${\rm{CH}}$≡${\rm{CH}}\xrightarrow{{{{\rm{O}}_3}/{\rm{NaOH}}}}X\xrightarrow{{{\rm{Zn}}/{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}}}Y$ is:

${\rm{C}}{{\rm{H}}_2}{\rm{OH}}$—${\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{OH}}$

JEE CHEMISTRY MCQ: ALCOHOLS & PHENOLS & ETHERS

1 Medium

1 of 1

Application +4 | -1

\mathrm{CH}

≡

\mathrm{CH}\xrightarrow{{{\mathrm{O}_3}/\mathrm{NaOH}}}X\xrightarrow{{\mathrm{Zn}/\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}}}Y

is:

View answer and explanation

Correct option: A

\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

—

\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

Option A (

\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

—

\mathrm{C}{\mathrm{H}_2}\mathrm{OH}

) is correct because the reaction sequence involves ozonolysis of ethyne followed by reduction, leading to the formation of ethylene glycol.