Chapter 25: Q26P (page 710)

Chromatography–mass spectrometry. HPLC separation of
enantiomers of the drug Ritalin on a chiral stationary phase was
shown in Problem 25-13.
(a) Detection is by atmospheric pressure chemical ionization with
selected reaction monitoring of the m/z 23484 transitions. Explain
how this detection works and propose structures for m/z 234 and m/z 84.
(b) For quantitative analysis, the internal standard ${}^{2}H_{3}$ -Ritalin with
a deuterated methyl group was added. Deuterated enantiomers have
the same retention times as unlabelled enantiomers. Which selected
reaction monitoring transition should be monitored to produce a
chromatogram of the internal standard in which unlabelled Ritalin
will be invisible?

Short Answer

Expert verified

The part (a), part (b) is

It is selected by mass filter Q1 in a triple quadrupole spectrometer and It is selected by mass filter Q3 in a triple quadrupole spectrometer
Internal standard has a nominal mass of 236, so the protonated molecule (right structure on picture) has m/z 237

Step by step solution

Propose structures for m/z 234 and m/z 84

Part (a)

At m/z 234, compound has prominent peak, so the structure is ${MH}^{+}$ :

It is selected by mass filter Q1 in a triple quadrupole spectrometer.

C5H10N+

At m/z 84 the fragment is: $C_{5} H_{10} N^{+}$

It is selected by mass filter Q3 in a triple quadrupole spectrometer.

Unlabelled Ritalin will be invisible

Part (b)

The selected reaction monitoring transition that should be monitored to produce a chromatogram of the internal standard in which unlabelled Ritalin will be invisible is m/z

237 84:

Internal standard has a nominal mass of 236, so the protonated molecule (right structure on picture) has m/z 237. Cleavage of bond give the sam fragment as in task (a)