Answer: (A)

$$\boxed{\text{III} < \text{I} < \text{II} < \text{IV}}$$

Reason:

• III → more steric hindrance → slowest $S_N2$
• I → secondary alkyl chloride
• II → secondary alkyl bromide; Br⁻ is a better leaving group than Cl⁻
• IV → $\alpha$-bromoketone; carbonyl group strongly activates the $S_N2$ reaction → fastest

Correct option: (A) III < I < II < I

V is highest because it is an α-bromo ketone:$$\mathrm{CH_3CO-CH(Br)-CH_2CH_3}$$

The key effect is the carbonyl group (C=O)

Why does $C=O$ increase $S_N2$ rate?

The carbonyl group has a strong −I (electron-withdrawing) effect:$$\mathrm{C=O \rightarrow CH(Br)}$$

It pulls electron density away from the carbon bearing Br, making that carbon more electrophilic.