${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{C}} = CHCO{\rm{C}}{{\rm{H}}_3}$ can be oxidised to ${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{C}} = CHCO{\rm{O}}{{\rm{H}}_3}$ by:

NEET UG CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

1 Hard

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Understanding +4 | -1

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C} = CHCO\mathrm{C}{\mathrm{H}_3}

can be oxidised to

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{C} = CHCO\mathrm{O}{\mathrm{H}_3}

by:

View answer and explanation

Correct option: D

NaOI

Option D (NaOI) is correct because NaOI (sodium hypoiodite) is used in the haloform reaction, which oxidizes methyl ketones to carboxylic acids.