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The compound which shows metamerism is:
A
$C_3H_8O$
B
$C_3H_6O$
C
$C_4H_{10}O$
D
$C_5H_{12}$
Detailed Solution
Metamerism requires different alkyl groups on either side of a polyvalent functional group.
$C_4H_{10}O$: $CH_3-CH_2-O-CH_2-CH_3$ and $CH_3-O-CH_2-CH_2-CH_3$ are metamers.
$C_3H_6O$ (acetone) and $C_3H_8O$ ($CH_3-O-CH_2-CH_3$) allow only one arrangement, and $C_5H_{12}$ has no polyvalent functional group – so they cannot show metamerism.