Question

The figure shows the separation of two enantiomers on a chiral stationary phase.

Sepation of enantiomers of Ritalin by HPLC with a chiral stationary phase.[Data from R.Bakhitar,L.Ramos,and F.L.S.Tse, “Quantification of methlylphenidate in plasma using chiral Liquid-chromatography/Tandem mass spectrometry: Application to Toxicokinetric studies,”Anal Chim Acta 2002,469,261.]

(a)From $t_r$and${\mathrm{w}}_{1/2}$ find N for each peak.

(b) From${\mathrm{t}}_{\mathrm{r}}$and${\mathrm{w}}_{1/2}$find the resolution.

(c)Given${\mathrm{t}}_{\mathrm{m}}=1.62$min, use Equation$23-23$with the average N to predict the resolution.

Short answer

The separation of two enantiomers on a chiral stationary phase.

$$\begin{gathered} {\mathrm{t}}_{\mathrm{r}}=470\min \\ {\mathrm{w}}_{1/2}=0.28\min \\ \mathrm{N}=5.55{\left(\frac{4.70\min }{0.28\min }\right)}^2 \end{gathered}$$

$$\begin{gathered} =\frac{0.589(5.37-4.70)\min }{0.28\min +0.35\min } \\ =0.626 \\ =\frac{\sqrt{1435.13}}{4}\frac{(1.22-1)}{1.22}\left(\frac{2.31}{1+2.31}\right) \\ =1.19 \\ \mathrm{a})1306.49 \\ \mathrm{b})0.626 \\ \mathrm{c})1.19 \end{gathered}$$

Step-by-step solution

definition of enantiomers

In chemistry, an enantiomer – also known as an optical isomer, antipode, or optical antipode – is one of two non-superposable stereoisomers that are mirror images of each other.

Find N for each peak

a) Use this formula to find the N by substituting the known value for both enantiomers

b)Find the resolution

The resolution equation goes:

$$\begin{gathered} \mathrm{Resolution}=\frac{0.589∆{\mathrm{t}}_{\mathrm{r}}}{{\mathrm{w}}_{1/2\mathrm{LD}}} \\ =\frac{0.589.(5.37-4.70)\min }{0.28\min +0.35\min } \\ =0.626 \end{gathered}$$

equation resolution

The average N to predict the resolution

c) This is the equation we are going to use for resolution:

$$\begin{gathered} \mathrm{Resolution}=\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+{\mathrm{k}}_2}\right) \\ =\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+{\mathrm{k}}_2}\right) \end{gathered}$$

The average N

$$\begin{gathered} \overline{\mathrm{N}}={\mathrm{N}}_{\mathrm{l}}+{\mathrm{N}}_{\mathrm{d}} \\ =\frac{1563.77+1306.49}{2} \\ =1435.13 \\ {\mathrm{t}}_{\mathrm{m}}=1.62\min \end{gathered}$$

Now to calculate the resolution find the retention factor for the second peak by first finded

$$\begin{gathered} =2.31 \\ \mathrm{Now}\mathrm{find}\mathrm{the}{\mathrm{t}}^{\mathrm{r}}\mathrm{for}\mathrm{the}\mathrm{L}\mathrm{enantiomer}: \\ {{\mathrm{t}}^{\mathrm{r}}}_{\mathrm{r},\mathrm{l}}={\mathrm{t}}_{\mathrm{r},\mathrm{l}}-{\mathrm{t}}_{\mathrm{m}} \\ =4.70\min -1.62\min \\ =3.08\min \\ \mathrm{And}\mathrm{substitute}\mathrm{all}\mathrm{the}\mathrm{known}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{resolution}\mathrm{equation}:\mathrm{Resolution} \\ \mathrm{Resolution}=\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+2.31}\right) \\ =\frac{\sqrt{1435.13}}{4}\frac{\left(1.22-1\right)}{1.22}\left(\frac{2.31}{1+2.31}\right) \\ =1.19 \\ \mathrm{Hence}, \\ \mathrm{a})1306.49 \\ \mathrm{b})0.626 \\ \mathrm{c})1.19 \end{gathered}$$

equation resolution

The average N to predict the resolution

c) This is the equation we are going to use for resolution:

$$\begin{gathered} \mathrm{Resolution}=\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+{\mathrm{k}}_2}\right) \\ =\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+{\mathrm{k}}_2}\right) \end{gathered}$$

The average N

$$\begin{gathered} \overline{\mathrm{N}}={\mathrm{N}}_{\mathrm{l}}+{\mathrm{N}}_{\mathrm{d}} \\ =\frac{1563.77+1306.49}{2} \\ =1435.13 \\ {\mathrm{t}}_{\mathrm{m}}=1.62\min \end{gathered}$$

Now to calculate the resolution find the retention factor for the second peak by first finded

$$\begin{gathered} {{\mathrm{t}}^{\text{'}}}_{\mathrm{r},\mathrm{D}}={\mathrm{t}}_{\mathrm{r},\mathrm{d}}-{\mathrm{t}}_{\mathrm{m}} \\ =5.37\min -1.62\min \\ =3.75\min \\ {\mathrm{k}}_2=\frac{{{\mathrm{t}}^{\mathrm{r}}}_{\mathrm{r},\mathrm{d}}}{{\mathrm{t}}_{\mathrm{m}}} \\ =\frac{3.75\min }{1.62\min } \\ =2.31 \\ \mathrm{Now}\mathrm{find}\mathrm{the}{\mathrm{t}}^{\mathrm{r}}\mathrm{for}\mathrm{the}\mathrm{L}\mathrm{enantiomer}: \\ {{\mathrm{t}}^{\mathrm{r}}}_{\mathrm{r},\mathrm{l}}={\mathrm{t}}_{\mathrm{r},\mathrm{l}}-{\mathrm{t}}_{\mathrm{m}} \\ =4.70\min -1.62\min \\ =3.08\min \\ \mathrm{And}\mathrm{substitute}\mathrm{all}\mathrm{the}\mathrm{known}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{resolution}\mathrm{equation}:\mathrm{Resolution} \\ \mathrm{Resolution}=\frac{\sqrt{\mathrm{N}}}{4}\frac{\left(\mathrm{\alpha }-1\right)}{\mathrm{\alpha }}\left(\frac{{\mathrm{k}}_2}{1+2.31}\right) \\ =\frac{\sqrt{1435.13}}{4}\frac{\left(1.22-1\right)}{1.22}\left(\frac{2.31}{1+2.31}\right) \\ =1.19 \\ \mathrm{Hence}, \\ \mathrm{a})1306.49 \\ \mathrm{b})0.626 \\ \mathrm{c})1.19 \end{gathered}$$