Chemistry Carboxylic Acids MDCAT 2013
PMDC Verified Question 83 of 94
Methyl cyanides, on boiling with mineral acids or alkalies, yield:
A
Butanoic acid
B
Formic acid
C
Acetic acid
D
Propanoic acid
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: Acetic acid
Concept:

The hydrolysis of a nitrile containing "n" carbon atoms always yields a carboxylic acid with the exact same number of "n" carbon atoms.

Formula:

$$ \text{CH}_3\text{C}\equiv\text{N} + 2\text{H}_2\text{O} \xrightarrow{\text{H}^+ / \text{OH}^-} \text{CH}_3\text{COOH} + \text{NH}_3 \text{ (or } \text{NH}_4^+ \text{)} $$

Solution:

  • Methyl cyanide (\( \text{CH}_3\text{CN} \)), also known as acetonitrile, contains 2 carbon atoms.


  • Upon complete hydrolysis, the \(-\text{CN}\) group transforms into a \(-\text{COOH}\) group.


  • The resulting 2-carbon carboxylic acid is acetic acid (ethanoic acid).


Why other options are incorrect:

Formic acid has 1 carbon, propanoic acid has 3 carbons, and butanoic acid has 4 carbons. None of these match the 2-carbon skeleton of methyl cyanide.

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