In the reaction, ${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_3}\xrightarrow{{{\rm{Oxidation}}}}A\xrightarrow{{{\rm{NaOH}}}}B\xrightarrow{{{\rm{Soda}}\;{\rm{lime}}}}C$ The product $C$ is

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{OH}}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{COONa}}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{ONa}}$

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JEE CHEMISTRY MCQ: HYDROCARBONS

1 Easy

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Application +4 | -1

In the reaction,

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_3}\xrightarrow{{\mathrm{Oxidation}}}A\xrightarrow{{\mathrm{NaOH}}}B\xrightarrow{{\mathrm{Soda}\;\mathrm{lime}}}C

The product

C

View answer and explanation

Correct option: B

{\mathrm{C}_6}{\mathrm{H}_6}

Option B (

{\mathrm{C}_6}{\mathrm{H}_6}

) is the correct answer because the reaction sequence involves oxidation of toluene to benzoic acid, followed by conversion to sodium benzoate, and finally decarboxylation to benzene.

JEE CHEMISTRY MCQ: HYDROCARBONS

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