Question

Mark the correct increasing order of reactivity of the following compounds with $HBr/HCl$.

• A. $\left(i\right)<\left(ii\right)<\left(iii\right)$
• B. $\left(ii\right)<\left(i\right)<\left(iii\right)$
• C. $\left(ii\right)<\left(iii\right)<\left(i\right)$
• D. $\left(iii\right)<\left(ii\right)<\left(i\right)$

Answer 1

The correct option is C $\left(ii\right)<\left(iii\right)<\left(i\right)$

The reaction is a nucleophilic substitution reaction. The reaction site i.e. $C{H}_2$ is highly hindered thus will follow $S_{N}1$ mechanism. In $S_{N}1$ mechanism, the reaction involves the first slow step of carbocation formation followed by an attack of the incoming nucleophile (Br^-/Cl^-$).

1. $R-OH+HX\to R-\stackrel{+}{O}{H}_2+X^{-}\to R^{+}+H_{2}O+X^{-}$: slow step

2. $R^{+}+X^{-}\to R-X$: fast

since slow step controls the reactivity, therefore stability and ease of formation of the carbocation will decide the reactivity of the molecule.

In case $p-N{O}_2$ group it is strongly electron-withdrawing group by -Inductive and -Mesomeric effect and thus will destabilize the carbocation and will therefore not favour the reaction.

In case $p-Cl$ group it is strongly electron-withdrawing group by -Inductive and also shows +Mesomeric effect due to the presence of its lone pair of electrons. As a result, the electron-withdrawing tendency is relatively lower than $N{O}_2$ group by is still decreases the electron density in the ring. Thus will destabilize the carbocation and will therefore not favour the reaction.

In case of no such destabilizing factor $C_6{H}_{5}C{H}_{2}OH$ will undergo substitution with relative ease.

order of reactivity will be: $p-\left(N{O}_2\right)C_6{H}_{4}C{H}_{2}OH<p-\left(Cl\right)C_6{H}_{4}C{H}_{2}OH<C_6{H}_{5}C{H}_{2}OH$

Therefore the correct increasing order of reactivity of the given compounds with $HBr/HCl$ is $\left(ii\right)<\left(iii\right)<\left(i\right)$