$A$ halide $C_5H_{11}Br$ on treatment with alc. $KOH$ gives $2$-pentene only. The halide will be

Which reaction is faster $(i)$ or $(ii)$ and what is the mechanism?
$(i) \ CH_3CH_2CH_2Br + CH_3ONa \xrightarrow{DMF} CH_3CH_2CH_2OCH_3 + Br^-$
$(ii) \ CH_3CH_2CH_2Br + CH_3SNa \xrightarrow{DMF} CH_3CH_2CH_2SCH_3 + Br^-$

Among the following organic halides,arrange them in the increasing order of their dehydrohalogenation reactions in the presence of alcoholic $KOH$:
$(A)$ $CH_3-CH_2-Br$
$(B)$ $CH_3-CH_2-CH_2-Br$
$(C)$ $CH_3-CH(Br)-CH_3$
$(D)$ $(CH_3)_3-CBr$

Identify $A$ and $B$ in the following reaction:
$C_2H_5Cl \xrightarrow{A} C_2H_5OH \xleftarrow{B} C_2H_5Cl$

The reaction of chloroform with concentrated $HNO_3$ yields ........

$\xrightarrow[\Delta]{KOH \text{ (alc.) }}$ major product $P$
Product $P$ is