Chlorination of substance $A$ gives substance $B$. Substance $B$ reacts with alcoholic $KOH$ to give substance $C$,which decolorizes Baeyer's reagent. Ozonolysis of substance $C$ gives $HCHO$. What is substance $A$?

Match the reactions in List-$I$ with the major products in List-$II$:
| List-$I$ | List-$II$ (Major Product) |
| :--- | :--- |
| $(A)$ $CH_3-CHBr-CH_2Br \xrightarrow{KOH/C_2H_5OH}$ | $(I)$ $1^{\circ}$-alkyl bromide |
| $(B)$ $CH_3-CH_2-CH=CH_2 \xrightarrow{HBr, (C_6H_5CO)_2O_2, \Delta}$ | $(II)$ $2^{\circ}$-alkyl bromide |
| $(C)$ $CH_3CH_2CH_3 \xrightarrow{Br_2, h\nu}$ | $(III)$ Allyl bromide |
| $(D)$ $CH_3-CH=CH_2 \xrightarrow{NBS, \Delta}$ | $(IV)$ Alkenyl bromide |

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,

The compound which is $NOT$ formed when a mixture of $n$-butyl bromide and ethyl bromide is treated with sodium metal in the presence of dry ether is:

The products $(A)$ and $(B)$ respectively 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: