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The hydrocarbon which can react with sodium in liquid ammonia is

$\begin{array}{1 1}(a)\;CH_3CH_2C\equiv CCH_2CH_3\\(b)\;CH_3CH_2CH_2C\equiv CCH_2CH_2CH_3\\(c)\;CH_3CH_2C\equiv CH\\(d)\;CH_3CH=CHCH_3\end{array}$

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Terminal alkynes react with sodium in liquid ammonia to yield ionic compounds (ie) sodium alkylides
Hence (c) $CH_3CH_2C\equiv CH$ is the correct option.
answered Mar 21, 2014 by sreemathi.v

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