The reaction of the compound (CH_3)_2C = CH-C | vedclass

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(D) The reaction of an alkene with $KMnO_4$ and $NaIO_4$ (Lemieux-Johnson reagent or oxidative cleavage conditions) results in the cleavage of the dou

Vedclass · Accepted Answer

$CH_3COCH_3 + CH_3CHO$

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(D) The reaction of an alkene with $KMnO_4$ and $NaIO_4$ (Lemieux-Johnson reagent or oxidative cleavage conditions) results in the cleavage of the double bond.For the alkene $(CH_3)_2C = CH-CH_3$:$1$. The double bond breaks between the two carbons.$2$. The $(CH_3)_2C=$ part is oxidized to acetone,$CH_3COCH_3$.$3$. The $=CH-CH_3$ part is oxidized to acetaldehyde,$CH_3CHO$.Therefore,the products are $CH_3COCH_3$ and $CH_3CHO$.

The reaction of the compound $(CH_3)_2C = CH-CH_3$ with $NaIO_4$ in the presence of $KMnO_4$ yields:

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