The compound $CH_3-C(CH_3)=CH-CH_3$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives :
A
$CH_3COCH_3$
B
$CH_3COCH_3 + CH_3COOH$
C
$CH_3COCH_3 + CH_3CHO$
D
$CH_3CHO + CO_2$
The answer is B.
Solution
$NaIO_4$ with a small amount of $KMnO_4$ (Lemieux–von Rudloff reagent) cleaves a C=C double bond. $KMnO_4$ first forms a diol, which $NaIO_4$ cleaves; $NaIO_4$ also regenerates the permanganate. The compound is 2-methylbut-2-ene, $(CH_3)_2C=CH-CH_3$. Cleavage at the double bond gives $(CH_3)_2C=O$ from the left part and $CH_3CH=O$ from the right part. The ketone (acetone) is not oxidised further, but the aldehyde fragment is oxidised by permanganate to a carboxylic acid: $CH_3CHO \rightarrow CH_3COOH$. $(CH_3)_2C=CHCH_3 \xrightarrow{NaIO_4/KMnO_4} CH_3COCH_3 + CH_3COOH$ So the products are $CH_3COCH_3 + CH_3COOH$.
ConceptLemieux-von Rudloff oxidation
Hydrocarbons in past papers
75 questions from this chapter have appeared across 26 exam years.
