The addition of $HI$ to the double bond of propene yields isopropyl iodide and not $n$-propyl iodide because the addition proceeds through the formation of:

Consider compounds $A$, $B$, and $C$ with the following structural formulae:
$A = CH_3 - CH_2 - CH_2 - CH_2 - CH_2 - OH$
$B = CH_2 = CH - CH_2 - CH_2 - CH_3$
$C = HO - CH_2 - CH_2 - CH(OH) - CH_3$
For the conversion of $B$ from $A$, the reagent $(D)$ required is . . . . . . and the structural formula of the product $(E)$ obtained when $C$ undergoes the same reaction using excess reagent $(D)$ is . . . . . . .

The percentage composition of an organic compound $A$ is: carbon = $85.71 \%$ and hydrogen = $14.29 \%$. Its vapour density is $14$. Consider the following reaction sequence:
$A$ $\xrightarrow{Cl_2/H_2O} B$ $\xrightarrow[(ii) H_3O^+]{(i) KCN/EtOH} C$
Identify $C$.

The metal used for the de-bromination reaction of $1, 2$-dibromoethane is:

What is the product of the following reaction?

The product of the reaction between propene and $HBr$ in the presence of a peroxide is: