The reagents required for the following two-step transformation are:

The following reaction is supposed to take place through the $S_N1$ mechanism: $CH_3-C(Ph)(OH)-CH_2-CH_3 \xrightarrow{HI} \text{products}$. If the configuration of the substrate is $D$,then the configuration of the products will be:

Toluene reacts with excess of $Cl_2$ in the presence of sunlight to give a product which on hydrolysis followed by reaction with $NaOH$ gives:

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

What is the major product obtained in the reaction of $1-$phenyl$-2-$bromobutane with $NaOMe$?

Which among the following compounds will undergo the fastest $S_{N}2$ reaction?