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JEE CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

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In the given reaction,

(C6H5CO)2O⟶H2OI\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{I}

C6H5COOCOCH3⟶H2OII\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOCOCH}_3 \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow} \mathrm{II}
Identify the product(s) formed in the given reaction.
View answer and explanation
Correct option: B
2 molecules of benzoic acid 1 molecules of benzoic acid and 1 molecule of ethanoic acid
Option D (Al3+\mathrm{Al}^{3+}) is correct because, according to the Hardy-Schulze rule, ions with higher charges have greater coagulating power for colloids.

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