Identify the pair of reactants that upon reaction, with elimination of HCl will give rise to the dipeptide Gly-Ala.

The dipeptide Gly-Ala is written as
$$\text{NH}_2{-}CH_2{-}CO{-}NH{-}CH(CH_3){-}COOH$$

Key observations about this structure:

1. The N-terminal (left-hand) residue is Gly (glycine); its $$\text{NH}_2$$ group is free.
2. The peptide (amide) bond is between the carboxyl group of Gly and the amino group of Ala.
3. The carboxyl group of Ala remains free at the C-terminal end.

To prepare this dipeptide in one step with elimination of $$HCl$$, we need the following.

• The carboxyl function of Gly must be activated as an acyl chloride so it can supply the $$\text{CO}$$ part of the peptide bond and furnish the $$\text{Cl}$$ that leaves as $$HCl$$.
• The amino group of Ala must be available as a nucleophile (unprotected $$NH_2$$) so it can attack the activated acyl chloride and form the $$CONH$$ linkage.

Let us inspect each option in this light.

Case A: $$NH_2{-}CH_2{-}COCl$$ (Gly, acyl chloride) and $$NH_2{-}CH(CH_3){-}COOH$$ (Ala, free amine)
• Gly provides the acyl chloride; Ala provides the free $$NH_2$$.
• Nucleophilic acyl substitution gives
$$NH_2{-}CH_2{-}COCl + NH_2{-}CH(CH_3){-}COOH \;\longrightarrow\; NH_2{-}CH_2{-}CO{-}NH{-}CH(CH_3){-}COOH + HCl$$
This product is exactly Gly-Ala.

Case B: $$NH_2{-}CH_2{-}COCl$$ (Gly, acyl chloride) and $$NH_3^+{-}CH(CH_3){-}COCl$$ (Ala, acyl chloride)
• Both molecules are acyl chlorides; there is no free $$NH_2$$ to act as nucleophile.
• Peptide bond formation cannot occur directly, so this pair is unsuitable.

Case C: $$NH_2{-}CH_2{-}COOH$$ (Gly, free amine) and $$NH_2{-}CH(CH_3){-}COCl$$ (Ala, acyl chloride)
• Here the acyl part comes from Ala and the amine from Gly, giving
$$NH_2{-}CH(CH_3){-}CO{-}NH{-}CH_2{-}COOH$$
This is the dipeptide Ala-Gly, not Gly-Ala.

Case D: Both reactants are simple amino acids with carboxyl groups (no acyl chloride).
• Their direct condensation would eliminate $$H_2O$$, not $$HCl$$, and the reaction is not favoured under the stated condition.

Only Case A fulfils all requirements and yields the desired Gly-Ala dipeptide with elimination of $$HCl$$.

Hence, the correct choice is Option A.