Consider the following reactions of $\mathrm{CH}{}_3\mathrm{Br}$ and $(\mathrm{CH}{}_3){}_3\mathrm{CBr}$ with aqueous $\mathrm{NaOH}$ at fixed temperature. Assume $\mathrm{CH}{}_3\mathrm{Br}$ reacts only by the SN2 path and $(\mathrm{CH}{}_3){}_3\mathrm{CBr}$ only by the SN1 path. Select the correct statements.

1. ADoubling $[\mathrm{OH}{}^{-}]$ leaves the rate of hydrolysis of $(\mathrm{CH}{}_3){}_3\mathrm{CBr}$ unchanged.
2. BHydrolysis of $\mathrm{CH}{}_3\mathrm{Br}$ proceeds through a free $\mathrm{CH}{}_3{}^{+}$ carbocation intermediate.
3. CDoubling $[\mathrm{OH}{}^{-}]$ doubles the rate of hydrolysis of $\mathrm{CH}{}_3\mathrm{Br}$.
4. DDoubling the concentration of the halide doubles the rate in both cases.