1. Tardigrade
  2. Question
  3. Chemistry
  4. A hydrocarbon C 6 H 12 decolourizes bromine solution and yields n-hexane on hydrogenation. On oxidation with KMnO 4 it forms two different monobasic acids of the type RCOOH. The compound is

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Q. A hydrocarbon $C _{6} H _{12}$ decolourizes bromine solution and yields $n$-hexane on hydrogenation. On oxidation with $KMnO _{4}$ it forms two different monobasic acids of the type $RCOOH$. The compound is

Hydrocarbons

Solution:

This hydrocarbon cannot be a cycloalkene, because $C _{6} H _{12}$ has only one double bond and gives $n$-hexane on hydrogenation. Hex-1-ene with $KMnO _{4}$ gives pentanoic acid and $1 mol$ of $CO _{2}$.