Ethyl alcohol exhibits acidic character on reacting with$:$
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$\ce{CH_3CH_2−OH + Na \rightarrow CH_3−CH_{2}−O^−Na^{+ } + \frac{1}{2} H_2}$ Acid reacts with metal to give hydrogen gas.
Hence, in above reaction $\ce{CH_{3}CH_2OH}$ behaves as an acid.
Hence, in above reaction $\ce{CH_{3}CH_2OH}$ behaves as an acid.
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