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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Hydrocarbons
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1-Butyne on oxidation with hot alkaline $KMnO_4$ would yield

$\begin{array}{1 1}(a)\;CH_3CH_2CH_2COOH\\(b)\;CH_3CH_2COOH\\(c)\;CH_3CH_2COOH+CO_2+H_2O\\(d)\;CH_3CH_2COOH+HCOOH\end{array}$

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1 Answer

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$CH_3CH_2C\equiv CH\quad\underrightarrow{HOT\;KMnO_4}\quad CH_3CH_2COOH +[HCOOH]\quad\underrightarrow{[O]}\quad CO_2+H_2O$
Thus the ultimate products are $CH_3CH_2COOH,CO_2$ and $H_2O$ and as such option(c) correct.
Hence (c) is the correct option.
answered Mar 21, 2014 by sreemathi.v
 

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