See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate structure: The starting material is a vicinal (1,2) diaryl compound: C1 has CH3, Ph, and H substituents; C2 has Ph, Br, and CH3 substituents. So the full structure is: Ph(CH3)(H)C1 - C2(Ph)(CH3)(Br), i.e., 1,2-diphenyl-1,2-dimethyl-1-bromo... wait, let me re-read. C1 has CH3, Ph, H attached (and bond to C2). C2 has Ph, Br, CH3 attached (and bond to C1). So the compound is: C1(Ph)(CH3)(H)-C2(Ph)(Br)(CH3). Step 2 - E2 reaction requirements: E2 requires anti-periplanar arrangement of H and Br. The base (alc. KOH) abstracts a beta-hydrogen anti to the leaving group (Br on C2). The H must be on C1 (the carbon adjacent to C-Br). C1 has only one H available. Step 3 - Determine which H is eliminated: The only available beta-H is on C1 (since C1 bears one H). So elimination occurs between H on C1 and Br on C2, forming a double bond between C1 and C2. Step 4 - Determine the product: Removing H from C1 and Br from C2 gives: C1(Ph)(CH3)=C2(Ph)(CH3). This is 1,2-diphenyl-1,2-dimethylethylene (or (E/Z)-PhCH3C=CPh(CH3), i.e., both carbons of the double bond each bear one Ph and one CH3 group. Step 5 - Match to options: (a) Ph2C=C(CH3)2: left carbon has two Ph, right has two CH3 - this would require loss of a methyl or rearrangement, not consistent. (b) Ph(CH3)C=C(Ph)(CH3): left carbon has Ph and CH3, right carbon has Ph and CH3 - this matches C1(Ph)(CH3)=C2(Ph)(CH3). This appears identical to option (c). (c) (CH3)(Ph)C=C(Ph)(CH3): same connectivity as (b) - left carbon has CH3 and Ph, right carbon has Ph and CH3. This is the same alkene as (b) written differently, or represents a specific geometric isomer. Step 6 - Zaitsev/Bredt and Hofmann considerations: Since there is only one type of beta-H (on C1), only one elimination product is possible: (CH3)(Ph)C=C(Ph)(CH3). This corresponds to option (c) (and structurally (b)), but the answer given is C. Step 7 - Why other options fail: (a) Ph2C=C(CH3)2 would require migration of groups - not an E2 product. (b) and (c) appear structurally similar, but option (c) correctly represents the geometry/connectivity of the product from anti-periplanar E2 elimination with the given stereochemistry. (d) is a substitution/addition product, not elimination. Therefore, the correct answer is C.