Question

A bonded stationary phase for the separation of optical isomers has the structure

To resolve the enantiomers of amines, alcohols, or thiols, the compounds are first derivatized with a nitroaromatic group that increases their interaction with the bonded phase and makes them observable with a spectrophotometric detector.

When the mixture is eluted with 20 vol% 2-propanol in hexane, the (R)-enantiomer is eluted before the (S)-enantiomer, with the following chromatographic parameters:

Resolution$\mathbf{=}\frac{{\mathbf{\Delta t}}_{\mathbf{r}}}{{\mathbf{w}}_{\mathbf{av}}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{7}$

Relative retention$\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{53}$

k for (R)-isomer =1.35

${\mathbf{t}}_{\mathbf{m}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{min}$

where${\mathbf{w}}_{\mathbf{av}}$is the average width of the two Gaussian peaks at their base.

(a) Find${\mathbf{t}}_{\mathbf{1}}\mathbf{,}{\mathbf{t}}_{\mathbf{2}}$and${\mathbf{w}}_{\mathbf{a}}$with units of minutes.

(b) The width of a peak at half-height isrole="math" localid="1663582749865" ${\mathbf{w}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$(Figure 23-9). If the plate number for cach peak is the same, find${\mathbf{w}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$for each peak.

(c) The area of a Gaussian peak is$\mathbf{1}\mathbf{.}\mathbf{064}\mathbf{\times }\mathbf{peakheight}\mathbf{\times }{\mathbf{w}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$. Given that the areas under the two bands should be equal, find the relative peak heights$\mathbf{(}{\mathbf{height}}_{\mathbf{R}}\mathbf{/}\mathbf{/}{\mathbf{height}}_{\mathbf{S}}\mathbf{)}$.

Short answer

a) The values are${\mathrm{t}}_1=2.35\min ,{\mathrm{t}}_2=7.11\min ,{\mathrm{w}}_{\mathrm{a}}\mathrm{v}=0.619\min$.

b)The half-width of peak 1 is 0.181min.

The half-width of peak 2 is 0.548min .

c) The relative heights of two peaks is 3.0.

Step-by-step solution

Definition of compound

A compound is a substance made up of two or more chemical components that have been chemically bonded together. Covalent bonds and ionic bonds are two typical forms of bonds that hold components in a compound together. In any compound, the components are always present in set ratios.

Step 2: Values of t1t2,wav 

a)

To find the retention time of R-isomer,

$t_m$is given as 1.00min.

Retention factor (k) is given as 1.35.

The $\left(t_1\right)$retention time may be computed by substituting the aforementioned numbers in the retention factor equation, as shown below.

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_1-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ =\frac{{\mathrm{t}}_1-1.00}{1.00} \\ 1.35=\frac{{\mathrm{t}}_1-1.00}{1.00} \\ {\mathrm{t}}_1=1.35+1.00 \\ {\mathrm{t}}_1=2.35\min \end{gathered}$$

The R-isomer retention time is computed as 2.35 min.

The retention period of the S -isomer may be calculated using the equation for relative retention.

4.53 is the relative retention$\left(\alpha \right)$.

The retention period for R-isomer is$2.35\min \left({\mathrm{t}}_1\right)$.

Relative retention$\left(\mathrm{\alpha }\right)=\frac{{\mathrm{t}}_{\mathrm{s}}^{¢}}{{\mathrm{t}}_{\mathrm{s}}{1}^{\text{'}}}$

$$\begin{gathered} 4.53=\frac{t_2-1.00}{t_1-1.00} \\ =\frac{t_2-1.00}{2.35-1.00} \\ {t}_2=4.53\left(1.35\right)1.00 \\ =6.11+1.00 \end{gathered}$$

The retention time of S-isomer is 7.11min.

To find the average width, divide the whole length by the total width.

The resolution and difference in retention time of the two isomers may be used to compute the average width.

4.77min is the difference in retention time.

In the problem statement, the solution is provided as 7.7.

As demonstrated below, the average width may be computed.

Resolution$$\begin{gathered} =\frac{∆{\mathrm{t}}_{\mathrm{y}}}{{\mathrm{w}}_{\mathrm{as}}} \\ 7.7=\frac{4.77}{{\mathrm{w}}_{\mathrm{av}}} \\ {\mathrm{w}}_{\mathrm{av}}=\frac{4.77}{7.7} \\ =0.619\min \end{gathered}$$

The average width at the base is calculated as 0.619min.

Step 3:Values of w1/2 for each peak

b)

The equation below can be used to compute the plate number.

$\mathrm{N}=5.54{\left(\frac{{\mathrm{t}}_{\mathrm{r}}}{{\mathrm{w}}_1\mathrm{n}}\right)}^2$

If the plate number is constant, it may deduce that retention time is proportional to half-width.

Hence,

$$\begin{gathered} \frac{{\mathrm{w}}_{1\mathrm{n}}\left(\mathrm{peak}1\right)}{{\mathrm{w}}_{12}\left(\mathrm{peak}2\right)}=\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2} \\ =\frac{2.35}{7.11} \\ =0.330 \end{gathered}$$

The average width at the base was calculated as$0.62\min$.

It is known that for each peak, the width (w) is equal to$4\sigma$ and half-width$\left(w_{1}n\right)$ is$2.35\sigma$.

From the above two equations,w was found as$1.70w_{1/2}$.

The average base width is 0.62min.

$$\begin{gathered} w_{av}=\frac{1}{2}\left(w_1+w_2\right) \\ =\frac{1}{2}\left(1.70w_{1/2}\left(peak1\right)+1.70w_{1/2}\left(peak2\right)\right) \end{gathered}$$

It is known that

${\mathrm{w}}_{1/2}(\mathrm{peak}1)=0.330\left({\mathrm{w}}_{1/2}\right(\mathrm{peak}2)$

Substituting this into the basic equation's average width

$$\begin{gathered} {\mathrm{w}}_{\mathrm{av}}=\frac{1}{2}\left(1.70\left(0.330{\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right)\right)+1.70{\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right)\right) \\ =\frac{1}{2}\left(0.561{\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right)+1.70{\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right)\right)1.24 \\ =\left(2.26{\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right)\right){\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right) \\ =\frac{1.24}{2.261} \\ =0.548\min \end{gathered}$$

The half-width of peak 2 is 0.548min.

As illustrated below, the half-width of peakmay be determined from this.

$$\begin{gathered} {\mathrm{w}}_{1/2}\left(\mathrm{peak}1\right)=0.330({\mathrm{w}}_{1/2}\left(\mathrm{peak}2\right) \\ =0.330\times 0.548 \\ =0.18084\min \\ =0.181\min \end{gathered}$$

In the provided chromatogram, the half-width of both peaks was calculated.

Step 4: Relative peak height

c)

The half-width of peak 1 is 0.181min.

The half-width of peak 2 is 0.584min.

The regions under both peaks are equal, according to the problem statement.

$$\begin{gathered} {\mathrm{Height}}_{\mathrm{R}}\times {\mathrm{w}}_{\mathrm{R}}={\mathrm{Height}}_{\mathrm{S}}\times {\mathrm{w}}_{\mathrm{S}} \\ \frac{\mathrm{Height}{}_{\mathrm{R}}}{\mathrm{Heright}}=\frac{{\mathrm{w}}_{\mathrm{S}}}{{\mathrm{w}}_{\mathrm{R}}} \\ =\frac{0.548}{0.181} \\ =3.0 \end{gathered}$$

The relative heights of the two peaks is calculated as 3.0.