Which one among the following will undergo a faster $S_{N}1$ reaction?

For the following compounds,the correct statement$(s)$ with respect to nucleophilic substitution reactions is(are):
$I$: Benzyl bromide $(C_6H_5CH_2Br)$
$II$: Cyclohexylmethyl bromide $(C_6H_{11}CH_2Br)$
$III$: tert-Butyl bromide $((CH_3)_3CBr)$
$IV$: $1-$Phenylethyl bromide $(C_6H_5CH(CH_3)Br)$
$A$. $I$ and $III$ follow $S_{N}1$ mechanism
$B$. $I$ and $II$ follow $S_{N}2$ mechanism
$C$. Compound $IV$ undergoes inversion of configuration
$D$. The order of reactivity for $I$,$III$ and $IV$ is: $IV > I > III$

Reaction $(A)$:
$Ph-CH=CH-CH_3 \xrightarrow[H_2O_2 / OH^-]{B_2H_6 / THF}$ products.
Reaction $(B)$:
$CH_3-CH=CH-Ph \xrightarrow[NaBH_4]{Hg(OAc)_2 / H_2O}$ products.
Correct statement is:

Which of the following is most reactive toward $S_N2$ reaction?

The major product obtained in the following conversion is

$1-$chlorobutane reacts with alcoholic $KOH$ to form