Answer: C (the O–H bond). Ethanoic acid is a carboxylic acid, CH₃COOH, and it reacts with magnesium like any acid with a reactive metal: \(2\text{CH}_3\text{COOH} + \text{Mg} \to (\text{CH}_3\text{COO})_2\text{Mg} + \text{H}_2\), giving effervescence as hydrogen gas is released.
The acidity comes from the carboxyl group, –COOH. When the acid reacts, it is the O–H bond of this group that breaks: the hydrogen leaves as H⁺ (which pairs up to form H₂), and the remaining ethanoate ion, CH₃COO⁻, bonds to the Mg²⁺ ion to form magnesium ethanoate.
The C–H, C–C and C=O bonds all stay intact throughout, so none of those is broken. Only the O–H bond of the carboxylic acid is broken, which is option C. The key idea is that the acidic proton released by a carboxylic acid comes from the –O–H of the –COOH group.
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