NUCLEAR MAGNETIC RESONANCE SPECTROSCPY A guide for A level students KNOCKHARDY PUBLISHING 2008...

NUCLEAR MAGNETIC NUCLEAR MAGNETIC RESONANCE SPECTROSCPYRESONANCE SPECTROSCPY

A guide for A level studentsA guide for A level students

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING2008 2008

SPECIFICATIONSSPECIFICATIONS

NMR SPECTROSCOPYNMR SPECTROSCOPYKNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING

INTRODUCTION

This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.

Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.

Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...

www.knockhardy.org.uk/sci.htm

Navigation is achieved by...

either clicking on the grey arrows at the foot of each page

or using the left and right arrow keys on the keyboard

CONTENTS• Prior knowledge

• What is nmr?

• Origin of spectra

• Tetramethylsilane

• Chemical shift

• Resolution

• Multiplicity - splitting patterns

• Integration

• OH signals and the use of D2O

• Working out spectra

• Questions on proton nmr

• Carbon-13 nmr

NMR SPECTROSCOPYNMR SPECTROSCOPY

Before you start it would be helpful to…

• know the names and structures of organic functional groups

• find the structures of isomers given the molecular formula


WHAT IS NMR AND WHAT DOES AN NMR SPECTRUM TELL YOU?

Proton nuclear magnetic resonance spectroscopy provides...

• information about the hydrogen atoms in molecules

How does it work?

• involves the interaction of materials with the LOW ENERGY RADIO WAVESLOW ENERGY RADIO WAVES

It provides the information by...

spinning a sample of the compound in a magnetic field

• hydrogen atoms in different environments respond differently to the field

• each different environment of hydrogen produces a signal in a different position

• the area under each peak / signal is proportional to the number of hydrogens

• signal can be split according to how many H’s are on adjacent atoms

PREVIEW

NMR SPECTROSCOPY – ORIGIN OF SPECTRA

All nuclei possess charge and mass. Those with either an odd mass number or an odd atomic number also possess spin. This means they have angular momentum.

POSSESS SPIN 1 2 13 19 31 H H C F P

1 1 6 9 15

DON’ T POSSESS SPIN 12 C

6

h

aligned with the field

aligned against the fieldE

NE

RG

Y

A nucleus without spin cannot be detected by nuclear magnetic resonance spectroscopy.

A spinning nucleus such as 1H behaves as a spinning charge and generates a magnetic field. It can be likened to a bar magnet. When it is placed in an externally applied field it can align with, or against, the field. The energy difference between the two states () depends on the applied field.

The sample is spun round in the field of a large electromagnet and a radio-frequency (RF) field is applied. The magnetic field is increased and the excitation or “flipping” of nuclei from one orientation to another is detected as an induced voltage resulting from the absorption of energy from the RF field.

An nmr spectrum is the plot of the induced voltage against the sweep of the field. The area under a peak is proportional to the number of nuclei “flipping”

Not all hydrogen nuclei absorb energy at the same field strength at a given frequency; the field strength required depends on the environment of the hydrogen.

By observing the field strength at which protons absorb energy, one can deduce something about the structure of a molecule.

NMR SPECTROMETERS

RADIOFREQUENCY

OSCILLATOR

THE BASIC ELEMENTS OF AN NMR

SPECTROMETER

INTERPRETATION OF SPECTRA

NMR spectra provide information about the structure of organic molecules from the ...

• number of different signals in the spectrum• position of the signals (chemical shift)• intensity of the signals• splitting pattern of the signals

NMR SPECTROSCOPY




NMR SPECTROSCOPY




OBTAINING SPECTRA

• a liquid sample is placed in a tube which spins in a magnetic field

• solids are dissolved in deuterated solvents (CDCl3) or solvents without H’s (CCl4 ) [solvents with hydrogen atoms in them will produce peaks in the spectrum]

• TMS, tetramethylsilane, (CH3)4Si, is added to provide a reference signal

• when the spectrum is run, it can be integrated to find the relative peak areas

• spectrometers are now linked to computers to analyse data and store information

NMR SPECTROSCOPY

• non-toxic liquid - SAFE TO USE

• inert - DOESN’T REACT WITH COMPOUND BEING ANALYSED

• has a low boiling point - CAN BE DISTILLED OFF AND USED AGAIN

• all the hydrogen atoms are chemically equivalent - PRODUCES A SINGLE PEAK

• twelve hydrogens so it produces an intense peak - DON’T NEED TO USE MUCH

• signal is outside the range shown by most protons - WON’T OBSCURE MAIN SIGNALS

• given the chemical shift of = 0

• the position of all other signals is measured relative to TMS

TETRAMETHYLSILANE - TMS

The molecule contains four methyl groups attached to a silicon atom in a tetrahedral arrangement. All the hydrogen atoms are chemically equivalent.

PROVIDES THE REFERENCE SIGNAL

• each proton type is said to be chemically shifted relative to a standard (usually TMS)• the chemical shift is the difference between the field strength at which it absorbs and the field strength at which TMS protons absorb• the delta () scale is widely used as a means of reporting chemical shifts

Observed chemical shift (Hz) x 106

= ppm (parts per million) Spectrometer frequency (Hz)

• the chemical shift of a proton is constant under the same conditions (solvent, temperature) • the TMS peak is assigned a value of ZERO ( = 0.00)• all peaks of a sample under study are related to it and reported in parts per million• H’s near to an electronegative species are shifted “downfield” to higher values

CHEMICAL SHIFTCHEMICAL SHIFT

Approximatechemical shifts

The actual values

depend on the environment

13 12 11 10 9 8 7 6 5 4 3 2 1 0

DOWNFIELD - ‘deshielding’

-COOH

-CHO

-C=CH-

ROH

- C - X

H

- C - H

TMS

LOW RESOLUTION - HIGH RESOLUTIONLOW RESOLUTION - HIGH RESOLUTION

LOW RESOLUTION SPECTRUM OF 1-BROMOPROPANE

• low resolution nmr gives 1 peak for each environmentally different group of protons• high resolution gives more complex signals - doublets, triplets, quartets, multiplets• the signal produced indicates the number of protons on adjacent carbon atoms

LOW RESOLUTION - HIGH RESOLUTIONLOW RESOLUTION - HIGH RESOLUTION

HIGH RESOLUTION SPECTRUM OF 1-BROMOPROPANE

The broad peaks are split

into sharper signals

The splitting pattern depends on the number of hydrogen atoms on adjacent atoms


Number of peaks = number of chemically different H’s on adjacent atoms + 1

1 neighbouring H 2 peaks “doublet” 1:1

2 neighbouring H’s 3 peaks “triplet” 1:2:1

3 neighbouring H’s 4 peaks “quartet” 1:3:3:1

4 neighbouring H’s 5 peaks “quintet” 1:4:6:4:1

Signals for the H in an O-H bond are unaffected by hydrogens on adjacent atoms - get a singlet

MULTIPLICITY (Spin-spin splitting)MULTIPLICITY (Spin-spin splitting)



0 neighbouring H’s signal isn’t split 1 peak “singlet” 1 neighbouring H signal split into 2 peaks “doublet” ratio = 1:1 2 neighbouring H’s 3 peaks “triplet” 1:2:1 3 neighbouring H’s 4 peaks “quartet” 1:3:3:1 4 neighbouring H’s 5 peaks “quintet” 1:4:6:4:1




0 neighbouring H’s signal isn’t split 1 peak “singlet” 1 neighbouring H signal split into 2 peaks “doublet” ratio = 1:1 2 neighbouring H’s 3 peaks “triplet” 1:2:1 3 neighbouring H’s 4 peaks “quartet” 1:3:3:1 4 neighbouring H’s 5 peaks “quintet” 1:4:6:4:1


PASCAL’S TRIANGLE1

1 11 2 1

1 3 3 11 4 6 4 1

1 5 10 10 5 1

It is interesting to note the relationship between the successive peak ratios. It follows the pattern found in Pascal’s triangle.

Each number in the series is the sum of the two numbers above it in the triangle

What would be the pattern for 6 neighbouring hydrogens? 1 6 15 20 15 6 1

PRESS THE SPACE BAR FOR THE ANSWER


Splitting patterns are worked out by considering the effect adjacent, chemically different protons have on another signal in a given environment.

The spin of the proton producing the signal is affected by each of the two forms of the adjacent proton.

One orientation augments/enhances its field and the other opposes/reduces it. This is done by calculating the various possible combinations of alignment of adjacent protons.

HOWEVER Signals for the H in an O-H bond are not affectedby hydrogens on adjacent atoms so are not split


CONTENTSCONTENTS


ANALOGYImagine you had an opinion on something. If nobody influenced you, your opinion would be the same. However if another person had a view on the topic, they would either agree or disagree with you. Their ideas would either enhance what you thought or diminish it. There would be two possibilities of equal chance.

If there were two people offering views they could either be both for it (1 possibility) , both against (1 possibility) or one could be in favour and the other against (2 possibilities). There would be three possibilities of relative chance 1:2:1

FOR AGAINST

O adjacent H’sThere is no effect



1 adjacent Hcan be aligned either with or against the fieldthere are only two equally probable possibilitiesthe signal is split into 2 peaks of equal intensity




2 adjacent H’smore possible combinationsget 3 peaks in the ratio 1 : 2 : 1




2 adjacent H’smore possible combinationsget 3 peaks in the ratio 1 : 2 : 1

3 adjacent H’seven more possible combinationsget 4 peaks in the ratio 1 : 3 : 3 : 1

EXPLAIN THE THEORY BEHIND THE SPLITTING PATTERN FOR 4 ADJACENT H’s


4 adjacent H’sgives 5 peaks in the ratio 1 : 4 : 6 : 4 : 1


• the area under a signal is proportional to the number of hydrogen atoms present• an integration device scans the area under the peaks• lines on the spectrum show the relative abundance of each hydrogen type (old method)• computers now display the information as numerical data

By measuring the distances between the integration lines one canwork out the simple ratio between the various types of hydrogen.

before integration after integration

INTEGRATIONINTEGRATION

NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET(see later for an explanation of this)

INTEGRATIONINTEGRATION

HOW TO WORK OUT THE SIMPLE RATIOS (OLD METHOD)• Measure how much each integration line rises as it goes of a set of signals• Compare the relative values and work out the simple ratio between them• In the above spectrum the rises are in the ratio... 1:2:3

IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio

Measure the distance between the top and bottom lines.

Compare the heights from each signal and make them into a simple ratio.

Computers now do the integration automatically

• The signal due to the hydroxyl (OH) hydrogen is a singlet ... there is no splitting• H’s on OH groups do not couple with adjacent hydrogen atoms

Arises because the H on the OH, rapidly exchanges with protons on other molecules (such as water or acids) and is not attached to any particular oxygen long enough to register a splitting signal.

O-H bonds and splitting patternsO-H bonds and splitting patterns

• The signal due to the hydroxyl (OH) hydrogen is a singlet ... there is no splitting• H’s on OH groups do not couple with adjacent hydrogen atoms

Arises because the H on the OH, rapidly exchanges with protons on other molecules (such as water or acids) and is not attached to any particular oxygen long enough to register a splitting signal.

O-H bonds and splitting patternsO-H bonds and splitting patterns

OH hydrogens are always seen as a singlet ... there is no splitting

This is a quartet despite the fact that there are 4 H’s on adjacent atoms - the H on the OH doesn’t couple

As has been pointed out, the signal due to the hydroxyl (OH) hydrogen is a singlet.

It is possible to identify which signal is caused by the H of an O-H group by doing a ‘D2O shake’. A small amount of deuterium oxide D2O, a form of water, is added to the sample and the spectrum is re-run. Any signal due to O-H proton disappears.

The H in the O-H bond changes places with a deuterium atom, 2H or D, from D2O

Deuterium doesn’t exhibit nuclear magnetic resonance under the conditions used for proton nmr so the signal is removed to another part of the spectrum.

.

O-H bonds and the DO-H bonds and the D22O shakeO shake





before shaking with D2O






.

before shaking with D2O after shaking with D2O


H atoms attached to the N in amines also interchange with deuteriumH atoms attached to the N in amines also interchange with deuterium

When is a hydrogen chemically different?


TWO SIGNALSQuartet and triplet :- ratio of peak areas = 3 : 2

Carbons 1 & 4 are the similar and so are carbons 2 & 3 so there are only two different chemical environments.The signal for H’s on carbon 2 is a quartet - you ignore the two neighbours on carbon 3 because they are chemically identical.

BUTANE

1 2 3 4

When is a hydrogen chemically different?


TWO SIGNALSQuartet and triplet :- ratio of peak areas = 3 : 2

Carbons 1 & 4 are the similar and so are carbons 2 & 3 so there are only two different chemical environments.The signal for H’s on carbon 2 is a quartet - you ignore the two neighbours on carbon 3 because they are chemically identical.

TWO SIGNALSboth singlets :- ratio of peak areas = 2 : 1

Hydrogens on OH groups only give singlets. The signal for H’s on each carbon are not split, because- H’s on the neighbouring carbon are chemically identical... and- H’s on adjacent OH groups do not couple.

BUTANE

ETHANE-1,2-DIOL

1 2 3 4

An nmr spectrum provides several types of information :-

number of signal groups tells you the number of different proton environmentschemical shift the general environment of the protonspeak area (integration) the number of protons in each environment multiplicity how many protons are on adjacent atoms

In many cases this information is sufficient to deduce the structure of an organic molecule but other forms of spectroscopy are used in conjunction with nmr.

NMR SPECTROSCOPY - NMR SPECTROSCOPY - SUMMARYSUMMARY

HOW TO WORK OUT AN NMR SPECTRUM

1. Get the formula of the compound2. Draw out the structure3. Go to each atom in turn and ask the ‘census’ questions4. Work out what the spectrum would look like ... signals due to H’s nearer electronegative atoms (Cl,Br,O) are shifted downfield to higher values


THE BASIC “CENSUS” Ask each hydrogen atom to... - describe the position of the atom on which it lives - say how many hydrogen atoms live on that atom - say how many chemically different hydrogen atoms live on adjacent atoms

BUT, REMEMBER THAT

H atoms on OH groups - ONLY PRODUCE ONE PEAK

- DON’T COUNT AS A NEIGHBOUR

THE BASIC “CENSUS” Ask each hydrogen atom to... - describe the position of the atom on which it lives - say how many hydrogen atoms live on that atom - say how many chemically different hydrogen atoms live on adjacent atoms

BUT, REMEMBER THAT

H atoms on OH groups - ONLY PRODUCE ONE PEAK

- DON’T COUNT AS A NEIGHBOUR


ATOMUNIQUE DESCRIPTION OF THE POSITION

OF THE HYDROGEN ATOMSH’S ON THE

ATOMCHEMICALLY DIFFERENT

H’S ON ADJACENT ATOMSSIGNAL

SPLIT INTO

1On an end carbon, two away

from the carbon with the bromine atom on it

3 2 2+1 = 3

2On a carbon atom second from the end and one away from the carbon with the bromine atom

2 3+2 = 5 5+1 = 6

3On an end carbon atom which

also has the bromine atom on it 2 2 2+1 = 3

1 2 3

“CENSUS” QUESTIONS

- describe where each hydrogen lives - say how many hydrogens live on that atom - say how many chemically different hydrogen atoms live on adjacent atoms

“CENSUS” QUESTIONS

- describe where each hydrogen lives - say how many hydrogens live on that atom - say how many chemically different hydrogen atoms live on adjacent atoms

1-BROMOPROPANE

Spectrum of 1-bromopropaneSpectrum of 1-bromopropane

5 4 3 2 1 0

CHEMICAL SHIFTS

3 environments = 3 signals

Triplet = 3.4Sextet = 1.9Triplet = 1.0

Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen

1 2 3

TMS


5 4 3 2 1 0

INTEGRATION

Area ratio from relative heights of integration lines = 2 : 2 : 3

Carbon 1 3Carbon 2 2Carbon 3 2

1 2 3

2

2

3

TMS


5 4 3 2 1 0

SPLITTING

SPLITTING PATTERNCarbon 1

Chemically different hydrogen atoms on adjacent atoms = 2

2 + 1 = 3

The signal will be aTRIPLET

1 2 3

1

TMS


5 4 3 2 1 0

SPLITTING



5 + 1 = 6

The signal will be aSEXTET

1 2 3

2

TMS


5 4 3 2 1 0

SPLITTING



2 + 1 = 3

The signal will be aTRIPLET

1 2 3

3

TMS

The signal is shifted furthest away (downfield) from TMS as the hydrogen atoms are nearest the electronegative bromine atom.


5 4 3 2 1 0

SPLITTING

3 environments = 3 signals

1 Triplet = 1.0 3 H’s2 Sextet = 1.9 2 H’s3 Triplet = 3.4 2 H’s

Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen

1 2 3

3 2 1

TMS

Peaks Three different signals as there are three chemically different protons.Shift Signals are shifted away from TMS signal, are nearer to the halogen.Splitting Signals include a triplet ( = 1.0) sextet ( = 1.8) triplet ( = 3.4) Integration The integration lines show that the ratio of protons is 2:2:3

The signals due to the protons attached to carbon ...

C1 triplet ( = 1.0) coupled to the two protons on carbon C2 ( 2+1 = 3 ) C2 sextet ( = 1.8) coupled to five protons on carbons C1 and C3 ( 5+1 = 6 ) C3 triplet ( = 3.4) coupled to the two protons on carbon C2 ( 2+1 = 3 )

4 3 2 1 0 SUMMARY


TMS

1 2 3


SUPPLEMENTARY QUESTIONS

1. Why is proton nmr more useful for the investigation of organic compounds ?

2. What other nucleus found in organic compounds is investigated using nmr ?

3. What compound is used as the internal reference for proton nmr chemical shifts ? How many peaks does it produce and at what delta () value does it appear ?

4. What uses have been made of nuclear magnetic resonance in other scientific areas ?

Supplementary Questions - Answers

1. Because organic compounds tend to contain hydrogen atoms.

2. Carbon 13

3. Tetramethylsilane (TMS) gives a strong single peak at = 0

4. Magnetic resonance imaging in body scanners

SEE NEXT PAGE FOR ANSWERS


SUPPLEMENTARY QUESTIONS

1. Why is proton nmr more useful for the investigation of organic compounds ?

2. What other nucleus found in organic compounds is investigated using nmr ?

3. What compound is used as the internal reference for proton nmr chemical shifts ? How many peaks does it produce and at what delta () value does it appear ?

4. What uses have been made of nuclear magnetic resonance in other scientific areas ?

Supplementary Questions - Answers

1. Because organic compounds tend to contain hydrogen atoms.

2. Carbon 13

3. Tetramethylsilane (TMS) gives a strong single peak at = 0

4. Magnetic resonance imaging in body scanners

WHAT IS IT!WHAT IS IT!

C2H5Br


C2H3Br3


C2H4Br2


C6H12


C2H4O2


C4H8O2


C3H6O


C3H6O


C4H8O


C8H16O2


C11H16


C8H10


C8H10


C9H12


C6H10O3


C4H8Br2

CARBON-13CARBON-13


After hydrogen, the most useful atom providing information is carbon-13.

Natural carbon contains about 1% of this isotope so the instruments for its detection need to be sensitive and spectra will take longer to record.

Only the chemical shift is important as each spectrum gives only single lines for each chemically equivalent carbon.

CARBON-13 NMR SPECTROSCOPYCARBON-13 NMR SPECTROSCOPY

After hydrogen, the most useful atom providing information is carbon-13.

Natural carbon contains about 1% of this isotope so the instruments for its detection need to be sensitive and spectra will take longer to record.

Only the chemical shift is important as each spectrum gives only single lines for each chemically equivalent carbon.

Environment Chemical shift / C - C (alkanes) 10 - 35

C - C=O 10 - 35 C - Cl or C - Br 30 - 70 C - N (amines) 35 - 65 C - OH 50 - 65 C = C (alkenes) 115 - 140 aromatic C’s (benzene rings) 125 - 150 C=O (esters, acids, amides) 160 - 185 C=O (aldehydes, ketones) 190 – 220

Carbon-13 nmr has wide applications in the study of naturalproducts, biological molecules and polymers.

CARBON-13 NMR SPECTROSCOPYCARBON-13 NMR SPECTROSCOPY

IsomersIsomersof Cof C33HH77BrBr

3 peaks 2 peaks all three carbons are different the two outer carbons are similar

CARBON-13 NMR SPECTRACARBON-13 NMR SPECTRA

HCCCBr

H

H

H

H

H

H

HCCCH

H

H

Br

H

H

H

IsomersIsomersof Cof C33HH77BrBr

3 peaks 2 peaks all three carbons are different the two outer carbons are similar

EthanolEthanolCC22HH55OHOH


HCCCBr

H

H

H

H

H

H

HCCCH

H

H

Br

H

H

H

This is where the proton nmr spectrum of ethanol would be on the same scale.

H C C OH

H

H H

H

HC

H

H

COH

H

H

The carbon-13 spectrum of 2-methylbutane (CHThe carbon-13 spectrum of 2-methylbutane (CH33))22CHCHCHCH22CHCH33

Other isomers of C5H12

pentane CH3CH2CH2CH2CH3 3 peaks

2,3-dimethylpropane (CH3)4C 2 peaks


There are four chemically different carbon atoms in the molecule so there are four peaks in the C-13 nmr spectrum.

NO SPLITTING WITH C-13ONLY ONE PEAK FOR

EACH CARBON

NO SPLITTING WITH C-13ONLY ONE PEAK FOR

EACH CARBON

chemically equivalent

carbon atoms

H C C C C H

H

H

CH3

H

H

H H

H

How many peaks would you expect there to be in the carbon-13 spectrum of…

• butane CH3CH2CH2CH3

• 2-methylpropane CH3CH(CH3)CH3

• butanal CH3CH2CH2CHO

• butanone CH3COCH2CH3

• pentan-2-one CH3COCH2CH2CH3

• pentan-3-one CH3CH2COCH2CH3

• cyclohexane C6H12

CARBON-13 NMR SPECTRA - CARBON-13 NMR SPECTRA - QUESTIONSQUESTIONS

How many peaks would you expect there to be in the carbon-13 spectrum of…

• butane CH3CH2CH2CH3 2

• 2-methylpropane CH3CH(CH3)CH3 2

• butanal CH3CH2CH2CHO 4

• butanone CH3COCH2CH3 4

• pentan-2-one CH3COCH2CH2CH3 5

• pentan-3-one CH3CH2COCH2CH3 3

• cyclohexane C6H12 1


19

Identify the isomers of C4H8O


Identify the isomers of C4H8O


A butanal

B butanone

C 2-methylpropanal

Identify the isomers of C6H12


Identify the isomers of C6H12


X hex-1-ene or hex-2-ene or 2-methylpent-1-ene or 3-methylpent-1-ene or 2-methylpent-2-ene or 3-methylpent-2-ene or

Y cyclohexane

Z 2,3-dimethylbut-2-ene

REVISION CHECKREVISION CHECK

What should you be able to do?

Recall the how an nmr spectrum is produced

Explain and understand the origin of chemical shift

Explain and understand the purpose of integration

Explain and understand the purpose of shaking with D2O

Recall the differences between high and low resolution spectra

Explain and understand the origin of splitting patterns

Interpret and explain a simple proton or carbon-13 nmr spectrum

NUCLEAR MAGNETIC NUCLEAR MAGNETIC RESONANCE SPECTROSCPYRESONANCE SPECTROSCPY

THE ENDTHE END

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