IChO 1993 · Theory problem 2

HPLC determination of bile acids

Download PDFIChO 1993

The reflux of bile duodenal matter is suspected to be the major cause of gastritis and medical therapy is based on the treatment with antiacida that buffer the \(pH\) of gastric juices by binding bile acids and lysolecithin. Two bile acids, i. e. cholic acid (CA) and glycocholic acid (GCA), were chosen to study the properties of some antiacida commercial formulations. Separation and determination of the two acids in artificial and natural gastric juices were required and high performance liquid chromatography (HPLC) was used as analytical technique.

Two chromatographic columns (A and B, respectively) were selected on the base of published literature data and preliminary tests were carried out to choose the column (between A and B) which would yield the best separation of the aforementioned compounds. The retention times (\(t\)) of the two acids, of a substance not retained by the chromatographic columns and of a compound used as internal standard (i.s.) are shown in the first two columns of the Table. Both chromatographic columns are 25 cm long and show the efficiency of \(2.56\times10^{4}\) theoretical plates per meter (\(N\,\mathrm{m}^{-1}\)).

In actual analysis, an artificial gastric juice was extracted with an appropriate solvents mixture and then final solution (\(1\,\mathrm{cm}^{3}\)) contained 100 % of the two acids present in the original mixture. \(100\,\mu\mathrm{cm}^{3}\) of the extract with 2.7 µmoles of the internal standard were analysed by HPLC using the selected column. The response factors (F) of CA and GCA with respect to the i.s. and the chromatographic peak areas of the two compounds are reported in the Table.

Column A \(t\) (s) Column B \(t\) (s) F Area
Unretained compound 120 130 - -
Cholic Acid (CA) 380 350 0.5 2200
Glycocholic Acid (GCA) 399 395 0.2 3520
Internal standard 442 430 - 2304

In answering to the questions remember that:

\[ R=\frac{\sqrt{N}}{4}\times\frac{\alpha-1}{\alpha}\frac{K'}{K'+1} \tag{1} \]

\[ \alpha=\frac{t'_2}{t'_1}=\frac{t_2-t_0}{t_1-t_0} \tag{2} \]

\[ K'=K'_2=\frac{t'_2}{t_0}=\frac{t_2-t_0}{t_0} \tag{3} \]

Questions:

2.1 Using \(R\), \(\alpha\) and \(K'\) values, demonstrate which of the two chromatographic columns would you use, considering that an accurate determination requires the best possible (baseline) separation of all compounds (do not take into account the unretained compound)?

2.2 Calculate the amounts of each acid in the extract solution.

Official English text of the problem.

Answers

  • 2.1: Choose column B. Its adjacent-peak resolutions are \(R_{\mathrm{CA,GCA}}=2.28\) and \(R_{\mathrm{GCA,is}}=1.63\), both exceeding the usual baseline-separation criterion \(R\geq1.5\). Column A has \(R_{\mathrm{CA,GCA}}=0.952\), which is insufficient.
  • 2.2: The printed aliquot unit and the undefined response-factor convention require qualifications. If the intended aliquot is \(100\,\mu\mathrm L\), the amounts in the complete extract are:
Meaning assigned to \(F_X\) Cholic acid Glycocholic acid
Relative response: \(\displaystyle F_X=\frac{A_X/n_X}{A_{\mathrm{is}}/n_{\mathrm{is}}}\) \(51.6\,\mu\mathrm{mol}\), approximately \(52\,\mu\mathrm{mol}\) \(206\,\mu\mathrm{mol}\), approximately \(2.1\times10^2\,\mu\mathrm{mol}\)
Reciprocal correction factor: \(\displaystyle F_X=\frac{n_X/A_X}{n_{\mathrm{is}}/A_{\mathrm{is}}}\) \(12.9\,\mu\mathrm{mol}\), approximately \(13\,\mu\mathrm{mol}\) \(8.25\,\mu\mathrm{mol}\), approximately \(8.3\,\mu\mathrm{mol}\)

If the printed aliquot is instead read literally as \(100\times10^{-6}\,\mathrm{cm}^{3}\), all these amounts are 1,000 times larger: the same numerical values apply in mmol.

Solution

2.1 — Selection of the chromatographic column

The number of theoretical plates in each complete column is

\[ N=(2.56\times10^{4}\,\mathrm{m}^{-1})(0.25\,\mathrm m) =6400, \qquad \frac{\sqrt N}{4}=20. \]

The unretained compound determines the dead time:

\[ t_{0,A}=120\,\mathrm s,\qquad t_{0,B}=130\,\mathrm s. \]

In both columns, the elution order is CA, GCA, internal standard. Thus, the adjacent pairs that must be resolved are CA–GCA and GCA–internal standard.

For each pair, label the earlier peak 1 and the later peak 2. Then

\[ \alpha=\frac{t_2-t_0}{t_1-t_0}, \qquad K'=\frac{t_2-t_0}{t_0}. \]

Substitution into the supplied resolution equation gives a useful simplification:

\[ \begin{aligned} R &=20 \left(\frac{t_2-t_1}{t_2-t_0}\right) \left(\frac{t_2-t_0}{t_2}\right)\\ &=20\frac{t_2-t_1}{t_2}. \end{aligned} \]

The calculated parameters are all dimensionless:

Column Adjacent pair \(\alpha\) \(K'\), for the later peak \(R\)
A CA–GCA \(\dfrac{399-120}{380-120}=1.0731\) \(\dfrac{399-120}{120}=2.3250\) \(\dfrac{20(399-380)}{399}=0.9524\)
A GCA–i.s. \(\dfrac{442-120}{399-120}=1.1541\) \(\dfrac{442-120}{120}=2.6833\) \(\dfrac{20(442-399)}{442}=1.9457\)
B CA–GCA \(\dfrac{395-130}{350-130}=1.2045\) \(\dfrac{395-130}{130}=2.0385\) \(\dfrac{20(395-350)}{395}=2.2785\)
B GCA–i.s. \(\dfrac{430-130}{395-130}=1.1321\) \(\dfrac{430-130}{130}=2.3077\) \(\dfrac{20(430-395)}{430}=1.6279\)

For approximately Gaussian peaks, baseline separation conventionally requires \(R\gtrsim1.5\).

Column A separates GCA from the internal standard, but its two acid peaks overlap appreciably. Column B satisfies the criterion for both adjacent pairs, so all three retained compounds can be determined separately:

\[ \boxed{\text{Use column B.}} \]

2.2 — Amounts of the bile acids

The two analytes are:

  • Cholic acid (CA), \(\ce{C24H40O5}\): C[C@H](CCC(=O)O)[C@H]1CC[C@H]2[C@@H]3[C@H](O)C[C@@H]4C[C@H](O)CC[C@]4(C)[C@H]3C[C@H](O)[C@]12C
  • Glycocholic acid (GCA), the glycine amide of cholic acid, \(\ce{C26H43NO6}\): C[C@H](CCC(=O)NCC(=O)O)[C@H]1CC[C@H]2[C@@H]3[C@H](O)C[C@@H]4C[C@H](O)CC[C@]4(C)[C@H]3C[C@H](O)[C@]12C

These isomeric SMILES encode the natural bile-acid stereochemistry, including the \(3\alpha,7\alpha,12\alpha\) hydroxyl groups and \(5\beta\) configuration.

Let \(n_X\) denote the amount of an acid in the analysed aliquot. The internal-standard amount is

\[ n_{\mathrm{is}}=2.7\,\mu\mathrm{mol}. \]

With the relative-response definition

\[ F_X= \frac{A_X/n_X}{A_{\mathrm{is}}/n_{\mathrm{is}}}, \]

rearrangement gives

\[ n_X=\frac{A_X}{A_{\mathrm{is}}}\frac{n_{\mathrm{is}}}{F_X}. \]

Consequently,

\[ n_{\mathrm{CA,aliquot}} =\frac{2200}{2304}\frac{2.7}{0.5} =5.15625\,\mu\mathrm{mol}, \]

\[ n_{\mathrm{GCA,aliquot}} =\frac{3520}{2304}\frac{2.7}{0.2} =20.625\,\mu\mathrm{mol}. \]

If the intended aliquot is \(100\,\mu\mathrm L=0.100\,\mathrm{cm}^{3}\), the complete extract contains ten aliquots:

\[ D=\frac{1.00\,\mathrm{cm}^{3}}{0.100\,\mathrm{cm}^{3}}=10. \]

Therefore,

\[ \boxed{n_{\mathrm{CA,extract}} =51.5625\,\mu\mathrm{mol}\approx52\,\mu\mathrm{mol}}, \]

\[ \boxed{n_{\mathrm{GCA,extract}} =206.25\,\mu\mathrm{mol}\approx2.1\times10^2\,\mu\mathrm{mol}}. \]

If \(F\) instead denotes the reciprocal, amount-per-area correction factor,

\[ F_X= \frac{n_X/A_X}{n_{\mathrm{is}}/A_{\mathrm{is}}}, \qquad n_X=F_X\frac{A_X}{A_{\mathrm{is}}}n_{\mathrm{is}}, \]

the corresponding extract amounts for a \(100\,\mu\mathrm L\) aliquot are

\[ n_{\mathrm{CA,extract}} =10(0.5)\frac{2200}{2304}(2.7) =12.890625\,\mu\mathrm{mol} \approx13\,\mu\mathrm{mol}, \]

\[ n_{\mathrm{GCA,extract}} =10(0.2)\frac{3520}{2304}(2.7) =8.25\,\mu\mathrm{mol} \approx8.3\,\mu\mathrm{mol}. \]

Finally, taking the printed volume literally as

\[ V_{\mathrm{aliquot}} =100\times10^{-6}\,\mathrm{cm}^{3} =10^{-4}\,\mathrm{cm}^{3} \]

gives

\[ D=\frac{1\,\mathrm{cm}^{3}}{10^{-4}\,\mathrm{cm}^{3}}=10^4. \]

This increases each extract amount above by a factor of \(10^3\). Because extraction recovery is 100%, the calculated extract amounts also equal the amounts originally present in the gastric-juice sample.

The solution was checked against the official solution.