Question

For $C{H}_{3}Br+O{H}^{-}\to C{H}_{3}OH+B{r}^{-}$, the rate of reaction is given by the expression is:

• A. $rate=k\left[C{H}_{3}Br\right]$
• B. $rate=k\left[O{H}^{-}\right]$
• C. $rate=k\left[C{H}_{3}Br\right]\left[O{H}^{-}\right]$
• D. $rate=k\left[C{H}_{3}Br{]}^{\circ }\right[O{H}^{-}{]}^{\circ }$

Answer 1

Answer: (C)

$rate=k\left[C{H}_{3}Br\right]\left[O{H}^{-}\right]$

The reaction $C{H}_{3}Br+O{H}^{-}\to C{H}_{3}OH+B{r}^{-}$ proceeds through $S_{N}2$ mechanism. The rate of $S_{N}2$ reaction is given by the expression $r=k\left[S\right]\left[N\right]$. Here, S and N are the substrate and nucleophile respectively. The $S_{N}2$ reaction is second order reaction and its rate depends on the concentration of the substrate as well as the nucleophile. Thus, the rate of conversion of methyl bromide to methanol is given by the expression $rate=k\left[C{H}_{3}Br\right]\left[O{H}^{-}\right]$.