The correct sequence of the alkyl halides towards $S_{N} 2$ reactions is

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${CH}_{3} Br > {CH}_{3} Cl > {CH}_{3} F$

${CH}_{3} Br < {CH}_{3} Cl > {CH}_{3} F$

${CH}_{3} Br > {CH}_{3} Cl < {CH}_{3} F$

${CH}_{3} Br < {CH}_{3} Cl < {CH}_{3} F$

answer is A.

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Detailed Solution

Since the order of the leaving group is I>Br>Cl>F, the more reactive the alkyl halide will be in $S N_{2}$
Consequently, $C H_{3} I > C H_{3} B r > C H_{3} C l > C H_{3} F$ is the order of alkyl halide reactivity in the $S N_{2}$ reaction.

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