When acetic acid ($\mathrm{CH_3COOH}$) is treated with $\mathrm{LiAlH_4}$ followed by aqueous workup, the organic product is

1. NCERT §8.6 (Reactions of carboxylic acids — reduction): $\mathrm{LiAlH_4}$ (lithium aluminium hydride) is a strong reducing agent that reduces carboxylic acids ALL THE WAY to primary alcohols. 2. Mechanism overview: $\mathrm{LiAlH_4}$ delivers hydride ($\mathrm{H^-}$) twice — first converting the $\mathrm{C{=}O}$ to a carbonyl intermediate, then the intermediate aldehyde further to the primary alcohol. 3. For acetic acid: $\mathrm{CH_3COOH} \xrightarrow{(i)\,\mathrm{LiAlH_4}; (ii)\,\mathrm{H_2O}} \mathrm{CH_3CH_2OH}$ (ethanol). 4. The carbonyl carbon now has 2 H's and the original methyl → CH$_3$-CH$_2$-OH. Note: the $\alpha$ methyl group is preserved. 5. Option A (acetaldehyde) is the intermediate — but $\mathrm{LiAlH_4}$ does not stop there. Option B (methanol) would require loss of a carbon, which doesn't happen here. Option D (ethane) is hyper-reduction not achievable with LAH. _Source: NCERT Class 12 Chemistry Part 2, Ch 8, §8.6 (Reduction of carboxylic acids with $\mathrm{LiAlH_4}$), p. 20._