When $(\mathrm{CH}{}_3){}_3\mathrm{C}-\mathrm{Br}$ is dissolved in 50% aqueous ethanol it gives both $(\mathrm{CH}{}_3){}_3\mathrm{C}-\mathrm{OH}$ and $(\mathrm{CH}{}_3){}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{}_2\mathrm{H}{}_5$. The most direct inference is that

1. Awater attacks the bromide by ${\mathrm{S}}_{\mathrm{N}}2$ and ethanol attacks by ${\mathrm{S}}_{\mathrm{N}}1$
2. Bbackside attack by ethanol is faster than by water, giving a retention product
3. Cthe rate-determining ionisation gives a carbocation that is captured by either water or ethanol of the solvent
4. Dthe substrate first eliminates and then adds the solvent