Question

The antitumor drug gimatecan is available as nearly pure (S)-enantiomer. Neither pure (R)-enantiomer nor a racemic (equal) mixture of the two enantiomers is available. To measure small quantities of (R)-enantiomer in nearly pure (S)-gimatecan, a preparation was subjected to normal-phase chromatography on each of the enantiomers of a commercial, chiral stationary phase designated (S,S)- and (R,R)-DACH-DNB. Chromatography on the (R,R)-stationary phase gave a slightly asymmetric peak at tr 5 6.10 min with retention factor k 5 1.22. Chromatography on the (S,S)- stationary phase gave a slightly asymmetric peak at tr 5 6.96 min with k 5 1.50. With the (S,S) stationary phase, a small peak with 0.03% of the area of the main peak was observed at 6.10 min.

Chromatography of gimatecan on each enantiomer of a chiral stationary phase. Lower traces have enlarged vertical scale. [Data from E. Badaloni, W. Cabri, A. Ciogli, R. Deias, F. Gasparrini, F. Giorgi, A. Vigevani, and C. Villani, “Combination of HPLC ‘Inverted Chirality Columns Approach’ and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.” Anal. Chem. 2007, 79, 6013.]

(a) Explain the appearance of the upper chromatograms. Dashed lines are position markers, not part of the chromatogram. What Problems 709 would the chromatogram of pure (R)-gimatecan look like on the same two stationary phases?

(b) Explain the appearance of the two lower chromatograms and why it can be concluded that the gimatecan contained 0.03% of the (R)-enantiomer. Why is the (R)-enantiomer not observed with the (R,R)-stationary phase?

(c) Find the relative retention (a) for the two enantiomers on the (S,S)-stationary phase.

(d) The column provides N 5 6 800 plates. What would be the resolution between the two equal peaks in a racemic (equal) mixture of (R)- and (S)-gimatecan? If the peaks were symmetric, does this resolution provide baseline separation in which signal returns to baseline before the next peak begins?

Short answer

The part (a), part (b), part (c), part (d) is

1. Chiral stationary phase it will be eluted at 6.10 min.
2. Chromatography enables us to locate where enantiomer is eluted.
3. $\propto =1.26$
   
4. $\mathrm{resolution}=2.55$
   

Step-by-step solution

Two stationary phases

Part (a)

If we look (R, R) chiral stationary phase, we see that (S)-gimatecan is eluted at 6.1 min. If we look (S, S) chiral stationary phase, we see that (S)-gimatecan is eluted at 6.96 min which means that it is retained more strongly. (R)-enantiomer of gimatecan should have the opposite behaviour. So, at (R, R) chiral stationary phase, it will be eluted ate 6.96 min, while at (S, S) chiral stationary phase it will be eluted at 6.10 min.

Step 2: (R)-enantiomer not observed with the (R,R)-stationary phase

Part (b)

At (S, S) stationary phase, the small peak for (R)-gimatecan (at 6-. min) is separated from the big (S)-gimatecan peak (at 6.96 min). The area between the two peaks can be integrated, so we can compare with each other. At (R, R) stationary phase, we see (S)- gimatecan peak (6.1 min), while there is no (R)-gimatecan peak (it is lost beneath the tail of (S-gimatecan). Chromatography enables us to locate where enantiomer is eluted.

Relative retention (a) for the two enantiomers

Part (c)

First, we need to calculate ${\mathrm{t}}_{\mathrm{m}}$from retention factor of each enantiomer:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ {\mathrm{t}}_{\mathrm{m}}=\frac{{\mathrm{t}}_{\mathrm{r}}}{\mathrm{k}+1} \end{gathered}$$

For (S)-gimatecan:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{m}}=\frac{6.96\min }{1.50+1} \\ {\mathrm{t}}_{\mathrm{m}}=2.784\min \end{gathered}$$

For (R)-gimatecan:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{m}}=\frac{6.10\min }{1.22+1} \\ {\mathrm{t}}_{\mathrm{m}}=2.784\min \end{gathered}$$

The average ${\mathrm{t}}_{\mathrm{m}}$is:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{m}}=\frac{\left(2.784+2.784\right)\min }{2} \\ {\mathrm{t}}_{\mathrm{m}}=2.766\min \end{gathered}$$

The adjusted retention time is:

${\mathrm{t}}_{\mathrm{r}}^1={\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}$

For (R)-gimatecan:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{r}1}^1=\left(6.10-2.766\right)\min \\ {\mathrm{t}}_{\mathrm{r}1}^1=3.334\min \end{gathered}$$

For (S)-gimatecan:

$$\begin{gathered} {\mathrm{t}}_{\mathrm{r}2}^1=\left(6.96-2.766\right)\min \\ {\mathrm{t}}_{\mathrm{r}2}^1=4.194\min \end{gathered}$$

The relative retention can be calculated using the formula:

$\propto =\frac{{\mathrm{t}}_{\mathrm{r}2}^1}{{\mathrm{t}}_{\mathrm{r}1}^1}=\frac{4.194}{3.334}=1.26$

Signal returns to baseline before the next peak begins

Part (d)

The resolution can be calculated using the formula:

$$\begin{gathered} \mathrm{resolution}=\frac{\sqrt{\mathrm{N}}}{4}.\frac{\propto -1}{\propto }.\frac{\mathrm{k}}{1+\mathrm{k}} \\ \mathrm{resolution}=\frac{\sqrt{6800}}{4}.\frac{1.26-1}{1.26}.\frac{1.50}{1+1.50} \\ \mathrm{resolution}=2.55 \end{gathered}$$

If the peaks were symmetric, this resolution would provide adequate baseline separation.

Here, the final result of part(a), part(b), part(c), part(d) is

1. Chiral stationary phase it will be eluted at 6.10 min.
2. Chromatography enables us to locate where enantiomer is eluted.
3. $\propto =1.26$
   
4. $\mathrm{resolution}=2.55$