The reagent which can react with $2-$bromopropane to give mainly a substitution product is:

Arrange in decreasing order of the rate of ${\mathrm{S}}_{\mathrm{N}}2$ reaction.

The conversion of $\mathrm{I}$ to $\mathrm{II}$

The major product in the following reaction is :

${\mathrm{S}}_{\mathrm{N}}1$ reactivity of the following halides,

(i) ${\left({\mathrm{CH}}_3\right)}_3\mathrm{CBr}$

(ii) ${\left({\mathrm{C}}_6{\mathrm{H}}_5\right)}_2\mathrm{CHBr}$

(iii) ${\left({\mathrm{C}}_6{\mathrm{H}}_5\right)}_2\mathrm{C}\left({\mathrm{CH}}_3\right)\mathrm{Br}$

(iv) ${\left({\mathrm{CH}}_3\right)}_2\mathrm{CHBr}$

The compound $\mathrm{A}$ in the reaction is:

 $+2\mathrm{Na} \stackrel{\mathrm{dry} \mathrm{ether}}{\to }$ $\mathrm{A}+2\mathrm{NaBr}$ is: