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$\mathrm{CH} \equiv \mathrm{CH} \frac{\mathrm{HCl}}{\text { Excess }} A \frac{\text { Hydrolysis }}{\Delta} B \frac{\text { Reduction }}{\mathrm{Na}-\mathrm{Hg} / \mathrm{H}_{2} \mathrm{O}} C$
Compound $C$ is
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Compound $C$ is
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ethanol
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