Question

Match the following reactants in Column I with the corresponding reaction products in Column II.

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(A\right)& \text{Benzene}& \left(i\right)& \text{Benzoic acid}\\ & +C{l}_2\stackrel{AlC{l}_3}{\to }& & \\ \left(B\right)& \text{Benzene}& \left(ii\right)& \text{methyl}\\ & +C{H}_{3}Cl\stackrel{AlC{l}_3}{\to }& & \text{phenyl}\\ & & & \text{ketone}\\ \left(C\right)& \text{Benzene}& \left(iii\right)& \text{Toluene}\\ & +C{H}_{3}COCl\stackrel{AlC{l}_3}{\to }& & \\ \left(D\right)& +C{H}_{3}COCl\stackrel{KMn{O}_4/NaOH}{\to }& \left(iv\right)& \text{Chlorobenzene}\\ & & \left(v\right)& \text{Benzene}\\ & & & \text{hexachloride}\end{array}$

Answer 1

Correct answer:

$\left(A\right)\to \left(iv\right),\left(B\right)\to \left(iii\right),\left(C\right)\to \left(ii\right),\left(D\right)\to \left(i\right)$

* The first reaction shows benzene undergoes chlorination.

* The second reaction is Friedel-Craft alkylation.

* The third reaction exhibits friedel-Craft acylation of
benzene.

* In the fourth reaction shown below, toluene undergoes
oxidation with $KMn{O}_4$ to form benzoic acid.


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