See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: E2 elimination with EtOH/heat (acting as base/solvent under heating conditions, effectively an E1 or E2 elimination). The substrate is 2-bromo-2,3-dimethylbutane... let me re-read the structure. The starting material is: CH3-C(CH3)2-CH(Br)-CH3, which is 3-bromo-2,2-dimethylbutane (the carbon bearing Br is C3, adjacent to the neopentyl-like quaternary carbon). Actually, let me number carefully: - C1: CH3 - C2: C(CH3)2 (quaternary, two methyl groups) - C3: CH(Br) - C4: CH3 So the compound is 3-bromo-2,2-dimethylbutane. For elimination, the H can be removed from either: 1. C2: but C2 has no H (it is quaternary - attached to two CH3, C1, and C3). So elimination toward C2 is not possible. 2. C4: removing H from C4 (the CH3 at C4) gives the double bond between C3 and C4, producing 2,3-dimethylbut-2-ene: (CH3)2C=C... wait, C3-C4 double bond gives CH3-C(CH3)2-C(=CH2)... no. Wait: C3=C4 double bond: C2(CH3)2 - C3 = C4(H2), this gives (CH3)3C-CH=CH2... that would be 3,3-dimethylbut-1-ene. Alternatively, if we consider the structure again: the Br is on C3 (the CH bearing Br). Beta carbons are C2 (no H, quaternary) and C4 (CH3, three H's available). Elimination toward C4: gives C3=C4 double bond → (CH3)3C-CH=CH2... but C2 has three methyls attached making it (CH3)2C... The product from C3-C4 elimination is: CH3-C(CH3)2-CH=CH2... Wait, Zaitsev's rule states that the more substituted alkene is the major product. Since elimination can only go toward C4 (C2 has no beta-H), the only elimination product is: (CH3)2C(CH3)-CH=CH2 → neopentyl type → 3,3-dimethylbut-1-ene. But the answer is B, which is 2,3-dimethylbut-2-ene, a trisubstituted alkene. This suggests a carbocation rearrangement occurs. Under EtOH/heat, E1 conditions apply. The secondary carbocation at C3 can rearrange via a 1,2-methyl shift from C2 to C3, giving a tertiary carbocation at C2. From this tertiary carbocation at C2, elimination of H from C1 (CH3) or C3 gives the more substituted alkene. After 1,2-methyl shift: carbocation moves to C2, giving: CH3-C+(CH3)-CH(CH3)-CH3 (2-methylbutan-2-yl... with the extra methyl). This tertiary carbocation can lose H from C1 or C3: - Loss from C3: gives CH3-C(CH3)=C(CH3)-CH3 → 2,3-dimethylbut-2-ene (trisubstituted, more stable by Zaitsev) - Loss from C1: gives CH2=C(CH3)-CH(CH3)-CH3 → less substituted Zaitsev product is 2,3-dimethylbut-2-ene, which matches option (b), the trisubstituted internal alkene. Options (a), (c), (d) represent less substituted or non-rearranged products and are not the major product under E1 conditions with rearrangement. Therefore, the correct answer is B.