1. What is the target dipeptide?

   The dipeptide Gly-Ala means:

   • the N-terminal amino acid is glycine: $\mathrm{H_2N-CH_2-COOH}$
   • the C-terminal amino acid is alanine: $\mathrm{H_2N-CH(CH_3)-COOH}$

   So the peptide formed is: $\mathrm{H_2N-CH_2-CO-NH-CH(CH_3)-COOH}$

2. How is a peptide bond formed?

   A peptide bond is formed between:

   • the carboxyl group of the first amino acid, and
   • the amino group of the second amino acid.

   For Gly-Ala, the bond must form as: $\mathrm{Gly-CO-NH-Ala}$

3. Why does the question mention elimination of HCl?

   Elimination of $\mathrm{HCl}$ suggests that one reactant must provide:

   • an acid chloride group $(-\mathrm{COCl})$, and
   • the other reactant must provide an amino group $(-\mathrm{NH_2})$.

   Thus, to obtain Gly-Ala, we need:

   • glycine acid chloride at the carboxyl end, and
   • alanine with free amino group.

4. Required reactants

   The correct pair should therefore be: $\mathrm{H_2N-CH_2-COCl} \quad \text{and} \quad \mathrm{H_2N-CH(CH_3)-COOH}$

   On reaction: $\mathrm{H_2N-CH_2-COCl + H_2N-CH(CH_3)-COOH} \rightarrow \mathrm{H_2N-CH_2-CO-NH-CH(CH_3)-COOH + HCl}$

   This gives the dipeptide Gly-Ala.

5. Comparison with stored answer

   The stored correct answer is C. Since the correct choice must be the option containing glycine acid chloride + alanine, this matches C.

Final Answer: $\boxed{\text{C}}$