JEE Chemistry 2023 Question with Solution

Given: The oligopeptide is Alanylglycylphenylalanylisoleucine.

Find: The number of $sp^2$ hybridised carbon atoms.

From the given structure, identify carbon atoms that are $sp^2$ hybridised.

The following carbons are $sp^2$ hybridised:

1. The carbonyl carbon of the peptide bond between alanine and glycine.
2. The carbonyl carbon of the peptide bond between phenylalanine and alanine.
3. The carbonyl carbon of the carboxyl group at the C-terminal of isoleucine.
4. The carbons of the phenyl group in phenylalanine.

Hence, counting all the $sp^2$ hybridised carbons in the structure:

• One from each of the peptide bonds.
• One from the carboxyl group.
• The aromatic ring carbons in phenylalanine.

Thus, the total number of $sp^2$ hybridised carbons is $10$.

Therefore, the required answer is $10$.

Given: The peptide contains peptide carbonyl groups, one terminal carboxyl group, and one phenyl ring.

Find: The total number of $sp^2$ hybridised carbons.

A quick way is to count standard $sp^2$ carbon sources:

• Each peptide bond contributes one carbonyl carbon.
• The terminal carboxyl group contributes one carbonyl carbon.
• The phenyl ring contributes six $sp^2$ carbons.

Adding these contributions gives the final count as $10$.

Therefore, the correct numerical value is $10$.