Title: CHE2FCA Mass Spectrometry

Q: Coronene is a polycyclic aromatic hydrocarbon (PAH) containing six fused benzene rings with a molecular formula of C_24H_12.

The mass spectrum of caronene exhibits a molecular ion at m/z 300 with a relative ion intensity of 100% (the molecular ion is also the base peak in this instance).

True/false statement follows.

The signal due to M + 1 at m/z 301 (mostly due to contribution form ^13C) is expected to have a relative ion intensity of 26%.
- True (Well done. That's right, there's a contribution from each carbon atom in the molecule to the M + 1 signal due to the isotope abundance of ^13C and a smaller contribution from each hydrogen due to ^2H. (M + 1) intensity = (24 x 1.08) + (12 x 0.012) = 26.09% [Image])
- False (Refresh the isotope abundance of ^13C and the contribution of each carbon atom to the M + 1 signal. Remember, there's a contribution from each carbon atom in the molecule to the M + 1 signal due to the isotope abundance of ^13C and a smaller contribution from each hydrogen due to ^2H. [Image])
A: True

Q: The term ‘mean free path’ that is regularly used in mass spectrometry refers to the distance an ion travels from the ion source to the detector.
- True (Remember, in mass spectrometery, after we form a characteristic gaseous ion (such as a molecular ion) we want it to travel intact through the mass spectrometer until it is detected. That is, we do not want it to fragment after collision with other gaseous molecule or ion species.)
- False (That's right. The term ‘mean free path’ is actually the distance an ion travels before a collision with another gaseous molecule or ion species.)
A: False

Q: The term ‘mean free path’ that is regularly used in mass spectrometry refers to the distance an ion travels through the mass spectrometer before a collision with another gaseous species.
- True (Well done)
- False (Remember, in mass spectrometery, after we form a characteristic gaseous ion (such as a molecular ion) we want it to travel intact through the mass spectrometer until it is detected. That is, we do not want it to fragment after collision with other gaseous ion species.)
A: True

Q: Fragmentation of a molecule can occur at a radical site where an unpaired electron is donated to form a new bond.
- True (Well done. Fragmentation of radical cations by a structure rearrangement mechanism can occur at the radical site by donation of unpaired electron to form a new bond. (fragmentation also at charge site where positive charge attracts electron to form new bond ))
- False (Refresh fragmentation of a radical ion. Fragmentation of radical cationsby structure rearrangement mechanism can occur at the radical site by donation of unpaired electron to form a new bond. (fragmentation also at charge site where positive charge attracts electron to form new bond ))
A: True

Q: A fragment ion must always have a lower m/z than that of the molecular ion from which it is formed.
- True (Correct. A fragment ion (as opposed to a recombination ion) always has a lower m/z than the parent molecular ion.)
- False (Note: A fragment ion (as opposed to a recombination ion) always has a lower m/z than the parent molecular ion.)
A: True

Q: Electrospray ionization is an example of a “soft ionization” technique.
- True (Correct. • Electron ionisation (EI) (~ 70 eV) is a hard ionisation technique • Soft ionisation techniques include: chemical ionization (CI) photoionization (PI) fast atom bombardment (FAB) laser desorption (LD) electrospray ionization (ESI) atmospheric pressure ionization (API))
- False (Remember: • Electron ionisation (EI) (~ 70 eV) is a hard ionisation technique • Soft ionisation techniques include: chemical ionization (CI) photoionization (PI) fast atom bombardment (FAB) laser desorption (LD) electrospray ionization (ESI) atmospheric pressure ionization (API))
A: True

Q: A minimum electron energy of 70 eV is required for the electron ionisation of a molecule.
- True (Refresh ionisation energies needed to remove a electron (bonding and non-bonding) from a molecule. Ionisation of sigma- and pi-bonds typically requires only 7 - 13 eV. Electron ionisation (70 eV) is a standard energy for comparison of mass spectrums. Electron ionisation at 70 eV is normal practise and used to produce characteristic fragmentation ions from the molecular ion in the mass spectrum that can be compared to reference spectra for identifications purposes.)
- False (Well done. Ionisation of sigma- and pi-bonds typically requires only 7 - 13 eV. Electron ionisation at 70 eV is normal practise and used to produce characteristic fragmentation ions from the molecular ion in the mass spectrum that can be compared to reference spectra for identifications purposes.)
A: False

Q: Fast Atom Bombardment (FAB) is a “hard” ionisation method used in mass spectrometry.
- True (Refresh "soft and "hard" ionisation methods. Electron ionisation (EI) (~ 70 eV) is a hard ionisation technique. Soft ionisation techniques include: chemical ionization (CI), photoionization (PI), fast atom bombardment (FAB), laser desorption (LD), electrospray ionization (ESI), & atmospheric pressure ionization (A)
- False (Well done. Electron ionisation (EI) (~ 70 eV) is a hard ionisation technique. Soft ionisation techniques include: chemical ionization (CI), photoionization (PI), fast atom bombardment (FAB), laser desorption (LD), electrospray ionization (ESI), & atmospheric pressure ionization (API))
A: False

Q: The most favoured ionisation site in a molecule is associated with the removal of the electron with the highest ionisation energy.
- True (Refresh: Most favoured ionization site involves removal of the electron of lowest ionization energy. This is typically a non-bonding (lone pair) electron when elements such as oxygen, nitrogen or sulphur, etc, are present in the compound.)
- False (Correct. Most favoured ionization site involves removal of the electron of lowest ionization energy. This is typically a non-bonding (lone pair) electron when elements such as oxygen, nitrogen or sulphur, etc, are present in the compound.)
A: False

Q: The number of 70 eV electrons in an ion source can be increased by using an electron multiplier.
- True (Note: An electron multiplier is used to measure the detector signal not to generate more electrons in the ion source.)
- False (Well done. An electron multiplier is used to measure detector signal not to generate more electrons in the ion source.)
A: False

Q: Chlorine has two isotopes with masses of 35 and 37 Da. The molecular ion for Cl_2 occurs at m/z 70.
- True (Correct. The molecular ion is the peak of highest intensity containing the most abundant isotopes in group of molecule peaks at the highest m/z values in mass spectrum associated with the intact molecule. Chlorine isotopes: ^35Cl 34.968 85 Da, 75.78% ^37Cl 36.965 90 Da, 24.22% Chlorine molecular ion is ^35Cl-^35Cl = m/z 70 Th)
- False (Refresh the definition of a molecular ion. Refresh the identification of the molecular ion from a group of molecule ion m/z signals. Chlorine isotopes: ^35Cl 34.968 85 Da, 75.78% ^37Cl 36.965 90 Da, 24.22%)
A: True

Q: A single carbon-12 cation has a m/z of exactly 12.0000 Th.
- True (Correct. • one dalton is 1/12 the mass of a single ^12C atom • one ^12C atom has a mass of 12.0000 Da • one ^12C cation has a m/z of 12.0000 Th (thomson))
- False (Note: • one dalton is 1/12 the mass of a single ^12C atom • one ^12C atom has a mass of 12.0000 Da • one ^12C cation has a m/z of 12.0000 Th (thomson))
A: True

Q: A quadrupole ion trap can be used as a mass analyser in a mass spectrometer.
- True (Well done. Separation of ions according to their mass-to-charge ratio (m/z) occurs in the mass analyser. The mass analyser can be: • magnetic sector • quadrupole filter • quadrupole ion trap • time-of-flight)
- False (Refresh the different types of mass analysers used in mass spectrometry. Separation of ions according to their mass-to-charge ratio (m/z) occurs in the mass analyser. The mass analyser can be: • magnetic sector • quadrupole filter • quadrupole ion trap • time-of-flight)
A: True

Q: Mass analyzers are usually maintained at a typical vacuum of 10 kPa
- True (Remember, a mass spectrometer needs to operate under low pressure (= high vacuum). Typically, a mass spectrometer operates at a low vacuum < 6.6 x 10^-3 Pa (< 6.6 mPa) so as to remove atmospheric gaseous molecules and maintain a long 'mean free path'. Note: 10 kPa is about 1/10 of an atmosphere (101 kPa), ie, not a high vacuum (= low pressure).)
- False (Well done! A mass spectrometer requires a high vacuum (= low pressure) to remove atmospheric gaseous molecules and maintain a 'long mean free path'. Typically, a mass spectrometer vacuum is < 6.6 x 10^-3 Pa.)
A: False

Q: The width of a mass spectrum peak at m/z 180 Th is found to be 0.6 Th when measured at 10% peak height. The 10% mass resolution of the mass spectrometer is 300.
- True (Well done.)
- False
A: True
Feedback: Check your calculation.

Q: The width of a mass spectral peak at m/z 150 is found to be 0.4 Th when measured at 50% peak height. The 50% mass resolution of the mass spectrometer is 60.
- True (Check your calculation. Rs = m / Δm = 150 / 0.4 = 375)
- False (Correct.)
A: False

Q: Samples of approximately 10^-3 – 10^-6 gram are required to produce a mass spectrum.
- True (Refresh required sample sizes in mass spectrometry. Mass Spectrometry can identify compounds using ~1 ng – 1 pg (10^-9 – 10^-12 g) of sample)
- False (Correct. Mass Spectrometry can identify compounds using ~1 ng – 1 pg (10^-9 – 10^-12 g) of sample)
A: False

Q: There are five “rings + double bonds” associated with a C_7H_5NOF_2^+• molecular ion.
- True (Correct. RDB = IV – ½ I + ½ III + 1 (for A_I B_II C_III D_IV) = 7 – ½ (7) + ½ (1) + 1 = 5)
- False (Check your calculation. Remember to use RDB = IV – ½ I + ½ III + 1 (for A_I B_II C_III D_IV))
A: True

Q: Not all molecules produce a base peak in their mass spectrum.
- True (Note: All mass spectrums have a base peak = most intense peak.)
- False (Correct. All mass spectrums have a base peak = most intense peak.)
A: False

Q: All molecules produce a base peak in their mass spectrum.
- True (Correct. Base peak in a mass spectrum is the commonly observed most abundant peak that is assigned a peak intensity of 100% to which all other peak intensities are referred.)
- False (Refresh the difference between a base peak and molecular ion peak in a mass spectrum.)
A: True

Q: There are four “rings + double bonds” associated with a C_9H_11BrN_2O_5^•+ molecular ion.
- True (Check your calculation. Remember to use RDB = IV – ½ I + ½ III + 1 (for A_I B_II C_III D_IV) where A = H + Br = 11 + 1 = 12 B = O = 5 (not needed) C = N = 2 D = C = 9)
- False (Correct. RDB = IV – ½ I + ½ III + 1 (for A_I B_II C_III D_IV) = 9 – ½ (12) + ½ (2) + 1 = 5)
A: False

Q: Bromine has two isotopes with masses of 79 and 81 Da. The molecular ion for Br_2 occurs at m/z 162.
- True (Refresh the definition of a molecular ion Remember, the molecular ion is the signal of highest intensity containing the most abundant isotopes in the group of molecule ion signals at the highest m/z values in mass spectrum associated with the intact molecule. Bromine isotopes: ^79Br 78.918 34Da, 50.69% ^81Br 80.916 29Da, 49.31% Bromine molecular ion is ^79Br-^79Br = m/z 158 Th)
- False (Correct. The molecular ion is the signal of highest intensity containing the most abundant isotopes in the group of molecule ion signals at the highest m/z values in mass spectrum associated with the intact molecule. Bromine isotopes: ^79Br 78.918 34Da, 50.69% ^81Br 80.916 29Da, 49.31% Bromine molecular ion is ^79Br-^79Br = m/z 158 Th)
A: False

Q: The McLafferty rearrangement of a carbonyl compound (aldehyde or ketone) involves the transfer of a hydrogen atom from a carbon three atoms away to the oxygen atom that results in the elimination of a neutral water molecule and formation of a 'new' cation radical.
- True (Note the following. The McLafferty rearrangement of a carbonyl compound (aldehyde or ketone) involves the transfer of a hydrogen atom from a carbon three atoms away to the oxygen atom that results in the elimination of a neutral alkene molecule and formation of a 'new' cation radical.)
- False (Correct, well done. Correct, well done. The McLafferty rearrangement of a carbonyl compound (aldehyde or ketone) involves the transfer of a hydrogen atom from a carbon three atoms away to the oxygen atom that results in the elimination of a neutral alkene molecule and formation of a 'new' cation radical.)
A: False

Q: Electron ionisation involves firing a beam of high energy electrons, typically 70 eV, at a gaseous molecule to remove an electron.
- True (Well done. That's right, electron ionization (EI) is a hard ionisation technique that involves bombardment of vaporized gaseous molecules with a beam of electrons with, typically, energy of 70 eV. This knocks an electron from the molecule to produce a cation radical. Whereas, chemical ionisation is a soft ionisation technique that uses a reagent gas, such as methane or ammonia, to transfer a proton to the gaseous molecule to produce an intact protonated molecule ion (MH^+) in the mass spectrum.)
- False (Remember the difference between electron ionisation and chemical ionisation. Electron ionization (EI) is a hard ionisation technique that involves bombardment of vaporized gaseous molecules with a beam of electrons with, typically, energy of 70 eV. This knocks an electron from the molecule to produce a cation radical. Whereas, chemical ionisation is a soft ionisation technique that uses a reagent gas, such as methane or ammonia, to transfer a proton to the gaseous molecule to produce an intact protonated molecule ion (MH^+) in the mass spectrum.)
A: True

Q: Coronene is a polycyclic aromatic hydrocarbon (PAH) containing six fused benzene rings with a molecular formula of C_24H_12.

The mass spectrum of caronene exhibits a molecular ion at m/z 300 with a relative ion intensity of 100% (the molecular ion is also the base peak in this instance).

True/false statement follows.

The signal due to M + 1 at m/z 301 (mostly due to contribution form ^13C) is expected to have a relative ion intensity of 24%.
- True (Well done. That's right, there's a contribution from each carbon atom in the molecule to the M + 1 signal due to the isotope abundance of ^13C [Image])
- False (Refresh the isotope abundance of ^13C and the contribution of each carbon atom to the M + 1 signal. Remember, there's a contribution from each carbon atom in the molecule to the M + 1 signal due to the isotope abundance of ^13C [Image])
A: True