(i) The following compound is an example of a peptide: H2N-CH(CH2COOH)-C(=O)-NH-CH(CH2-C6H5)-COOCH3 (a dipeptide with a free -CH2COOH side chain on the first residue and a benzyl side chain + methyl ester on the second residue). (r) Write the Zwitterion structure of the compound.

Question

(i) The following compound is an example of a peptide: H2N-CH(CH2COOH)-C(=O)-NH-CH(CH2-C6H5)-COOCH3 (a dipeptide with a free -CH2COOH side chain on the first residue and a benzyl side chain + methyl ester on the second residue). (r) Write the Zwitterion structure of the compound. (s) How many amino acids will be obtained on hydrolysis of the compound? (2 marks total for (r)+(s)) (ii) Write the equation for the reaction of glucose with periodic acid. (1 mark)

Accepted Answer

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> [!TLDR]
> The zwitterion forms between the free N-terminal amino group and the free side-chain carboxyl group (since the C-terminus is blocked as a methyl ester); this dipeptide hydrolyses to 2 amino acids; and glucose consumes 5 moles of periodic acid, cleaving all its C-C bonds bearing adjacent -OH/=O groups.
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> **(r) Zwitterion structure:** The compound $H_2N$-$CH(CH_2COOH)$-$CO$-$NH$-$CH(CH_2C_6H_5)$-$COOCH_3$ has a free N-terminal $-NH_2$ and a free side-chain $-CH_2COOH$ (the C-terminus is esterified as $-COOCH_3$ and so is not free). Proton transfer occurs internally from the side-chain acid to the terminal amine:
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> Zwitterion: $^+H_3N$-$CH(CH_2COO^-)$-$CO$-$NH$-$CH(CH_2C_6H_5)$-$COOCH_3$
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> **(s) Number of amino acids on hydrolysis:** This is a **dipeptide** (one peptide/-CONH- bond), so complete hydrolysis of the peptide bond gives **2 amino acid units**: aspartic acid (from residue 1) and phenylalanine (from residue 2, whose methyl ester also hydrolyses to free -COOH + methanol under full hydrolysis).
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