Question

From each of the following pairs select the compound that will react faster with sodium iodide in acetone
(a) 2-Chloropropane or 2-bromopropane
I II
(b) 1-Bromobutane or 2-bromobutane
I II

• A. (a) - I, (b) - I
• B. (a) - I, (b) - I
• C. (a) - I, (b) - I
• D. (a) - II, (b) - I

Answer 1

The correct option is D (a) - II, (b) - I

$NaI$ in polar aprotic solvent (acetone) favours $S_{N}2$ reaction mechanism.
The $S_{N}2$ reaction involves displacement of a leaving group (usually a halide or a tosylate), by a nucleophile. This reaction works the best with methyl and primary halides because bulky alkyl groups block the backside attack of the nucleophile, but the reaction does work with secondary halides, and will react very slowly with tertiary halides. Also better will be leaving group more will be rate of $S_{N}2$ reaction.
In the following example, the bromine and chlorine are acting as leaving group.

Order of rate $\to r_2>r_1$ because $B{r}^{\ominus }$ is better leaving group than $C{l}^{\ominus }$
Order of rate $\to r_1>r_2$ primary alkylhalide react faster than secondary alkylhalide.