The second-order rate constant for alkene formation from $(\mathrm{CH}{}_3){}_2\mathrm{CHBr}$ with $\mathrm{NaOEt}$ in ethanol at $25{}^{\circ }\mathrm{C}$ is $4.02\times 10^{-4}\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. For $(\mathrm{CD}{}_3){}_2\mathrm{CHBr}$ under the same conditions it is $6.0\times 10^{-5}\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. The best conclusion is

1. A$k_H/k_D=6.7$: the C-H bond is broken in the rate-determining step, together with loss of bromide in one concerted step (E2)
2. B$k_H/k_D=6.7$: ionisation of the C-Br bond is rate determining and a carbocation is formed (E1)
3. C$k_H/k_D=6.7$: the reaction is SN2 at the alpha carbon, with no involvement of the beta C-H bond
4. D$k_H/k_D=1.1$: no C-H bond cleavage occurs in the rate-determining step