To determine which compound shows metamerism, we need to understand what metamerism is.$\\$Metamerism is a type of isomerism where compounds have the same molecular formula but different alkyl groups on either side of a functional group.$\\$Let's analyze each option:$\\$Option 1: $$C$_5$H$_{12} \\$• This is an alkane with no functional group, so it cannot exhibit metamerism.$\\$Option 2: $$C$_3$H$_8$O$\\$• This is an alcohol or ether. Possible structures are $$CH$_3$CH$_2$CH$_2$OH$$ (propanol) or $$CH$_3$OCH$_2$CH$_3$ (methoxyethane).$\\$• Both structures have no different alkyl groups on either side of the functional group, so no metamerism.$\\$Option 3: $$C$_3$H$_6$O$\\$• This could be an aldehyde or ketone. Possible structures are $$CH$_3$CH$_2$CHO$$ (propanal) or $$CH$_3$COCH$_3$ (acetone).$\\$• Again, no different alkyl groups on either side of the functional group, so no metamerism.$\\$Option 4: $$C$_4$H$_{10}$O$\\$• This could be an ether. Possible structures are $$CH$_3$OCH$_2$CH$_2$CH$_3$ (methoxypropane) and $$CH$_3$CH$_2$OCH$_2$CH$_3$ (ethoxyethane).$\\$• These structures have different alkyl groups on either side of the oxygen atom, showing metamerism.$\\$Therefore, the compound that shows metamerism is $$C$_4$H$_{10}$O$. \\$This corresponds to Option 4.