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Chapter 25: Q6P (page 707)

a. Why are HPLC particles porous?

b. Why are the particles with60-120A掳pores used for small molecules but wide pore300-A掳stationary phases used to separate polypeptides and proteins?s

Short Answer

Expert verified

The porous HPLC particles increase the surface area, resulting in greater sample capacity.

Smaller pore stationary phases are used for small molecules, but wide pore stationary phases are used to separate polypeptides and proteins because the pores must be large enough for the dissolved substances to enter. Larger pores take up more surface area, so they should be avoided if possible.

Step by step solution

01

Step 1:Explanation

For part a, Because porous HPLC particles increase surface area, they have a higher sample capacity.

02

Explanation

For part b, Smaller pore stationary phases are used to separate small molecules, whereas wide pore stationary phases are used to separate polypeptides and proteins because the pores must be large enough for the dissolved substances to enter. Because larger pores take up more surface area, they should be avoided if possible.

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Most popular questions from this chapter

The rate at which heat is generated inside a chromatographycolumn from friction of flowing liquid is power (watts, W=J/s)= volume flow raterole="math" localid="1656474760665" (m3/s)脳pressure drop (pascals, role="math" localid="1656474828044" Pa=kg/[m?s2]).

(a) Explain the analogy between heat generated in a chromatography column and heat generated in an electric circuit (power = current脳voltage).

(b) At what rate (watts = J/s) is heat generated for a flow of 1 mL/minwith a pressure difference of3500 bar between the inlet and outlet?

You will need to convert mL/min tom3/s.Also 1 bar=105Pa.

Question:The graph shows retention data from aC8silica column with an acetonitrile/water mobile phase.


(a) What mobile phase composition provides greatest retention(k)for the components? Least retention? Coelution (equalk)of two components?

(b) Predict the retention time of each peak at 40%and60%acetonitrite. Draw a chromatogram (a "stick diagram" representing each peak as a vertical line) of the separation at each mobile phase composition.

(c) Would60%acetonitrile yield adequate resolution?

(d) Assuming Gaussian peaks, does the separation at60%acetonitrile have the attributes of a good separation?

a. Why is high pressure needed in HPLC?

b. For a given column length , why do smaller particles give a higher plate number?

c. What is bonded phase in liquid chromatography?

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, 鈥淐ombination of HPLC 鈥業nverted Chirality Columns Approach鈥� and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.鈥� 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?

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=螖迟rwav=7.7

Relative retention(伪)=4.53

k for (R)-isomer =1.35

tm=1.00min

wherewavis the average width of the two Gaussian peaks at their base.

(a) Findt1,t2andwawith units of minutes.

(b) The width of a peak at half-height isrole="math" localid="1663582749865" w1/2(Figure 23-9). If the plate number for cach peak is the same, findw1/2for each peak.

(c) The area of a Gaussian peak is1.064脳peakheight脳w1/2. Given that the areas under the two bands should be equal, find the relative peak heights(heightR//heightS).
See all solutions

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