CH_3-C(CH_3)=CH-CH_3 xrightarrow[KMnO_4/Delta | vedclass

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(D) The reaction of $2$-methylbut-$2$-ene $(CH_3-C(CH_3)=CH-CH_3)$ with $NaIO_4/KMnO_4$ involves the oxidative cleavage of the double bond.In this pro

Vedclass · Accepted Answer

$CH_3-COOH + CH_3-C(=O)-CH_3$

Vedclass · Answer

(D) The reaction of $2$-methylbut-$2$-ene $(CH_3-C(CH_3)=CH-CH_3)$ with $NaIO_4/KMnO_4$ involves the oxidative cleavage of the double bond.In this process,the $=CH-CH_3$ part is oxidized to ethanoic acid $(CH_3-COOH)$,and the $CH_3-C(CH_3)=$ part is oxidized to propanone $(CH_3-C(=O)-CH_3)$.Therefore,the products are acetic acid and acetone.

$CH_3-C(CH_3)=CH-CH_3 \xrightarrow[KMnO_4/\Delta]{NaIO_4}$ Product will be

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