© 2011 Pearson Education, Inc. 1 Chapter 13 Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible Spectroscopy Organic Chemistry 6 th Edition

© 2011 Pearson Education, Inc.1

Chapter 13

Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible

Spectroscopy

Organic Chemistry 6th Edition

Paula Yurkanis Bruice


Spectrally Identifiable Functional Groups


The Mass SpectrometerA mass spectrum records only positively charged fragments, either cations or radical cations

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


Information obtained from a mass spectrum:

• The molecular ion (M): measured to the nearest whole number or up to four decimal places (high-resolution mass spectrometry).

• Isotope peaks (M + 1, M + 2 etc.).

Typically M and the isotope peaks are the highest masses in the spectrum

• The high-resolution mass of the molecular ion provides the molecular formula directly.

• The whole-number mass of the molecular ion and the relative intensities of M + 1, M + 2, etc., can also provide the molecular formula.

…M + 1

M

Exception: a compound whose molecular ion completely fragments


…

• Fragment masses and intensities together provide structural information.

M

M-15

• The base peak has the greatest intensity in the spectrum.

• Intense peaks correspond to relatively stable cationic and/or relatively stable radical species lost.

• The fragments lost also provide structural information.

M-29Base peak

For example:

• Fragment m/z 57 resulted from the loss of methyl (m/z = 15) from the molecular ion.

• Given its intensity, m/z 57 must be the sec-butyl carbocation (not the primary butyl carbocation).


The Mass Spectrum of Pentane

Note weak m/z = 57 peak, primary butyl carbocation


The base peak of m/z 43 in the mass spectrum of pentane indicates the preference for C-2 to C-3 fragmentation:

The mass of the radical species lost in a fragmentation is the difference between the m/z values of the fragment ion and the molecular ion

All fragments originate from the molecular ion


Note strong m/z = 57 peak, secondary butyl carbocation

The Mass Spectrum of Isopentane


2-Methylbutane is more likely than pentane to lose a methyl radical because a secondary carbocation can be formed:


What are the structures of m/z 42 and 41?These ions arise from loss of the ethyl radical and either hydrogen atom or H2 from the pentane molecular ion:

Note: All fragments originate from the molecular ion.Exception: Tandem Mass spectrometry where fragments of fragments are observed.

Two-Fragment Loss from the Molecular Ion


Isotopes in Mass Spectrometry

• M + 1 peak: a contribution from 2H or 13C.

• M + 2 peak: a contribution from 18O or from two heavy isotopes (2H or 13C) in the same molecule.

• A large M + 2 peak suggests a compound containing either chlorine or bromine: a Cl if M + 2 is one-third the

intensity of M; a Br if M + 2 is the same intensity as M.

• To calculate the molecular masses of molecular ions and fragments, the atom mass of a single isotope of an atom must be used.


Fragmentation Patterns of Alkyl Halides

79Br 81Br


The Mass Spectrum of 2-Chloropropane

35Cl

37Cl

35Cl

37Cl


-Cleavage results from the homolytic cleavage of aC—C bond at the carbon:


-Cleavage occurs because the C—Cl and C—C bonds have similar strengths, and the species that is formed is a relatively stable cation:

-Cleavage is less likely to occur in alkyl bromide because C—C bond is stronger than C—Br bond


Fragmentation Patterns of Ethers


A C—O bond is cleaved heterolytically, with the electrons going to the more electronegative atom:


A C—C bond is cleaved homolytically at an -position because it leads to a relatively stable cation:


Fragmentation Patterns of Alcohols

Because they fragment, molecular ions obtained from alcohols usually are not observed


Like alkyl halides and ethers, alcohols undergo -cleavage:

In alcohols, loss of water results in a fragmentation peakat m/z = M-18:


Common Fragmentation Behavior in Alkyl Halides,

Ethers, and Alcohols

1. A bond between carbon and a more electronegative atom breaks heterolytically

2. A bond between carbon and an atom of similar electronegativity breaks homolytically

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


Fragmentation Patterns of Ketones

An intense molecular ion peak:


McLafferty rearrangement may occur:


Spectroscopy and the Electromagnetic Spectrum

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


Electromagnetic radiation has wave-like properties

High frequencies and short wavelengths are associated with high energy


Vibrational Transitions Observed in IR Spectroscopy

Functional groups stretch at different frequencies, and IR spectroscopy is used to identify functional groups


C H h

C H

Higher EnergyVibrational State

Infrared transitions require a bond dipole to occur:

The more polar the bond, the more intense the absorptions:

The intensity of an absorption band also depends on the number of bonds responsible for the absorption


1-butene — infrared active

2,3-dimethyl-2-butene — infrared inactive

2,3-dimethyl-2-heptene — infrared active, but very weak absorption band

Influence of symmetry on IR activity of the alkene stretch:


The Vibrating Bond as a Quantized Harmonic Oscillator

Ball-and-spring model:Quantum levels for a stretching vibration:

Fundamental transition: o 1

Overtone: o 2

Overtones are twice the frequency of the fundamental transition and are always weak


The approximate wavenumber of an absorption can be calculated from Hooke’s law:

v 1

2cK

Reduced MassM1M 2

M1 M 2

= wavenumberc = speed of lightK = force constantM1 and M2 = masses of atoms


Hooke’s law predicts that lighter atoms will vibrate at a higher frequency than heavy atoms:

C—H ~3000 cm–1

C—D ~2200 cm–1

C—O ~1100 cm–1

C—Cl ~700 cm–1

Increasing the s character of a bond (higher K value) increases the stretching frequency:

spsp2

sp2

sp3


Note the influence of mass and s character on stretching frequency:


An Infrared SpectrumThe functional group

region (4000–1400 cm–1)The fingerprint

region (1400–600 cm–1)

The functional group, or diagnostic region, is used to determine the functional group presentThe fingerprint region is used for structure elucidation by spectral comparison

High energy Low energy


Functional group regions: Both compounds are alcohols

Fingerprint regions: Compounds are different alcohols


The exact position of the absorption band depends on electron delocalization, the electronic effect of neighboring substituents, and hydrogen bonding:


Esters have a carbonyl and a C—O stretch Ketones have only a carbonyl stretch

Carbonyl overtone


Putting an atom other than carbon next to the carbonylgroup causes the position of the carbonyl absorptionband to shift:

The predominant effect of the oxygen of an ester is inductive electron withdrawal

The predominant effect of the nitrogen of an amide iselectron donation by resonance



The position of a C—O absorption varies because of resonance release in acids and esters:

~1050 cm–1

~1050 cm–1

~1250 cm–1

~1250 cm–1 and 1050 cm–1


Acids are readily distinguished from alcohols

BroadOH stretch

C═O stretch

Higher-frequencyC─O stretch


The position and the breadth of the O—H absorptionband depend on the concentration of the solution

It is easier to stretch an O—H bond if it is hydrogen bonded


The strength of a C—H bond depends on the hybridizationof the carbon


Examine the absorption bands in the vicinity of 3000 cm–1


• Sharp absorption bands at ~1600 cm–1 and 1500–1430 cm–1.• Overtones at 1700–1900 cm–1 for the in-plane and out-of-

plane benzene C—H bends. • The benzene overtones in the diagnostic region are readily

recognized.

Benzene ring:Benzene in-plane and out-of-plane C—H bends


Stretch of C—H Bond in an Aldehyde

The stretch of the C—H bond of an aldehyde shows one absorption band at ~2820 cm–1 and another one at ~2720 cm–1


• Primary amine: two N—H stretches at 3350 cm–1.• Amine: N—H bend.• “Isopropyl split” at 1380 cm–1 indicates the presence of an

isopropyl group.

Identifying a functional group by the bending vibrations:


The absence of absorption bands can be useful inidentifying a compound in IR spectroscopy

Bonds in molecules lacking dipole moments will not bedetected

Analyzing Infrared Spectra

The position, intensity, and shape of an absorption band are helpful in identifying functional groups


wavenumber (cm–1) assignment

30752950

1650 and 890absence of bands

1500–1430 and 720

sp2 CHsp3 CH

a terminal alkene with two substituentshas less than four adjacent CH2 groups



30502810 and 27301600 and 1460

1700

sp2 CHan aldehydebenzene ring

a partial single-bondcharacter carbonyl



330029502100

OH group sp3 CHalkyne



3300295016601560

N—Hsp3 CH

amide carbonylN—H Bend



>3000<3000

1605 and 150017201380

sp2 CHsp3 CH

a benzene ringa ketone carbonyl

a methyl group


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


UV and Vis light cause only two kinds of electronic transition:

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

• A visible spectrum is obtained if visible light is absorbed.

• A UV spectrum is obtained if UV light is absorbed.

Forbidden transition: lone pair orthogonal to system

Symmetry: allowed transition


A chromophore is the part of a molecule that absorbs UV or visible light

Only compounds with electrons can produce UV/Vis spectra

Allowed

Forbidden


The Beer–Lambert Law

The molar absorptivity of a compound is a constant that is characteristic of the compound at a particular wavelength

A = log(I0/I)c = concentration of substance in solutionl = length of the cell in cm = molar absorptivity, a measure of the probability of the transition

A = c l = ~10,000 M–1cm–1,

Allowed

= <100 M–1cm–1, Forbidden


Effect of Conjugation on max

The max and values increase as the number of conjugated double bonds increases




If a compound has enough conjugated double bonds, it will absorb visible light (max >400 nm), and the compound will be colored


An auxochrome is a substituent in a chromophore thatalters the max and the intensity of the absorption:


The Visible Spectrum and Color


Uses of UV/Vis Spectroscopy

• Measure the rates of a reaction

• Determine the pKa of a compound

• Estimate the nucleotide composition of DNA

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© 2011 Pearson Education, Inc. 1 Chapter 13 Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible Spectroscopy Organic Chemistry 6 th Edition