13C NMR
NMR spectroscopy There are two main types of NMR 1. C13 NMR 2. H (proton) NMR There is only around 1% C13 in organic molecules but modern NMR machines are sensitive enough to give a full spectra for C13 The C13 spectra is a simpler spectrum than the H NMR Equivalent Carbon atoms. In a C13 NMR spectrum, there is one signal (peak) for each set of equivalent C atoms. C O C H H C H H H H C C O O C C H H H H H H H H CH3 C O O CH3 C O H C C C H H H H H H C C H H H H H 3 peaks 4 peaks 4 peaks 2 peaks a a a a a a b b b b c b c c d d 1,2 dinitrobenzene 1,3 dinitrobenzene 1,4 dinitrobenzene 3 peaks 4 peaks 2 peaks NO2 NO2 1 1 2 2 2 2 H3C C CH3 CH3 Cl a a a b 2 peaks CH3 NO2 a b c c d d e 5 peaks NMR spectroscopy involves interaction of materials with the lowenergy radiowave region of the electromagnetic spectrum NMR spectroscopy is the same technology as that used in ‘magnetic resonance imaging’ (MRI) to obtain diagnostic information about internal structures in body scanners. Solvents Samples are dissolved in solvents without any 1H atoms, e.g. CCl4 , CDCl3 . This means that in the H NMR the solvent will not give any peaks The same solvent is used in C13 NMR and in this case there will be one peak due to the solvent that will appear on the spectrum. However, it is known where this peak is so it can be ignored. In the exam it is likely this peak will not occur on the spectra. Calibration and shift A small amount of TMS (tetramethylsilane) is added to the sample to calibrate the spectrum The same calibration compound is used for both H and C NMR TMS is used because: •its signal is away from all the others •it only gives one signal •it is non-toxic •it is inert •it has a low boiling point and so can be removed from sample easily The spectra are recorded on a scale known as the chemical shift (δ), which is how much the field has shifted away from the field for TMS.. The δ is a measure in parts per million (ppm) is a relative scale of how far the frequency of the proton signal has shifted away from that for TMS. Si CH3 H3C CH3 CH3 tetramethylsilane δ chemical shift (ppm). C13 NMR shift
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6.3.2 Spectroscopy
NMR Spectroscopy (a) analysis of a carbon-13 NMR spectrum of an organic molecule to make predictions about: (i) the number of carbon environments in the molecule (ii) the different types of carbon environment present, from chemical shift values (iii) possible structures for the molecule M3.1 All carbon-13 NMR spectra that are assessed will be proton decoupled. In examinations, NMR chemical shift values will be provided on the Data Sheet. Restricted to functional groups studied in the A level specification. HSW3,5 Interpretation of spectra to analyse organic compounds.(c) prediction of a carbon-13 or proton NMR spectrum for a given molecule M3.1 (d) (i) the use of tetramethylsilane, TMS, as the standard for chemical shift measurements (ii) the need for deuterated solvents, e.g. CDCl 3, when running an NMR spectrum (iii) the identification of O–H and N–H protons by proton exchange using D2O
H NMR
Equivalent Hydrogen atoms. In an H NMR spectrum, there is one signal for each set of equivalent H atoms. In addition the intensity (integration value) of each signal is proportional to the number of equivalent H atoms it represents. Ethanol has 3 groups of different hydrogen atoms 3 2 1 CH3 CH CH2 CH3 Br a b c d 4 sets of equivalent H’s: ratio 3:1:2:3 H3C C CH3 O a a 1 signal CH2 C O O H3C CH3 3 sets of equivalent H’s: ratio 3:2:3 a b c CH2 C O O H3C C CH3 CH3 CH3 3 sets of equivalent H’s: ratio 3:2:9 a b c c c H3C CH CH2 a b c 3 sets of equivalent H’s: ratio 3:1:2 H C C C H H H H H H C C H H H H a H a b c d a a a a c d d 4sets of equivalent H’s: ratio 6:1:2:3 H C C H H H H O H a a a b b c N Goalby chemrevise.org Solvents Samples are dissolved in solvents without any 1H atoms, e.g. CCl4 , CDCl3 . This means that in the H NMR the solvent will not give any peaks The same solvent is used in C13 NMR and in this case there will be one peak due to the solvent that will appear on the spectrum. However, it is known where this peak is so it can be ignored. In the exam it is likely this peak will not occur on the spectra. Calibration and shift A small amount of TMS (tetramethylsilane) is added to the sample to calibrate the spectrum The same calibration compound is used for both H and C NMR TMS is used because: •its signal is away from all the others •it only gives one signal •it is non-toxic •it is inert •it has a low boiling point and so can be removed from sample easily The spectra are recorded on a scale known as the chemical shift (δ), which is how much the field has shifted away from the field for TMS.. The δ is a measure in parts per million (ppm) is a relative scale of how far the frequency of the proton signal has shifted away from that for TMS. Si CH3 H3C CH3 CH3 tetramethylsilane δ chemical shift (ppm). H NMR shift CH2 C O O H3C CH3 3 3 2 C C O H O H C H H H H 3 2 1 δ ppm δ ppm N Goalby chemrevise.org If D2O is added to a sample then a process of proton exchange happens with the H in any O-H and N-H bonds. This has the effect of removing the peaks from the HNMR spectra. This can help with the identification of O–H and N–H peaks on the spectra. Addition of D2O to the sample of Propanoic acid would make the peak at δ = 11.7 (ppm) in the above spectrum disappear Proton exchange using D2O. The δ depends on what other atoms/groups are near the H – more electronegative groups gives a greater shift. Spin-Spin coupling in H NMR Splitting of peak = number of inequivalent H’s on neighbouring C atoms + 1 In high resolution H NMR each signal in the spectrum can be split into further lines due to inequivalent H’s on neighbouring C atoms. Nuclei in identical chemical environments do not show coupling amongst themselves! signal singlet doublet triplet quartet quintet appearance Split number of peaks 1 2 3 4 5 number of neighbouring inequivalent H atoms 0 1 2 3 4 relative size 1:1 1:2:1 1:3:3:1 1:4:6:4:1 CH2 C O O H3C CH3 a b c The peak due to group a will be a triplet as it is next to b (a carbon with 2 H’s) The peak due to group b will be a quartet as it is next to a (a carbon with 3H’s) The peak due to group c will be a singlet as it is next to a carbon with no H’s) C C O O C C H H H H H H H H a b c The peak due to group a will be a singlet as it is next to a carbon with 0 H’s Shift 2.1-2.6 Integration trace 3 The peak due to group b will be a quartet as it is next to a carbon with 3 H’s Shift 3.7 -4.1 Integration trace 2 The peak due to group c will be a triplet as it is next to a carbon with 2 H’s Shift 0.7-1.2 Integration trace 3 a b c ppm ppm N Goalby chemrevise.org For 6 split peaks use the term hextet or multiplet
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6.3.2 Spectroscopy
(b) analysis of a high resolution proton NMR spectrum of an organic molecule to make predictions about: (i) the number of proton environments in the molecule (ii) the different types of proton environment present, from chemical shift values (iii) the relative numbers of each type of proton present from relative peak areas, using integration traces or ratio numbers, when required (iv) the number of non-equivalent protons adjacent to a given proton from the spin– spin splitting pattern, using the n + 1 rule (v) possible structures for the molecule M3.1 In examinations, NMR chemical shift values will be provided on the Data Sheet. Restricted to functional groups studied in the A level specification. Learners will be expected to identify aromatic protons from chemical shift values but will not be expected to analyse their splitting patterns. HSW3,5 Interpretation of spectra to analyse organic compounds. (c) prediction of a carbon-13 or proton NMR spectrum for a given molecule M3.1 (d) (i) the use of tetramethylsilane, TMS, as the standard for chemical shift measurements (ii) the need for deuterated solvents, e.g. CDCl 3, when running an NMR spectrum (iii) the identification of O–H and N–H protons by proton exchange using D2O
Elemental analysis
Please refer to Empirical and molecular formula under Amount of substance
Infrared spectroscopy
Please refer to Infrared spectroscopy under Analytical techniques
Mass spectroscopy
Please refer to Mass spectroscopy under Analytical techniques
Combined techniques
Bringing it all together Elemental analysis C 66.63% H 11.18% O 22.19% 66.63/12 1. Work out empirical formula 11.18/1 22.19/16 C H O =5.5525 =11.18 =1.386875 =4 =8 =1 2. Using molecular ion peak m/z value from mass spectrum calculate Molecular formula molecular ion peak m/z value= 144 Mr empirical formula C4H8O = 72 If Mr molecular formula 144 then compound is C8H16O2 3. Use IR spectra to identify main bonds/functional group C8H16O2 could be an ester, carboxylic acid or combination of alcohol and carbonyl. Look for IR spectra for C=O and O-H bonds 4 3 2 1 5 δ ppm 2 2 3 9 C=O C-H There is a C=O but no O-H absorptions, so must be an ester. 4. Use NMR spectra to give details of carbon chain 4 peaks – only 4 different environments. singlet of area 9 At δ =0.9 Means 3 CH3 groups C CH3 H3C CH3 Peak at δ 4 shows H–C–O Area 2 suggests CH2 Quartet means next to a CH3 O C H H H Peak at δ 2.2 shows H–C=O Area 2 suggests CH2 Singlet means adjacent to C with no hydrogens O C H H C O H Peak at δ 1.2 shows R-CH3 Area 3 means CH3 Triplet means next to a CH2 CH3 Put all together to give final structure
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6.3.2 Spectroscopy
Combined techniques (h) deduction of the structures of organic compounds from different analytical data including: (i) elemental analysis (see also 2.1.3c) (ii) mass spectra (iii) IR spectra. M3.1 Limited to functional groups encountered in this specification. Learners will not be expected to interpret mass spectra of organic halogen compounds. HSW3,5,6 Analysis and interpretation of different analytical data.
Credits:Neil Goalby