See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: This is an E2 elimination reaction of a cyclopropane ring bearing a quaternary ammonium leaving group (NMe3+). In E2 eliminations the base (NaOEt) abstracts a beta-hydrogen that is anti-periplanar to the leaving group, and the Hofmann-type or Zaitsev selectivity is governed by stereochemistry and the available beta-hydrogens. Step 1 – Identify the substrate. The molecule is a 1,2,3-trisubstituted cyclopropane (actually the ring bears four substituents making it a 1,2,3,3- or 1,1,2,3-substituted cyclopropane). Reading the Newman/circle projection: one carbon (C1) carries H and H; one carbon (C2) carries Ph and Me; one carbon (C3) carries Ph and NMe3+. So the ring is: C1(H,H)–C2(Ph,Me)–C3(Ph,NMe3+) with C1 also bonded to C3. Step 2 – Determine the leaving group and available beta-hydrogens. The leaving group is NMe3+ on C3. The beta-carbons are C2 and C1. Beta-hydrogens available: C2 has no H (it bears Ph and Me); C1 bears two H atoms. Therefore elimination must occur by abstracting one of the H atoms on C1, giving a double bond between C1 and C3 (since NMe3+ is on C3). Step 3 – Apply anti-periplanar (E2) requirement. For E2 on a cyclopropane ring, the H and the leaving group must be anti to each other. From the Newman-style drawing, the H on C1 that is anti to NMe3+ on C3 is available, making C1–C3 the bond that becomes the double bond. Step 4 – Determine the alkene product. The double bond forms between C1 and C3. C1 originally had two H atoms; after losing one H it bears one H. C3 originally had Ph and NMe3+; after losing NMe3+ it bears Ph. C2 is still attached to C1 via ring opening... Wait – in cyclopropane ring-opening E2, the ring opens. The three-membered ring undergoes elimination to give an alkene: C2–C1=C3 connectivity. C1 bears (H, and bond to C2's Ph/Me side), and C3 bears Ph. So the product alkene is: C2(Ph)(Me)=C1(H) with a bond from C1 to... Actually for cyclopropane E2: the C–C bond of the ring between C2 and C3 breaks (the one beta to leaving group and alpha to the H being abstracted), giving an allyl-type alkene. The product is: (Ph)(Me)C=C(H)(Ph), which matches option (c): Ph and Me on one carbon, H and Ph on the other carbon of the double bond. Step 5 – Why other options fail. Option (a) requires four Ph groups total, but the substrate only has two Ph groups. Option (b) requires three Ph groups. Option (d) introduces an OEt group from the solvent which does not occur in simple E2. Only option (c) accounts for both Ph groups and the Me group in the correct arrangement. Therefore, the correct answer is C.