(CH3)3C-CH2Br (neopentyl bromide) is heated in ethanol and, in a separate experiment, treated with NaOEt in ethanol and with NaI in acetone.

Neopentyl bromide has a quaternary carbon next to the $\mathrm{CH_2Br}$ group, which governs everything. (A) is wrong: E2 needs a $\beta$-hydrogen, but the $\beta$-carbon is quaternary and bears none, and the $\mathrm{S_N2}$ route is also blocked, so with ethoxide the reaction is extremely slow and cannot give 2-methylbut-2-ene by E2. (B) is right: under solvolysis the leaving of bromide is accompanied by migration of a methyl group (a $1,2$-methyl shift) which converts the primary centre into the tertiary $\mathrm{(CH_3)_2C^+CH_2CH_3}$ cation; ethanol then traps it to give 2-ethoxy-2-methylbutane. (C) is wrong because the back-side approach is shielded by the three methyl groups on the neighbouring carbon, and neopentyl bromide reacts about $10^5$ times slower than $n$-propyl bromide in $\mathrm{S_N2}$. (D) is wrong because the adjacent carbon carries no hydrogen, so no hydride shift is possible; only a methyl shift can occur.