12. Structure Determination: Mass Spectrometry and Infrared Spectroscopy

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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)

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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

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Sample vaporized and subjected to bombardment by electrons that remove an electron creating a cation radical

Bonds in cation radicals begin to break (fragment)

Mass to charge (m/z) ratio is measured

Molecular ion (M+)shows molecular weight

12.1 Mass Spectrometry of Small Molecules:Magnetic-Sector Instruments

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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+)

C

H

H

H

C

H

H

C

H

H

H

Propane MW = M+ = 44

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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+)

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Predict possible molecular formulas containing C, H, and maybe O if :

M+ = 86 M+ = 156

M+ peak Example:

Possible formulas can be predicted from the mass. (M+)

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Predict possible molecular formulas containing C, H, and maybe O if :

M+ = 86 M+ = 156

C6H14

C5H10O C4H6O2

C3H2O3

C11H24

C12H12 C11H8OC10H20O

C10H4O2 C9H16O2

C8H12O3

C7H8O4

C6H4O5

(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

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M+ peak :

Peaks above the molecular weight appear as a result of naturally occurring heavier isotopes in the sample

C

H

H

H

C

H

H

C

H

H

H

Propane MW = M+ = 44

M+1 = 45From 1.1% 13C

(M+1) from 1.1% 13C in nature

M+ peak: Halides

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Click on image to enlarge

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

Br

CH3Cl

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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

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MW=72; M+ peak not seen

CH3=15

M -15 = 57

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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:

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Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

CH3

MW=98 MW=98

Solution:

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Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

CH3

MW=98 MW=98

98-69=29 loss of (CH3CH2)

98-83=15 (loss of CH3)

M+ =98

Learning Check:

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Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

CH3

MW=98 MW=98

Solution:

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Identify the following MS as that of methylcyclohexane or ethylcyclopentane.

CH3

MW=98 MW=98

98-83=15 (loss of CH3)

M+ =98

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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

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CH3

CHH3C

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Mass Spec: Amines

Nitrogen Rule: Nitrogen Rule: Amines with odd # of N’s have Odd M+

Amines undergo -cleavage, generating radicals

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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

•Carbonyl compounds can also undergo cleavage

Loss of 28

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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

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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

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Low frequency (u) Long wavelength (l)= Low energy

High frequency (u) Short wavelength (l)= High energy

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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

Infrared Absorption of Ethyl Alcohol CH3CH2OH

Low frequency (u) Long wavelength (l)= Low energy

High frequency (u) Short wavelength (l)= High energy

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12.6 Infrared Spectroscopy

IR region lower energy than visible light (< red – produces heating as with a heat lamp)

2.5 106 m to 2.5 105 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

High frequency = High E Low frequency = Low E

Infrared Spectroscopy

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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

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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)

12.7 Interpreting Infrared Spectra Most functional groups absorb at a characteristic energy

Infrared Spectra: Functional Grps Characteristic higher energy IR absorptions used to confirm the

existence of the presence of functional groups

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12.8 IR Spectra: Functional Grps

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Alkane

Alkene

Alkyne

-C-H C-C

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IR: Aromatic Compounds

(Subsituted benzene “teeth”)

C≡C

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IR: Alcohols and Amines

CH3CH2OH

Amines similar to OH

O-H broadens with Hydrogen bonding

N-H broadens with Hydrogen bonding

C-O

IR: Alcohols: O-H stretch

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Gas phase(no H-bonding)

Liquid Film(Lots of H-bonding)

CCl4 sln (0.25M)(some H-bonding)

IR: Alcohols

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C-O

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.

IR: Amines

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Amines similar to OHN-H broadens with Hydrogen bonding

IR: Amines Examples

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IR: Carbonyls: C=O Aldehydes

Carbonyls in general: Strong, sharp C=O peak 1670 to 1780 cm1

Conjugation lowers stretching frequency

IR: C=O: Aldehydes

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IR: C=O: Ketones

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Conjugation with a double bond or benzene ring lowers the stretching frequency by 30 to 40 cm-1.

IR: C=O: Ketones

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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 cm1 in saturated esters Electron donating O increased the frequency

1715 cm1 in esters next to aromatic ring or a double bond Conjugation decreases the frequency

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IR: Carboxylic Acids

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Learning Check:

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

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

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-C-H=C- H

-C=C-

Learning Check:

41.Propose a structure for the following unknown hydrocarbon:

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Solution:

41.Propose a structure for the following unknown hydrocarbon:

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-C-H

M=68

68-53 =15 loss of CH3

≡C- H

C5H8

-C≡C-

M-1=67

2 deg of unsat

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Learning Check:

42. Propose a structure for the following unknown hydrocarbon:

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Solution:

42. Propose a structure for the following unknown hydrocarbon:

-C-H

M=70

70-55 =15 loss of CH3

=C-H

C5H10

-C=C-

In mass spectrometry, what term is used to describe the ion that results from the ejection of one electron from a molecule?

1 2 3 4 5

20% 20% 20%20%20%

1. base peak

2. parent peak

3. fragment

4. analyte

5. none of these

What quantity is detected by mass spectrometry?

1 2 3 4 5

20% 20% 20%20%20%1. the energy of a molecule

2. the number of electrons ejected from a molecule

3. the number of ions of a particular mass to charge ratio

4. the number of electrons needed to ionize a molecule

5. the number of hydrogen atoms in a molecule

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

1 2

50%50%

1. True

2. False

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

1 2

50%50%

1. True

2. False

BrBr

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

1 2 3 4

25% 25%25%25%

O

1.O

2.

OH

3.

O

4.

Which of the following reasons explains why some mass spectrum peaks are larger than others?

1 2 3 4 5

20% 20% 20%20%20%1. The larger peaks represent fragments that are more stable.

2. The larger peaks represent fragments that are less stable.

3. Cations tend to give larger peaks than radical cations.

4. Radical cations tend to give larger peaks than cations.

5. None of these

Which of the following compounds is most likely to undergo a McLafferty rearrangement?

1 2 3 4 5

20% 20% 20%20%20%

NH2

1.O

2.

O

3.

OH

4.OH

5.

Frequency is commonly reported in units of:

1 2 3 4 5

20% 20% 20%20%20%

1. Joules

2. nm

3. amu

4. Hz

5. m/z

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

1 2

50%50%

1. True

2. False

Which type of electromagnetic radiation possesses the highest energy?

1 2 3 4 5

20% 20% 20%20%20%

1. IR light

2. UV light

3. microwaves

4. visible light

5. FM radio waves

Infrared spectroscopy is based on _____ excitation.

1 2 3 4

25% 25%25%25%

1. electronic

2. rotational

3. nuclear

4. vibrational

What functional groups are most likely present in a compound whose IR spectrum shows absorbances at 2217 and 1648 cm-1?

1 2 3 4 5

20% 20% 20%20%20%

1. aldehyde and alkene

2. alkene and alcohol

3. alcohol and nitrile

4. alkyne and ketone

5. nitrile and alkene

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

1 2

50%50%

1. True

2. False

Which of the following compounds will have its carbonyl absorb at the lowest frequency in IR spectroscopy?

1 2 3 4 5

20% 20% 20%20%20%

O

1.

O

O

2.

H

O

3.

O

O

4.O

5.

What approximate frequency range is considered the fingerprint region in infrared spectroscopy?

1 2 3 4 5

20% 20% 20%20%20%

1. 4000 – 3500 cm-1

2. 3500 – 3000 cm-1

3. 3000 – 2000 cm-1

4. 2000 – 1500 cm-1

5. 1500 – 500 cm-1

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

1 2

50%50%1. True

2. False