${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}\xrightarrow[{{\rm{step}} - 1}]{{{\rm{C}}{{\rm{l}}_2}}}{\rm{C}}{{\rm{H}}_3}{\rm{CHO}}\xrightarrow[{{\rm{step}} - 2}]{{3{\rm{C}}{{\rm{l}}_2}}}{\rm{C}}{{\rm{l}}_3}{\rm{CCHO}}$ In above reactions the role of ${\rm{C}}{{\rm{l}}_2}$ in step-1 and step-2 respectively is

JEE CHEMISTRY MCQ: ALCOHOLS & PHENOLS & ETHERS

1 Easy

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Understanding +4 | -1

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}\xrightarrow[{\mathrm{step} - 1}]{{\mathrm{C}{\mathrm{l}_2}}}\mathrm{C}{\mathrm{H}_3}\mathrm{CHO}\xrightarrow[{\mathrm{step} - 2}]{{3\mathrm{C}{\mathrm{l}_2}}}\mathrm{C}{\mathrm{l}_3}\mathrm{CCHO}

In above reactions the role of

\mathrm{C}{\mathrm{l}_2}

in step-1 and step-2 respectively is

View answer and explanation

Correct option: A

Oxidation, chlorination

Option A (Oxidation, chlorination) is correct because in step-1, chlorine acts as an oxidizing agent, and in step-2, it acts as a chlorinating agent.