Only two isomeric monochloro derivatives are possible for (only structural)

n-Butane (C₄H₁₀)

Structure: CH₃–CH₂–CH₂–CH₃

Possible Cl substitutions:

CH₃–CH₂–CH₂–CH₂Cl → chlorine on end carbon

CH₃–CHCl–CH₂–CH₃ → chlorine on second carbon

Only two different structures are possible because putting Cl on the 3rd carbon gives the same as putting it on the 2nd (mirror image).

So n-butane → 2 isomers

2-Methylpropane (C₄H₁₀)

Structure:

Only the central carbon is attached to three methyl groups, the other three carbons are equivalent.

Cl can only attach to a methyl group → All three options are the same structurally.

Actually, you can only have one unique Cl substitution, but in some books, they count two possibilities if considering end vs central (technically only one structural isomer).

2,4-Dimethylpentane

More complex, many positions → more than two monochloro derivatives.

Benzene

Only one unique H position for Cl → C₆H₅Cl → only one derivative

Only n-butane and 2-methylpropane give exactly two structural monochloro derivatives.

That’s why answer is A and D.