CH_3-C equiv C-CH_3 xrightarrow{H^{+}/KMnO_4, | vedclass

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Question

(B) The reaction of $CH_3-C \equiv C-CH_3$ with acidic $KMnO_4$ under heating conditions causes oxidative cleavage of the alkyne to form two molecules

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$CH_3-CH_2-OH$

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(B) The reaction of $CH_3-C \equiv C-CH_3$ with acidic $KMnO_4$ under heating conditions causes oxidative cleavage of the alkyne to form two molecules of acetic acid $(A)$: $CH_3-C \equiv C-CH_3 \xrightarrow{H^{+}/KMnO_4, \Delta} 2 CH_3-COOH$ Next,the reduction of acetic acid $(CH_3-COOH)$ using $LiAlH_4$ followed by acidic workup $(H^{+})$ yields ethanol $(B)$: $CH_3-COOH \xrightarrow{(1) LiAlH_4, (2) H^{+}} CH_3-CH_2-OH$

$CH_3-C \equiv C-CH_3$ $\xrightarrow{H^{+}/KMnO_4, \Delta} A$ $\xrightarrow{(1) LiAlH_4, (2) H^{+}} B$
$B$ will be

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$CH_3-C \equiv C-CH_3$ $\xrightarrow{H^{+}/KMnO_4, \Delta} A$ $\xrightarrow{(1) LiAlH_4, (2) H^{+}} B$ $B$ will be