Consider the following reactions, which are carried out at the same temperature.

${\mathrm{CH}}_3-\mathrm{Br}+{\mathrm{OH}}^{\ominus }\stackrel{\mathrm{EtOH}}{⟶}{\mathrm{CH}}_3-\mathrm{OH}+{\mathrm{Br}}^{⊝} \left(\mathrm{i}\right)$

${\mathrm{CH}}_3-\mathrm{Br}+{\mathrm{OH}}^{\ominus }\stackrel{\mathrm{DMSO}}{⟶}{\mathrm{CH}}_3-\mathrm{OH}+{\mathrm{Br}}^{\ominus } \left(\mathrm{ii}\right)$

Which of the following statement is correct about these reactions?

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+\mathrm{HCl}\stackrel{{\mathrm{ZnCl}}_2}{⟶}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}$

In the above reaction, the leaving group is:

What are the products of the following reaction?

In the following reaction, product is:

Which of the following reaction is incorrect?

${\mathrm{S}}_{\mathrm{N}}1$ reaction undergoes through a carbocation intermediate as follows:

$\mathrm{R}-\mathrm{X}\left(\text{ aq. }\right)\stackrel{\text{ Slow }}{\rightleftharpoons }{\mathrm{R}}^{*}\left(\text{ aq. }\right)+{\mathrm{X}}^{-}\left(\text{aq. }\right)\underset{\mathrm{fast}}{\overset{{\mathrm{H}}_2\mathrm{O}}{\to }}\mathrm{ROH}(\mathrm{aq}.)+{\mathrm{H}}^{+}(\mathrm{aq}.)$

$[\mathrm{R}=\mathrm{t}-\mathrm{Bu},\text{ iso-Pr, }\mathrm{Et},\mathrm{Me}](\mathrm{X}=\mathrm{Cl},\mathrm{Br},\mathrm{I})$

The correct statements are,

(I) The decreasing order of rate of ${\mathrm{S}}_{\mathrm{N}}1$ reaction is $\mathrm{t}-\mathrm{BuX}>\mathrm{iso}-\mathrm{PrX}>\mathrm{EtX}>\mathrm{MeX}$.

(II) The decreasing order of ionization energy is $\mathrm{MeX}>\mathrm{EtX}>\mathrm{iso}-\mathrm{PrX}>\mathrm{t}-\mathrm{BuX}$.

(III) The decreasing order of energy of activation is $\mathrm{t}-\mathrm{BuX}>\text{ iso-PrX }>\mathrm{EtX}>\mathrm{MeX}$.

The insecticide chlordane is warmed with dilute $\mathrm{NaOH}$ solution for some time. The expected product would be: