An alkane with molecular formula $\mathrm{C}_5\mathrm{H}_{12}$ can exist in the following isomeric forms : $(I) \underset{\text{n-pentane}}{\mathrm{H_3CCH_2CH_2CH_2CH_3}}$ $(II) \underset{\text{2-methylbutane}}{\mathrm{H_3C-\overset{\overset{\mathrm{CH}_3}{\vert}}{\mathrm{C}}-\mathrm{HCH_2CH_3}}}$ $(III) \underset{\text{2,2-dimethylpropane}}{\mathrm{H_3C-\overset{\overset{\mathrm{CH}_3}{\vert}}{\underset{\underset{\mathrm{CH}_3}{\vert}}{\mathrm{C}}}-\mathrm{CH}_3}}$ Upon photochemical chlorination, the H atom in the alkane is replaced by halogen atom. Looking at the structures we find that 3 different halogen derivatives are possible for (I), while 4 are possible for (II) and only I for (III). (i) All the H atoms in (III) are equivalent. Therefore, we get only a single monochloride from (III). (ii) (I) yields 3 isomeric products due to presence of 3 different H atoms. (iii) (II) yields 4 isomeric products due to presence of 4 different H atoms.