In the S_N2 reaction CH3Br + OH- -> CH3OH + Br- the rate law is rate = k[CH3Br][OH-]. If the concentrations of both CH3Br and OH- are doubled at constant temperature, the rate becomes

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In the $S_{N} 2$ reaction $CH X_{3} Br + OH X^{-} CH X_{3} OH + Br X^{-}$ the rate law is rate $= k [CH X_{3} Br] [OH X^{-}]$ . If the concentrations of both $CH X_{3} Br$ and $OH X^{-}$ are doubled at constant temperature, the rate becomes

The SN2 step is bimolecular and the rate is first order in each reactant. Doubling both gives a factor

2 \times 2 = 4