In S$$_N$$2 reactions, the correct order of reactivity for the following compounds: CH$$_3$$Cl, CH$$_3$$CH$$_2$$Cl, (CH$$_3$$)$$_2$$CHCl and (CH$$_3$$)$$_3$$CCl is:

In bimolecular nucleophilic substitution, written as $$S_N2$$, a nucleophile attacks the electrophilic carbon from the side opposite to the leaving group and forms the transition state in a single concerted step. Because both the nucleophile and the substrate appear in the rate-law expression, the experimentally observed rate is

$$\text{Rate}=k\,[\text{Substrate}]\,[\text{Nucleophile}].$$

This one-step mechanism requires that the nucleophile approach the carbon atom easily; therefore, any crowding (steric hindrance) around that carbon slows the reaction. The guiding principle is:

$$\text{Less steric hindrance}\;\Rightarrow\;\text{Faster }S_N2\;\text{reaction}.$$

For alkyl chlorides we classify the carbon bearing chlorine as

$$\begin{aligned} &\text{Methyl} &:&\; CH_3Cl \\ &\text{Primary} &:&\; CH_3CH_2Cl \\ &\text{Secondary} &:&\; (CH_3)_2CHCl \\ &\text{Tertiary} &:&\; (CH_3)_3CCl \end{aligned}$$

The steric bulk increases in the same sequence, so the ease of backside attack decreases accordingly. Translating this qualitative idea into a quantitative rate sequence, we have

$$\text{Methyl} > \text{Primary} > \text{Secondary} \gg \text{Tertiary}.$$

Substituting the given compounds into this general order yields

$$CH_3Cl \; \big( \text{methyl} \big) \; > \; CH_3CH_2Cl \; \big( \text{primary} \big) \; > \; (CH_3)_2CHCl \; \big( \text{secondary} \big) \; > \; (CH_3)_3CCl \; \big( \text{tertiary} \big).$$

Comparing our derived sequence with the four options, only Option B lists the order exactly as obtained.

Hence, the correct answer is Option B.