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Question: The compound on reaction with \( NaI{O_4} \) in the presence of \( \;KMn{O_4} \) gives ![](https:/...

The compound on reaction with NaIO4NaI{O_4} in the presence of   KMnO4\;KMn{O_4} gives

A) CH3COCH3C{H_3}COC{H_3}
B) CH3COCH3+CH3COOH        C{H_3}COC{H_3} + C{H_3}COOH\;\;\;\;
C) CH3COCH3+CH3CHO  C{H_3}COC{H_3} + C{H_3}CHO\;
D) CH3CHO+CO2C{H_3}CHO + C{O_2}

Explanation

Solution

Hint : Here the reactant is alkene which reacts with Lemieux reagents.   KMnO4\;KMn{O_4} is a strong oxidizing agent which oxidizes alkene to cis-diol and NaIO4NaI{O_4} cleaves the bond between two carbon atoms of the obtained product.

Complete Step By Step Answer:
The given molecular formula is- CH3−C(CH3)=CH−CH3C{H_3} - C\left( {C{H_3}} \right) = CH - C{H_3}
It is also called is isopentane and its IUPAC name is 3−methylbut−2−ene3 - methylbut - 2 - ene
This compound reacts with given reagents- sodium per-iodate and potassium permanganate. . The aqueous solution of both the reagent is known as Lemieux reagent
We have to find the product.
Here,   KMnO4\;KMn{O_4} is oxidizing agent so it oxidizes the alkene and cis-diol product is obtained. The reaction is given as-
CH3−C(CH3)=CH−CH3→KMnO4CH3−C(CH3)(OH)−CH(OH)−CH3C{H_3} - C\left( {C{H_3}} \right) = CH - C{H_3}\xrightarrow{{KMn{O_4}}}C{H_3} - C\left( {C{H_3}} \right)\left( {OH} \right) - CH\left( {OH} \right) - C{H_3}
This obtained product is then cleaved by sodium periodate into aldehyde and ketone. The reaction is given as-
CH3−C(CH3)(OH)−CH(OH)−CH3→−H2ONaIO4CH3COCH3+CH3CHOC{H_3} - C\left( {C{H_3}} \right)\left( {OH} \right) - CH\left( {OH} \right) - C{H_3}\xrightarrow[{ - {H_2}O}]{{NaI{O_4}}}C{H_3}COC{H_3} + C{H_3}CHO
Now, since aldehyde is more reactive than ketone it is further oxidized by KMnO4KMn{O_4} to carboxylic acid. The reaction is given as-
CH3CHO→KMnO4CH3COOHC{H_3}CHO\xrightarrow{{KMn{O_4}}}C{H_3}COOH
The whole reaction can be summarized as-

Hence correct answer is option B.

Note :
Students may go wrong if they think that the reaction stops when aldehyde is formed and choose the option C which is incorrect.
We know that KMnO4KMn{O_4} is a strong oxidizing agent which can oxidize the reagent in any medium whether it is acidic, basic or neutral so it further oxidizes aldehyde to carboxylic acid. KMnO4KMn{O_4} Has following properties-
It is odourless and purple in colour.
It’s aqueous solution, a sweet taste.
It is soluble in water, pyridine, methanol and other organic solvents.
It has a strong oxidizing property so it is used as an oxidant in many reactions.