13C NMR Spectroscopy

13 C NMR Spectroscopy

Introduction

• Nuclear magnetic resonance spectroscopy (NMR) is the most powerful tool available for structural determination.

• A nucleus with an odd number of protons, an odd number of

neutrons, or both, has a nuclear spin that can be observed by the NMR spectrometer.

• NMR active nuclei include: ¹H, ¹³C, ¹⁹F, and ³¹P.

• Remember a spinning nucleus generates a magnetic field (magnetic moment).

• In the absence of an external magnetic field, proton magnetic moments have random orientations.

• However, in the presence of an external magnetic field, the

magnetic moment is aligned either with or against the external field.

Introduction

• The stronger the magnetic field, the greater the energy difference between the two spin states, resulting in a greater population

Nuclear Spin Energy Levels

• A photon of light with the right amount of energy (radiofrequency, rf) can be absorbed and cause the spinning proton to flip.

• The nuclei undergo a “spin flip”, and the nuclei are said to be “in resonance”. This absorption of energy leads to the NMR signal

External magnetic field

E₁ E₂

E₁ E₂

h absorption

of energy

Nuclear Spin Energy Levels

• If the two states become equally populated, then no net spin transitions occur and no signal is produced. This is called saturation.

• The frequency of EM radiation necessary for resonance depends on the strength of the magnetic field and on the chemical environment of the nucleus.

• Fortunately, protons (in

H NMR) in molecules usually

experience different chemical environments (i.e. are

shielded to varying extents).

1 H NMR Spectroscopy

• Therefore, different frequencies are required to bring different protons into resonance.

• Consider CH

OH:

O C H H

H

H Deshielded, senses higher

effective magnetic field so comes into resonance at a higher frequency.

Shielded, senses a smaller effective magnetic field so comes into resonance at a lower frequency.

1 H NMR Spectroscopy

• A ¹H NMR spectrum provides the following information:

1. The # of different types of H – number of basic groups of signals.

2. The relative numbers of different types of H –

3. The electronic environment of the different types of H – 4. The number of hydrogen “neighbors” a proton has –

Simple Correlation Table of ¹ H

Chemical Shifts

Why Carbon ( ¹³ C) NMR Spectroscopy

• Some organic compounds have few C-H bonds:

• Others have very similar ¹H NMR spectra:

C C C

C C

O O

O HO

HO

O

O

CH₃ H₃C

H H

O

O H H₃C

CH₃ H

Carbon ( ¹³ C) NMR vs ¹ H NMR

• The ¹³C nucleus can also undergo nuclear magnetic resonance.

• ¹³C NMR vs ¹H NMR :

– ¹²C, the most abundant isotope of carbon, does NOT exhibit NMR behavior. Why?

– ¹³C, only _____ natural abundance, does exhibit NMR behavior.

– Due to low abundance, ¹³C-¹³C coupling is usually not observed.

– Chemical shift ranges are much larger –

– Integration in ¹³C NMR is NOT reliable due to variable relaxation times from C to C. Also Nuclear Overhauser Effect - the

intensity of the C signal increases as the number of attached protons increases. Not uniform however.

Fourier Transform (FT) spectroscopy

• The magnetic moment of the ¹³C nucleus is about 1/4 that of the H nucleus resulting in lower sensitivity.

• The low natural abudance and small magnetic moment of the ¹³C isotope results in the ¹³C nucleus being about ______ less sensitive than the ¹H nucleus to NMR phenomena.

• Consequently, much longer acquisition times were required.

• The development of Fourier transform (FT) spectroscopy has made ¹³C NMR acquisition routine.

• The old way of acquiring NMR was to apply a constant magnetic field to the sample and scan the range of frequencies = continuous wave (CW) NMR.

• With FT-NMR the data is collected all at once by exciting the sample with an RF pulse (typically only a few microseconds long) which

covers all the resonance frequencies, and thus changes the

Fourier Transform (FT) spectroscopy

• After the pulse has stopped, the decay of the signal from the sample is measured. The decaying sine wave called a free induction

decay (FID):

• A Fourier transform converts the intensity vs time data into intensity vs frequency information.

Time (s) Frequency

Fourier Transform

C

H₃C CH₃ O

Intensity of signal

Fourier Transform (FT) spectroscopy

Fourier Transform

Chemical Shifts in ¹³ C NMR

• Two simple ideas will make interpretation of

C NMR spectra easier:

1. Hybridization of the C atom determines the chemical shift:

 sp

hybridized carbons have chemical shift values _________.

 sp

hybridized carbons have chemical shift values _________.

2. The presence of an EN element near a C atom will

cause its chemical shift to move _____________.

Simple Correlation Table of ¹³ C chemical shifts

See text (p. 593) and Lab manual (p. 60) for more extensive tables

Coupling in Carbon NMR

• The low abundance of ¹³C makes C-C coupling very rare.

• However, ¹³C-H coupling is common. N+1 rule still applies:

Coupling in Carbon NMR

• Spectra which show

C-H coupling are called proton- coupled spectra.

• However, extensive

C-H coupling often produces splitting patterns that are difficult to interpret.

• To simply

C NMR spectra, often recorded using broad band proton decoupling.

• Therefore each carbon signal appears as a singlet, because C-H splitting has been eliminated.

• Spectra recorded in the broad band proton decoupling

mode give the number of unique carbon atoms in a

molecule.

Proton-coupled vs Proton-decoupled

13 C NMR Spectra

Interpreting ¹³ C NMR Spectra

H₃C C CH₂ CH₃ CH₃

CH₃

TMS CDCl₃

solvent

Interpreting ¹³ C NMR Spectra

CH₃ C O O

CH₂ CH₃

Interpreting ¹³ C NMR Spectra

CH₃ O

Interpreting ¹³ C NMR Spectra

O

O

CH₃ H₃C

H H

O

O

H H₃C

CH₃ H

Mass Spectrometry

Basic Principles

• Mass Spectroscopy (MS) is a destructive analytical technique for measuring the ______________ (____) of ions in the gas phase.

 This allows accurate determination of the _______________ of a molecule.

 Structural information is also gained.

 Molecular Formula determination is sometimes possible.

• While the method is destructive, only very small amounts (1 mg or less) is required.

• MS does not involve the absorption or emission of light.

• A mass spectrometer is designed to do 3 things:

1. Convert a neutral molecule, M, into positive (or negative) ions usually by bombardment with a beam of high energy

electrons.

2. Separate the ions based on mass (mass-to-charge ratio, ___).

Basic Principles

M + e M + 2e 10-70 eV

in energy 1 eV = 23 kcal/mol

Schematic of Mass Spectrometer

• First the sample is vaporized under vacuum.

• A beam of electrons bombards the molecules in the gas phase causing ionization and formation of radical cations.

~70 Volts

+

_ +

_

e^- e^- e^-

+ + + + + +

_

Electron Collector (Trap)

Repeller

Extraction Plate Filament

to Analyzer Inlet

Electrons Neutral

Molecules

Positive Ions

Electron Impact Ionization Source

Electron impact Ionization source

Schematic of Mass Spectrometer

• The radical cations fragment further after ionization owing to the large amount of energy transferred by the electron beam.

• Some fragments carry a positive charge, others are neutral:

• Only the positively charged fragments are accelerated into the

CH₄ - e C

H H H

H

m/z =

[CH₃] + H m/z =

[CH₂] + 2H m/z =

Molecular ion

Schematic of Mass Spectrometer

• The analyzer tube is surrounded by a magnet whose magnetic field deflects the positively charge fragments in a curved path.

• The amount of deflection depends on m/z.

ion trajectory not in register (too heavy) Ion

Source

Detector ion trajectory

not in register (too light)

ion trajectory in register

S

N

Magnetic Sector Mass Analyzer

Electromagnet

Basis of Fragment Separation

• Fragments with smaller m/z value are deflected ______ than a larger m/z value.

• Since z is usually _____, the fragments are sorted by mass.

• By varying the magnetic field, cations of different masses are sorted and counted by a detector.

• The more stable the fragment the more likely it will make it to the detector.

• The masses are graphed or tabulated according to their relative abundance = The Mass Spectrum.

The Mass Spectrum of Methane

12 13 14 15 16 17

M⁺ = 15 C¹²H₃⁺

M⁺ = 16 Molecular ion

[C¹²H₄]^+.

[C¹²H₂]^+.

[C¹²]^+.

C¹²H⁺

m/z Intensity

1 3.4

2 0.2

12 2.8

13 8.0

14 16.0

15 86.0

16 100.0

17 1.11

Base peak

Isotopes

• Most elements common to organic compounds are mixtures of isotopes.

• The existence of atomic isotopes in nature accounts for the appearance of M+1 and M+2 peaks in a mass spectrum.

• Organic compounds containing only C, H, O, and N usually have relatively small M+1 and M+2 peaks.

Relative abundance, %

M+

M+1⁺

M+

M+1⁺ M+2⁺

Isotopes

Element Most abundant isotope Less abundant isotope Relative abundance

Hydrogen ¹H ²H 0.016

Carbon ¹²C ¹³C 1.08

Nitrogen ¹⁴N ¹⁵N 0.38

Oxygen ¹⁶O ¹⁸O 0.20

Sulfur ³²S ³⁴S 4.4

Chlorine ³⁵Cl ³⁷Cl 32.5

Bromine ⁷⁹Br ⁸¹Br 98.0

Isotopes

• MS is particularly valuable for compounds which contain Cl and Br:

 If one S atom is present, M + 2 is ~ 4% of M⁺.

 If one Cl atom is present, M + 2 is ~ 33% of M⁺.

 If one Br atom is present, M + 2 is ~ to M⁺.

Relative abundance, % M+

M M+2

24 1

M M+2

3 1

M M+2

1 1

M+

M+2⁺

M+ M+2⁺

Mass Spectrum with Chlorine

M+

M+2⁺

CH

H₃C CH₃ Cl

Mass Spectrum with Bromine

M+ M+2⁺

CH₂

H₃C CH₂ Br

Isotopes

• Carbon Rule – For compounds containing only C, H, and O, the following formula can be used to determine the number of carbons in the molecule:

1.1

peak 1

+ M of intensity relative

= s C no.

Determine the molecular formula of the unknown organic compound whose mass spectral data is given in the table below:

Peak Mass (m/z)

Relative intensity

M 86 100.0

M+1 87 5.6

M+2 88 0.4

Isotopes

• Nitrogen Rule: if a compound has:

– An odd number of nitrogen atoms, its molecular ion, M+, will be odd.

– Zero or an even number of nitrogen atoms, its molecular ion, M+, will be even.

Resolution

• Resolution: a measure of how well a mass spectrometer separates ions of different mass.

 Low resolution – capable of distinguishing among ions of different nominal mass, i.e. different by at least one or more amu.

 High resolution – capable of distinguishing among ions that differ in mass by as little as 0.0001 amu.

• For example: CO, N₂, and ethene all have a nominal mass of 28 amu. High resolution MS can distinguish these molecules.

CO 27.9949 amu

N₂ 28.0061 amu

CH₂=CH₂ 28.0314 amu

Fragmentation Pathways

• Structural information is available from analysis of fragments formed by bond cleavages in the molecular ion, M⁺.

• In general, the molecular ion, M⁺, will fragment so as to form the most stable cationic fragment (usually a carbocation).

• In some cases, the M⁺ peak is very small or absent. Occurs if the fragments are considerably more stable M⁺.

Mass Spectrum-Fragmentation

• Consider the mass spectrum of pentane: p. 516-517 text

• Fragmentation of the molecular ion often results:

Mass Spectrum of Pentane

Fragmentation of Alkanes

CH₃ CH CH₂ CH₂ CH₃

CH₃

CH₂ CH₂

CH₃ HC CH₃

CH₃ +

CH CH₂ CH₂ CH₃

CH₃

+ CH₃ 1

2 m/z 86 3

m/z 43

m/z 71

86 M⁺ 71

57

CH₃ CH CH₂ CH₂ CH₃

CH₃ 43

Compounds with Heteroatoms

• Molecules containing O, N, halogens, or other heteroatoms often undergo ___________ (adjacent to heteroatom).

• Driving force is resonance stabilized cations.

Fragmentation of Alcohols

• Alcohols common fragmentation is -cleavage and loss of “H₂O” to give an M-18 peak.

M⁺= 88 (not observed) M-15

M-18 M-29

M-18 -15

C OH

H₃C CH₂ CH₃ CH₃

Fragmentation of Amines

M⁺ = 73 CH₂=NH+ ₂

m/z=30

NH₂ CH₂ CH H₃C

H₃C

Fragmentation of Ketones

M⁺ = 72 57

43

C O

H₃C CH₂ CH₃

McLafferty Rearrangement

• If one of the alkyl groups attached to the carbonyl carbon of an aldehyde or ketone has a hydrogen, a cleavage known as a McLafferty rearrangement can occur.

Mass spectrum of butyraldehyde

M⁺=72 M–28

O

H

Aromatic Compounds

• Usually strong M+ peak.

• m/z =91 for tropylium ion and methylene spacings above 91 (105, 119, etc. for alkyl chains) often observed.

• m/z = 65 (C₅H₅⁺), 77 (C₆H₅⁺) are sometimes observed.

-R

m/z 91

CH₂ R CH₂

Aromatic Compounds

Common Fragments

CH₂ CH₃ CH₃CH₂

O H C

O H₃C C

O HO C

C O m/z = 15 29 29 43

m/z = 45 91 105