Recall the trend for $S_N2$ reactions

In an $S_N2$ reaction, the nucleophile attacks the carbon atom from the backside in a single step.

Therefore, the reaction is highly sensitive to steric hindrance around the carbon attached to the leaving group.

Compare the substrates

• $RCH_2X$ = primary alkyl halide
• $R_2CHX$ = secondary alkyl halide
• $R_3CX$ = tertiary alkyl halide

Effect of steric hindrance on $S_N2$ reactivity

Since backside attack becomes more difficult as substitution increases:

$\text{Primary} > \text{Secondary} > \text{Tertiary}$

Hence,

$RCH_2X > R_2CHX > R_3CX$

Match with the given options

• A: $RCH_2X > R_3CX > R_2CHX$ ❌
• B: $RCH_2X > R_2CHX > R_3CX$ ✅
• C: $R_3CX > R_2CHX > RCH_2X$ ❌
• D: $R_2CHX > R_3CX > RCH_2X$ ❌

Final answer

The correct order of decreasing $S_N2$ reactivity is:

$RCH_2X > R_2CHX > R_3CX$

So, Option B is correct.