1. Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Chapter 12
Structure Determination: Mass Spectrometry and Infrared Spectroscopy
Suggested Problems – 1-11,14-16,18,23,26,30-34,41-2

2. Determining the Structure of an Organic Compound
The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants
In the 19th and early 20th centuries, structures were determined by synthesis and chemical degradation that related compounds to each other
Physical methods now permit structures to be determined directly. We will examine:
mass spectrometry (MS)
infrared (IR) spectroscopy
nuclear magnetic resonance spectroscopy (NMR)
ultraviolet-visible spectroscopy (UV-VIS)

3. Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments
Mass spectrometry (MS) determines molecular weight by measuring the mass of a molecule
Components of a mass spectrometer:
Ionization source - Electrical charge assigned to sample molecules
Mass analyzer - Ions are separated based on their mass-to-charge ratio
Detector - Separated ions are observed and counted
Mass spectrometry is a technique for measuring the mass, and therefore the molecular weight of a molecule.
In addition, it’s often possible to gain structural information about a molecule by measuring the masses of the fragments produced when molecules are broken apart.

4. Electron-Ionization, Magnetic-Sector Mass Spectrometer
Small amount of sample undergoes vaporization at the ionization source to form cation radicals
Amount of energy transferred causes fragmentation of most cation radicals into positive and neutral pieces
While many different kinds of mass spectrometers are available, the electron impact, magnetic-sector instruments are the most common.
In these a beam of electrons knocks an electron from a molecule in the gas phase. Electron bombardment transfers so much energy that most of the cation radicals fragment after formation. They break apart into smaller pieces, some of which retain the positive charge and some of which are neutral.

5. Electron-Ionization, Magnetic-Sector Mass Spectrometer
Fragments pass through a strong magnetic field in a curved pipe that segregates them according to their mass-to-charge ratio
Positive fragments are sorted into a detector and are recorded as peaks at the various m/z ratios
Mass of the ion is the m/z value
The fragments then flow through a curved pipe in a strong magnetic field, which deflects them into different paths based on their mass-to-charge ratio, m/z. Neutral fragments are not deflected by the magnetic field and are lost on the walls of the pipe. Positively charged fragments are sorted by the mass spectrometer onto a detector, which records them as peaks at the various m/z ratios. Since the typical molecule has only one positive charge (loses only one electron), the m/z ratio corresponds to the particle’s mass.

6. The electron-ionization, magnetic-sector mass spectrometer

7. Quadrupole Mass Analyzer
Comprises four iron rods arranged parallel to the direction of the ion beam
Specific oscillating electrostatic field is created in the space between the four rods
Only the corresponding m/z value is able to pass through and reach the detector
Other values are deflected and crash into the rods or the walls of the instrument
In a quadrupole mass analyzer, a set of four solid rods is arranged parallel to the direction of the ion beam. An oscillating electrostatic field is generated in the space between the rods.

8. The Quadrupole Mass Analyzer
Only one m/z value will make it through the quadrupole region – the others will crash into the quadrupole rods or the walls of the instrument and never reach the detector.

9. Representing the Mass Spectrum
Plot mass of ions (m/z) (x-axis) versus the intensity of the signal (roughly corresponding to the number of ions) (y-axis)
Tallest peak is base peak (Intensity of 100%)
Peak that corresponds to the unfragmented radical cation is parent peak or molecular ion (M+)
The mass spectrum of a compound is typically presented as a bar graph, with masses (m/z values) on the x axis and intensity, or relative abundance of ions of a given m/z striking the detector, on the y axis. The molecular ion is often not the base peak.

10. Interpreting Mass Spectra
Provides the molecular weight from the mass of the molecular ion
Double-focusing mass spectrometers have a high accuracy rate
In compounds that do not exhibit molecular ions, soft ionization methods are used
Double focusing instruments have such high resolution that they provide mass measurements accurate to amu, making it possible to distinguish between two formulas with the same nominal mass. For example, both C5H12 and C4H8O have MW=72, but they differ slightly beyond the decimal point: C5H12 has an exact mass of amu, whereas C4H8O has an exact mass of amu. A high resolution mass spectrometer can easily distinguish between them.
It is important to note that exact mass measurements refer to molecules with specific isotopic compositions. Thus, the sum of the exact atomic masses of the specific isotopes in a molecule is measured – amu for 1H, amu for 12C, amu for 14N, amu for 16O, and so on – rather than the sum of the average atomic masses of eleements, as found on a periodic table.
Because some compounds fragment so easily, it is often difficult to observe the molecular ion. In these cases, soft ionization methods that do not involve electron bombardment can be used to minimize or prevent fragmentation. The above mass spectrum is for 2,2-dimethylpropane which has a MW of 72. The molecular ion is not observed because it fragments too easily with electron impact instruments.
There is also a small peak at M + 1 due to the presence of different isotopes in the molecule. The peak at 58 above would be due to the small presence of 13C in the molecule.

11. High Resolution Mass Spectrometry Can Distinguish
Between Compound with the Same Molecular Mass
Exact Masses of Isotopes

12. Natural Abundance of Isotopes

13. Other Mass Spectral Features
Mass spectrum provides the molecular fingerprint of a compound
The way molecular ions break down, can produce characteristic fragments that help in identification
Interpretation of molecular fragmentation pattern assists in the derivation of structural information
The mass spectrum serves as a kind of molecular fingerprint for the a compound. Each organic compound fragments in a unique way.
The positive charge, not surprisingly, often remains with the fragment that is best able to stabilize it.

14. Mass Spectral Fragmentation of Hexane
Hexane (m/z = 86 for parent) has peaks at m/z = 71, 57, 43, 29
Most hydrocarbons fragment in many ways. The hexane molecule shows an abundance of different peaks. Since all the carbon-carbon bonds of hexane are electcronically similar, all break to a similar extent, giving rise to the observed mixture of ions.
In the above spectrum, the loss of a methyl radical from the hexane cation radical leads to a peak at 71. The loss of an ethyl radical leads to the peak at 75, etc…
With practice, one can often analyze the fragmentation pattern of an unknown compound and work backward to a structure that is compatible with the data.

15. Worked Example
The male sex hormone testosterone contains only C, H, and O and has a mass of amu, as determined by high-resolution mass spectrometry
Determine the possible molecular formula of testosterone

16. Worked Example Solution:
Assume that hydrogen contributes to the mass of
Dividing by ( difference between the atomic weight of one H atom and 1) gives 26.67
Approximate number of H in testosterone
Determine the maximum number of carbons by dividing 288 by 12
List reasonable molecular formulas containing C,H, and O that contain hydrogens and whose mass is 288
Remember the following exact mass account for the major isotopes of C, H, O, and N: amu for 1H, amu for 12C, amu for 14N, amu for 16O.
Since the sex hormone only contains C, H, and O, we can see that C will not contribute to any digits to the right of the decimal point. The contribution from O will be quite minor. H on the other hand makes the major contribution to the mass to the right of the decimal point. The assumption that all of the values for the mass to the right of the decimal point are contributed by H is erroneous, because O contributes some, but it is a very good first approximation.

17. Worked Example The possible formula for testosterone is C19H28O2
Knowing a mass of 288, one can consult a table of possible compounds whose C, H, and O atoms total 288. Finding the exact mass corresponding to a particular formula allows identification of the compound.

18. Mass Spectrometry of Some Common Functional Groups
Alcohols
Fragment through alpha cleavage and dehydration
Alcohols undergo alpha cleavage on either side of the carbon being the hydroxyl group. The oxygen containing moiety typically bears the positive charge due to resonance stabilization of the positive charge by the oxygen atom.
Alcohols also typically will undergo dehydration – the loss of water – with the loss of 18 mass units.

19. Mass Spectrometry of Some Common Functional Groups
Amines
Nitrogen rule of mass spectrometry
A compound with an odd number of nitrogen atoms has an odd-numbered molecular weight
Amines undergo -cleavage, generating alkyl radicals and a resonance-stabilized, nitrogen-containing cation
The nitrogen rule is a result of the fact that N is trivalent, thus requiring an odd number of hydrogen atoms.
An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms.
Aliphatic amines undergo alpha cleavage in a similar fashion to alcohols. Resonance stabilization by the nitrogen is a driving force in the nitrogen species being the radical cation.

20. Mass Spectrometry of Some Common Functional Groups
Halides
Elements comprising two common isotopes possess a distinctive appearance as a mass spectra
Since both chlorine and bromine have two common isotopes gives their mass spectra a distinctive appearance.
Chlorine, for example, exists as two isotopes, 35C and 37C, in a roughly 3:1 ratio. In any peak that contains chlorine, there will be a peak corresponding to the 35C as well as a peak of 1/3 the intensity two units higher corresponding to 37C. The mass spectrum above shows the chlorine substitution pattern for the peak at 64 and at 49 (loss of a methyl group).
In the case of bromine, the isotopic distribution is 50.7% 79Br and 49.3% 81Br, a ratio of roughly 1:1. Consequently any peak that contains bromine will exhibit a peak two mass units higher of nearly equal intensity. The absence of an M+2 peak in a molecule that contains bromine suggests the fragment does not contain a bromine atom.

21. Fragmentation of Carbonyl Compounds
A C–H that is three atoms away leads to an internal transfer of a proton to the C=O called the McLafferty rearrangement
Carbonyl compounds can also undergo -cleavage
Ketones and aldehydes that have a hydrogen on a carbon three atoms away from the carbonyl group undergo a characteristic mass-spectral cleavage called the McLafferty rearrangment.
The hydrogen atom is transferred to the carbonyl oxygen, a C-C bond is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment.
Ketones and aldehydes also undergo alpha cleavage to generate acylium ions that bear the positive charge.

22. Worked Example
List the masses of the parent ion and of several fragments that can be found in the mass spectrum of the following molecule
2-methyl-2-pentanol

23. Worked Example Solution: The molecule is 2-methyl-2-pentanol
It produces fragments resulting from dehydration and alpha cleavage
Peaks may appear at M+=102(molecular ion), 87, 84, 59
One would expect to see alpha cleavage on both sides of the carbon bearing the hydroxyl group as well as dehydration.

24. Mass Spectroscopy in Biological Chemistry: Time-of-Flight (TOF) Instruments
Most biochemical analyses by MS use soft ionization methods that charge molecules with minimal fragmentation
Electrospray ionization (ESI)
High voltage is passed through the solution sample
Sample molecule gains one or more protons from the volatile solvent, which evaporates quickly
Matrix-assisted laser desorption ionization (MALDI)
Sample is absorbed onto a suitable matrix compound
Upon brief exposure to laser light, energy is transferred from the matrix compound to the sample molecule
Because large molecules do not vaporize easily, methods such as soft ionization can be used to charge molecules without fragmenting them to too great a degree.
Time of flight instruments are very sensitive because they separate masses based on how long it takes them to migrate down a magnetically charged tube. Smaller molecules move move rapidly while larger molecules move very slowly. Molecules up to 100,000 dalton (amu) can give mass spectra with these instruments.

25. MALDI–TOF Mass Spectrum of Chicken Egg-White Lysozyme

26. Spectroscopy and the Electromagnetic Spectrum
Waves are classified by frequency or wavelength ranges

27. Spectroscopy and the Electromagnetic Spectrum
Electromagnetic radiation seems to have dual behavior
Possesses the properties of a photon
Behaves as an energy wave

28. Spectroscopy and the Electromagnetic Spectrum
Speed of the wave
The unit of electromagnetic energy is called quanta

29. Spectroscopy and the Electromagnetic Spectrum
Considering the plank equation and multiplying ɛ by Avogadro’s number NA:

30. Absorption Spectrum
Organic compounds exposed to electromagnetic radiation can absorb energy of only certain wavelengths (unit of energy)
Transmit energy of other wavelengths
Changing wavelengths to determine which are absorbed and which are transmitted produces an absorption spectrum
In infrared radiation, absorbed energy causes bonds to stretch and bend more vigorously
In ultraviolet radiation, absorbed energy causes electrons to jump to a higher-energy orbital
When an organic compound is exposed to a beam of electromagnetic radiation, it absorbs energy of some wavelengths but passes, or transmits, energy of other wavelengths. If we irradiate the sample with energy of many different wavelengths and determine which are absorbed and which are transmitted, we can measure the absorption spectrum of the compound.

31. Worked Example
Calculate the energy in kJ/mol for a gamma ray with λ = 5.0×10-11m
Solution:

32. Infrared Spectroscopy
IR region has lower energy than visible light (below red - produces heating as with a heat lamp)
Wavenumber:
Infrared spectroscopy

33. Infrared Energy Modes
Molecules possess a certain amount of energy that causes them to vibrate
Molecule absorbs energy upon electromagnetic radiation only if the radiation frequency and the vibration frequency match
The energy a molecule gains when it absorbs radiation must be distributed over the molecule in some way. With infrared radiation, the absorbed energy causes bonds to stretch and bend more vigorously.
The useful IR region is from 4000 to 400 cm-1.
The amount of energy a molecule contains is not continuously variable but is quantized. That is, a molecule can stretch or bend only at specific frequencies corresponding to specific energy levels.
When a molecule is irradiated with electromagnetic radiation, energy is absorbed if the frequency of the radiation matches the frequency of the vibration. The result of this energy absorptionis an increased amplitude for the vibration.
Since each frequency absorbed by a molecule corresponds to a specific molecular motion, we can find what kinds of motions a molecule has by measuring itsIR spectrum. By interpreting these motions, we can find out what kinds of bonds (functional groups) are present in the molecule.

34. Interpreting Infrared Spectra
IR spectrum interpretation is difficult as the arrangement of organic molecules is complex
Disadvantage - Generally used only in pure samples of fairly small molecules
Advantage - Provides a unique identification of compounds
Fingerprint region cm-1 to 400 cm-1 (approx)
Complete interpretation of the IR spectrum is not necessary to gain useful structural information
IR absorption bands are similar among compounds
Most IR spectra contain many peaks because most organic molecules have dozens of different bond stretching and bending motions, and thus have dozens of different absorptions.
Neverthless, the IR is very useful to identify the presence, or absence, of a multitude of different kinds of functional groups.
The region from 1500 to 4000 cm-1 is referred to as the functional group region and can be utilized to ascertain the presence of a variety of functional groups. The region from 400 to 1500 cm-1 is the fingerprint region which serves as a fingerprint for an individual compound. As such, if two IR are identical in the fingerprint region, they are most likely the same compound.

35. Characteristic IR Absorptions of Some Functional Groups
The absorptions of these functional groups are consistent from one molecule to the next. That is, if they appear in an IR of a molecule, one can usually deduce the presence of that functional group.

36. IR Spectra of Hexane, 1-Hexene, and 1-Hexyne
In comparing the spectra of hexane, hexene, and hexyne, two features stand out. One is the presence of a C-C double bond or triple bond stretch which is indicative of the presence of a double or triple bond. The second is the presence of C-H stretches in the 3000 cm-1 region. Sp3 C-H stretches occur around 2900 cm-1. An sp2 C-H stretch occurs at higher frequency (3050 cm-1) and an sp C-H stretch occurs higher still (3300 cm-1).

37. Regions of the Infrared Spectrum
Region from 4000 to 2500 cm-1 can be divided into areas characterized by:
Single-bond stretching motions
Triple-bond stretching motions
Absorption by double bonds
Fingerprint portion of the IR spectrum
Single bond stretching regions occur above 2850 cm-1.
Triple bond stretching regions occur between 2850 cm-1 and 2000 cm-1.
Double bond stretching regions occur from 2000 cm-1 to aboug 1500 cm-1.
The fingerprint region usually involves single bond stretching vibrations of the C-C, C-O, and C-N type.
Note the stronger triple bond stretches at a higher frequency than does the double bond which stretches at a higher frequency than the single bond, indicative of the bond strengths involved and the energies needed to effect their stretch.

38. Harmonic Oscillator System comprising two atoms connected by a bond
In a vibrating bond, the vibrating energy is in a constant state of change from kinetic to potential energy and vice versa
Equation for natural frequency of vibration for a bond -
Interpreting Infrared Spectra

39. Worked Example
Using IR spectroscopy, distinguish between the following isomers:
CH3CH2OH and CH3OCH3
Solution:
CH3CH2OH is a strong hydroxyl bond at 3400–3640 cm-1
CH3OCH3 does not possess a band in the region 3400–3640 cm-1

40. Infrared Spectra of Some Common Functional Groups
Alkanes
No functional groups
C–H and C–C bonds are responsible for absorption
C–H bond absorption ranges from 2850 to 2960 cm-1
C–C bonds show bands between 800 to cm-1

41. Infrared Spectra of Some Common Functional Groups
Alkenes
Vinylic =C–H bonds are responsible for absorption from 3020 to 3011cm-1
Alkene C=C bonds are responsible for absorption close to 1650cm-1
Alkenes possess =C–H out-of-plane bending absorptions in the 700 to 1000 cm-1 range

42. Infrared Spectra of Some Common Functional Groups
Alkynes
C≡C stretching absorption exhibited at 2100 to 2260 cm-1
Similar bonds in 3-hexyne show no absorption
Terminal alkynes such as 1-hexyne possess ≡C–H stretching absorption at 3300 cm-1
The reason that 3-hexyne does not show a C-C triple bond absorption is because the molecule is symmetrical. The stretch of this bond results in no change to the dipole moment of this molecule as a result of the symmetry.

43. Some Vibrations are Infrared Inactive
A bond absorbs IR radiation only if its dipole moment changes when it vibrates.

44. Aromatic Compounds Weak C–H stretch at 3030 cm1
Weak absorptions at 1660 to 2000 cm1 range
Medium-intensity absorptions at 1450 to 1600 cm1

45. Alcohols and Amines Alcohols Amines O–H 3400 to 3650 cm1
Usually broad and intense
Amines
N–H 3300 to 3500 cm1
Sharper and less intense than an O–H
Hydrogen bonding is responsible for making the O-H stretch broad.
Primary amines tend to exhibit two N-H stretches between 3300 and 3500 cm-1. Secondary amines only exhibit one. This is due to two N-H stretches which are possible for primary amines.

46. The IR Spectrum of an Alcohol

47. The IR Spectrum of an Amine

48. Carbonyl Compounds
Strong, sharp C=O peak in the range of 1670 to 1780 cm1
Exact absorption is characteristic of type of carbonyl compound
Principles of resonance, inductive electronic effects, and hydrogen bonding provides a better understanding of IR radiation frequencies
Carbonyl C=0 stretches represent one of the most easily identified absorptions.
The exact position of the absorption is characteristic of the specific type of carbonyl compound – for example, aldehyde, ketone, ester, amide.

49. Carbonyl Compounds Aldehydes 1730 cm1 in saturated aldehydes
1705 cm1 in aldehydes next to double bond or aromatic ring
Low absorbance frequency is due to the resonance delocalization of electron density from the C=C into the carbonyl
Conjugation of the carbonyl group in the aldehyde lowers the frequency of the C=O stretch.
Resonance delocalization of electron density introduced by the conjugation serves to increase the C=O bond length slightly. This results in a lower vibrational frequency for the carbonyl group.

50. The IR Spectrum of an Aldehyde
The two aldehyde C-H stretches at 2820 and 2720 cm-1 are commonly referred to as aldehyde fangs.
The carbon—hydrogen stretch of an aldehyde hydrogen occurs
at 2820 cm–1 and at 2720 cm–1.

51. Ketones
Saturated open-chain ketones and six-membered cyclic ketones absorb at 1715cm-1
Five-membered ketones absorb at 1750cm-1
Stiffening of C=O bond due to ring strain
Four members absorb at 1780cm-1
As a general rule, a conjugated carbonyl group absorbs IR radiation 30 to 30 cm-1 lower than the corresponding saturated carbonyl compound.

52. This C═O Bond Is Essentially a Pure Double Bond

53. This C═O Bond Has Significant Single Bond Character
The less double bond character, the lower the frequency.

54. Carbonyl Compounds Esters Saturated esters absorb at 1735 cm-1
Esters possess two strong absorbances within the range of 1300 to 1000 cm-1
Esters adjacent to an aromatic ring or a double bond absorb at 1715 cm-1
The electron withdrawing effect of the non-carbonyl oxygen raises the frequency of absorption of the carbonyl group relative to a ketone to a small degree.

55. The IR Spectrum of an Ester

56. The IR Spectrum of a Carboxylic Acid
The broadening of the O-H resonance in a carboxylic acid is due to hydrogen bonding that results from dimer formation with another carboxylic acid molecule.

57. Hydrogen Bonded OH Groups Stretch at a Lower Frequency
It is easier to stretch a hydrogen bonded OH group.

58. The IR Spectrum of an Amide
Why does the amide carbonyl resonate at a lower frequency than the traditional carbonyl group?
Answer: the nitrogen donates electrons into the carbonyl group by resonance which means the C=O bond has more single bond character.

59. Worked Example
Identify the possible location of IR absorptions in the compound below

60. Worked Example Solution:
The compound possesses nitrile and ketone groups as well as a carbon–carbon double bond
Nitrile absorption occurs at 2210–2260 cm-1
Ketone exhibits an absorption bond at 1690 cm-1
Double bond absorption occurs at 1640–1680 cm-1
Infrared Spectra of Some Common Functional Groups