See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The starting material is a cyclopentane bearing a gem-dimethyl group at C1, an N(CN3)2 (diazide amine, i.e., a tertiary amine equivalent) at C2 (wedge), H at C3 (wedge), and D at C3 (dash). The two reaction pathways represent Hofmann Elimination and Cope Elimination respectively, both of which are stereospecific eliminations. Step 1 - Identify the amine and stereochemistry: The nitrogen substituent N(CN3)2 is actually better read as N(CH3)2 (a dimethylamino group, common in Hofmann/Cope elimination problems). The nitrogen at C2 is on a wedge (beta/up face). At C3, H is on a wedge and D is on a dash. Step 2 - Pathway A: Hofmann Elimination (1) CH3I quaternizes the amine to give a quaternary ammonium salt. (2) Ag2O converts the iodide to hydroxide. (3) Heat (Delta) causes E2 elimination. Hofmann elimination proceeds via an E2 mechanism, which requires anti-periplanar geometry between the leaving group (NMe3, at C2, wedge/up) and the beta-hydrogen being removed. For anti-periplanar elimination, the H or D at C3 must be anti to the nitrogen leaving group. The nitrogen is on the wedge (up) at C2; therefore the anti-periplanar hydrogen at C3 must be on the dash (down) face, which is D. Thus, D is eliminated anti to the leaving group, giving a product where H remains on the double bond carbon (the vinyl H is H, not D). Product A has H at the vinylic position (no D in the alkene). Step 3 - Pathway B: Cope Elimination (1) H2O2 oxidizes the amine to an N-oxide. (2) Heat (Delta) causes syn (cis) elimination via a 5-membered cyclic transition state. Cope elimination is a syn elimination. The N-oxide at C2 (wedge, up) removes a syn-periplanar hydrogen from C3. The syn hydrogen at C3 is on the wedge (up) face, which is H (not D). Thus, H is eliminated syn to give product B, and D remains on the double bond carbon. Product B has D at the vinylic position. Step 4 - Match to answer choices: Product A: alkene with H (no D) at the new vinylic position -> the structure shows H and H at the double bond end. Product B: alkene with D at the new vinylic position -> the structure shows H and D at the double bond end. This matches option (a): A = 1,1-dimethylcyclopentene with H,H and B = 1,1-dimethylcyclopentene with H,D. Why other options fail: (b) reverses A and B incorrectly. (c) both show H,H - ignores stereospecificity of Cope. (d) both show H,D - ignores stereospecificity of Hofmann. Therefore, the correct answer is A.