1. Identify the peptide structure

The dipeptide Gln-Gly means:

• N-terminal amino acid = Glutamine (Gln)
• C-terminal amino acid = Glycine (Gly)

So the peptide skeleton is:

$H_2N-CH(CH_2CH_2CONH_2)-CO-NH-CH_2-COOH$

2. Reagent used: $CH_3COCl$ followed by aqueous work-up

Acetyl chloride reacts readily with free amino groups to give N-acetyl derivatives.

In this dipeptide, the functional groups are:

• free N-terminal amino group: $-NH_2$
• one peptide linkage: $-CO-NH-$
• side-chain amide of glutamine: $-CONH_2$
• terminal carboxylic acid: $-COOH$

3. Which group will react?

The most reactive site toward acetyl chloride here is the free amino group at the N-terminus.

Thus,

$-NH_2 \xrightarrow{CH_3COCl} -NHCOCH_3$

The peptide amide nitrogen and glutamine side-chain amide are not acetylated under this simple treatment because amide nitrogens are much less nucleophilic.

4. Product formed

Therefore the product is the N-acetyl derivative of Gln-Gly:

$CH_3CO-NH-CH(CH_2CH_2CONH_2)-CO-NH-CH_2-COOH$

5. Conclusion

So the correct option must be the one showing acetylation of only the terminal amino group.

Hence, the answer is Option C.