Chapter 13 Spectroscopy Infrared spectroscopy

Chapter 13Chapter 13SpectroscopySpectroscopy

Infrared spectroscopyInfrared spectroscopy

Ultraviolet-visible spectroscopyUltraviolet-visible spectroscopy

Nuclear magnetic resonance spectroscopyNuclear magnetic resonance spectroscopy

Mass spectrometryMass spectrometry

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

13.113.1Principles of Molecular Principles of Molecular

Spectroscopy:Spectroscopy:Electromagnetic RadiationElectromagnetic Radiation

Cosmic rays

Rays

X-rays

Ultraviolet light

Visible light

Infrared radiation

Microwaves

Radio waves

Figure 13.1: The Electromagnetic Spectrum

Energy

Hi , Short

Lo , Long

Is propagated at the speed of light,

has properties of particles and waves,

and the energy of a photon is proportional to its frequency.

Electromagnetic Radiation

E = h

Figure 13.1: The Electromagnetic Spectrum

400 nm 750 nm

Visible Light

Longer Wavelength ()Shorter Wavelength ()

Higher Frequency () Lower Frequency ()

Higher Energy (E) Lower Energy (E)

Ultraviolet Infrared

The UV, VIS, IR region.

13.213.2Principles of Molecular Spectroscopy: Principles of Molecular Spectroscopy:

Quantized Energy StatesQuantized Energy States

Electromagnetic radiation is absorbed when theenergy of photon corresponds to difference in energy between two states.

E = h

Effect

Electron excitation

Vibration of bonds

Rotation of molecules

Nuclear spin states

Energy

UV-Vis

infrared

microwave

radiofrequency

Energy and its Effect

13.313.3Introduction to Introduction to

11H NMR SpectroscopyH NMR Spectroscopy(Proton NMR)(Proton NMR)

1H and 13C

Both have nuclear spin = ±1/2.1H is 99% at natural abundance.13C is 1.1% at natural abundance.12C is 98.9% at natural abundance )

but is not NMR active.

The Nuclei that are Most Useful toOrganic Chemists are:

Nuclear Spin

A spinning charge, such as the nucleus of 1H or 13C, generates a magnetic field. The magnetic field generated by a nucleus of spin +1/2 is opposite in direction from that generated by a nucleus of spin –1/2.

+ +

++

+

+

+

The distribution of nuclear spins is random in the absence of an external magnetic field B0.

+

+

+

+

+

B0 = field strength

An external magnetic field (B0) causes nuclear magnetic moments to align parallel or antiparallel to applied field.

+

+

+

+

+

B0

There is a slight excess of nuclear magnetic moments aligned parallel to the applied field.

There is no difference in energy in the two spin states absence of a magnetic field. ΔE (the energy required to flip states) is proportional to strength of external magnetic field.

Energy Differences Between Nuclear Spin States

+

+

E E '

increasing field strength

Energy Required to Flip Spin States

H1 C13

Radiofreq. Ho Ho

60MHz 14091.6 56025.0 90MHz 21137.4 84037.4100MHz 23486.0 93374.9200MHz 46971.9 186749.9300MHz 70457.9 280124.8600MHz 140915.8 560249.7

Ho = B0 (Terms for the applied field, in gauss below).

Some Important Relationships in NMR

The frequency of absorbedelectromagnetic radiationis proportional to

the energy difference betweentwo nuclear spin stateswhich is proportional to

the applied magnetic field.

Units

Hz

kJ/mol(kcal/mol)

tesla (T)


The frequency of absorbed electromagneticradiation is different for different elements, and for different isotopes of the same element.

For a field strength of 4.7 T:1H absorbs radiation having a frequencyof 200 MHz (200 x 106 s-1)13C absorbs radiation having a frequencyof 50.4 MHz (50.4 x 106 s-1)


The frequency of the electromagnetic radiation absorbed to flip spin states for a particular nucleus (such as 1H) depends on its molecular environment.

So, hydrogens in different environments in the same compound will absorb energy of different frequencies.

This makes NMR a very useful tool for structure determination.

13.4Nuclear Shielding

and1H Chemical Shifts

What do we mean by "shielding"?What do we mean by "shielding"?What do we mean by "chemical shift"?What do we mean by "chemical shift"?

Shielding

An external magnetic field affects the motion of the electrons in a molecule, inducing a magnetic field within the molecule.

The direction of the induced magnetic field, Bi, is opposite to that of the applied field.

C H

B0

Bi

Shielding

The induced field shields the nuclei (in this case, C and H) from the applied field.

So, a stronger external field is needed in order for the energy difference between spin states to match energy of radiofrequency (rf) radiation.

C H

B0

Bi

Chemical Shift

Chemical shift is a measure of the degree to which a nucleus in a molecule is shielded.

Protons in different environments are shielded to greater or lesser degrees which results in different chemical shifts.

C H

B0

Bi

Chemical Shift

Chemical shifts () are measured relative to the protons in tetramethylsilane (TMS) as a standard.

TMS has highly shielded hydrogens and the peak for these Hs in a scan is set at 0 .

Si CH3

CH3

CH3

H3C

=position of signal - position of TMS peak

spectrometer frequencyx 106

01.02.03.04.05.06.07.08.09.010.0

Chemical shift (, ppm)measured relative to TMS

UpfieldIncreased shielding

DownfieldDecreased shielding

(CH3)4Si (TMS)

Chemical Shift

Example: The signal for the proton in chloroform (HCCl3) appears 1456 Hz downfield from TMS at a spectrometer frequency of 200 MHz.

=position of signal - position of TMS peak

spectrometer frequencyx 106

=1456 Hz - 0 Hz

200 x 106 Hxx 106

= 7.28

Chemical shift (, ppm)

7.28 ppm

H C

Cl

Cl

Cl

01.02.03.04.05.06.07.08.09.010.0

The more electron electron density that is withdrawn from a hydrogen results in a smaller field induced from the remaining electrons.

13.5Effects of Molecular Structure

on1H Chemical Shifts

Protons in different environments experience different degrees of shielding and have

different chemical shifts.

Electronegative Substituents Decreasethe Shielding of Methyl Groups

least shielded H most shielded H CH3F CH3OCH3 (CH3)3N CH3CH3 (CH3)4Si

4.3 3.2 2.2 0.9 0.0

A more electronegative atom will remove more electron density from hydrdogen.

Electronegative Substituents Decrease Shielding

H3C—CH2—CH3

O2N—CH2—CH2—CH3

0.9 0.9 1.3

1.0 4.3 2.0

CHCl3 7.3

CH2Cl2 5.3

CH3Cl 3.1

Methyl, Methylene, and Methine

CH3 more shielded than CH2 ; CH2 more shielded than CH

H3C C

CH3

CH3

H

0.9

1.6 0.8

H3C C

CH3

CH3

CH2

0.9

CH3

1.2

Protons Attached to sp2 Hybridized Carbonare Less Shielded than those Attached

to sp3 Hybridized Carbon H H

HH

H

H

C C

HH

H H

CH3CH3

7.3 5.3 0.9

But Protons Attached to sp Hybridized Carbonare More Shielded than those Attached

to sp2 Hybridized Carbon

2.4CH2OCH3C CHC C

HH

H H

5.3

Protons Attached to Benzylic and AllylicCarbons are Somewhat Less Shielded than Usual

1.5 0.8

H3C CH3

1.2

H3C CH2

2.6

H3C—CH2—CH3

0.9 0.9 1.3

Allylic

Benzylic

Alkyl

Proton Attached to C=O of Aldehydeis Most Deshielded C—H

2.4

9.7

1.1

C C

O

H

H

CH3

H3C

Table 13.1

Type of proton Chemical shift (),ppm


CH R 0.9-1.8

1.5-2.6CH CC

2.0-2.5CH C

O

2.1-2.3CH NC

CH Ar 2.3-2.8

Table 13.1



CH Br 2.7-4.1

9-10C

O

H

2.2-2.9CH NR

3.1-4.1CH Cl

6.5-8.5H Ar

C C

H

4.5-6.5

3.3-3.7CH O

Table 13.1


1-3H NR

0.5-5H OR

6-8H OAr

10-13C

O

HO

13.6Interpreting 1H NMR Spectra

1. Number of signals.

2. Position of signals.

3. Signal intensity (measured by area under peak).

4. Splitting pattern (multiplicity).

Information Contained in an NMRSpectrum Includes:

Number of Signals

Protons that have different chemical shifts are chemically nonequivalent.

They exist in different molecular environments.

Protons that have the same chemical shift are chemically equivalent.


CCH2OCH3N

OCH3

NCCH2O

Figure 13.12

01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0

chemically nonequivalent

Are in identical environments

Have same chemical shift

Replacement test: replacement by some arbitrary "test group" generates same compound

H3CCH2CH3

chemically equivalent

Chemically Equivalent Protons

H3CCH2CH3

chemically equivalent

CH3CH2CH2ClClCH2CH2CH3

Chemically Equivalent Protons

Replacing protons at C-1 and C-3 gives same compound (1-chloropropane).C-1 and C-3 protons are chemically equivalent and have the same chemical shift.

Diastereotopic protons are those whoswe replacement by some arbitrary test group generates diastereomers.

Diastereotopic protons can have differentchemical shifts.

Diastereotopic Protons

C C

Br

H3C

H

H

5.3 ppm

5.5 ppm

Enantiotopic protons are those whose replacement by some arbitrary test group generates enantiomers.

They are in mirror-image environments.

Enantiotopic protons have the samechemical shift.

Enantiotopic Protons

C CH2OH

H3C

HH

EnantiotopicProtons

C CH2OH

H3C

ClH

C CH2OH

H3C

HCl

R S

13.7Spin-Spin Splitting in 1H NMR

Spectroscopy

Not all peaks are singlets.Not all peaks are singlets.

Signals can be split by coupling of Signals can be split by coupling of nuclear spins.nuclear spins.


Cl2CHCH3Figure 13.13

4 lines;quartet

2 lines;doublet

CH3CH

01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0

Two-bond and Three-bond Coupling

C C

H

H

C C HH

protons separated bytwo bonds

(geminal relationship)

protons separated bythree bonds

(vicinal relationship)

In order to observe splitting, protons cannot have same chemical shift.

Coupling constant (2J or 3J) is independent of field strength.

Two-bond and Three-bond Coupling

C C

H

H

C C HH

01.02.03.04.05.06.07.08.09.010.0


Cl2CHCH3Figure 13.13

4 lines;quartet

2 lines;doublet

CH3CH

coupled protons are vicinal (three-bond coupling)CH splits CH3 into a doublet

CH3 splits CH into a quartet

Why Do the Methyl Protons of1,1-Dichloroethane Appear as a Doublet?

C C HH

Cl

Cl

H

Hsignal for methyl protons is split into a doublet

To explain the splitting of the protons at C-2, we first focus on the two possible spin orientations of the proton at C-1.


A result of spin-spin spliting.

There are two orientations of the nuclear spin forthe proton at C-1. One orientation shields theprotons at C-2; the other deshields the C-2 protons.The protons at C-2 "feel" the effect of both theapplied magnetic field and the local field resultingfrom the spin of the C-1 proton.

C C HH

Cl

Cl

H

H


"true" chemicalshift of methylprotons (no coupling)

This line correspondsto molecules in which

the nuclear spin of the proton at C-1 reinforcesthe applied field.

This line correspondsto molecules in which

the nuclear spin of the proton at C-1 opposesthe applied field.

C C HH

Cl

Cl

H

H

Why Does the Methine Proton of1,1-Dichloroethane Appear as a Quartet?

signal for methine proton is split into a quartet

The proton at C-1 "feels" the effect of the applied magnetic field and the local fields resulting from the spin states of the three methyl protons. The possible combinations are shown on the next slide.

C C HH

Cl

Cl

H

H

There are eight combinations of nuclear spins for the three methyl protons.These 8 combinations split the signal into a 1:3:3:1 quartet.

Why Does the Methine Proton of1,1-Dichloroethane Appear as a Quartet?

C C HH

Cl

Cl

H

H

For simple cases, the multiplicity of a signalfor a particular proton is equal to the number of equivalent vicinal protons + 1.

Called the N + 1 Rule.

The Splitting Rule for 1H NMR

13.8Splitting Patterns:The Ethyl Group

CHCH33CHCH22X is characterized by a triplet-quartet X is characterized by a triplet-quartet pattern (quartet at lower field than the triplet).pattern (quartet at lower field than the triplet).


BrCH2CH3Figure 13.16

4 lines;quartet

3 lines;tripletCH3

CH2

01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0

Splitting Patterns of Common Multiplets

Number of equivalent Appearance Intensities of linesprotons to which H of multiplet in multipletis coupled

1 Doublet 1:12 Triplet 1:2:13 Quartet 1:3:3:14 Pentet 1:4:6:4:15 Sextet 1:5:10:10:5:16 Septet 1:6:15:20:15:6:1

Table 13.2

13.9Splitting Patterns:

The Isopropyl Group

(CH(CH33))22CHX is characterized by a doublet-septet CHX is characterized by a doublet-septet pattern (septet at lower field than the doublet).pattern (septet at lower field than the doublet).


ClCH(CH3)2Figure 13.18

7 lines;septet

2 lines;doublet

CH3

CH

01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0

13.10Splitting Patterns:Pairs of Doublets

Splitting patterns are not always symmetrical, Splitting patterns are not always symmetrical, but lean in one direction or the other.but lean in one direction or the other.

Pairs of Doublets

Consider coupling between two vicinal protons.

If the protons have different chemical shifts, each will split the signal of the other into a doublet.

C CH H

Pairs of Doublets

Let be the difference in chemical shift in Hz between the two hydrogens.

Let J be the coupling constant between them in Hz.

C CH H

AX

When is much larger than J the signal for each proton is a doublet, the doublet is symmetrical, and the spin system is called AX.

J J

C CH H

AM

As /J decreases, the signal for each proton remains a doublet, but becomes skewed. The outer lines decrease while the inner lines increase, causing the doublets to "lean" toward each other.

J J

C CH H

AB

When and J are similar, the spin system is called AB. Skewing is quite pronounced. It is easy to mistake an AB system of two doublets for a quartet.

J J

C CH H

A2

When = 0, the two protons have the same chemical shift and don't split each other. A single line is observed. The two doublets have collapsed to a singlet.

C CH H


Figure 13.20

OCH3

skewed doublets

H H

HH

Cl OCH3

01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0

13.11Complex Splitting Patterns:

Multiplets of multipletsMultiplets of multiplets

m-Nitrostyrene

Consider the proton shown in red.

It is unequally coupled to the protons shown in blue and white.

Jcis = 12 Hz; Jtrans = 16 Hz

H

HO2N

H

m-Nitrostyrene

16 Hz16 Hz

12 Hz 12 Hz

The signal for the proton shown in red appears as a doublet of doublets.

H

HO2N

H

Figure 13.21 H

HO2N

H

doublet of doublets

doublet doublet

13.121H NMR Spectra of Aldohols

What about H bonded to O ?What about H bonded to O ?

O—H

The chemical shift for O—H is variable ( 0.5-5 ppm) and depends on temperature and concentration.

Splitting of the O—H proton is sometimes observed, but often is not. It usually appears as a broad peak.

Adding D2O converts O—H to O—D. The O—H peak disappears.

C OH H

13.13NMR and Conformations

NMR is “Slow”

Most conformational changes occur faster than NMR can detect them.

An NMR spectrum shows the weighted average of the conformations.

For example: Cyclohexane gives a single peak for its H atoms in NMR. Half of the time a single proton is axial and half of the time it is equatorial. The observed chemical shift is halfway between the axial chemical shift and the equatorial chemical shift.

13.1413C NMR Spectroscopy

1H and 13C NMR Compared:

Both give us information about the number of chemically nonequivalent nuclei (nonequivalent hydrogens or nonequivalent carbons).

Both give us information about the environment of the nuclei (hybridization state, attached atoms, etc.).

It is convenient to use FT-NMR techniques for 1H; it is standard practice for 13C NMR.


13C requires FT-NMR because the signal for a carbon atom is 10-4 times weaker than the signal for a hydrogen atom.

A signal for a 13C nucleus is only about 1% as intense as that for 1H because of the magnetic properties of the nuclei, and

at the "natural abundance" level only 1.1% of all the C atoms in a sample are 13C (most are 12C).


13C signals are spread over a much wider range than 1H signals, making it easier to identify and count individual nuclei.

Figure 13.26 (a) shows the 1H NMR spectrum of 1-chloropentane; Figure 13.26 (b) shows the 13C spectrum. It is much easier to identify the compound as 1-chloropentane by its 13C spectrum than by its 1H spectrum.


ClCH2

Figure 13.26(a)

CH3ClCH2CH2CH2CH2CH3

1H

01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0


Figure 13.26(b)

ClCH2CH2CH2CH2CH3

020406080100120140160180200

13C

CDCl3

A separate, distinct peak appears for each of the 5 carbons.

13.1513C Chemical Shifts

Are measured in ppm (Are measured in ppm ())from the carbons of TMSfrom the carbons of TMS

13C Chemical Shifts are Most Affected By:

• Electronegativity of groups attached to carbon • Hybridization state of carbon

Electronegativity has an even greater effect on 13C chemical shifts than it does on 1H chemical shifts.

Types of Carbons

(CH3)3CH

CH4

CH3CH3

CH3CH2CH3

(CH3)4C

primary

secondary

tertiary

quaternary

Classification Chemical shift, 1H 13C

0.2

0.9

1.3

1.7

-2

8

16

25

28

Replacing H with C (more electronegative) deshieldsC to which it is attached.

Electronegativity Effects on CH3

CH3F

CH4

CH3NH2

CH3OH

Chemical shift,

1H

0.2

2.5

3.4

4.3

13C

-2

27

50

75

Electronegativity Effects and Chain Length

Chemicalshift,

Cl CH2 CH2 CH2 CH2 CH3

45 33 29 22 14

Deshielding effect of Cl decreases as number of bonds between Cl and C increases.

Hybridization Effects

sp3 hybridized carbon is more shielded than sp2.

114

138

36

36 126-142sp hybridized carbon is more shielded than sp2, but less shielded than sp3.

CH3H C C CH2 CH2

68 84 22 20 13

Carbonyl Carbons are Especially Deshielded O

CH2 C O CH2 CH3

127-13441 1461171

Table 13.3

Type of carbon Chemical shift (),ppm


RCH3 0-35

CR2R2C

65-90CRRC

R2CH2 15-40

R3CH 25-50

R4C 30-40

100-150 110-175

Table 13.3



RCH2Br 20-40

RCH2Cl 25-50

35-50RCH2NH2

50-65RCH2OH

RCH2OR 50-65

RCOR

O

160-185

RCR

O

190-220

RC N 110-125

13.1613C NMR and Peak Intensities

Pulse-FT NMR distorts intensities of signals. Pulse-FT NMR distorts intensities of signals. Therefore, peak heights and areas can be Therefore, peak heights and areas can be deceptive.deceptive.

CH3

OH

Figure 13.27


7 carbons give 7 signals, but intensities are not equal.

020406080100120140160180200 020406080100120140160180200

13.1713C-H Coupling

13C—13C splitting is not seen because theprobability of two 13C nuclei being in the samemolecule is very small.13C—1H splitting is not seen because spectrumis measured under conditions that suppress this splitting (broadband decoupling).

Peaks in a 13C NMR Spectrum are TypicallySinglets

13.18Using DEPT to Count the Hydrogens

Attached to 13C

DDistortionless istortionless EEnhancement nhancement of of PPolarization olarization TTransferransfer

1. Equilibration of the nuclei between the lower and higher spin states under the influence ofa magnetic field

2. Application of a radiofrequency pulse to givean excess of nuclei in the higher spin state

3. Acquisition of free-induction decay dataduring the time interval in which the equilibriumdistribution of nuclear spins is restored

4. Mathematical manipulation (Fourier transform) of the data to plot a spectrum

Measuring a 13C NMR Spectrum Involves

In DEPT, a second transmitter irradiates 1H during the sequence, which affects the appearanceof the 13C spectrum.

Some 13C signals stay the same.Some 13C signals disappear.Some 13C signals are inverted.

Measuring a 13C NMR Spectrum Involves


Figure 13.29 (a)

O

C

C

CH

CH CH

CH2

CH2

CH2

CH3

CCH2CH2CH2CH3

O

020406080100120140160180200

DEPT 45


Figure 13.29 (b)

CH

CH CH

CH2 CH2

CH2

CH3

CCH2CH2CH2CH3

O

CH and CH3 unaffected

C and C=O nulledCH2 inverted

020406080100120140160180200 020406080100120140160180200

DEPT 135

13.20Introduction to Infrared Spectroscopy

Gives information about the functional groups in a molecule

The region of infrared that is most useful lies between 2.5-16 m (4000-625 cm-1).

IR absorption depends on transitions between vibrational energy states (bond stretching and bending).

Bond stretching vibrations require more energy than bond bending vibrations.

Infrared Spectroscopy

Stretching Vibrations of a CH2 Group

Symmetric Antisymmetric

These vibrations are analogous to the stretching motion of two springs.

In-plane Bending Vibrations of a CH2 Group

Antisymmetric,“rocking”

Symmetric,“sissoring”

These vibrations are analogous to the in-plane bending motion of two springs.

Out-of-plane Bending Vibrations of a CH2 Group

Antisymmetric,“twisting”

Symmetric,“wagging”

These vibrations are analogous to the out-of-plane bending motion of two springs.

13.21Infrared Specta

Characteristic functional groups usually found between 4000-1600 cm-1. The fingerprint region is from 1300-625 cm-1.

Infrared Spectroscopy

Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

Figure 13.35(a): Infrared Spectrum of Hexane

Figure 13.35(b): Infrared Spectrum of 1-Hexene


Figure 13.35(c): Infrared Spectrum of Benzene


Figure 13.35(d): Infrared Spectrum of Hexylbenzene


13.22Characteristic Absorption Frequencies

Stretching vibrations (single bonds)

Structural unit Frequency, cm-1

sp C—H 3310-3320

sp2 C—H 3000-3100

sp3 C—H 2850-2950

sp2 C—O 1200

sp3 C—O 1025-1200

Table 13.4 Infrared Absorption Frequencies

Figure 13.36(f): Infrared Spectrum of Dihexyl Ether


Stretching vibrations (multiple bonds)


Table 13.4Infrared Absorption Frequencies

C C 1620-1680

—C N

—C C— 2100-2200

2240-2280

Figure 13.36(b): Infrared Spectrum of Hexanenitrile


Stretching vibrations (carbonyl groups)


Aldehydes and ketones 1710-1750

Carboxylic acids 1700-1725

Acid anhydrides 1800-1850 and 1740-1790

Esters 1730-1750

Amides 1680-1700


C O

Figure 13.36(c): Infrared Spectrum of Hexanoic Acid


Figure 13.36(d): Infrared Spectrum of 2-Hexanone


Figure 13.36(e): Infrared Spectrum of Methyl Hexanoate


Bending vibrations of alkenes



CH2RCH

CH2R2C

CHR'cis-RCH

CHR'trans-RCH

CHR'R2C

910-990

890

665-730

960-980

790-840

Bending vibrations of derivatives of benzene


Monosubstituted 730-770 and 690-710

Ortho-disubstituted 735-770

Meta-disubstituted 750-810 and 680-730

Para-disubstituted 790-840


Stretching vibrations (single bonds)


O—H (alcohols) 3200-3600

O—H (carboxylic acids) 3000-3100

N—H 3350-3500


Figure 13.36(a): Infrared Spectrum of 1-Hexanol


Figure 13.36(g): Infrared Spectrum of Hexylamine


Figure 13.36(h): Infrared Spectrum of Hexanamide


13.23Ultraviolet-Visible (UV-VIS) Spectroscopy

Gives information about conjugated electron systems.

Gaps between electronic energy levels are greater than thosebetween vibrational levels.

Gap corresponds to wavelengthsbetween 200 and 800 nm.

Transitions Between Electronic Energy States

E = h

X-axis is wavelength in nm (high energy at left, low energy at right).

max is the wavelength of maximum absorption and is related to electronic makeup of molecule, especially electron system.

Y axis is a measure of absorption of electromagnetic radiation expressed as molar absorptivity () or it may be absorbance.

Beer’s Law: A = cl

Conventions in UV-VIS

200 220 240 260 280

1000

2000

Wavelength, nm

max 230 nmmax 2630

Molarabsorptivity ()

UV Spectrum of cis,trans-1,3-Cyclooctadiene

-Electron configuration of the ground State.

Most stable

-Electron configuration of the excited state.

* Transition in cis,trans-1,3-Cyclooctadiene

HOMO

LUMO

E = h

* Transition in Alkenes

HOMO-LUMO energy gap is affected by substituents on double bond.

As HOMO-LUMO energy difference decreases (smaller E), max shifts to longer wavelengths.

Table 13.5

Methyl groups on double bond cause max to shift to longer wavelengths.

C C

H

H

H

H CH3

max 170 nm

CH3

max 188 nm

C C

H

H

Table 13.5

Extending conjugation has a larger effect on max; shift is again to longer wavelengths.

max 170 nm

max 217 nm

C C

H

H

H

H

C C

H

H

H

C C

H

H

H

Table 13.5

max 217 nm(conjugated diene)

max 263 nmconjugated triene plus

two methyl groups

C C

H

H

H

C C

H

H

H

C C

H

H

C C

H

H

C C

H

H

CH3

CH3

Lycopene

max 505 nm

Orange-red pigment in tomatoes(Lycopene)

13.24Mass Spectrometry

Atom or molecule is hit by high-energy electron.

Principles of Electron-Impact Mass Spectrometry

e–

Electron is deflected but transfers much of its energy to the molecule.

e–


This energy-rich species ejects an electron,


forming a positively charged, odd-electron species called the molecular ion.The molecular ion represents the molecular weight of the compound.

e–++••

The molecular ion passes between poles of a magnet and is deflected by magnetic field.

Amount of deflection depends on mass-to-charge ratio.

Highest m/z isdeflected least.

Lowest m/z is deflected most.


++••


If the only ion that is present is the molecular ion, mass spectrometry provides a way to measure the molecular weight of a compound and is often used for this purpose.

However, the molecular ion commonly fragments to a mixture of species of lower m/z.

The molecular ion dissociates to a cationand a radical.

Fragmentation

++ •

Usually several fragmentation pathways are available and a mixture of ions is produced.

cation radical

++••

Only the cation fragments are deflected by the magnet.

A mixture of ions of different mass gives a separate peak for each m/z.

Intensity of a peak is proportional to thepercentage of each ion of different mass in mixture.

Separation of peaks depends on relative mass.

+

++

+

+

+

Fragmentation

After passing through the magnet each ion of different mass will be detected.

Also all ions of the same mass will be detected and will reflect the intensity of that mass peak.

+ + + +

+ +

Fragmentation

20 40 60 80 100 120 m/z

m/z = 78

100

80

60

40

200

Relative intensity

Some Molecules Undergo Very Little Fragmentation

Benzene is an example. The major peak corresponds to the molecular ion.

H

H H

HH

H

H

H H

HH

H

H

H H

HH

H

all H are 1H and all C are 12C

one C is 13C one H is 2H

Isotopic Clusters

78 79 79

93.4% 6.5% 0.1%

Isotopic Abundances

M+ % M + 1 % M + 2 %1H 100

12C 98.9 13C 1.114N 99.6 15N 0.416O 99.8 18O 0.232S 95 33S 0.8 34S 4.235Cl 75.5 37Cl 24.579Br 50.5 81Br 49.5127I 100

20 40 60 80 100 120

m/z

100

80

60

40

20

0

Relative intensity

112

114

Isotopic Clustersin Chlorobenzene

Visible in peaks for molecular ion

35Cl 37Cl

20 40 60 80 100 120

m/z

Relative intensity

77

Isotopic Clustersin Chlorobenzene

No m/z 77, 79 pair; therefore ion responsible form/z 77 peak does not contain Cl.

H

H

H

H

H +

100

80

60

40

20

0

Alkanes Undergo Extensive Fragmentation

m/z

Decane

142

4357

71

85

99

CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH3

Relative intensity

100

80

60

40

20

0

20 40 60 80 100 120

Propylbenzene Fragments Mostlyat the Benzylic Position

20 40 60 80 100 120

m/z

Relative intensity

120

91 CH2—CH2CH3

100

80

60

40

20

0

13.25Molecular Formula as a Clue to Structure

Molecular Weights

One of the first pieces of information we try to obtain when determining a molecular structure is the molecular formula.

However, we can gain some information even from the molecular weight. Mass spectrometry makes it relatively easy to determine molecular weights.

The Nitrogen Rule

A molecule with an odd number of nitrogens has an odd molecular weight.

A molecule that contains only C, H, and O or which has an even number of nitrogens has an even molecular weight.

NH2 93

138

NH2O2N

183

NH2O2N

NO2

Exact Molecular Weights

CH3(CH2)5CH3

Heptane

CH3CO

O Cyclopropyl acetate

Molecular formula

Molecular weight

C7H16 C5H8O2

100 100

Exact mass 100.1253 100.0524

High Resolution Mass spectrometry can measure exact masses. Therefore, mass spectrometry can give molecular formulas.

Molecular Formulas

Knowing that the molecular formula of a substance is C7H16 tells us immediately that is an alkane because it corresponds to CnH2n+2.

C7H14 lacks two hydrogens of an alkane, therefore contains either a ring or a double bond.

Index of Hydrogen Deficiency

Relates molecular formulas to multiple bonds and rings and is also referred to as the number of unsaturations.

Index of hydrogen deficiency = ½ (number of hydrogens in the saturated formula – number of hydrogens in the compound).

Example 1

Index of hydrogen deficiency

C7H14

21 (molecular formula of alkane –

molecular formula of compound)=

21 (C7H16 – C7H14)=

21 (2) = 1=

Therefore, one ring or one double bond.

Example 2

C7H12

21 (C7H16 – C7H12)=

21 (4) = 2=

Therefore, two rings, one triple bond,two double bonds, or one double bond + one ringor two rings.

Oxygen Has No Effect

CH3(CH2)5CH2OH (1-heptanol, C7H16O) has same number of H atoms as heptane.

Neglect the oxygen in the formula below.

2

Index of hydrogen deficiency =

1 (C7H16 – C7H16O) = 0

No rings or double bonds.

2

Oxygen Has No Effect

Index of hydrogen deficiency =

1 (C5H12 – C5H8O2) = 2

One ring plus one double bond.

CH3CO

O Cyclopropyl acetate

If Halogen is Present

Treat a halogen as if it were hydrogen.

C C

CH3

ClH

H

C3H5Cl

Same index of hydrogendeficiency as for C3H6.

Rings versus Multiple Bonds

Index of hydrogen deficiency tells us the sum of the number of rings plus multiple (pi) bonds.

Catalytic hydrogenation tells us how many multiple (pi) bonds there are.

The difference between these indicates the number of rings.

End of Chapter 13End of Chapter 13SpectroscopySpectroscopy

Chapter 13 Spectroscopy Infrared spectroscopy

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