Morphine and morphine 3-b-d-glucuronide were separated on two different 50 3 4.6 mm columns with 3-mm particles.61 Column A was C18-silica run at 1.4 mL/min and column B was bare silica run at 2.0 mL/min.

(a) Estimate the volume,Vm, and time,tm, at which unretained solute would emerge from each column. The observed times are 0.65 min for column A and 0.50 min for column B.

(b) Column A was eluted with 2 vol% acetonitrile in water containing 10 mM ammonium formate at pH 3. Morphine 3-β-d-glucuro-nide emerged at 1.5 min and morphine at 2.8 min. Explain the order of elution.

(c) Find the retention factor k for each solute on column A, usingtm5 0.65 min.

(d) Column B was eluted with a 5.0-min gradient beginning at 90 vol% acetonitrile in water and ending at 50 vol% acetonitrile in water. Both solvents contained 10 mM ammonium formate, pH 3. Morphine emerged at 1.3 min and morphine 3-b-d-glucuronide emerged at 2.7 min. Explain the order of elution. Why does the gradient go from high to low acetonitrile volume fraction?

(e) From Equation 25-12 in Box 25-4, estimate k* on Column B assuming S = 4 and withtm5 0.50 min.

Short Answer

Expert verified

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

aVm=0.529cm3tm=0.378mintm=0.265min

(b) The polar compound would less be retained

ck1=1.31k2=3.31

(d) The more polar compound can be removed.

ek*=6.25

Step by step solution

01

Column A and 0.50 min for column B

Part (a)

First, we calculate the volume at which solvent front appears:

Then we calculatedtm:

tm=L.dc22.F

Or:

tm=VmF

For column A:

tm=0.529cm31.4cm3/mintm=0.378min

For column B:

tm=0.529cm32cm3/mintm=0.265min

02

Order of elution

Part (b)

The morphine-3-β-D-glucuronide would elute before morphine because it is more polar (it has more hydroxy groups and one carboxylic acid). So, because the reversed phase column is nonpolar the polar compound would less be retained.

03

Retention factor k for each solute on column A

Part (c)

The retention factor can be calculated using the formula:

k=tr-tmtm

For morphine-3-β-D-glucuronide, the retention factor is:

k1=1.5-0.65min0.65mink1=1.31

For morphine, the retention factor is:

k2=2.8-0.65min0.65mink2=3.31

04

Gradient go from high to low acetonitrile volume fraction

Part (d)

Column B has bare silica which is polar. Morphine should be retained less than morphine-3-β-D-glucuronide because it is less polar. The gradient goes from high to low acetonitrile volume fraction because it tends to increase concentration of water, so that the polarity can be increased. Because of that, the more polar compound can be removed.

05

k* on Column B

Part (e)

The average retention factor is calculated using the equation:

k*=tG.Fɸ.Vm.S

First, we need to calculate the volume of mobile phase in the column:

Vm=F.tmVm=2mL/min.0.5minVm=1mL

The k*is:

k*=5min.2mL/min0.4.1mL.4k*=6.25

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

a) Vm=0.529cm3tm=0.378mintm=0.265minb) Thepolarcompoundwouldlessberetainedc)k1=1.31k2=3.31d) Themorepolarcompoundcanberemoved.e) k*=6.25

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Most popular questions from this chapter

(a) You wish to use a wide gradient from 5 vol % to 95 vol %B for the first separation of a mixture of small molecules to decide whether to use gradient or isocratic elution. What should be the gradient time, tG,for a 15×0.46-cmcolumn containing 3-μmparticles with a flow of 1.0mL/min

You optimized the gradient separation going from 20vol%to 34 vol% Bin 11.5min at 1.0mLFindk*for this optimized

separation. To scale up to a 15×1.0cmcolumn, what should be the gradient time and the volume flow rate? If the sample load on

the small column was 1mgwhat sample load can be applied to the large column? Verify that k*is unchanged.

(a) use figure 25-28a select a tetrahydrofurn/water mobile phase strength of equivalent strength to 80%methanol.

(b)Describe how to prepare 1 L of this tetrahydrofurn mobile phase.

(c)what limitation would be imposed by the use tetrahydrofurn.

HPLC peak should generally not have an asymmetry factor, B/A in figure 23-14,outside the range0.9-1.5

  1. Sketch the shape of a peak with an asymmetry of 1.8
  2. What might you do to correct the asymmetry?

Two peaks emerge from a reversed-phase chromatography column as sketched in the illustration.

According to Equation 23-33, resolution is given by

Resolution=N4(α-1)α(k21+k2)

where Nis plate number, αis relative retention (Equation23-20), and k2 is the retention factor for the more retained component (Equation 23-16).

(a) If you decrease the amount of organic solvent in the mobile phase, you will increase retention. Sketch the chromatogram if retention factors increase but Nand αare constant.

(b) If you change the solvent type or the stationary phase, you will change the relative retention. Sketch the chromatogram ifαincreases but Nandk1are constant.

(c) If you decrease particle size or increase column length, you can increase the plate number. Sketch the chromatogram if Nincreases by (i) decreasing particle size and (ii) increasing column length. Assume αand k2are constant.

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?

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