IChO 2002 · Theory problem 7

Glutathione, an Essential Mini-Peptide

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Glutathione, abbreviated as GSH, is a small peptide that is present in almost all tissues of animals. GSH fulfils important biological functions, such as detoxification of electrophilic chemicals and reduction of (organic) peroxides in blood. An electrophilic compound reacts irreversibly with GSH, especially in the liver, to give a primary product that is converted by a series of biotransformations into a so-called mercapturic acid, which is excreted via the urine. Oxidants react with GSH to give the disulfide GSSG, which can be enzymatically reverted to GSH with reductases. The ratio GSH/GSSG in most cells is ≥ 500.

Figure 1

7.1 (a) How many amino acid residues are present in GSH?

(b) Draw the structures of the corresponding amino acids and mark the chiral centers with an asterisk.

A mercapturic acid A isolated from urine of a person who has been exposed to acrylonitrile (\(\ce{H2C=CH-CN}\)) has the molecular formula \(\ce{C8H12N2O3S}\). The \({}^{1}\mathrm{H}\)-NMR spectrum of A in \(\ce{(CD3)2SO}\) is shown in Figure 1. When the product is pretreated with \(\ce{D2O}\), the signals at \(\delta\) 12.8 and \(\delta\) 6.8 are no longer present and the signal 3 is simplified.

Figure 2
Figure 3

Figure 1

7.2 (a) The NMR-signals correspond with protons in the following groups: \(\ce{CH}\), \(\ce{CH2}\), \(\ce{CH3}\), \(\ce{OH}\) and \(\ce{NH}\). Indicate the appropriate proton group in the boxes for the signals 1 – 7.

Signals 1 2 3 4/5 6 7
Protons

(b) How many carbon atoms are present in compound A that do not carry any protons?

(c) Draw the structure of compound A.

Vitamin C (ascorbic acid) reacts with oxidants to give dehydroascorbic acid D

Figure 4

7.3 Eating fresh fruit and vegetables is healthy

  • [ ] because vitamin C forms a complex with GSH.
  • [ ] because vitamin C reacts with electrophilic compounds.
  • [ ] because vitamin C removes oxidants and prevents undesired depletion of GSH.
  • [ ] for many reasons, but none of them has anything to do with GSH.

Official English text of the problem.

Answers

  • 7.1(a): Three amino acid residues.
  • 7.1(b):
  • L-Glutamic acid: N[C@@H](CCC(=O)O)C(=O)O. Mark its α-carbon with an asterisk; it has \(S\) configuration.
  • L-Cysteine: N[C@@H](CS)C(=O)O. Mark its α-carbon with an asterisk; it has \(R\) configuration.
  • Glycine: NCC(=O)O. It has no chiral center.
  • 7.2(a): \(1:\ce{OH};\ 2:\ce{NH};\ 3:\ce{CH};\ 4/5:\) two \(\ce{CH2}\) groups; \(6:\ce{CH2};\ 7:\ce{CH3}\).
  • 7.2(b): Three: the carboxyl carbon, the amide carbonyl carbon, and the nitrile carbon.
  • 7.2(c): N-Acetyl-S-(2-cyanoethyl)-L-cysteine, CC(=O)N[C@@H](CSCCC#N)C(=O)O. Its α-carbon has \(R\) configuration.
  • 7.3: Select the third statement: “because vitamin C removes oxidants and prevents undesired depletion of GSH.”

Solution

7.1(a)

GSH is γ-L-glutamyl-L-cysteinylglycine, with SMILES N[C@@H](CCC(=O)N[C@@H](CS)C(=O)NCC(=O)O)C(=O)O.

The two amide linkages connect one glutamic acid residue, one cysteine residue, and one glycine residue. Thus, GSH contains three amino acid residues. The glutamyl linkage uses glutamic acid’s side-chain carboxyl group.

7.1(b)

The free amino acids have the following structures. An asterisk above a \(\ce{CH}\) group marks its carbon atom.

Amino acid Structure with chiral center marked Configuration
L-Glutamic acid \(\ce{HOOC}-\overset{*}{\ce{CH}}(\ce{NH2})-\ce{CH2-CH2-COOH}\) \(S\)
L-Cysteine \(\ce{HOOC}-\overset{*}{\ce{CH}}(\ce{NH2})-\ce{CH2-SH}\) \(R\)
Glycine \(\ce{H2N-CH2-COOH}\) Achiral

The α-carbon of glutamic acid and that of cysteine each bear four different substituents. Glycine’s α-carbon bears two identical hydrogen atoms and therefore is not chiral.

L-Cysteine has \(R\) configuration because its sulfur-containing side chain has higher CIP priority than its carboxyl group; this does not change its designation as an L-amino acid.

7.2(a)

The integral increases give the proton counts:

Signal Approximate \(\delta\) / ppm Number of protons Assignment
1 12.8 1 Carboxylic-acid \(\ce{OH}\)
2 6.8 1 Amide \(\ce{NH}\)
3 4.8 1 \(\ce{CH}\)
4/5 2.7–3.1 4 combined Two \(\ce{CH2}\) groups
6 2.55 2 \(\ce{CH2}\)
7 2.0 3 \(\ce{CH3}\)

Signals 1 and 2 disappear after treatment with \(\ce{D2O}\), identifying exchangeable hydrogens. The very downfield signal at \(\delta 12.8\) is characteristic of a carboxylic-acid proton. Signal 2 is consequently the \(\ce{NH}\) proton.

Signal 3 integrates to one proton and is assigned to the amino-acid α-\(\ce{CH}\). Its simplification after deuterium exchange shows that part of its original splitting arose from coupling to the neighboring \(\ce{NH}\) proton.

The remaining integrals identify three methylene groups and one methyl group. The total is

\[ 1+1+1+4+2+3=12 \]

hydrogens, agreeing with \(\ce{C8H12N2O3S}\).

7.2(b)

The carbon-bound hydrogens belong to one \(\ce{CH}\), three \(\ce{CH2}\), and one \(\ce{CH3}\) group. These account for

\[ 1+3+1=5 \]

carbon atoms. Since A contains eight carbon atoms, the number without attached hydrogens is

\[ \boxed{8-5=3}. \]

In the identified structure, these are the two carbonyl carbons and the nitrile carbon.

7.2(c)

Acrylonitrile, C=CC#N, is an activated alkene. Conjugate addition of the cysteine sulfur of GSH to its terminal alkene carbon produces the fragment

\[ \ce{-S-CH2-CH2-C#N}. \]

The cysteine portion supplies the carboxylic-acid group, the α-\(\ce{CH}\), and another methylene group attached to sulfur. Removal of the glutamyl and glycyl portions and acetylation of the remaining amino group produce the mercapturic acid.

The methyl signal near \(\delta 2.0\), together with the amide \(\ce{NH}\), supports an N-acetyl group, \(\ce{-NH-C(=O)-CH3}\). Combining these fragments gives

\[ \boxed{\ce{HOOC-CH(NH-C(=O)-CH3)-CH2-S-CH2-CH2-C#N}}. \]

Thus, A is N-acetyl-S-(2-cyanoethyl)-L-cysteine, CC(=O)N[C@@H](CSCCC#N)C(=O)O. Draw the α-carbon in the \(R\) configuration retained from the cysteine residue of GSH.

The structure accounts for every signal:

  • The carboxylic-acid and amide hydrogens give signals 1 and 2.
  • The α-\(\ce{CH}\) gives signal 3.
  • The two sulfur-adjacent methylene groups give the overlapping signals 4/5.
  • The methylene adjacent to the nitrile gives signal 6.
  • The acetyl methyl group gives signal 7.

Its hydrogen count is \(1_{\mathrm{OH}}+1_{\mathrm{NH}}+1_{\mathrm{CH}}+6_{\mathrm{CH_2}}+3_{\mathrm{CH_3}}=12\). Its three oxygen atoms belong to the carboxylic acid and amide, and its two nitrogen atoms belong to the amide and nitrile.

As a further check, the degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(8)+2+2-12}{2}=4. \]

The two carbonyl groups contribute two units, and the nitrile triple bond contributes two more.

7.3

Select the third statement.

Vitamin C, L-ascorbic acid, OC[C@H](O)[C@H]1OC(=O)C(O)=C1O, reduces oxidants while being oxidized to dehydroascorbic acid D, OC[C@H](O)[C@H]1OC(=O)C(=O)C1=O, as depicted in the problem. The corresponding balanced oxidation half-reaction is

\[ \ce{C6H8O6 -> C6H6O6 + 2H+ + 2e-}. \]

By donating electrons to oxidants, vitamin C reduces the amount of GSH that must be consumed in removing them, helping preserve reduced glutathione.

The solution was checked against the official solution.