Q. One mole of hydrocarbon (A) reacts with one mole of bromine giving a dibromo compound, $C_5H_{10}Br_2$. Substance (A) on treatment with cold dilute alkaline $KMnO_4$ solution forms a compound $C_5H_{12}O_2$. On ozonolysis (A) gives equimolar quantities of propanone and ethanal. (A) is
Hydrocarbons
Solution:
One mole of the hydrocarbon (A) adds on one mole of bromine to form $C_5H_{10}Br_2$ therefore, (A) must be an alkene having molecular formula $C_5H_{10}$. The position
$\therefore $ The compound (A) is 2-methylbut-2-ene. With alkaline $KMnO_4$, it forms a compound $C_5H_{12}O_2$. of double bond is indicated by ozonolysis as
