2016 Org Chem Chap 14

© 2014 Pearson Education, Inc.

Mass Spectrometry, Infrared Spectroscopy,

and Ultraviolet/Visible

Spectroscopy

Paula Yurkanis Bruice University of California,

Santa Barbara

Chapter 14


2

Identification of Organic Compounds: Instrumental Techniques/information

Instrumental techniques Compound information Mass spectrometry: Mol. Mass, Mol. Formula, structural feature Infrared spectroscopy: functional group(s) UV/Vis spectroscopy: conjugated double bonds NMR spectroscopy: carbon-hydrogen framework


3

14.1 Mass spectrometry

70 ev = 1615 kcal/mol or 6754 kJ/mol

electron beam

An electron is ejected from the compound, thereby forming a molecular ion.


Classes of Organic Compounds

[Insert Table 14.1]


A Mass Spectrometer

Only positively charged species reach the recorder.


The Mass Spectrum of Pentane

m/z = mass-to-charge ratio of the fragment because z = 1

14.2 The Mass Spectrum: Fragmentation

Fragment ion peaks- positively charged fragments of the molecule.


The Molecular Ion

Pentane forms a molecular ion with m/z = 72.


Fragmentation of the Molecular Ion

The more stable the fragments, the more abundant they will be.

C-2—C-3 fragmentation forms more stable fragments.


Loss of H2 From a Fragment


More Stable Fragments are More Abundant

The peak at m/z = 57 is more abundant for isopentane than for pentane because a secondary carbocation is more stable than a primary carbocation.


Secondary Carbocations are More Stable Than Primary Carbocations


12

To differentiate Cycloalkanes: C7H14


14.3 Using the m/z value of the molecular ion to calculate the molecular formula

Rule of 13: Base value/13(C+H) = number of Carbons in this molecule Example: For a molecule with base peak m/z = 142 142/13 = 10…..12 C10H(10+12) = C10H22

Contains 1 “O” remove 1C and 4H C9H18O


14

relative intensity of M + 1 peak Number of carbon atoms = ————— 0.011 X ( relative intensity of M peak )

14.4 Isotopes in Mass Spectrometry


Natural Abundance of Isotopes


14.5 High-Resolution Mass Spectrometry Can Distinguish Between Compound with the Same Molecular Mass

Exact Masses of Isotopes


14.6 The Fragmentation Patterns of Functional Groups

The Carbon—Bromine Bond Breaks Heterolytically


The Carbon—Chlorine Bond Breaks Heterolytically The Carbon—Carbon Bond Breaks Homolytically


α-Cleavage in an Alkyl Chloride

The homolytic cleavage of the carbon—carbon bond is called α-cleavage.

The bonds that break are • the weakest bonds, and • the bonds that form the most stable fragments.


The Mass Spectrum of 2-Chloropropane

- ·CH3

- ·Cl


More Cl atoms


With Br and Cl atoms


α-Cleavage Occurs in Alkyl Chlorides but is Less Likely to Occur in Alkyl Bromides

The carbon—carbon bond and the carbon—chlorine bond have similar strengths.

The carbon—carbon bond is much stronger than the carbon—bromine bond.


Fragmentation in Ethers: Path a: The C—O Bond Breaks Heterolytically


Fragmentation in Ethers: Path b: α-Cleavage of a C-C bond

Same m/z but different ionic structures


α-Cleavage in an Alcohol


Loss of a Hydrogen from a γ-Carbon

Lost an even number of m/z indicates lost of a neutral molecule.

Two bonds break because of a stable water molecule is formed as a result of fragmentation.


28

1-methoxybutane 2-methoxybutane 2-methoxy-2-methylpropane

(A)

(B)

(C)


Common behavior of alkyl halides, ethers, and alcohols after forming a molecular ion by losing a lone-pair electron:

1. A bond between carbon and a more electronegative atom (halogen or oxygen) breaks heterolytically and the electrons stay with the more electronegative atom.

2. A bond between carbon and an atom of similar electronegativity (carbon, chlorine, or hydrogen) breaks homolytically

3. The bonds most likely to break are the weakest bonds and those that lead to formation of the most stable products(cation).


α-Cleavage or loss of a neutral molecule in a Ketone

Release a neutral molecule



32

Nuclear spin Electron spin

14.9 Spectroscopy and the Electromagnetic Spectrum

low frequency

high frequency

short wavelength

long wavelength


33

電磁波

波長(λ, m) 種類能量(kcal/mol) 對分子的影響

< 10-10 Gamma rays 106

~10-8 X rays 104 游離化

真空UV 102

近UV 電子躍遷

10-6 可見光 10

10-4 紅外光 1 分子振動

10-2 微波 10-4 分子轉動

1 ~ 100 無線電波 10-6 核自旋變化


34

ν = c / λ c = 3 X 1010 cm/s E = hν =hc / λ

wavenumber ν (cm-1) = 1 /λ (cm) = 104/λ(µm) ~

The Greater the Energy, the Greater the Frequency The Greater the Energy, the Shorter the Wavelength


35

14.10 Infrared Spectroscopy: Stretching and Bending Vibrations

Infrared active vibrational mode: Change of bond dipole moment

A stretching vibration occurs along the line of the bond.


Stretching and Bending Vibrations


37 Functional group region: 4000-1400 cm-1 Fingerprint region: 1400-600 cm-1

Each Stretching and Bending Vibration Occurs at a Characteristic Wavenumber


The Functional Group Region (4000–1400 cm–1) The Fingerprint Region (1400–600 cm–1)

Functional group regions: Both compounds are alcohols

Fingerprint regions: Compounds are different alcohols



14.12 The Intensity of Absorption Bands

The More Polar the Bond, the More Intense the Absorption

Intensity of absorption band is proportional to the number of bonds


41

14.13 The Position of Absorption Bands

The amount of energy required to stretch a bond depends on the strength of the bond and the masses of the bonded atoms.

Hooke’s Law For a harmonic oscillator: f = force constant = 5 x 105 dynes/cm / bond

ν =1

2πcf (m1 + m2)

m1 m2

1 / 2~

ν (cm-1) = f (m1 + m2)m1 m2

1 / 2~ 4.12


42

C-H bond: C-D bond:

ν (cm-1) = 5 x 105 (12 + 1)12 x 1

1 / 2~ 4.12

= 3032 cm-1 (calculated)(experimental = 3000 cm-1)

ν (cm-1) = f (m1 + m2)m1 m2

1 / 2~ 4.12

Hooke’s Law

ν (cm-1) = 5 x 105 (12 + 2 )12 x 2

1 / 2~ 4.12



43

ν (cm-1) = 10 x 105 (12 + 12 )12 x 12

1 / 2~ 4.12


C=C bond:

Lighter atoms show absorption bands at larger wavenumbers. C—H ~3000 cm-1

C—D ~2200 cm-1

C—O ~1100 cm-1

C—Cl ~700 cm-1


44

The Effect of Bond Order

• C ≡ C (~2100 cm-1) C = C (~1650 cm-1) C―C (~ 1200-800 cm-1) • C = O (~1700 cm-1) C―O (~1100 cm-1) • C ≡ N (~2200 cm-1) C = N (~1600 cm-1) C―N (~1100 cm-1)

The Greater the Bond Order, the Larger the Wavenumber.


14.14 The Position of an Absorption Band is Affected by Electron Delocalization, Electron Donation and Withdrawal,

and Hydrogen Bonding

The more double bond character, the greater the frequency (wavenumber).

Electron Delocalization (Resonance) Affects the Frequency of the Absorption


This C═O Bond Is Essentially a Pure Double Bond

1720 cm-1


This C═O Bond Has Significant Single Bond Character

The less double bond character, the lower the frequency.

1680 cm-1


48

1720 cm-1

1680 cm-1


Resonance Electron Donation Decreases the Frequency Inductive Electron Withdrawal Increases the Frequency


The IR Spectrum of an Ester

1740 cm-1


The IR Spectrum of an Amide

1650 cm-1


Carbon—Oxygen Bonds

The carbon—oxygen bond in an alcohol is a pure single bond.

The carbon—oxygen bond in an ether is a pure single bond. The carbon—oxygen single bond in a carboxylic acid has partial double bond character.

One carbon—oxygen single bond in an ester is a pure single bond and one has partial double bond character.


The IR Spectrum of an Alcohol and a Carboxylic Acid


Hydrogen Bonded OH Groups Stretch at a Lower Frequency

It is easier to stretch a hydrogen bonded OH group.



The Strength of a Carbon—Hydrogen Bond Depends on the Hybridization of the Carbon

An sp3-carbon—hydrogen bond is the weakest, so its stretch occurs at the shortest wavenumber (< 3000 cm–1).


An sp3-carbon—hydrogen stretch occurs at < 3000 cm–1. An sp2-carbon—hydrogen stretch occurs at > 3000 cm–1.

Where Carbon—Hydrogen Bonds Stretch and Bend

Stretching vibrations require more energy than bending vibrations.


Where Carbon—Hydrogen Bonds Bend

An sp3-carbon—hydrogen bend of a methyl occurs at < 1400 cm–1.

An sp2-carbon—hydrogen bend of a methyl and/or a methylene occurs at > 1400 cm–1.


The IR Spectrum of an Aldehyde

The carbon—hydrogen stretch of an aldehyde hydrogen occurs at 2820 cm–1 and at 2720 cm–1.


The IR Spectrum of an Amine


The IR Spectrum of Diethyl Ether


62

C―H absorption Bands Bending vibrations

Csp3—H ~1400 cm-1

CH3 appear on both sides of 1400 cm-1


63

14.15 The Absence of Absorption Bands

C-O

CH3CH2OCH2CH3


14.16 Some Vibrations are Infrared Inactive

A bond absorbs IR radiation only if its dipole moment changes when it vibrates.


14.17 How to Interpret an Infrared Spectrum

Wavenumber (cm–1) Assignment

3075 2950 1650 and 890 absence of bands 1500–1430 and 720

sp2 CH sp3 CH A terminal alkene with two substituents has less than four adjacent CH2 groups.

The IR Spectrum of 2-Methyl-1-pentene

Information for substituted benzene


The IR Spectrum of Benzaldehyde


3050 2810 and 2730 1600 and 1460 1700

sp2 CH an aldehyde benzene ring a partial single-bond character carbonyl


The IR Spectrum of 2-Propyn-1-ol


3300 2950 2100

OH group sp3 CH alkyne

C≡C-H

OH

C≡C

C-O


The IR spectrum of N-Methylethanamide


3300 2950 1660 1560

N—H sp3 CH amide carbonyl N—H bend

N-H C=O


The IR Spectrum of Ethyl Benzyl Ketone

Wavenumber (cm–1) Assignment >3000

<3000 1605 and 1500 1720 1380

sp2 CH sp3 CH a benzene ring a ketone carbonyl a methyl group

C=O


70

How to Identify an Infrared Spectrum

1. Check if C=O present; In general, C=O stretching frequency appeared

around 1820 ~ 1660 cm-1 with strong absorption intensity.

2. If C=O present: additional band will be observed Acid COOH 3400-2400 cm-1 strong

broad absorption Amide CO-NHR ~3500 cm-1 Ester CO-OR 1300-1000 cm-1 Anhydride O=C-O-C=O 1810, 1760 cm-1 Aldehyde O=C-H 2850, 2750 cm-1 Ketone -C=O no extra band


How to Identify an Infrared Spectrum

3. If no C=O: Alcohol, phenol 3600-3300 cm-1 (O-H) 1300-1000 cm-1 (C-O) Amine 3500 cm-1 (N-H) ether 1300-1000 cm-1 (C-O) 4. C=C or aromatic ring C=C 1650 cm-1 (w) Ar 1650-1450 cm-1 >3000 cm-1 (C-H) 5. Triple bond? C≡C 2150 cm-1 C≡N 2250 cm-1


Which compound has the following IR spectrum?

72


14.18 Ultraviolet and Visible Spectroscopy

• Spectroscopy is the study of the interaction between matter and electromagnetic radiation

• UV/Vis spectroscopy provides information about compounds with conjugated double bonds


74

Ultraviolet and Visible Spectroscopy

HOMO (Highest Occupied MO)

LUMO (Lowest Unoccupied MO)

UV 180-400 nm, visible 400-780 nm

Electronic Transition

nπ*

π π*


An Electronic Transition

Only organic compounds with π electrons can produce UV/Vis spectra.

A UV spectrum is obtained if UV light is absorbed.

A visible spectrum is obtained if visible light is absorbed.


76

π→π* transition

n→π* transition

A UV Spectrum

λmax 195 nm(ε = 9000/M cm, hexane)

λmax 270 nm(ε = 15/M cm, hexane)


UV/Vis Absorption Bands are Broad

UV/Vis absorption bands are broad because an electronic state has vibrational sublevels.


Chromophore

A chromophore is that part of a molecule that is responsible for a UV/Vis spectrum.


14.19 The Beer–Lambert Law

A = ε c l

A = absorbance of the sample

c = concentration of substance in solution

l = length of the cell in cm

ε = molar absorptivity of the sample (a measure of the probability of the transition)

A = log I0/I


Cells Used for Taking UV/Vis Spectra


The More Conjugated Double Bonds, the Longer the Wavelength


14.20 Effect of Conjugation on λmax

Conjugation Makes the Electronic Transition Easier




84

Highest Occupied Molecular Orbital

Lowest Unoccupied Molecular Orbital



Colored Compounds Absorb Visible Light (> 400 nm)


86

Auxochrome: a substituent that, when attached to a chromophore, alters the λmax and the intensity of the absorption, usually increasing both of them.

Red shift blue shift

Hypsochromic shift Bathochromic shift

Bathochromic shift Auxochrome


14.21 The Visible Spectrum and Color

The Color Observed Depends on the Color Absorbed


Common Dyes


Anthocyanins


UV/Vis Spectroscopy Can Be Used to Measure the Rate of a Reaction


UV/Vis Spectroscopy Can Be Used to Measure the Rate of a Reaction


UV/Vis Spectroscopy Can Be Used to Determine a pKa Value


UV/Vis Spectroscopy Can Be Used to Determine the Melting Temperature of DNA

Documents

2016 Org Chem Chap 14