One mole of hydrocarbon (A) reacts with one mole of bromine giving a dibromo compound, C5H10Br2. Substance (A) on treatment with cold dilute alkaline KMnO4 solution forms a compound C5H12O2. On ozonolysis (A) gives equimolar quantities of propanone and ethanal. (A) is

Question

One mole of hydrocarbon (A) reacts with one mole of bromine giving a dibromo compound, C5H10Br2. Substance (A) on treatment with cold dilute alkaline KMnO4 solution forms a compound C5H12O2. On ozonolysis (A) gives equimolar quantities of propanone and ethanal. (A) is (A) 1-pentene (B) 2-pentene (C) 2-methylbut-2-ene (D) 3-methylbut-2-ene

Tardigrade Admin · Accepted Answer

One mole of the hydrocarbon (A) adds on one mole of bromine to form

C_{5} H_{10} B r_{2}

therefore, (A) must be an alkene having molecular formula

C_{5} H_{10}

. The position

∴

The compound (A) is 2-methylbut-2-ene. With alkaline

K M n O_{4}

, it forms a compound

C_{5} H_{12} O_{2}

. of double bond is indicated by ozonolysis as

Q. One mole of hydrocarbon (A) reacts with one mole of bromine giving a dibromo compound, C5​H10​Br2​. Substance (A) on treatment with cold dilute alkaline KMnO4​ solution forms a compound C5​H12​O2​. On ozonolysis (A) gives equimolar quantities of propanone and ethanal. (A) is

1-pentene

2-pentene

2-methylbut-2-ene

3-methylbut-2-ene