See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When an alkyl halide reacts with a strong base like NaOMe (sodium methoxide), elimination (E2) is favored over substitution, especially when the base is bulky or when there are multiple beta-hydrogens available. Step 1: Identify the substrate. 1-phenyl-2-bromobutane has the structure: Ph-CH2-CHBr-CH2-CH3. The bromine is at C2, with a phenyl group at C1. Step 2: Identify possible elimination pathways. There are two sets of beta-hydrogens: - Beta to C2 on the C1 side: the CH2 adjacent to phenyl (C1 hydrogens) — removal gives Ph-CH=CH-CH2-CH3, i.e., 1-phenylbut-2-ene (a trisubstituted or disubstituted alkene conjugated... wait, actually not conjugated with phenyl) - Beta to C2 on the C3 side: CH2 of the propyl chain (C3 hydrogens) — removal gives Ph-CH2-CH=CH2... no, that would be Ph-CH2-CH=CHCH3... wait, let me re-examine. Step 3: Re-examine structure. 1-phenyl-2-bromobutane: C1 is PhCH2-, C2 is CHBr, C3 is CH2, C4 is CH3. So structure is Ph-CH2-CHBr-CH2-CH3. - Elimination toward C1: removes H from C1 (PhCH2-), giving Ph-CH=CH-CH2-CH3 = 1-phenylbut-2-ene. This alkene is conjugated with the phenyl ring, making it more stable. - Elimination toward C3: removes H from C3, giving Ph-CH2-CH=CH-CH3 = 1-phenylbut-2-ene... no, that gives Ph-CH2-CH=CHCH3 which is 1-phenylbut-2-ene from C3 side. Actually removing H from C1 gives Ph-CH=CH-CH2CH3 (double bond between C1-C2), and removing H from C3 gives Ph-CH2-CH=CH-CH3 (double bond between C2-C3). Step 4: Stability comparison. Ph-CH=CH-CH2CH3 (double bond conjugated with phenyl, between C1-C2) is a styrene-type conjugated alkene, making it significantly more stable than the non-conjugated Ph-CH2-CH=CH-CH3. Step 5: Stereochemistry. The conjugated product Ph-CH=CH-CH2CH3 is (E)-1-phenylbut-1-ene (option a), where the phenyl and ethyl groups are trans. E2 elimination with anti-periplanar geometry and the conjugation with phenyl makes (E)-1-phenylbut-1-ene the major product. Step 6: Why other options fail: - Option (b) and (d): 1-phenylbut-2-ene (double bond at C2-C3) is not conjugated with phenyl, less stable, minor product. - Option (c): Substitution product (SN2) is disfavored with NaOMe (strong base, moderate nucleophile), and elimination dominates. Therefore, the correct answer is A.