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

Question

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 (A) hex-2-ene (B) hex-3-ene (C) cyclohexene (D) hex-1-ene

Tardigrade Admin · Accepted Answer

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.

Q. A hydrocarbon $C_{6} H_{12}$ decolourizes bromine solution and yields $n$ -hexane on hydrogenation. On oxidation with $K M n O_{4}$ it forms two different monobasic acids of the type $RCOO H$ . The compound is

hex-2-ene

hex-3-ene

cyclohexene

hex-1-ene

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 $K M n O_{4}$ gives pentanoic acid and $1 m o l$ of $C O_{2}$ .

Q. A hydrocarbon C6​H12​ decolourizes bromine solution and yields n-hexane on hydrogenation. On oxidation with KMnO4​ it forms two different monobasic acids of the type RCOOH. The compound is