Question

Explain the mechanism of alkaline hydrolysis of tert-butyl bromide with energy profile diagram.
Define carbolic acid.
How carbolic acid is prepared from benzene sulphonic acid?

Answer 1

(1) The mechanism of alkaline hydrolysis of $tert-butylbromide$:
First step is the dissociation of $C-Br$ bond. It is slow step or rate determining step.
$(C{H}_3{)}_{3}C-Br\underset{slowstep}{\overset{ratedeterminingstep}{\to }}(C{H}_3{)}_3{C}^{+}+B{r}^{-}$
Second step is formation of $C-OH$ bond. It is fast step.
$(C{H}_3{)}_3{C}^{+}+O{H}^{-}\to (C{H}_3{)}_{3}C-OH$
The energy profile diagram is as shown.
(2) $Carbolicacid$ is also known as $phenol$. It is aromatic hydroxy compound in which one or more hydroxyl group(s) is/are directly attached to the aromatic nucleus (i.e $benzene$ ring).
(3) Preparation of $carbolicacid$from $benzenesulphonicacid$: Neutralisation of $benzenesulphonicacid$ with aqueous $NaOH$ gives $sodiumbenzenesulphonate$.
$Ph-S{O}_{3}H+NaOH\stackrel{neutralisation}{\to }Ph-S{O}_{3}Na+H_{2}O$
$sodiumbenzenesulphonate$ is fused with excess $NaOH$ to form $sodiumphenoxide$
$Ph-S{O}_{3}Na+2NaOH\underset{573K}{\overset{fused}{\to }}Ph-ONa+N{a}_{2}S{O}_3+H_{2}O$
Treatment of $sodiumphenoxide$ with dil $H_{2}S{O}_4$ gives $phenol$
$2Ph-ONa+H_{2}S{O}_4\stackrel{\Delta }{\to }2Ph-OH+N{a}_{2}S{O}_4$