Which product would be expected to predominate in the given reaction?

Match the following lists:

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

What will be the final product in the given reaction?

Complete the following reactions:
$(1)$ $1$-Chloropropane $\xrightarrow[{KOH}]{{\text{ethanol}}} (B)$ $\xrightarrow[{CCl_4}]{{Br_2}} (C)$
$(2)$ Ethane $\xrightarrow[{hv}]{{Cl_2}} (B)$ $\xrightarrow[{Na, \text{ether}}]{} (C)$

In the reaction shown below,the structural formula of $(A)$,$X$ and $Y$ respectively are:
$C_7H_7N_2OCl + C_2H_5OH \rightarrow \text{Anisole} + N_2 + X + Y$

Correct statements regarding alkyl halides ($R-X$) among the following are: A. Alcohol being less polar solvent as compared to water, alcoholic $KOH$ favours elimination reaction with $R-X$. B. Order of reactivity towards $S_N1$ mechanism is $C_6H_5-CH_2-Cl > C_6H_5-CHCl-C_6H_5$. C. Non substituted aryl halides exhibit properties similar to alkyl halides. D. Vinyl chloride is an example of haloalkene and allyl chloride is an example of haloalkyne. E. $R-Cl$ can be prepared by reacting $R-OH$ with $SOCl_2$ but $Ar-Cl$ cannot be prepared by reacting $Ar-OH$ with $SOCl_2$. Choose the correct answer from the options given below: