${\rm{CH}} \equiv {\rm{CH}}\xrightarrow{{{{\rm{O}}_3}/{\rm{NaOH}}}}{\rm{X}}\xrightarrow{{{\rm{Zn}}/{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}}}{\rm{Y}}{\rm{.Y}}\;{\rm{is}}$

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

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

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

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

NEET UG CHEMISTRY MCQ: HYDROCARBONS

1 Medium

1 of 1

Application +4 | -1

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

View answer and explanation

Correct option: A

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

The correct answer is

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

because ozonolysis of ethyne followed by reduction yields ethylene glycol.