(4) C₄H₁₀O

Explanation:

(4) C₄H₁₀O will have different alkyl group attached with polyvalent functional group that's why show metamerism

CH₃–CH₂–O–CH₂–CH₃

CH₃–O–CH₂–CH₂–CH₃

(3) C₃H₆O

⇒ \(CH_3-\underset{O}{\underset{||}C}-CH_3\)

Only one arrangement possible so can not show metamerism.

(2) C₃H₈O

⇒ CH₃–O–CH₂–CH₃

Only one arrangement possible so can not show metamerism.

(1) No polyvalent functional group in C₅H₁₂, so can not show metamerism.