Which of the following will react with $$CHCl_3$$ + alc. KOH?

We have the reagent $$CHCl_3$$ in the presence of alcoholic $$KOH$$. In organic chemistry this specific combination is called the carbylamine test. The underlying reaction is written in general form as

$$R{-}NH_2 \;+\; CHCl_3 \;+\; 3KOH \;\longrightarrow\; R{-}NC \;+\; 3KCl \;+\; 3H_2O$$

It is important to note that the reaction proceeds only when the substrate contains a primary amine group, i.e., a nitrogen atom bonded to exactly one carbon atom and two hydrogen atoms. Secondary and tertiary amines, amides, imides and other nitrogen-containing functional groups do not undergo this reaction.

Now we examine each substance mentioned in the options in terms of the presence or absence of a primary $$NH_2$$ group.

Adenine: In the purine ring of adenine there is an exocyclic $$NH_2$$ group attached to carbon-6. Since this nitrogen is bonded to two hydrogens and one carbon, it is a primary amine and therefore adenine will give the carbylamine reaction.

Thymine: Thymine is a pyrimidine derivative that contains two carbonyl groups and an $$NH$$ in the ring, but no exocyclic $$NH_2$$ group. Hence it lacks a primary amine and will not respond to the carbylamine reagent.

Proline: Proline is a cyclic amino acid in which the nitrogen is incorporated into a five-membered ring (pyrrolidine). That nitrogen is bonded to two carbons and one hydrogen, making it a secondary amine, so the carbylamine reaction is negative for proline.

Lysine: Lysine possesses two separate $$NH_2$$ groups: one attached to the α-carbon (the normal amino acid amino group) and another on the ε-carbon in its side chain. Both are primary amines, so lysine readily undergoes the carbylamine reaction.

Putting these observations together, we see that the compounds giving a positive test are adenine and lysine.

Option C lists exactly these two compounds. None of the other options contains the complete and correct set.

Hence, the correct answer is Option C.