Question

In a $S_{N}2$ substitution reaction of the type

$R-Br+C{l}^{-}\stackrel{DMF}{\to }R-Cl+B{r}^{-}$

Which one of the following has the highest relative rate?

• A.
• B. $C{H}_{2}C{H}_{2}C{H}_{2}C{H}_{2}Br$
• C. $(C{H}_3{)}_{3}C-Br$
• D.

Answer 1

The correct option is B

$C{H}_{2}C{H}_{2}C{H}_{2}C{H}_{2}Br$

For $S_{N}2$ reaction, as the steric crowding decreases, rate of reaction increases. So reactivity order of alkyl halides follow the order:
$1^{\circ }\text{alkyl halide}>2^{\circ }\text{alkyl halide}>3^{\circ }\text{alkyl halide}$$C{H}_{2}C{H}_{2}C{H}_{2}C{H}_{2}Br$ is the least crowded and $1^{\circ }\text{alkyl halide}$, it is the most reactive in given options.
Hence, (b) is the correct option.