One mole of a hydrocarbon `'A'` reacts with `1 mil` of brominne giving a dibromo compound, `C_(5)H_(10)Br_(2)`. Compound `'A'` on treatment with cold dilute alkaline potassium permanganate solution forms a compound, `C_(5)H_(12)O_(2)`. On ozonolysis, `'A'` gives equimolar quantities of propanone and ethanal. Deduce the structural formula of `'A'`

Explanation

One mole of the hydrocarbxon (A) adds on one mole of bromine to form `C_(5)H_(10)Br_(2)`. therefore, (A) must be an alkene having molecular formula `C_(5)H_(10)`. The position of double bond is indicated by ozonolysis as
`underset("Propanone")(CH_(3)- overset(CH_(3))overset(|)C=O)+ underset("Ethanal")(O= overset(H)overset(|)C- CH_(3) to CH_(3))- underset("2-Methylbut-2-enc") (overset(CH_(3))overset(|)C= overset(H)overset(|)C-CH_(3))`
The compound (A) is 2-Methylbut-2-ene. With alkaline `KMnO_(4)` it forms a compound `C_(5)H_(12)O_(2)`
`CH_(3)- overset(CH_(3))overset(|)C= overset(H)overset(|)C-CH_(3) overset("Alk. KMMnO_(4)")to underset(C_(5)H_(12)O_(2))(CH_(3)-underset(OH)underset(|)overset(CH_(3))overset(|)C-underset(OH)underset(|)overset(H)overset(|)C- CH_(3))`