Ethanoic acid undergoes Hell-Volhard Zelinsky reaction but methanoic acid does not, because of

Question

presence of $\alpha$-$\mathrm{H}$ atom in ethanoic acid

absence of $\alpha$-$\mathrm{H}$ atom in ethanoic acid

higher acidic strength of ethanoic acid than methanoic acid

presence of $\alpha$-$\mathrm{H}$ atom in methanoic acid

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