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

The number of all types of isomers of chlorobutane $(C_4H_9Cl)$ is:

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:

The possible products formed in the following reaction are:

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

Assertion : Alkylbenzene is not prepared by Friedel-Crafts alkylation of benzene.
Reason : Alkyl halides are less reactive than acyl halides.