The compound that will undergo $S_N 1$ reaction with the fastest rate is

The product formed on the reaction of ethyl alcohol with bleaching powder is:

The major product obtained in the following conversion is

What is the major product of the following reaction?
$CH_2=CH-CH_2-CH(Br)-CH_3 + KOH \xrightarrow[\text{heat}]{\text{alcohol}} \dots$

Which compound in each of the following pairs will react faster in $S_N2$ reaction with $OH^{-}$?
$(i)$ $CH_3-Br$ or $CH_3-I$
$(ii)$ $CH_3-C(CH_3)_2-Cl$ or $CH_3-Cl$

Which of the following haloalkanes reacts with aqueous $KOH$ most easily? Explain giving reason.
$(i)$ $1$-Bromobutane
$(ii)$ $2$-Bromobutane
$(iii)$ $2$-Bromo-$2$-methylpropane
$(iv)$ $2$-Chlorobutane