Question

Consider the following reaction,
$Phenol\stackrel{Zndust}{\to }X\underset{anhyd.Alc{l}_3}{\overset{C{H}_{3}Cl}{\to }}Y\stackrel{alkalineKMn{O}_4}{⟶}Z$
The product Z is:

• A. benzaldehyde
• B. benzoic acid
• C. benzene
• D. toluene

Answer 1

The correct option is B benzoic acid

$Phenol\stackrel{Zndust}{\to }X\underset{anhyd.Alc{l}_3}{\overset{C{H}_{3}Cl}{\to }}Y\stackrel{alkalineKMn{O}_4}{⟶}Z$

Phenol undergoes reduction in the presence of $Zn$ dust and gives benzene i.e X.

Benzene in the presence of Alkyl halide and anhy. $AlC{l}_3$ undergoes Friedel-craft alkylation and gives methyl substituted benzene i.e. toluene as Y.

Toluene in the presence of alkaline $KMn{O}_4$ undergoes oxidation to COOH group resulting in the formation of Benzoic acid.

Overall reaction is represented as:

$\underset{Phenol}{C_6{H}_{5}OH}\stackrel{Zndust}{\to }\underset{benzene}{C_6{H}_6}\underset{anhyd.AlC{l}_3}{\overset{C{H}_{3}Cl}{\to }}\underset{toluene}{C_6{H}_{5}C{H}_3}\stackrel{alkalineKMn{O}_4}{⟶}\underset{BenzoicAcid}{C_6{H}_{5}COOH}$

the product is Benzoic acid.