NMR Spectroscopy PDF - A Level Chemistry

A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY 1H-N.M.R N.M.R. stands for Nuclear Magnetic Resonance and it is a form of spectroscopy that provides detail of the actual structure of an organic molecule usi...

A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY 1H-N.M.R N.M.R. stands for Nuclear Magnetic Resonance and it is a form of spectroscopy that provides detail of the actual structure of an organic molecule using “hydrogen environments”. It is used in conjunction with: Mass Spectroscopy - Gives you the Mr of the molecule via the M+ peak Infrared Spectroscopy - Gives you information about which covalent bonds are preset oil the molecule. Chemical Tests: Give you information about which functional groups are present. 1 H-N.M.R. spectra provide information about the specific positions of H atoms in an organic molecule. In other words: How many H atoms are bonded to each carbon atom What other atoms may also be bonded to the carbon atoms (O, N, Cl etc) How many of each type of atom are bonded to neighbouring carbons. 1H-N.M.R ENVIRONMENTS All of this information combines to provide what are called “environments”. These are specific positions that H atoms can occupy within a molecule. Here’s a few examples of H environments, where “R” represents a carbon chain. R CH3 R CH2 R R OH H’s in a CH3 group H’s in a CH2 group H’s in an alcohol group H H O R H R C Cl C C R C C H H H’s in an alkene group H’s next to a H’s next to a C=O group halogenoalkane group A full list of every environment you need to use is always provided in your data sheet for the exam. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY 1H-N.M.R SPECTRA 1 H-N.M.R. spectra give us four pieces of information that we can use to deduce the structure of a molecule. Here’s the spectra for ethanol as an example: 3.0 H H H C C OH H H 2.0 1.0 𝜎 /ppm 4 3 2 1 0 SPECTRUM FEATURE MOLECULAR FEATURE Each peak is a different H environment in the molecule. So, the number of Number of Peaks peaks = the number of hydrogen environments present. In ethanol there are three. The heights of the peaks tell us, Relative Peak Height / Integration relatively, how many H atoms are in Value each environment. These are the numbers above the peaks. You can see that two of the peaks above are “split”. This splitting gives us Splitting Pattern information about how many H’s are bonded to the neighbouring carbon atom to that environment. This is their position on the x axis of the Chemical Shift (𝜎) in ppm spectrum. You can use this position and (parts per million) your data book to look up the type of environment that caused the peak. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY NUMBER OF PEAKS This is the most straight forward. The number of peaks that you see is literally the number of different carbon environments that the molecule has. However, watch out for equivalent environments. This is where two environments are the same as each other and have the same neighbouring group. e.g. H O H H O H H H C C C H H C C C C H H H H H H Propanone Butanone You can see that all 6 H’s exist within This also has two -CH3 groups. the two -CH3 groups. Both of these However, each -CH3 group has a -CH3 groups have the same different neighbour. neighbour (C=O), so they are The one on the left has a C=O “equivalent” and only cause one neighbour, and the one on the right peak in the spectrum. has a -CH2 neighbour. This means that, despite both being -CH3 groups, they are not equivalent and will cause two different peaks on the spectrum. How To Spot H-NMR Environments AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY INTEGRATION VALUES These relate directly the relative number of H’s in each environment. If the values given in the spectrum are not integers, you need to use the values to find the simplest whole number ratio between them to deduce the number of H’s in each environment. e.g. 1.8 H H H C C OH 1.2 H H 0.6 𝜎 /ppm 4 3 2 1 0 Here, we can see that the values are not integers, so we find the simplest whole number ration to find the number of H’s in each environment. 0.6 : 1.2 : 1.8 = 1 : 2 : 3 So, the number of H’s in each environment = 1, 2 and 3 respectively. You can see that in ethanol we have a -CH3, a -CH2 and an -OH, so it literally tells us the number in each environment. However, bear in mind that once you have found the whole number ratio, these are relative numbers of H’s. It could be that that the molecule that created this spectrum actually contains 6 H's in one environment, 3 in another and 2 in another. So, the same ratio, but the actual numbers of H’s is different. We can use other information given in questions to help us decide this, such as the molecular formula deduced from a mass spectrum. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY SPLITTING PATTERNS These are probably the trickiest part of NMR spectra to master. The splitting pattern for a peak could be: singlet doublet triplet quartet Multiplet (5 or more) The split for an environment related directly to the number of H’s bonded to the neighbouring carbons of the environment that causes the peak. What’s known as the “n+1” rule applies here. i.e. the split number is one more than the number of H’s bonded to the neighbouring carbons. Number of H’s Bonded to Splitting Pattern Neighbouring C to the Environment Singlet 0 (zero) Doublet One Triplet Two Quartet Three Top Tip: H’s in -OH groups and -NH2 groups are always singlets and do not cause any splitting in neighbouring environments. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY H H H C C O H -CH3 H H -CH2 -OH 𝜎 /ppm 4 3 2 1 0 The -CH3 has a -CH2 environment as a neighbour. In other words there are 2 H’s bonded to the neighbouring carbon. Using n+1, this is why the peak for the -CH3 group is a triplet. 2 H’s on the neighbouring carbon + 1. The -CH2 has a -CH3 environment as a neighbour. In other words there are 3 H’s bonded to the neighbouring carbon. Using n+1, this is why the peak for the -CH3 group is a quartet. 3 H’s on the neighbouring carbon + 1. The H on the neighbouring oxygen atom does not cause splitting. The -OH is a singlet as the oxygen atom prevents any splitting signals. IN other words, no neighbouring H’s are picked up. 0 + 1 = 1. How To Deal With Splitting Patterns AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY CHEMICAL SHIFT The position of the peaks on the x axis related directly to a table that you will find in your data sheet. Simply make a note of the chemical shift value where the peak is on the spectrum and look it up in the table. This tells you what type of H environment you have. There may be overlaps in the ranges provided in the data sheet, but you can use other data at your disposal, such as the molecular formula, in order to make an informed decision. T.M.S. & CCl4 / CDCl3 TMS (tetramethyl silane) Si(CH3)4 is an inert standard that is used in NMR spectroscopy CH3 that serves two purposes. 1. It serves as a reference point for the chemical shift scale, as its peak is at 0, H 3C Si CH3 away from the other peaks. All other chemical shifts are relative to this. CH3 2. Its peak is a singlet and has a height of “12”. This serves as a reference point for the other peak heights in the spectrum. 12 𝜎 /ppm 4 3 2 1 0 AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY Both CCl4 and CDCl3 are solvents used to dissolve the test substance. CCl4 is a non-polar solvent used for non-polar organic molecules. CDCl3 is a polar solvent used for polar molecules. Both of these are inert (unlikely to react with the test substance) and neither create peaks as neither contain hydrogen. How To Answer H-NMR Questions AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.3.15 N.M.R. SPECTROSCOPY 13C-N.M.R. SPECTROSCOPY 13 C-NMR spectroscopy is very similar to H-NMR spectroscopy, except that you are only ever required to work with two pieces of information. 1. The number of peaks = the number of C environments 2. The chemical shift values relate the to environment type and can be looked up in your data sheet. So, in other words, you won’t have to deal with relative peak heights or splitting patterns. e.g. H H H C C O H H H C C 𝜎 /ppm 60 40 20 0 Ethanol has two C environments, therefore two pets on the C-NMr spectrum. The chemical shifts relate directly to the type of environment. The blue C is just a carbon in an “alkyl” group. The red C is a C bonded to an an oxygen with a single bond (C-O) How To Answer C-NMR Questions AQA www.chemistrycoach.co.uk © scidekick ltd 2024