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

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

2 NMR SPECTROSCOPY KNOCKHARDY 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... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard

3 CONTENTS Prior knowledge What is nmr? Origin of spectra Tetramethylsilane Chemical shift Resolution Multiplicity - splitting patterns Integration OH signals and the use of D 2 O Working out spectra Questions on proton nmr Carbon-13 nmr NMR SPECTROSCOPY

4 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 NMR SPECTROSCOPY

5 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? LOW ENERGY RADIO WAVES 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 PREVIEW

6 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 12 C 6  h aligned with the field   aligned against the field ENERGY A nucleus without spin cannot be detected by nuclear magnetic resonance spectroscopy. A spinning nucleus such as 1 H 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.

7 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

8 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

9 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

10 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 (CDCl 3 ) or solvents without H’s (CCl 4 ) [solvents with hydrogen atoms in them will produce peaks in the spectrum] TMS, tetramethylsilane, (CH 3 ) 4 Si, 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

11 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

12 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 10 6  = 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 SHIFT Approximate chemical shifts The actual values depend on the environment  DOWNFIELD - ‘deshielding’ -COOH -CHO -C=CH- ROH - C - X H - C - H TMS

13 LOW 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

14 LOW 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 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

15 Number of peaks = number of chemically different H’s on adjacent atoms neighbouring H2 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) 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

16 Number of peaks = number of chemically different H’s on adjacent atoms neighbouring H’s signal isn’t split1 peak“singlet” 1 neighbouring H signal split into2 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 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

17 Number of peaks = number of chemically different H’s on adjacent atoms neighbouring H’s signal isn’t split1 peak“singlet” 1 neighbouring H signal split into2 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 MULTIPLICITY (Spin-spin splitting) PASCAL’S TRIANGLE 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 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

18 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. HOWEVERSignals for the H in an O-H bond are not affected by hydrogens on adjacent atoms so are not split MULTIPLICITY (Spin-spin splitting)

19 CONTENTS MULTIPLICITY (Spin-spin splitting) 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

20 O adjacent H’s There is no effect MULTIPLICITY (Spin-spin splitting)

21 O adjacent H’s There is no effect 1 adjacent H can be aligned either with  or against  the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity MULTIPLICITY (Spin-spin splitting)

22 O adjacent H’s There is no effect 1 adjacent H can be aligned either with  or against  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 MULTIPLICITY (Spin-spin splitting)

23 O adjacent H’s There is no effect 1 adjacent H can be aligned either with  or against  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 MULTIPLICITY (Spin-spin splitting)

24 4 adjacent H’s gives 5 peaks in the ratio 1 : 4 : 6 : 4 : 1 MULTIPLICITY (Spin-spin splitting)

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

26 INTEGRATION 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

27 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

28 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 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

29 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 ‘D 2 O shake’. A small amount of deuterium oxide D 2 O, 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, 2 H or D, from D 2 O 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 D 2 O shake

30 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 ‘D 2 O shake’. A small amount of deuterium oxide D 2 O, 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, 2 H or D, from D 2 O 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 D 2 O O-H bonds and the D 2 O shake

31 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 ‘D 2 O shake’. A small amount of deuterium oxide D 2 O, 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, 2 H or D, from D 2 O 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 D 2 O after shaking with D 2 O O-H bonds and the D 2 O shake H atoms attached to the N in amines also interchange with deuterium

32 When is a hydrogen chemically different? NMR SPECTROSCOPY 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. BUTANE 12 34

33 When is a hydrogen chemically different? NMR SPECTROSCOPY 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. TWO SIGNALS 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. BUTANE ETHANE-1,2-DIOL 12 34

34 An nmr spectrum provides several types of information :- number of signal groups tells you the number of different proton environments chemical shiftthe general environment of the protons peak area (integration)the number of protons in each environment multiplicityhow 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 - SUMMARY

35 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  values NMR SPECTROSCOPY 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

36 NMR SPECTROSCOPY 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 On a carbon atom second from the end and one away from the carbon with the bromine atom 23+2 = = 6 3 On an end carbon atom which also has the bromine atom on it = “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

37 Spectrum of 1-bromopropane  CHEMICAL SHIFTS 3 environments = 3 signals Triplet  = 3.4 Sextet  = 1.9 Triplet  = 1.0 Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen 123 TMS

38 Spectrum of 1-bromopropane  INTEGRATION Area ratio from relative heights of integration lines = 2 : 2 : 3 Carbon 13 Carbon 22 Carbon TMS

39 Spectrum of 1-bromopropane  SPLITTING SPLITTING PATTERN Carbon 1 Chemically different hydrogen atoms on adjacent atoms = = 3 The signal will be a TRIPLET TMS

40 Spectrum of 1-bromopropane  SPLITTING SPLITTING PATTERN Carbon 2 Chemically different hydrogen atoms on adjacent atoms = = 6 The signal will be a SEXTET TMS

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

42 Spectrum of 1-bromopropane  SPLITTING 3 environments = 3 signals 1 Triplet  = 1.03 H’s 2 Sextet  = 1.92 H’s 3 Triplet  = 3.42 H’s Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen TMS

43 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 )  SUMMARY Spectrum of 1-bromopropane TMS 123

44 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 (  ) 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 Tetramethylsilane (TMS) gives a strong single peak at  = 0 4. Magnetic resonance imaging in body scanners SEE NEXT PAGE FOR ANSWERS

45 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 (  ) 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 Tetramethylsilane (TMS) gives a strong single peak at  = 0 4. Magnetic resonance imaging in body scanners

46 WHAT IS IT! C 2 H 5 Br

47 WHAT IS IT! C 2 H 3 Br 3

48 WHAT IS IT! C 2 H 4 Br 2

49 WHAT IS IT! C 6 H 12

50 WHAT IS IT! C2H4O2C2H4O2

51 C4H8O2C4H8O2

52 C3H6OC3H6O

53 C3H6OC3H6O

54 C4H8OC4H8O

55 C 8 H 16 O 2

56 WHAT IS IT! C 11 H 16

57 WHAT IS IT! C 8 H 10

58 WHAT IS IT! C 8 H 10

59 WHAT IS IT! C 9 H 12

60 WHAT IS IT! C 6 H 10 O 3

61 WHAT IS IT! C 4 H 8 Br 2

62 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

63 CARBON-13 NMR SPECTROSCOPY

64 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

65 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) 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 SPECTROSCOPY

66 Isomers of C 3 H 7 Br 3 peaks 2 peaks all three carbons are different the two outer carbons are similar CARBON-13 NMR SPECTRA H  C  C  C  Br H H H H H H HCCCHHCCCH H H Br H H H

67 Isomers of C 3 H 7 Br 3 peaks 2 peaks all three carbons are different the two outer carbons are similarEthanol C 2 H 5 OH CARBON-13 NMR SPECTRA H  C  C  C  Br H H H H H H HCCCHHCCCH 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 HCHC H H  C  OH H H

68 The carbon-13 spectrum of 2-methylbutane (CH 3 ) 2 CHCH 2 CH 3 Other isomers of C 5 H 12 pentane CH 3 CH 2 CH 2 CH 2 CH 3 3 peaks 2,3-dimethylpropane(CH 3 ) 4 C2 peaks CARBON-13 NMR SPECTRA 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-13 ONLY ONE PEAK FOR EACH CARBON NO SPLITTING WITH C-13 ONLY ONE PEAK FOR EACH CARBON chemically equivalent carbon atoms H  C  C  C  C  H H H CH3CH3 H H H H H

69 How many peaks would you expect there to be in the carbon-13 spectrum of… butaneCH 3 CH 2 CH 2 CH 3 2-methylpropaneCH 3 CH(CH 3 )CH 3 butanalCH 3 CH 2 CH 2 CHO butanoneCH 3 COCH 2 CH 3 pentan-2-oneCH 3 COCH 2 CH 2 CH 3 pentan-3-oneCH 3 CH 2 COCH 2 CH 3 cyclohexaneC 6 H 12 CARBON-13 NMR SPECTRA - QUESTIONS

70 How many peaks would you expect there to be in the carbon-13 spectrum of… butaneCH 3 CH 2 CH 2 CH methylpropaneCH 3 CH(CH 3 )CH 3 2 butanalCH 3 CH 2 CH 2 CHO4 butanoneCH 3 COCH 2 CH 3 4 pentan-2-oneCH 3 COCH 2 CH 2 CH 3 5 pentan-3-oneCH 3 CH 2 COCH 2 CH 3 3 cyclohexaneC 6 H 12 1 CARBON-13 NMR SPECTRA - QUESTIONS 19

71 Identify the isomers of C 4 H 8 O CARBON-13 NMR SPECTRA - QUESTIONS

72 Identify the isomers of C 4 H 8 O CARBON-13 NMR SPECTRA - QUESTIONS A butanal B butanone C 2-methylpropanal

73 Identify the isomers of C 6 H 12 CARBON-13 NMR SPECTRA - QUESTIONS

74 Identify the isomers of C 6 H 12 CARBON-13 NMR SPECTRA - QUESTIONS 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

75 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 D 2 O 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

76 NUCLEAR MAGNETIC RESONANCE SPECTROSCPY THE END KNOCKHARDY PUBLISHING


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