In the following reaction sequence,
$Br-CH_2-CH(Br)-Ph$ $\xrightarrow[2. NaNH_2]{1. Alc. KOH} X$ $\xrightarrow[4. Conc. HNO_3/H_2SO_4]{3. HgSO_4/dil. H_2SO_4, Heat} Y$
$X$ and $Y$ respectively are,

Reaction of $o$-toluic acid with $Br_2/Fe$ gives:

Match the reactions in Column-$A$ with the products in Column-$B$ and select the correct option:

Column-$A$	Column-$B$
$(1)$ $CH_3-CH_2-Cl + KOH_{(aq)} \rightarrow$	$(A)$ $1,2-\text{dichloroethane}$
$(2)$ $CH_3-CH_2-CH_2-CH_3 \xrightarrow{AlCl_3/HCl}$	$(B)$ $\text{Chloromethane}$
$(3)$ $CH_3-CH_2-Br + \text{alc. } KOH \rightarrow$	$(C)$ $\text{But-2-ene}$
$(4)$ $CH_2=CH_2 + Cl_2 \rightarrow$	$(D)$ $\text{Ethanol}$
	$(E)$ $\text{Chloroethane}$
	$(F)$ $\text{Ethene}$
	$(G)$ $\text{Isobutane}$

The products obtained in the above reactions $(1)$,$(2)$,and $(3)$ are:

Given below are two statements:
Statement $I$: Due to increase in van der Waals forces, the order of boiling points is $CH_3CH_2CH_2I > CH_3CH_2I > CH_3I$.
Statement $II$: As $para$-dichlorobenzene is more symmetric, its melting point is higher than $ortho$-dichlorobenzene, however its boiling point is lower than $ortho$-dichlorobenzene.
In the light of the above statements, choose the correct answer from the options given below:

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