1. Interpret the peptide name

The oligopeptide is: $\text{Alanylglycylphenyl alanyl isoleucine}$

This corresponds to the sequence: $\text{Ala} - \text{Gly} - \text{Phe} - \text{Ala} - \text{Ile}$

So, it is a pentapeptide containing 5 amino acid residues:

• Alanine
• Glycine
• Phenylalanine
• Alanine
• Isoleucine

2. Count the carbonyl carbons in the peptide

Each amino acid residue contributes one carbonyl carbon in the peptide backbone.

A pentapeptide has:

• 4 peptide bonds, and
• 1 terminal carboxyl group

But total carbonyl carbons remain equal to the number of residues: $5$

Each carbonyl carbon $(C=O)$ is $sp^2$ hybridised.

So from the backbone, number of $sp^2$ carbons = $5$

3. Check side chains for additional $sp^2$ carbons

Now inspect each side chain:

• Alanine side chain: $-CH_3$
  No $sp^2$ carbon.

• Glycine side chain: $-H$
  No carbon.

• Phenylalanine side chain: $-CH_2-C_6H_5$
  The benzene ring has 6 carbons, and all are $sp^2$ hybridised.

• Alanine side chain: $-CH_3$
  No $sp^2$ carbon.

• Isoleucine side chain: alkyl group
  No $sp^2$ carbon.

So additional $sp^2$ carbons from side chains = $6$

4. Total number of $sp^2$ hybridised carbons

$\text{Total } sp^2 \text{ carbons} = 5 + 6 = 11$

5. Compare with stored answer

Stored correct answer is: $10$

But the correct count is: $11$

Reason: all 5 carbonyl carbons are $sp^2$, and the phenyl ring contributes 6 more $sp^2$ carbons.

Hence the stored answer appears to be incorrect.