Question

Arrange the following compounds in the decreasing order of rate of reaction with $KSH$ in acetone.

• A. (III) > (I) > (II) > (IV)
• B. (II) > (IV) > (I) > (III)
• C. (IV) > (II) > (I) > (III)
• D. (I) > (III) > (II) > (IV)

Answer 1

The correct option is A (III) > (I) > (II) > (IV)

In presence of aprotic polar solvent, the reaction will go by $S_{N}2$ mechanism. In this, the nucleophile will attack the substrate from the back to form a transition state and then the bond between C and leaving group breaks to form the product.

Good leaving group favours the rate of $S_{N}2$ reaction. A good leaving group must have a weak C-X bond.
The bond between $C$ and $I$ is so weak due to the larger size of $I$ and it is highly reactive for $S_{N}2$ reaction.
The $C-F$ bond is too strong to break during $S_{N}2$ reaction.
Therefore, based on the bond strength between $C-X$, the increasing order of preference of alkyl halides for $S_{N}2$ reaction is
$R-F<R-Cl<R-Br<RI$

Hence, the decreasing order of rate of reaction is
(III) > (I) > (II) > (IV)