In S_N2 reaction, the correct order of reactivity for the following compounds:

CH₃Cl,CH3CH₂Cl, (CH₃)₂CHCl and (CH₃)₃CCl is

CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl

The correct answer is option 1. -- CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl

The correct order of reactivity in an $S_{N}2$ (Bimolecular Nucleophilic Substitution) reaction is governed primarily by steric hindrance.

In an $S_{N}2$ mechanism, the nucleophile attacks the electrophilic carbon from the backside (opposite to the leaving group). For this attack to be successful, the nucleophile must be able to approach the carbon atom easily.

The Role of Steric Hindrance

As the number of bulky alkyl groups attached to the reaction center increases, the space available for the nucleophile to attack decreases. This "crowding" significantly slows down the reaction rate.

Reactivity Comparison

1. $CH_{3}Cl$ (Methyl Chloride): The carbon is bonded to three small hydrogen atoms. It offers the least resistance to the incoming nucleophile. (Most Reactive)
2. $CH_{3}CH_{2}Cl$ (Primary Ethyl Chloride - 1°): The carbon is bonded to one relatively bulky methyl group.
3. $(CH_{3})_{2}CHCl$ (Secondary Isopropyl Chloride - 2°): The carbon is bonded to two methyl groups, making the backside attack much more difficult.
4. $(CH_{3})_{3}CCl$ (Tertiary t-Butyl Chloride - 3°): The carbon is surrounded by three bulky methyl groups. The steric hindrance is so high that $S_{N}2$ reactions generally do not occur at all for tertiary halides. (Least Reactive)

The Final Order

The reactivity decreases as the carbon becomes more substituted:

$\text{Methyl} > \text{Primary (1°)} > \text{Secondary (2°)} > \text{Tertiary (3°)}$

Therefore, the correct sequence is:

$CH_{3}Cl > CH_{3}CH_{2}Cl > (CH_{3})_{2}CHCl > (CH_{3})_{3}CCl$