See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Identify the substrate structure. The compound shown is 3-bromo-3-methylheptane. The carbon bearing Br (C3) is quaternary in terms of substitution: it has a methyl group, a propyl group on one side (C4-C5-C6-C7, i.e., a butyl chain), and an ethyl group on the other side (C2-C1), plus Br. More precisely, looking at the structure: the central carbon has Br, one methyl branch, and is part of a chain. Let me re-read: the structure shows a carbon with Br that has a two-carbon chain on the left (ethyl: CH2CH3) and a three-carbon chain on the right (propyl: CH2CH2CH3), plus a methyl group. So the compound is 3-bromo-3-methylhexane. Actually, re-examining: the left side shows ~2 carbons (making it an ethyl group from the Br carbon) and the right side shows ~3 carbons (propyl). With the methyl branch, this is 3-bromo-3-methylhexane. Step 2: Identify beta-carbons for E2 elimination. For E2, the base abstracts a beta-hydrogen (anti-periplanar to Br). The beta-carbons are those adjacent to the carbon bearing Br (C3): - C2 (part of the ethyl group on the left): CH2 with 2 H's - C4 (part of the propyl group on the right): CH2 with 2 H's - The methyl group carbon: CH3 with 3 H's Step 3: Determine E2 products from each beta-carbon. Elimination toward C2 gives: 3-methylhex-2-ene. C2=C3 double bond. C3 is trisubstituted (methyl, propyl on C3 side; ethyl remnant). The double bond C2=C3 has substituents: on C3 side = methyl and propyl (n-propyl); on C2 side = H and H (from the original CH2). Wait - C2 originally has 2H's and connects to C1 (CH3). So C2=C3: C2 bears H and CH3 (C1), C3 bears methyl and n-propyl. This gives E and Z isomers (cis/trans). That's 2 products. Elimination toward C4 gives: 3-methylhex-3-ene. C3=C4 double bond. C3 bears methyl and ethyl; C4 bears H and ethyl (C5-C6). C3 side: methyl and CH2CH3; C4 side: H and CH2CH2... wait C4 connects to C5-C6 (ethyl if hexane). So C4 bears H and n-propyl minus one = CH2CH3 (if hexane) or CH2CH2CH3. For 3-methylhexane: C4 has CH2, C5 has CH2, C6 has CH3. So C4=C3: C3 has methyl and ethyl (CH2CH3 from C2-C1), C4 has H and propyl (C5C6... no, C4-C5-C6 = ethyl from C4). This alkene also has E and Z isomers. That's 2 products. Elimination toward the methyl group gives: 3-methylenehexane (exocyclic-type, terminal =CH2). This is CH2=C(chain)(chain) — specifically methylenehexane. The product is 3-methylenehexane: CH2=C(CH2CH3)(CH2CH2CH3). This has no E/Z isomerism (one substituent is =CH2). That's 1 product. Step 4: Total E2 products = 2 + 2 + 1 = 5. Step 5: Why other options fail. - (a) 3: undercounts stereoisomers - (b) 4: misses one stereoisomeric pair or one structural product - (d) 6: overcounts Therefore, the correct answer is C.