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Hydrocarbons
Oxidative Cleavage of Alkenes
1 question from 1 paper, each with its answer.
- question
- 1
- paper
- 1
- of Hydrocarbons
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Concepts tested
- Lemieux-von Rudloff oxidation
All Questions
2003 AIPMT 1 question
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The compound $CH_3-C(CH_3)=CH-CH_3$ on reaction with $NaIO_4$ in the presence of $KMnO_4$ gives :$NaIO_4$ 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 $CH_3COCH_3 + CH_3COOH$.