Carbon NMR

1

NMR SPECTROSCOPY

Felipe Garcia

[email protected]

2

Determination of molecular structure

There are a range of techniques that can be used to determine and/or support structures of new compounds. These include X-ray diffraction, nuclear magnetic resonance (NMR), IR spectroscopy, UV-visible spectroscopy, microwave spectroscopy and mass spectrometry. Each technique provides different information and each of them have their own particular advantages and disadvantages.

During this course we are going to focus on: nuclear magnetic resonance (NMR)

Infrared spectroscopy (IR spectroscopy)

Introduction

Spectroscopy has been used for years in the search of an univocal way of identifying compounds once they are synthesised. The introduction of UV in 1930s and IR in 1940 provided techniques capable of elucidating from different functional groups. Using these techniques, for the first time, structural information could be obtained by small scale, non-destructive methods. Mass spectrometry introduced in 1950s, provided insight into the molecular formulae of compounds and their structures from molecular weights and fragmentation patterns. However, the analytical method that had the greatest was nuclear magnetic resonance (NMR). NMR provides information about the molecular skeleton; and in combination will all the techniques previously mentioned, it provides us the final piece of the ‘structural puzzle’ for the univocal structural determination of novel compounds.

Molecules are small, too small to be observed and studied directly; therefore we need methods capable of gathering information on the structures, motion and chemical reactions without significantly modifying the properties to be measured from the molecules and versatile enough to be applied in a wade range of situations and conditions.

Electrons in molecules and atoms may only posses certain energies; in simple terms, the energy in atoms and molecules is quantisized. Therefore to promote one electron from one level to another we need a fixed amount of energy, which depends on the initial (E1) and final energy (E2) levels.

Apart from the total energy associated with the system, (in our case molecules), is also convenient to separate the total energy into separate components like electronic, vibrational, rotational and

3

translational. However, regardless of the amount electronic energy there is always vibrational and rotational energies.

It is possible for an atom or a molecule to change their energy by moving from one energy level to another. For example: promote one electron from a low energy orbital to a higher energy one, or promoting the molecule from an specific lower vibrational energy level to a higher one by absorbing light energy or other forms of electromagnetic radiation (microwaves, infrared, ultraviolet, etc). Therefore, the vibrational and rotational energy levels are also quantisized.

The separation between the two energy levels, ΔE, is proportional to the frequency of the light absorbed, ν.

ΔE =hν

where (h is Planck’s constant =6.626 × 10− 34 J s).

• Moving from a lower energy level to a higher one requires an energy imput or absortion. • Moving from a higher energy level to a lower one produces an energy release or emission.

ABSORTION

∆E∆E= hν

E1

E2 E2

E1

EMISSIONE1

E2

E1

E2

The exact frequency at which energy is absorbed depends on the particular system. Different molecules will absorb at different frequencies. The frequency and therefore the energy of an electromagnetic radiation spans over many orders of magnitude. The energy required for a transition between to energy levels depends on the type of transition. For example, the energy separation between two energy electronic energy levels is much greater than the separation between two vibrational energy levels which is greater the separation for rotational levels.

4

Transition between electronic energy levels occur in the X-ray, ultraviolet or visible regions of the spectrum. Transitions between vibrational energy levels occur in the infrared, etc (see figure above)

NMR spectroscopy:

Every time a reaction takes place with me compounds being produces and/or isolated, chemists have to first identify the their molecular structures. In this chapter we will have a look at main two spectroscopic techniques such as NMR and IR. Being NMR the most important on a day to day basis for the vast majority of synthetic chemists as is an extremely sensitive technique with a wide range of uses and applications.

Electron spin:

electrons possess a property called ‘spin’. The electron has an angular momentum, which is a property associated with rotating objects and is denoted by the quantum number (S). For an electron in always 1/2.

In a magnetic field there are two different orientations of the angular momentum. The new orientation is denoted by a new quantum number (ms) which takes a value of +1/2 or -1/2 and denote electrons with opposite spin with up and down arrows, ↑ and ↓. Spin becomes important, when, for example, working out the lowest energy arrangement of electrons in atoms, as you have seen in the past.

Nuclear spin (I)

In a similar manner, certain nuclei possess spin - called nuclear spin and denoted by the quantum number I - resulting in very weak magnetic field associated with it.

This magnetic field comes from the incomplete quenching of all the spin angular momentum from individual neutrons and protons composing the nucleus.

5

Nucleous Number of protons and neutrons (p, n)

% Natural abundance I Number of energy levels (2I+1)

1H, single proton 1p, 0n 99.985 ½ 2 2H, deuterium 1p, 1n 0.015 1 3

10B 5p, 5n 20 3 7 11B 5p, 6n 80 3/2 4 12C 6p, 6n 98.9 0 1 13C 6p, 7n 1.1 ½ 2 14N 7p, 7n 99.6 1 3 16O 8p, 8p 99.8 0 19F 9p, 10n 100 ½ 2

When protons and neutrons are all not paired up, is not easy to predict the total angular momentum.

When a nucleus with I≠0 is placed in magnetic field there is an interaction between this nuclear spin and the applied field which induces 2I+1 number of orientations of the spin angular momentum. These orientations have different energies and are differentiated by the quantum number ml. which takes values from +I to -I in integer steps

For example:

In the case of 1H

1H (I=1/2) 2I+1 = 2 x ½ + 1 = 2

2 different orientations ml = +1/2 and -1/2

In the case of 13C

14N (I=1) 2I+1 = 2 x 1 + 1 = 3

3 different orientations ml = +1, 0 and -1

mmaaggnneettiicc ff iieelldd aapppplliieedd

eenneerrggyy ooffssppiinn ssttaatteess

nnoo mmaaggnneettiicc ff iieelldd aapppplliieedd ((II ==11//22)) ssppiinn ssttaatteess aarree ddeeggeenneerraatteedd

mmaaggnneettiicc ff iieelldd aapppplliieedd

((22II++11==22)) ssppiinn ssttaatteess ff oorr 11HH

BB

eenneerrggyy ddiiff ff eerreenncceeooff 11HHssppiinn ssttaatteess aatt aa mmaaggnneettiicc ff iieelldd ooff iinntteennssiittyy BB

11HH mmll== ++11// 22

11HH mmll== ++11// 22

ssttrreennggtthh ooff mmaaggnneettiicc ff iieelldd ((BB)) aapppplliieedd

eenneerrggyy ooffssppiinn ssttaatteess

nnoo mmaaggnneettiiccff iieelldd aapppplliieedd ((II ==11))

mmll == --11

mmll == 00

mmll == 00

mmaaggnneettiicc ff iieelldd ((BB)) aapppplliieedd ((22II ++11==33))

6

The difference in energy between the different spin states depends on the nucleus of study and strength of the magnetic field experienced by it.

magnetic f ield applied

energy ofspin states

1H ml= +1/2

1H ml= +1/2



13C ml= +1/2

13C ml= +1/2

It turns out that the exact frequency (remember ΔE=hν) of resonance (energy required for the transition to take place) is directly proportional to the strength of the magnetic field applied.

For example, For 1H



no magnetic f ield applied (I=1/2) spin states are degenerated


(2I+1=2) spin states for 1H

B1

energy differenceof 1Hspin states at a magnetic f ield of intensity B

1H ml= +1/2

1H ml= +1/2





(2I+1=2) spin states for 1H

B2

energy differenceof 1Hspin states at a magnetic f ield of intensity B2

1H ml= +1/2

1H ml= +1/2

For 13C





(2I+1=2) spin states for 13C

B1

energy differenceof 13Cspin states at a magnetic f ield of intensity B2

13C ml= +1/2

13C ml= +1/2





(2I+1=2) spin states for 13C

B2

energy differenceof 13Cspin states at a magnetic f ield of intensity B2

1H ml= +1/2

1H ml= +1/2

7

A few examples to illustrate these differences are shown in the tables below.

Bo (T) 1H Freq (MHz) 1.41 60 2.35 100 4.70 200 7.05 300 9.40 400 11.75 500 18.80 800

Frequency for different nuclei at fixed magnetic fileds:

Nucleus Frequency in a 4.7T magnetic field (MHz)

Frequency in a 18.8TT magnetic field (MHz)

13C 50 200 31P 81 324

195Pt 43.3 173.2

As we mentioned before when an active nucleus (I≠0) is placed in a strong magnetic field there is an interaction between this nuclear spin and the applied field which produces a set of nuclear spin energy levels (ml) with a certain energy separation. Once the spin nuclear levels split a radiowave of the appropriate frequency will cause transitions between these energy levels. The transitions between these energy levels gives rise to the NMR signal.

Nucleus

% Natural abundance

I

Number of energy levels

1H, a single proton 99.985 1/2 2 2H, a deuterium nucleus (D) 0.015 1 3

10B 20 3 7 11B 80 3/2 4 12C 98.9 0 1 13C 1.1 1/2 2 14N 99.6 1 3 15N 0.368 1/2 2 16O 99.8 0 1 17O 0.038 5/2 6 19F 100 1/2 2 31 100 1/2 2

Local Magnetic environment:

NMR is extremely sensitive as not all the 1H and 13C nucleus resonate at the same frequency. The exact frequency at which each nuclei of the same kind resonate depends on the local magnetic field experienced by each nuclei. The local magnetic field depends on the electron density. In a magnetic field, electrons move in such a way that they set up their own magnetic field which opposes the applied field.

8

As a result of this opposition, the nucleus experiences a weaker magnetic field than than it would have done had the electrons not been there. The weaker the magnetic field, the smaller the difference in energy between the nucleus’ spin states and hence the smaller the frequency of radio waves absorbed.

Electrons are said to shield the nucleus – the more electrons, the more shielded the nucleus is and hence the lower the frequency that it resonates at.

In an analogous way, anything which removes electron density from around a nucleus, deshields it and shifts the resonance to a higher frequency. Since electronegative elements withdraw electrons towards themselves, they will deshield atoms attached to them.

Increasing frequency for 13C

CH3F CH3OH CH3Cl CH4

The more electronegative the atom attached to a 13C nucleus, the higher the frequency

From the previous examples we may deduce that one factor contributing to chemical shift differences in resonance is the inductive effect (electron withdrawing effect produced by the electronegative atoms).

As we mentioned earlier, if the electron density about a nucleus is relatively high, the induced field due to electron motions will be stronger than if the electron density is relatively low. The shielding effect of the high electron density will be larger, and a higher external field (Bo) will be needed to excite the nuclear spin. Since the electronegativity decreases from left to right (F>O>Cl) (figure above), the local electron density around the C is expected to be greater, and the characteristic resonance signal from the C lie at a higher magnetic field. Such nuclei are said to be shielded (lower chemical shifts)

On the other hand elements that are more electronegative than carbon should produce the opposite effect (reduce the electron density); so groups bonded to such elements display lower field signals (they are deshielded). The deshielding effect of electron withdrawing groups is roughly proportional to their electronegativity. Furthermore, if more than one such group is present, the deshielding is additive (table on the right), and proton resonance is shifted even further downfield.

For example, Proton Chemical Shifts of Methyl Derivatives

9

Compound (CH3)4C (CH3)3N (CH3)2O CH3F 0.9 2.1 3.2 4.1

Compound (CH3)4Si (CH3)3P (CH3)2S CH3Cl 0.0 0.9 2.1 3.0

As seen from the tables, increasing the electronegative of the groups attached to the CH3 system decrease the electron density around the nuclei of study (in this case protons) protons, and there is less shielding (i.e. deshielding) so the chemical shift increases

Proton chemical shifts (ppm) with different number of substituents.

Compound\substituent Cl Br I Or Sr CH3X 3.0 2.7 2.1 3.1 2.1 CH2X2 5.3 5.0 3.9 4.4 3.7 CHX3 7.3 6.8 4.9 5.0

Carbon chemical shifts (ppm) with different substituents.

Compound\substituent F Cl Br I Or CH3X 75 25 10 -21 60 CH2X2 110 54 22 -54 103 CHX3 119 77.5 12 -140 115 CX4 94 97 -29 -293 121

The electronegativity effect (inductive effects) is cumulative, so the presence of a higher number of electronegative groups produce more deshielding and therefore, larger chemical shifts

Chemical shifts

No frequency is given in the previous picture as since the frequency will depend on the strength of the magnetic field (as was mentioned earlier); therefore the values for a given compound will be different in different magnetic fields.

To overcome this issue, the frequencies are expressed as shifts from an agreed reference - chemical shifts (δ) - which are independent of the magnetic field strength.

𝐶ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑠ℎ𝑖𝑓𝑡 (𝛿) 𝑖𝑛 𝑝𝑝𝑚 (𝑝𝑎𝑟𝑡𝑠 𝑝𝑒𝑟 𝑚𝑖𝑙𝑙𝑖𝑜𝑛) = 106 𝑥 𝜐𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒 − 𝜐𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒

𝜐𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒

νresonance = frequency of the nucleus of interest

νresonance = frequency of the nucleus of interest

peak from reference compoundPeak of interest

INCREASING FREQUENCY

10

Since the frequency depends on the strength of the magnet is common use to express the streng of the magnet using the frequency at which protons resonate. For example in instrument with a magnet of 18.80 Tesla protons resonate at 800MHz. So is commonly called an 800MHz instrument.

The frequency chances depending on the magnet but the chemical shift is a relative measure. So in an 800Mhz instrument 1ppm is equal to 800Hz,

Bo (T) 1H Freq (MHz) pmm – Hz equivalence 1.41 60 60 2.35 100 100 4.70 200 200 7.05 300 300 9.40 400 400 11.75 500 500 18.80 800 800

The frequency at which 13C resonates is approximately one quarter of the one for 1H.

The reference compound.

One method of solving this problem is to report the location of an nmr signal in a spectrum relative to a reference signal from a standard compound added to the sample. Such a reference standard should be chemically unreactive, and easily removed from the sample after the measurement. Also, it should give a single sharp nmr signal that does not interfere with the resonances normally observed for organic compounds. Tetramethylsilane, (CH3)4Si, usually referred to as TMS, meets all these characteristics, and has become the reference compound of choice for proton and carbon nmr.

All protons in TMS are equivalent so this compound only shows one signal in its proton NMR. Silicon is less electronegative than carbon so the carbons are more shielded relative to most organic compounds containing just C–C bonds. The net result is that in most organic compounds, the TMS reference signal is over to the right, out of the way.

Si

CH3

H3CH3C

CH3

tetremethyl silane, TMS

Some examples of chemical swifts respect to TMS are shown in the table bellow

Compound Shift of methyl carbon / ppm

Shift of methyl hydrogens

CH3Li –14 - Si(CH3)4 0 by definition 0 by definition CH3CH3 7 0.9 CH3Cl 26 3.1

CH3NH2 28 2.5 CH3OH 50 3.4 CH3NO2 61 4.3

CH3F 72 4.3 Note: CH3Li is primarily an ionic compound with CH3

- Li+. Therefore the carbon is very shielded giving negative value for its chemical shift

11

Number of peaks – looking for symmetry

The frequency at which at nuclei will resonate depends on the local electron density which will then “define” the local magnetic field. Each different environment will give raise to a signal in the NMR spectrum. (with different energy separations and therefore different frequencies for the transition).

For example:

In the case of 3-nonanone:

Every carbon nucleus is in a different environment; therefore nine different environments give raise to nine separate peaks in the 13C NMR spectrum.

In the case of 3-nonanone there are only 5 different environments giving rise to 5 distinct signals in the 13C NMR spectrum.

In the case of the 5-nonanone there is a plane of symmetry that interrelated on side of the molecule with the other. Therefore, the left and the right halves are equivalent.

Another example: in the case of the compound with formulae C4H9OH.

nBuOH

12

So there are 4 carbons environments – every carbon environment is slightly different - giving rise to four different signals.

However in the case of tBuOH, only two peaks are observed

All the CH3 groups in tert-butanol are equivalent due to the existence of a rotational axis

Nuclei that can be related one another through symmetry operations (mirror planes, rotation axis) are in the same environment.

For example in the case of the tert-butyl alcohol, there is three fold rotation axis around the HO-C1 making all the methyl groups equivalent. Therefore, just one signal on the NMR spectrum for the C2

C1

C2

C3

C2

C2

C1

C3

C1

C2

C1

C2

C3

Therefore each nuclei within the exact same environment will have the same chemical shift.

So NMR helps distinguish between isomers.

Interpreting NMR data

Well start first having a look at 13C NMR interpretation followed by 1H nmr.

13C NMR resonances usually fall within the range of 0 to 240ppm. In general terms the NMR spectrum may be divided into four regions

• Single bonded carbon (sp3 carbon atoms; C-X) atoms to strongly electronegative elements (e.g., O, F) or more than one less strongly withdrawing group: 50 – 100 ppm

• Single bonded carbon atoms (sp3 carbon atoms) with no such strongly electron withdrawing atoms (e.g., N and Cl): 0 – 50 ppm.

13

• Doubly bonded carbon atoms (C=C, sp2 carbon atoms); their shift varies depending on the

substituents: 100 – 150 ppm.

• Doubly bonded carbon atoms bonded to oxygen (sp2 carbon atoms; C=O); their shift varies depending on the substituents: 150 –200 ppm.

Dividing the NMR scale:

The carbon NMR spectrum is usually recorded between 0 to slightly over 200ppm since most of the organic compounds fall within this range.

These divisions act as guide lines only. For example, given enough electron withdrawing groups, an sp3 hybridised carbon will move up to the range of 100-150 ppm.

Compound\substituent F Cl Br I Or CH3X 75 25 10 -21 60 CH2X2 110 54 22 -54 103 CHX3 119 77.5 12 -140 115 CX4 94 97 -29 -293 121

H

CClH

H

Cl

CClH

H

Cl

CClCl

H

Cl

CClCl

Cl

9725 54 77.5δ/ppm

As we increase the number of Cl atoms the chemical shift increases; carbon atoms are more deshielded (i.e., they resonate at higher chemical shifts).

The compound on the As we increase the numer

OOCCHH33

CCOOCCHH33HH33CCOO

HH

111144ppppmm

5511ppppmm

14

A simplified version of the table above can be described as follows

C C

C N

C CC O

C O

sp2 carbons

sp2 carbons with very

electrongroups attachedsp3 carbons with very

electrongroups attached

sp3 carbons 0 ppm50150200 100

Coupling in NMR

The frequency at which a nucleus resonates depends on the local magnetic field that it experiences. We saw that local magnetic fields depends on:

(i) External magnetic field

(ii) Electron density around the nucleus

A further contribution may arise from close by nuclei with non-zero spin.

For example: in the 13C NMR of acetic acid, there are two different carbon environments giving rise to two different signals

Coupling to one equivalent nuclei

However if we have a look at the spectrum of fluoroacetic acid; each carbon environment gives rise to a separate signal. As expected, the chemical shifts are influenced by the electronegativity of the fluorine atom attached to C2 (c.f. ~δ21ppm for acetic acid and ~δ78ppm for the fluroroacetic acid)

15

However, if we look closely the signal for both carbon environments are doubled up. There are pairs of signals at 78ppm and 174ppm. If we look at the spacing between the pairs of signals; the signals for C2 - the carbon with the F directly attached - are more separated then the ones for C1.

Each of the “doubled” carbons is said to be coupled the fluorine atom (19F, 100% natural abundance, I= ½). Therefore, there are 2I+1 spin states possible for fluorine; half are fluorine “spin up” and the other half are “spin down”.

One spin state slightly increases the local magnetic field, the other slightly reduces it (see figure below)

If the fluorine had spin zero, there would be no coupling as show in red. However the spin is ½, so when the fluorine is “spin up” it sets a magnetic field that will reinforce the external magnetic field (left on figure, a), therefore this carbon when the fluorine is “spin up” experiences a slightly stronger magnetic field that would have if the spin of fluorine would have been zero. The slightly stronger magnetic field means that the resonance for the carbon is at a slightly higher chemical shift.

Conversely, when the fluorine is “spin down” it sets a magnetic field that will oppose the external magnetic field (right on figure, b), therefore this carbon when the fluorine is “spin down” experiences a slightly weaker magnetic field that would have if the spin of fluorine would have been zero. The slightly weaker magnetic field means that the resonance for the carbon is at a slightly lower chemical shift.

The signal of the carbons are split in two peaks of equal height, shifted equal amount either side of the original peak. This called a doublet. The effect of the coupling in a signal can be determined using what is called a ‘tree diagram’, which is a diagrammatic representation on how the original signal is successively split by the interaction with the different spin states of each nucleus one at a time.

16

J

original signal

spliting of the original signal intoa doublet caused by a nuclei with I=1/2

J = coupling constant(meassured in HZ)

1 1

The separation of the peaks is independent of the magnetic field applied; is related to the separation of the energy levels once the nuclei interact with each other. The size of the interaction is called coupling constant (J) and is measured in Hz – never in ppm as they depend on the strength of the external magnetic field – as they represent energy (which is directly proportional to frequency).

The interaction between the nuclei is transmitted through bonds. It depends on the nature of the nuclei involves and they are bonded together. Generally the coupling constant decreases as the number of connecting bonds increases. However there are other factors such as angles, rotation around the bonds, etc.

Coupling constants are denoted by the symbol J

1JA-B coupling through one bond between A and B

2JA-B coupling through two bonds between A and B

3JA-B coupling through three bonds between A and B

….

nJA-B coupling through n bonds between A and B

17

Coupling to two equivalent nuclei

when the coupling to two equivalent nuclei is considered. For example in the case of difluoro acetic acid. In the molecule, there are two equivalent atoms of fluorine (related by a plane of symmetry). There are three different ways the spins of the two fluorine atoms can be aligned:

Possible arrangements of the spin for two nuclei with I=1/2

Therefore this three possible arrangements for the spins of two equivalent nuclei with I=1/2 will split the original signal into three

With both spins up the carbon experiences a slightly stronger local magnetic field and resonate at higher chemical shifts. With both down they resonate at slightly lower chemical shift. With the combination of spins up-down and down-up neither has any effect on the magnetic field - they cancel one another – so the chemical shift remains the same

both spins down, thecarbon resonates at

lower chemical shif t

one spin up and one down,the carbon resonates at thesame chemical shif t

both spins up, thecarbon resonates at

higher chemical shif t

Both nuclei are indistinguisableboht spin configurations aremagnetically equivalent

The tree diagram shows how the original signal splits into two and subsequently each of these signals are split into two again. The coupling constants are the same every time each signal splits in two; as a consequence of this, the central signal is doubled in intensity as two of the signals overlap.

18

J

original signal

spliting of the original signal intoa triplet caused by two nuclei with I=1/2

J = coupling constant(meassured in HZ)

2 1

JJ

1

original signal is split due to the f irst f luorine nuclei

each signal is split due to the second f luorine nuclei

The 13C NMR of difluoro acetic show each of the carbon signals a triplet. As seen previously the coupling constant decreases over the number of bonds.

Coupling to three equivalent nuclei

In trifloro acetic acid, each carbon couple to three equivalent fluorine atoms. There are four different ways

Possible arrangements of the spins for three nuclei with I=1/2

Therefore this three possible arrangements for the spins of three equivalent nuclei with I=1/2 will split the original signal into four, called quartet.

19

Therefore in the carbon 13C NMR of trifluoroacetic acid each signal will be split into a quartet.

Predicting the intensity ratios using pascal’s triangle (Tartaglia’s triangle)

In general:

A resonance signal is split into 2I+1 signals by coupling with one nucleus with spin I

A resonance signal is split into 2nI+1 signals by coupling with n equivalent nuclei with spin I

Therefore we can simplify this when it comes to coupling with a nucleus with spin ½ by saying that a resonance signal is split into n +1 number of signals by coupling to n nuclei with spin ½ (note: 2nI+1 (when I= ½) is 2 x n x ½ +1 => n + 1.

For coupling with niclei with spin half, the intensities after each successive splitting are given by Pascal’s triangle.

20

Splitting exceptions

• Coupling to H in 13C NMR

In all the examples previously used to illustrate coupling patters splitting with hydrogen atoms (I= ½) was not observed in the same way it was seen for fluorine atom (I= ½). In general most carbon NMR spectra are in a way that any coupling to protons is removed. This process is known as broadband proton decoupling.

Proton coupled spectra can be recorded, they provide more information about the number of protons attached to the carbon atoms. Therefore we could distinguish between CH3, CH2, CH and quaternary (i.e., no protons attached). However it makes the spectrum difficult to interpret. Also coupling reduces the intensity of the peaks, making it more likely for weak peaks to be lost in the noise.

• Carbon – carbon coupling in 13C NMR

The natural abundance of carbon 13 is 1.1%, therefore – looking back into acetic acid – the majority of the sample will be A, there will be 1.1% of the molecules containing a 13C on carbon 1, there will be another 1.1% containing a 13C on carbon 2 and there will be 0.012% containing two 13C atoms.

21

O

12CHO 12CH3

O

13CHO 12CH3

O

12CHO 13CH3

O

13CHO 13CH3

1.1% 1.1% 0.012%

A B C D

• No coupling between equivalent nuclei

No splitting aer observed in an spectructum due to coupling between nuclei within the same environment. The reason for it is beyond the scope of this course.

Coupling between A and B and B and A

As we saw in pague 17 the coupling constants between A and B in the NMR spectrum of A is the same as the coupling between B and A in the NMR spectrum of B.

Couplings with spins greater than 1/2

many nuclei has spins higher than ½, the splitting should be equal to 2I+1. However, in most cases this splitting is not observed because the 2I+1 spin states rapidly interconvert – process called relaxation – making all the splitting due to the coupling with that nucleus to average to zero.

However there are a few examples of nuclei with spin greater than ½ which relaxation is slow enough for coupling to be observed.

A typical example is deuterium (2H, D, I= 1). For example CDCl3, CD2Cl2 and CD3SO are shown below

The splitting of patterns of CDCl3, CD2Cl2 are explained bellow

22

Documents

Carbon NMR