The compound CH_3-C(CH_3)=CH-CH_3 on reaction | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The reaction of an alkene with $NaIO_4$ in the presence of $KMnO_4$ (Lemieux-von Rudloff reagent) results in the oxidative cleavage of the $C=C$ d

Vedclass · Accepted Answer

$CH_3COCH_3 + CH_3COOH$

Vedclass · Answer

(D) The reaction of an alkene with $NaIO_4$ in the presence of $KMnO_4$ (Lemieux-von Rudloff reagent) results in the oxidative cleavage of the $C=C$ double bond.In the compound $CH_3-C(CH_3)=CH-CH_3$,the double bond is cleaved.The substituted carbon $CH_3-C(CH_3)=$ is oxidized to a ketone,$CH_3COCH_3$ (acetone).The $=CH-CH_3$ part is oxidized to an aldehyde,which is further oxidized to a carboxylic acid,$CH_3COOH$ (acetic acid),under these conditions.Therefore,the final products are $CH_3COCH_3 + CH_3COOH$.

The compound $CH_3-C(CH_3)=CH-CH_3$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives:

Similar Questions

The reaction of propene with diazomethane in the presence of $UV$ radiation yields ........................

Identify the major product $X$ in the following dehydration reaction: $(CH_3)_3C-CH(OH)-CH_3 \xrightarrow{H_2SO_4} X$.

The product of the following reaction: $CH_2=CH-CCl_3 + HBr$ is:

Which reaction among the following is correct?

Propane-$1$-ol can be prepared from propene by its reaction with

Vedclass Test Series

Exam Paper Generator

Online Exam Module