Chromatography–mass spectrometry. Cocaine metabolism in rats can be studied by injecting the drug and periodically with drawing blood to measure levels of metabolites by HPLC–mass spectrometry. For quantitative analysis, isotopically labelled internal standards are mixed with the blood sample. Blood was analysed by reversed-phase chromatography with an acidic eluent and atmospheric pressure chemical ionization mass spectrometry for detection. The mass spectrum of the collisionally activated dissociation products from the m/z 304 positive ion is shown in the figure on the next page. Selected reaction monitoring (m/z 304 from mass filter Q1 and m/z 182 from Q3 in Figure 22-33) gave a single chromatographic peak at 9.22 min for cocaine. The internal standard H52-cocaine gave a single peak at 9.19 min for m/z 309 (Q1) 182(Q3).

(a) Draw the structure of the ion at m/z 304.

(b) Suggest a structure for the ion at m/z 182.

(c) The intense peaks at m/z 182 and 304 do not have C2isotopic partners at m/z 183 and 305. Explain why.

(d) Rat plasma is exceedingly complex. Why does the chromatogram show just one clean peak?

(e) Given that H52-cocaine has only two major mass spectral peaks at m/z 309 and 182, which atoms are labelled with deuterium?

(f) Explain how you would use H52-cocaine for measuring cocaine in blood.

Spectrum for Problem 25-25.

Left: Mass spectrum of collisionally activated dissociation products from m/z 304 positive ion from atmospheric pressure chemical ionization mass spectrum of cocaine.

Right: Chromatograms obtained by selected reaction monitoring. [Data from G. Singh, V. Arora, P. T. Fenn, B. Mets, and I. A. Blair, “Isotope Dilution Liquid Chromatography Tandem Mass Spectrometry Assay for Trace Analysis of Cocaine and Its Metabolites in Plasma,” Anal. Chem. 1999, 71, 2021.]

Short Answer

Expert verified

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

  1. The given structure of cocaine in the problem has an m/z of 303, and is not charged

  2. The dissociation of this substituent and the hydrogen atom would leave the original structure with a m/z of 182

  3. The ions containing the C13isotopes, having a m/z of 305

  4. The ions with the C13isotopes, the ions that did not have a m/z of 304 or m/z of 182

  5. The 5 deuterium ions should be located in this fragment so that when it gets dissociated

  6. H52-cocaine is determined and quantified into the response factor

Step by step solution

01

Cocaine

Cocaine was analysed using the selected reaction monitoring method of chromatography-mass spectrometry. In this problem, the structure of fragments of the sample, and the structure of the ions that went through the first, second, and third quadrupoles were elucidated.

The reason why certain ions and certain isotopes were not detected, and how the chromatogram came to be were also studied. Also, the role of H52-cocaine in the measurement of cocaine from a blood sample was also explored.

02

Structure of the ion at m/z 304

Part (a)

The given structure of cocaine in the problem has an m/z of 303, and is not charged. To get the structure of the positively charged ion with a m/z of 304 that was detected by the mass spectrometer, we simply bond a hydrogen atom to the nitrogen with a methyl group bonded to it:

03

Structure for the ion at m/z 182

Part (b)

A possible structure of the fragment with a m/z of 182 that the third quadrupole allows passage to the detector is:

After going through collisionally-activated dissociation, the ester substituent (-COOC6H5 -) and one hydrogen atom in cocaine could have been removed. Having a m/z of 122 altogether, the dissociation of this substituent and the hydrogen atom would leave the original structure with a m/z of 182.

04

Intense peaks

Part (c)

The absence of M+1 peaks from the chromatogram is due to selected reaction monitoring. Because the first quadrupole and the third quadrupole only allowed structures with m/z 304 and m/z 182 respectively, the ions with these m/z were the only ones that reached the detector. The ions containing the C13isotopes, having a m/z of 305, were filtered out and not allowed access to the detector.

05

Chromatogram show just one clean peak

Part (d)

Like the ions with the C13isotopes, the ions that did not have a m/z of 304 or m/z of 182 were not able to reach the detector because they were filtered out by the quadrupoles used in the selected ion monitoring system.

06

Which atoms are labelled with deuterium

Part (e)

Because, after fragmentation cocaine should still have a m/z of 182 despite having 5 deuterium ions (H2 ),H52 should be bonded to the fragment that would be dissociated from cocaine to form an ion with an m/z of 182. Because the ester group fragment that would be dissociated from cocaine in (b) has 5 hydrogen atoms, the 5 deuterium ions should be located in this fragment so that when it gets dissociated, the remaining ion would have a m/z of 182.

07

Use \(^2{H_5}\)-cocaine for measuring cocaine in blood

Part (f)

To determine the concentration of cocaine in blood, what can be done is quantitative analysis based on the area of a chromatographic peak. Treating H52-cocaine as an internal standard, the relationship of cocaine and H52-cocaine is determined and quantified into the response factor. After determining the response factor using standard mixtures, the sample, which usually contains both the internal standard and the analyst, is fed into the chromatograph. Because the area of a peak is proportional to the quantity of the component it represents, equation 1 can be used to determine the amount of cocaine in a blood sample:

Ax[X]=FAs[S] (Equation 1)

where: Axis the area of the analyte signal,

[X]is the concentration of the analyte,

Fis the response factor,

Asis the area of the internal standard peak,

[S]is the concentration of the standard.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a). Sketch a graph of the van Deemnter equation (plate height versus linear flow rate).what would the curve look like if the multiple path term were 0? If the longtitundinal diffusinal diffusion term were 0?

(b). Explain why the van Deemter curve for 1.8μmparticles in figure 25-3is nearly flat at high flow rate.what can you say about each of the terms in the van Deemeter equation for 1.8μmparticles.

(c). Explain why the 2.7μmsuperficially porous particle enables separations similar to those achieved by 1.8μmtotally porous particles,but the superficially porous particle requires lower pressure.

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.

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?

A mixture of 14compounds was subjected to a reversed-phase gradient separation going from 5%to 100%acetonitrile with

a gradient time of 60min. The sample was injected at t =time. All peaks were eluted between 22and 50min.

(a) Is the mixture more suitable for isocratic or gradient elution?

(b) If the next run is a gradient, select the starting and ending %acetonitrile

and the gradient time.

To which kinds of analytes do these liquid chromatography detectors respond?

(a) ultraviolet

(b) refractive index

(c) evaporative light scattering

(d) charged aerosol

(e) electrochemical

(f) fluorescence

(g) nitrogen chemiluminescence

(h) conductivity

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free