Correct statement(s) about nucleophilic substitution is(are)

Option A is correct: $\mathrm{NaI}$ in acetone is a strong nucleophile in a polar aprotic solvent, so the secondary bromide reacts by $\mathrm{S_N2}$ with inversion; because I keeps the same CIP priority order as Br (I > ethyl > methyl > H), inversion converts $(S)$ into $(R)$. Option B is wrong: a tertiary halide in a polar protic solvent ionises ($\mathrm{S_N1}$) to a planar carbocation, giving a racemic alcohol, not complete inversion. Option C is wrong: a tertiary bromide treated with a strong base such as ethoxide undergoes E2 elimination (back-side $\mathrm{S_N2}$ attack at the tertiary carbon is blocked), giving 2-methylpropene as the main product. Option D is wrong: the neopentyl substrate has a quaternary carbon next to the reaction centre, so back-side attack is severely hindered and it reacts far more slowly than bromoethane.