12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

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 (VIS)

Why this Chapter? Finding structures of new molecules synthesized is critical To get a good idea of the range of structural techniques available and how they should be used

12.1 Mass Spectrometry of Small Molecules:Magnetic-Sector Instruments Bonds in cation radicals begin to break (fragment) Sample vaporized and subjected to bombardment by electrons that remove an electron creating a cation radical Mass to charge (m/z) ratio is measured Molecular ion (M+) shows molecular weight

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 (100%) Other peaks listed as the % of that peak Peak that corresponds to the unfragmented radical cation is parent peak or molecular ion (M+) Propane MW = M+ = 44

12.2 Interpreting Mass Spectra Molecular weight from the mass of the molecular ion Double-focusing instruments provide high-resolution “exact mass” 0.0001 atomic mass units – distinguishing specific atoms Example MW “72” is ambiguous: C5H12 and C4H8O but: C5H12 72.0939 amu exact mass C4H8O 72.0575 amu exact mass Result from fractional mass differences of atoms 16O = 15.99491, 12C = 12.0000, 1H = 1.00783 Instruments include computation of formulas for each peak If parent ion not present due to electron bombardment causing breakdown, “softer” methods such as chemical ionization are used

M+ peak Example: Possible formulas can be determined from the mass. (M+) Predict possible molecular formulas containing C, H, and maybe O if : M+ = 86 M+ = 156

M+ peak Example: C6H14 C5H10O C4H6O2 C3H2O3 C11H24 C12H12 C11H8O Possible formulas can be predicted from the mass. (M+) Predict possible molecular formulas containing C, H, and maybe O if : M+ = 86 M+ = 156 C6H14 C5H10O C4H6O2 C3H2O3 C11H24 C12H12 C11H8O C10H20O C10H4O2 C9H16O2 C8H12O3 C7H8O4 C6H4O5

M+ peak : Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample (M+1) from 1.1% 13C in nature (M and M+2) in 75.8%/24.2% ratio = 35Cl and 37Cl (M and M+2) in 50.7%/49.3% ratio = 79Br and 81Br Propane MW = M+ = 44 M+1 = 45 From 1.1% 13C

M+ peak: Halides M+ and M+2 in 75.8%:24.2% (~ 3:1) ratio Click on image to enlarge M+ peak: Halides M+ and M+2 in 75.8%:24.2% (~ 3:1) ratio = 35Cl and 37Cl M+ and M+2 in 50.7%:49.3% (~ 1:1) ratio = 79Br and 81Br CH3Cl

Mass-Spec Fragmentation Patterns Molecular ions break into characteristic fragments that can be identifed Serves as a “fingerprint” for comparison with known materials in analysis (used in forensics) Positive charge goes to fragments that best can stabilize it

Mass-Spec Fragmentation Patterns CH3=15 MW=72; M+ peak not seen

Mass Spec: Fragmentation of Hexane Hexane (m/z = 86 for parent) has peaks at m/z = 71, 57, 43, 29 71=Loss of CH3 (15) 57=Loss of CH3CH2 (29)

Learning Check: Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

Solution: Identify the following MS as that of methylcyclohexane or ethylcyclopentane. 98-69=29 loss of (CH3CH2) 98-83=15 (loss of CH3) M+ =98

Learning Check: Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

Solution: Identify the following MS as that of methylcyclohexane or ethylcyclopentane. 98-83=15 (loss of CH3) M+ =98

12.3 Mass Spec: Common Functional Groups Alcohols: Alcohols undergo -cleavage (at bond next to the C-OH) as well as loss of H-OH (18) to give C=C Loss of 18

Mass Spec: Alcohols

Mass Spec: Amines Nitrogen Rule: Amines with odd # of N’s have Odd M+ Amines undergo -cleavage, generating radicals

Mass Spec: 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 Loss of 28 Carbonyl compounds can also undergo  cleavage

12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments Most biochemical analyses by MS use: electrospray ionization (ESI) Matrix-assisted laser desorption ionization (MALDI) • Linked to a time-of-flight mass analyzer MALDI-TOF MS of chicken eg-white lysozyme

12.5 Spectroscopy and the Electromagnetic Spectrum Radiant energy is proportional to its frequency (cycles/s = Hz) as a wave (Amplitude is its height) Different types are classified by frequency or wavelength ranges

Electromagnetic Spectrum High frequency (u) Short wavelength (l) = High energy Low frequency (u) Long wavelength (l) = Low energy

Absorption Spectra Organic compound exposed to electromagnetic radiation, can absorb energy of certain wavelengths. Changing wavelengths to determine which are absorbed and which are transmitted produces an absorption spectrum Energy absorbed is shown as dips in spectrum High frequency (u) Short wavelength (l) = High energy Infrared Absorption of Ethyl Alcohol CH3CH2OH Low frequency (u) Long wavelength (l) = Low energy

12.6 Infrared Spectroscopy High frequency = High E Low frequency = Low E IR region lower energy than visible light (< red – produces heating as with a heat lamp) 2.5  106 m to 2.5  105 m region used by organic chemists for structural analysis IR energy in a spectrum is usually measured as wavenumber (cm-1), the inverse of wavelength and proportional to frequency Specific IR absorbed by organic molecule related to its structure

Infrared Spectroscopy IR energy absorption corresponds to atomic movements, such as vibrations and rotations from bending and stretching of bonds between groups of atoms Energy is characteristic of the bonding of atoms in a functional group Bond stretching dominates higher energy modes Light objects connected to heavy objects vibrate fastest: C-H, N-H, O-H For two heavy atoms, stronger bond requires more energy: C º C, C º N > C=C, C=O, C=N > C-C, C-O, C-N, C-halogen

12.7 Interpreting Infrared Spectra Most functional groups absorb at a characteristic energy 4000-2500 cm-1 N-H, C-H, O-H (stretching) 3300-3600 N-H, O-H 3000 C-H 2000-1500 cm-1 double bonds C=C, C=O, C=N (stretching) C=O 1680-1750 C=C 1640-1680 cm-1 Below 1500 cm-1 “fingerprint” Single bonds C-C, C-O, C-N, C-X (vibrations) 2500-2000 cm-1 CºC and C º N (stretching)

Infrared Spectra: Functional Grps Characteristic higher energy IR absorptions used to confirm the existence of the presence of functional groups

12.8 IR Spectra: Functional Grps Alkane C-C -C-H Alkene Alkyne

IR: Aromatic Compounds (Subsituted benzene “teeth”) C≡C

IR: Alcohols and Amines O-H broadens with Hydrogen bonding CH3CH2OH C-O N-H broadens with Hydrogen bonding Amines similar to OH

IR: Alcohols: O-H stretch Gas phase (no H-bonding) CCl4 sln (0.25M) (some H-bonding) Liquid Film (Lots of H-bonding)

IR: Alcohols C-O

IR: Amines Amines similar to OH N-H broadens with Hydrogen bonding Primary and secondary amines exhibit a characteristic broad IR N-H stretching absorption between 3250 and 3500 cm-1. Primary amines show two strong peaks in this range, whereas secondary amines show only one. Primary amines also show a band near 1600 cm-1 due to a scissoring motion of the NH2 group. Tertiary amines do not show any of these signals since they do not have a hydrogen bound to nitrogen. Amines similar to OH N-H broadens with Hydrogen bonding

IR: Amines Examples

IR: Carbonyls: C=O Aldehydes Carbonyls in general: Strong, sharp C=O peak 1670 to 1780 cm1 Conjugation lowers stretching frequency

IR: C=O: Aldehydes

IR: C=O: Ketones Conjugation with a double bond or benzene ring lowers the stretching frequency by 30 to 40 cm-1.

IR: C=O: Ketones Ring strain increases frequency: Incorporation of the carbonyl group in a small ring (5, 4 or 3-membered), raises the stretching frequency.

IR: C=O: Esters 1735 cm1 in saturated esters Electron donating O increased the frequency 1715 cm1 in esters next to aromatic ring or a double bond Conjugation decreases the frequency

IR: Carboxylic Acids

Learning Check: 32. Which of the following represents cyclohexane and which cyclohexene?

Solution: 32. Which of the following represents cyclohexane and which cyclohexene? =C- H -C-H -C=C-

Learning Check: 41.Propose a structure for the following unknown hydrocarbon:

Solution: 68-53 =15 loss of CH3 41.Propose a structure for the following unknown hydrocarbon: M-1=67 C5H8 2 deg of unsat M=68 ≡C- H -C≡C- -C-H

Learning Check: 42. Propose a structure for the following unknown hydrocarbon:

Solution: 42. Propose a structure for the following unknown hydrocarbon: 70-55 =15 loss of CH3 C5H10 M=70 -C-H =C-H -C=C-

base peak parent peak fragment analyte none of these In mass spectrometry, what term is used to describe the ion that results from the ejection of one electron from a molecule? base peak parent peak fragment analyte none of these

What quantity is detected by mass spectrometry? the energy of a molecule the number of electrons ejected from a molecule the number of ions of a particular mass to charge ratio the number of electrons needed to ionize a molecule the number of hydrogen atoms in a molecule

High-resolution mass spectrometry enables one to determine the molecular formula of a molecule. True False

High-resolution mass spectrometry would allow one to distinguish between the following molecules. True False

Which of the given C6H12O isomers would be expected to produce an m/z peak at M+-18? 3 4 2. 1. 4. 3.

Which of the following reasons explains why some mass spectrum peaks are larger than others? The larger peaks represent fragments that are more stable. The larger peaks represent fragments that are less stable. Cations tend to give larger peaks than radical cations. Radical cations tend to give larger peaks than cations. None of these

Which of the following compounds is most likely to undergo a McLafferty rearrangement? 1. 2. 1 2 3 4 Picture Choice 5 3. 4. 5.

Frequency is commonly reported in units of: Joules nm amu Hz m/z

The higher the wavenumber of a molecular vibration, the lower the energy of the infrared radiation needed to stimulate it. True False

Which type of electromagnetic radiation possesses the highest energy? IR light UV light microwaves visible light FM radio waves

Infrared spectroscopy is based on _____ excitation. electronic rotational nuclear vibrational

aldehyde and alkene alkene and alcohol alcohol and nitrile What functional groups are most likely present in a compound whose IR spectrum shows absorbances at 2217 and 1648 cm-1? aldehyde and alkene alkene and alcohol alcohol and nitrile alkyne and ketone nitrile and alkene

(R)-2-pentanol and (S)-2-pentanol give identical IR spectra. True False

Which of the following compounds will have its carbonyl absorb at the lowest frequency in IR spectroscopy? 1. 2. 1 2 3 4 5 3. 4. 5.

What approximate frequency range is considered the fingerprint region in infrared spectroscopy? 4000 – 3500 cm-1 3500 – 3000 cm-1 3000 – 2000 cm-1 2000 – 1500 cm-1 1500 – 500 cm-1

(E)-2-Butene and (Z)-2-butene give identical IR spectra. True False