The total number of monohalogenated organic products (including stereoisomers) formed in the following reaction is:
$A$ (simplest optically active alkene) $\xrightarrow[(ii) X_2/\Delta ]{(i) H_2/Ni/\Delta }$

In the given reaction,the products are:

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

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 |

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

Complete the following reactions:
$(1)$ Isopropyl bromide $\xrightarrow[{KOH, \Delta}]{{\text{alcoholic}}} X$ $\xrightarrow[{peroxide}]{{HBr}} Y$
$(2)$ $n$-propyl alcohol $\xrightarrow[{443 \ K}]{{\text{Conc. } H_2SO_4}} X$ $\xrightarrow{{HBr}} Y$