In the given reaction,$[X]$ will be:

Given below are two statements:
Statement $I$: $C-Cl$ bond is stronger in $CH_2=CH-Cl$ than $CH_3-CH_2-Cl$.
Statement $II$: The given optically active molecule on hydrolysis gives a solution that can rotate the plane polarized light.
In the light of the above statements,choose the correct answer from the options given below.

The product $(X)$ is:

Identify the correct order of reactivity for the following pairs towards the respective mechanism:
$(A)$ $S_{N}2$: Isobutyl bromide > tert-Butyl bromide
$(B)$ $S_{N}1$: Benzyl bromide > $2-$Phenylethyl bromide
$(C)$ Electrophilic substitution: Chlorobenzene > $1-$Chloro$-4-$nitrobenzene
$(D)$ Nucleophilic substitution: Bromobenzene < $1-$Bromo$-4-$nitrobenzene
Choose the correct answer from the options given below:

What happens when
$(i)$ $n$-butyl chloride is treated with alcoholic $KOH$.
$(ii)$ bromobenzene is treated with $Mg$ in the presence of dry ether.
$(iii)$ chlorobenzene is subjected to hydrolysis.
$(iv)$ ethyl chloride is treated with aqueous $KOH$.
$(v)$ methyl bromide is treated with sodium in the presence of dry ether.
$(vi)$ methyl chloride is treated with $KCN$?

Match the reactions in Column-$I$ with their mechanisms in Column-$II$.

Column-$I$ (Reaction)	Column-$II$ (Mechanism)
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$	$(1)$ $S_N2$ mechanism
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$	$(2)$ $S_E2$ aromatic
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$	$(3)$ $S_N1$ mechanism
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$	$(4)$ $\beta$-elimination (dehydrohalogenation)