What is the effect of doubling the concentration of alkyl halide on the rate of reaction for the given reaction?

$CH_3Br+OH^-→ CH_3OH + Br^-$

Double the rate

The correct answer is Option (1) → Double the rate

The reaction

$\mathrm{CH_3Br + OH^- \rightarrow CH_3OH + Br^-}$

proceeds via an $S_N2$ mechanism.

For an $S_N2$ reaction, the rate law is:

$\text{Rate} = k[\mathrm{CH_3Br}][\mathrm{OH^-}]$

So, if the concentration of alkyl halide (CH₃Br) is doubled while keeping $[\mathrm{OH^-}]$ constant, the rate becomes twice the original value.