Question

Hydrocarbon A (MF is $C_{5}H_{8}$) gave a white precipitate with ammonical silver nitrate. Oxidation of A with hot alkaline $KMnO_{4}$ gave 2-methyl propanoic acid. What is the structural formula of A?

Answer 1

"Compound A is ${ C }_{ 5 }{ H }_{ 8 }$ on treatment with $Ag{ NO }_{ 3 }/{ NH }_{ 4 }OH$gives white precipitate with triple bond at its terminal position.  

Compound A on treatment with Hot alkaline $KMn{ O }_{ 4 }$ gives${ CH }_{ 3 }CH\left( { CH }_{ 3 } \right) COOH$, 2-,ethyl propionic acid

Therefore structure of A is ${ CH }_{ 3 }\quad -\quad CH\left( { CH }_{ 3 } \right) \quad =\quad CH (triple bond)$

The scheme of the reaction is given below,

Therefore, the structural formula of A is ${ CH }_{ 3 }\quad -\quad CH\left( { CH }_{ 3 } \right) \quad -\quad C\quad \equiv \quad CH\quad \left( 3-methyl-1-butyne \right)$"

Answer 2

Answer:

A has acidic H so it will give precipitate with ammoniacal silver nitrate and ammoniacal cuprous chloride. B is di-alkene and C is cyclic alkene.