching P). For the ANTI substrate (C1-NR2O- up, C2-Ph up, same face): - The amine oxide at C1 must abstract a syn-H. The H syn to the N-oxide at C1 must be on the up face. At C2, Ph is already up (same face), so the available syn-H for elimination from C2 is not accessible (Ph is there). Therefore elimination occurs toward C6 (the other beta carbon), removing an H from C6 that is syn to the N-oxide. This gives a double bond between C1 and C6, with Ph remaining on C2 (allylic position next to the new double bond). This corresponds to structure (Q): cyclohex-2-ene with Ph at C1 (Ph directly on the ring adjacent to double bond in the other regiochemical sense, matching Q). Step 3 - Match products: - Syn isomer → A = P (98%) - Anti isomer → B = Q (85%) This matches option (c): A = P, B = Q. Why other options fail: - (a) A=P, B=P: Anti isomer cannot give P because syn elimination from C1 toward C2 is blocked by Ph being syn to N-oxide. - (b) A=Q, B=Q: Syn isomer gives P not Q. - (d) A=Q, B=P: Reversed, incorrect stereochemical analysis. Therefore, the correct answer is C.