Question

A compound A with the formula C5H10O give a positive 2,4 DNP test and a negative tollen's test. It can be oxidising to carboxylic acid B of molecular formula C3H6O2, when treated with alk. KMnO4 under vigorous conditions. The salt of B on gives hydrocarbon C on kolbe's electrolytic decarboxylation. Identify A, B and C and write chemical equations.

Answer 1

Hey dear,

● Answer-
A is CH3-CH2-CO-CH2-CH3
B is CH3-CH2-COOH
C is CH3-CH2-CH2-CH3

● Explaination-
A) First compound gives positive DNP test suggests it's aldehyde/ketone. Negative tollens test indicates its a ketone i.e.pentanone.

Thus, compound A is Pentan-3-one (CH3-CH2-CO-CH2-CH3).

B) Pentan-3-one oxidizes to form propanoic acid.
CH3-CH2-CO-CH2-CH3 + [O] ---> CH3-CH2-COOH + CH3-COOH

Hence, compound B is propanoic acid (CH3-CH2-COOH).

C) Propanoic acid on kolbes reaction gives n-butane.
CH3-CH2-COONa ---> CH3-CH2-CH2-CH3

Therefore, Compound C is n-butane (CH3-CH2-CH2-CH3).

Hope this is useful...