2008 SPECIFICATIONS NUCLEAR MAGNETIC RESONANCE SPECTROSCPY

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

KNOCKHARDY PUBLISHING
NMR SPECTROSCOPY 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... 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

NMR SPECTROSCOPY 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

Before you start it would be helpful to…
NMR 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?
PREVIEW 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 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

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 H H C F P DON’ T POSSESS SPIN C 6 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 (DE) depends on the applied field. DE = h aligned with the field b a aligned against the field ENERGY

THE BASIC ELEMENTS OF AN NMR SPECTROMETER
NMR SPECTROMETERS 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. RADIOFREQUENCY OSCILLATOR THE BASIC ELEMENTS OF AN NMR SPECTROMETER

INTERPRETATION OF SPECTRA
NMR SPECTROSCOPY 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

INTERPRETATION OF SPECTRA
NMR SPECTROSCOPY 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 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

TETRAMETHYLSILANE - TMS PROVIDES THE REFERENCE SIGNAL
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 d = 0 the position of all other signals is measured relative to TMS The molecule contains four methyl groups attached to a silicon atom in a tetrahedral arrangement. All the hydrogen atoms are chemically equivalent.

The actual values depend on the environment DOWNFIELD - ‘deshielding’
CHEMICAL SHIFT 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 (d) scale is widely used as a means of reporting chemical shifts Observed chemical shift (Hz) x 106 d = 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 (d = 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 d values - C - X H Approximate chemical shifts The actual values depend on the environment ROH -CHO - C - H -COOH -C=CH- TMS d DOWNFIELD - ‘deshielding’

LOW RESOLUTION - HIGH RESOLUTION
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 SPECTRUM OF 1-BROMOPROPANE

LOW RESOLUTION - HIGH RESOLUTION
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 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

MULTIPLICITY (Spin-spin splitting)
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 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

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 Number of peaks = number of chemically different H’s on adjacent atoms + 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” :2:1 3 neighbouring H’s 4 peaks “quartet” :3:3:1 4 neighbouring H’s 5 peaks “quintet” :4:6:4:1

MULTIPLICITY (Spin-spin splitting) PRESS THE SPACE BAR FOR THE ANSWER
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 Number of peaks = number of chemically different H’s on adjacent atoms + 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” :2:1 3 neighbouring H’s 4 peaks “quartet” :3:3:1 4 neighbouring H’s 5 peaks “quintet” :4:6:4:1 PASCAL’S TRIANGLE 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? 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 affected by hydrogens on adjacent atoms so are not split

ANALOGY Imagine 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 CONTENTS

O adjacent H’s There is no effect

O adjacent H’s There is no effect 1 adjacent H can be aligned either with a or against b the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity

O adjacent H’s There is no effect 1 adjacent H can be aligned either with a or against b the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity 2 adjacent H’s more possible combinations get 3 peaks in the ratio 1 : 2 : 1

O adjacent H’s There is no effect 1 adjacent H can be aligned either with a or against b the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity 2 adjacent H’s more possible combinations get 3 peaks in the ratio 1 : 2 : 1 3 adjacent H’s even more possible combinations get 4 peaks in the ratio 1 : 3 : 3 : 1 EXPLAIN THE THEORY BEHIND THE SPLITTING PATTERN FOR 4 ADJACENT H’s

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

before integration after integration
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 can work out the simple ratio between the various types of hydrogen. before integration after integration NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET (see later for an explanation of this)

HOW TO WORK OUT THE SIMPLE RATIOS (OLD METHOD)
INTEGRATION 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 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

O-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 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. 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

O-H bonds and the D2O shake
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. .

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. before shaking with D2O

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. . before shaking with D2O after shaking with D2O H atoms attached to the N in amines also interchange with deuterium

When is a hydrogen chemically different?
NMR SPECTROSCOPY When is a hydrogen chemically different? TWO SIGNALS Quartet 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. 1 2 3 4 BUTANE

When is a hydrogen chemically different?
NMR SPECTROSCOPY When is a hydrogen chemically different? TWO SIGNALS Quartet 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. both 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. 1 2 3 4 BUTANE ETHANE-1,2-DIOL

NMR SPECTROSCOPY - SUMMARY
An nmr spectrum provides several types of information :- number of signal groups tells you the number of different proton environments chemical shift the general environment of the protons peak 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.

HOW TO WORK OUT AN NMR SPECTRUM
NMR SPECTROSCOPY HOW TO WORK OUT AN NMR SPECTRUM 1. Get the formula of the compound 2. Draw out the structure 3. Go to each atom in turn and ask the ‘census’ questions 4. Work out what the spectrum would look like ... signals due to H’s nearer electronegative atoms (Cl,Br,O) are shifted downfield to higher d 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

NMR SPECTROSCOPY 3 2 3+2 = 5 “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 2 3 1-BROMOPROPANE ATOM UNIQUE DESCRIPTION OF THE POSITION OF THE HYDROGEN ATOMS H’S ON THE ATOM CHEMICALLY DIFFERENT H’S ON ADJACENT ATOMS SIGNAL SPLIT INTO 1 On an end carbon, two away from the carbon with the bromine atom on it 3 2 2+1 = 3 On a carbon atom second from the end and one away from the carbon with the bromine atom 3+2 = 5 5+1 = 6 On an end carbon atom which also has the bromine atom on it

Spectrum of 1-bromopropane
CHEMICAL SHIFTS 3 environments = 3 signals Triplet d = 3.4 Sextet d = 1.9 Triplet d = 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 d

INTEGRATION Area ratio from relative heights of integration lines = 2 : 2 : 3 Carbon 1 3 Carbon 2 2 Carbon 3 2 1 2 3 2 3 TMS 2 d

SPLITTING SPLITTING PATTERN Carbon 1 Chemically different hydrogen atoms on adjacent atoms = 2 = 3 The signal will be a TRIPLET 1 1 2 3 TMS d

SPLITTING SPLITTING PATTERN Carbon 2 Chemically different hydrogen atoms on adjacent atoms = 5 = 6 The signal will be a SEXTET 2 1 2 3 TMS d

SPLITTING SPLITTING PATTERN Carbon 3 Chemically different hydrogen atoms on adjacent atoms = 2 = 3 The signal will be a TRIPLET 3 1 2 3 TMS d The signal is shifted furthest away (downfield) from TMS as the hydrogen atoms are nearest the electronegative bromine atom.

SPLITTING 3 environments = 3 signals 1 Triplet d = H’s 2 Sextet d = H’s 3 Triplet d = H’s Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen 3 2 1 1 2 3 TMS d

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

NMR SPECTROSCOPY 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 (d) 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 d = 0 4. Magnetic resonance imaging in body scanners SEE NEXT PAGE FOR ANSWERS

SUPPLEMENTARY QUESTIONS Supplementary Questions - Answers
NMR SPECTROSCOPY 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 (d) 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 d = 0 4. Magnetic resonance imaging in body scanners

WHAT IS IT! C2H5Br

WHAT IS IT! C2H3Br3

WHAT IS IT! C2H4Br2

WHAT IS IT! C6H12

WHAT IS IT! C2H4O2

WHAT IS IT! C4H8O2

WHAT IS IT! C3H6O

WHAT IS IT! C4H8O

WHAT IS IT! C8H16O2

WHAT IS IT! C11H16

WHAT IS IT! C8H10

WHAT IS IT! C9H12

WHAT IS IT! C6H10O3

WHAT IS IT! C4H8Br2

WHAT ARE THEY! Compound 2,2-dimethylbutanal 3-methylbutanal
3-methylbutan-2-one pentan-2-one pentanal Spectrum triplet (1.05); quartet (2.47) doublet (1.02); singlet (2.13) heptet (2.22) includes... doublet (0.93); triplet (9.74) two singlets

CARBON-13 NMR SPECTROSCOPY

CARBON-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.

CARBON-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 / d C - C (alkanes) C - C=O C - Cl or C - Br C - N (amines) C - OH C = C (alkenes) aromatic C’s (benzene rings) C=O (esters, acids, amides) C=O (aldehydes, ketones) 190 – 220 Carbon-13 nmr has wide applications in the study of natural products, biological molecules and polymers.

CARBON-13 NMR SPECTRA HCCCBr HCCCH
Isomers of C3H7Br 3 peaks peaks all three carbons are different the two outer carbons are similar HCCCBr H H Br H HCCCH H H H

CARBON-13 NMR SPECTRA HCCCBr HCCCH
Isomers of C3H7Br 3 peaks peaks all three carbons are different the two outer carbons are similar Ethanol C2H5OH HCCCBr H H Br H HCCCH H H H H H COH HC H H This is where the proton nmr spectrum of ethanol would be on the same scale. H H H  C  C  OH H H

ONLY ONE PEAK FOR EACH CARBON
CARBON-13 NMR SPECTRA The carbon-13 spectrum of 2-methylbutane (CH3)2CHCH2CH3 Other isomers of C5H12 pentane CH3CH2CH2CH2CH3 3 peaks 2,3-dimethylpropane (CH3)4C 2 peaks chemically equivalent carbon atoms There are four chemically different carbon atoms in the molecule so there are four peaks in the C-13 nmr spectrum. H  C  C  C  C  H H CH3 NO SPLITTING WITH C-13 ONLY ONE PEAK FOR EACH CARBON

CARBON-13 NMR SPECTRA - QUESTIONS
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

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 C6H 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

What should you be able to do?
REVISION 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 RESONANCE SPECTROSCPY KNOCKHARDY PUBLISHING
THE END KNOCKHARDY PUBLISHING

2008 SPECIFICATIONS NUCLEAR MAGNETIC RESONANCE SPECTROSCPY

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