The reaction of $C_2H_5OH$ with $H_2SO_4$ does not give

$A$ on heating isomerizes to $B$. What is the structure of $B$?

The product $(C)$ in the below mentioned reaction is:
$CH_3-CH_2-CH_2-Br$ $\xrightarrow[\Delta]{KOH_{(alc)}} A$ $\xrightarrow{HBr} B$ $\xrightarrow[KOH_{(aq)}]{\Delta} C$

In the following reaction sequence,the compound $I$ is:
$I (C_3H_6Cl_2)$ $\xrightarrow{KOH_{(aq)}} II$ $\xrightarrow{(i) CH_3MgBr, (ii) H_2O/H^{+}} III$ $\xrightarrow{Anhy. ZnCl_2 + Conc. HCl}$ gives turbidity immediately.

The product of the reaction given below is :

Match the reagents in Column-$I$ with the major products obtained from the reaction of $Butan-1-ol$ in Column-$II$.

Column-$I$ (Reagents)	Column-$II$ (Products)
$A$. $HCl + ZnCl_2$	$i$. $CH_3CH=CHCH_3$
$B$. $HBr$	$ii$. $CH_3CH_2CH_2CH_2Cl$
$C$. $SOCl_2$	$iii$. $CH_3CH_2CH_2CH_2Br$
$D$. $H_2SO_4, \Delta$	$iv$. $CH_3CH_2CH_2CH_2Cl$