2016 Org Chem Chap 14
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Transcript of 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
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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
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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.
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Classes of Organic Compounds
[Insert Table 14.1]
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A Mass Spectrometer
Only positively charged species reach the recorder.
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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.
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The Molecular Ion
Pentane forms a molecular ion with m/z = 72.
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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.
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Loss of H2 From a Fragment
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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.
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Secondary Carbocations are More Stable Than Primary Carbocations
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To differentiate Cycloalkanes: C7H14
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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
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relative intensity of M + 1 peak Number of carbon atoms = ————— 0.011 X ( relative intensity of M peak )
14.4 Isotopes in Mass Spectrometry
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Natural Abundance of Isotopes
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14.5 High-Resolution Mass Spectrometry Can Distinguish Between Compound with the Same Molecular Mass
Exact Masses of Isotopes
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14.6 The Fragmentation Patterns of Functional Groups
The Carbon—Bromine Bond Breaks Heterolytically
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The Carbon—Chlorine Bond Breaks Heterolytically The Carbon—Carbon Bond Breaks Homolytically
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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.
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The Mass Spectrum of 2-Chloropropane
- ·CH3
- ·Cl
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More Cl atoms
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With Br and Cl atoms
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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.
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Fragmentation in Ethers: Path a: The C—O Bond Breaks Heterolytically
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Fragmentation in Ethers: Path b: α-Cleavage of a C-C bond
Same m/z but different ionic structures
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α-Cleavage in an Alcohol
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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.
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1-methoxybutane 2-methoxybutane 2-methoxy-2-methylpropane
(A)
(B)
(C)
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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).
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α-Cleavage or loss of a neutral molecule in a Ketone
Release a neutral molecule
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Nuclear spin Electron spin
14.9 Spectroscopy and the Electromagnetic Spectrum
low frequency
high frequency
short wavelength
long wavelength
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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
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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
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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.
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Stretching and Bending Vibrations
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37 Functional group region: 4000-1400 cm-1 Fingerprint region: 1400-600 cm-1
Each Stretching and Bending Vibration Occurs at a Characteristic Wavenumber
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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
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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
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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
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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
= 2228 cm-1 (calculated)(experimental = 2206 cm-1)
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ν (cm-1) = 10 x 105 (12 + 12 )12 x 12
1 / 2~ 4.12
= 1682 cm-1 (calculated)(experimental = 1650 cm-1)
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
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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.
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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
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This C═O Bond Is Essentially a Pure Double Bond
1720 cm-1
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This C═O Bond Has Significant Single Bond Character
The less double bond character, the lower the frequency.
1680 cm-1
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1720 cm-1
1680 cm-1
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Resonance Electron Donation Decreases the Frequency Inductive Electron Withdrawal Increases the Frequency
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The IR Spectrum of an Ester
1740 cm-1
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The IR Spectrum of an Amide
1650 cm-1
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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.
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The IR Spectrum of an Alcohol and a Carboxylic Acid
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Hydrogen Bonded OH Groups Stretch at a Lower Frequency
It is easier to stretch a hydrogen bonded OH group.
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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).
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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.
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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.
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The IR Spectrum of an Aldehyde
The carbon—hydrogen stretch of an aldehyde hydrogen occurs at 2820 cm–1 and at 2720 cm–1.
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The IR Spectrum of an Amine
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The IR Spectrum of Diethyl Ether
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C―H absorption Bands Bending vibrations
Csp3—H ~1400 cm-1
CH3 appear on both sides of 1400 cm-1
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14.15 The Absence of Absorption Bands
C-O
CH3CH2OCH2CH3
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14.16 Some Vibrations are Infrared Inactive
A bond absorbs IR radiation only if its dipole moment changes when it vibrates.
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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
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The IR Spectrum of Benzaldehyde
Wavenumber (cm–1) Assignment
3050 2810 and 2730 1600 and 1460 1700
sp2 CH an aldehyde benzene ring a partial single-bond character carbonyl
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The IR Spectrum of 2-Propyn-1-ol
Wavenumber (cm–1) Assignment
3300 2950 2100
OH group sp3 CH alkyne
C≡C-H
OH
C≡C
C-O
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The IR spectrum of N-Methylethanamide
Wavenumber (cm–1) Assignment
3300 2950 1660 1560
N—H sp3 CH amide carbonyl N—H bend
N-H C=O
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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
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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
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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
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Which compound has the following IR spectrum?
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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
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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*
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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.
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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)
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UV/Vis Absorption Bands are Broad
UV/Vis absorption bands are broad because an electronic state has vibrational sublevels.
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Chromophore
A chromophore is that part of a molecule that is responsible for a UV/Vis spectrum.
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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
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Cells Used for Taking UV/Vis Spectra
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The More Conjugated Double Bonds, the Longer the Wavelength
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14.20 Effect of Conjugation on λmax
Conjugation Makes the Electronic Transition Easier
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Conjugation Makes the Electronic Transition Easier
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Highest Occupied Molecular Orbital
Lowest Unoccupied Molecular Orbital
Conjugation Makes the Electronic Transition Easier
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Colored Compounds Absorb Visible Light (> 400 nm)
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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
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14.21 The Visible Spectrum and Color
The Color Observed Depends on the Color Absorbed
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Common Dyes
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Anthocyanins
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UV/Vis Spectroscopy Can Be Used to Measure the Rate of a Reaction
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UV/Vis Spectroscopy Can Be Used to Measure the Rate of a Reaction
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UV/Vis Spectroscopy Can Be Used to Determine a pKa Value
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UV/Vis Spectroscopy Can Be Used to Determine the Melting Temperature of DNA