NMR Spectroscopy for Structural Determination

NMR Spectroscopy for Structural Determination ➢ What is NMR? - Nuclear Magnetic Resonance Spectroscopy - Molecules placed in a strong magnetic field - Radiofrequency radiation causes excitation in atomic nuclei - Gives information on nuclei in different environments - Information to piece together the molecular skeleton ➢ Sample Preparation - Sample dissolved in a suitable solvent and added to a glass NMR tube - Size DOES matter → the amount of sample and solvent in an NMR tube is very important! - Deuterated solvents are used – deuterium (2H) does not resonate in the spectral region that we are investigating - Typical solvents: o CDCl3 (chloroform-D) o D 2O o DMSO-D6 ((CH3)2S=O) ➢ 1H NMR Spectra - 1H NMR spectra gives you four key pieces of information that allow us to determine the structure of a molecule: 1. Number of 1H environments (“peaks”) 2. Integration (height = no. of protons) 3. Chemical shift (x-axis) 4. Multiplicity NMR Signals Chemical Shift (𝛿) in ppm  14 ppm ⟷ 0 ppm ➢ Proton Environments - Proton in different magnetic environments will resonate at a different position in the NMR spectrum - Magnetic environment can be simplified to chemical environment. It can be applied to any element that we can run NMR Spectroscopy for - Need to look at the atoms and groups surrounding protons → the following molecule has 3 different chemical environments: A – C - A, B and C are differentiated by their distance from the NO2 group ➢ Exchangeable Protons - Ethanol has 3 different 1H environments - However, sometimes we may only see 2 peaks on a 1H NMR spectrum - This is because the alcohol proton is an exchangeable proton (same applies to N-H and S-H protons too) - Exchangeable protons have a polarised bond due to the high electronegativity of the heteroatom that they are attached to (O, N, S). This means that they can exchange with the deuterium atom in the solvent and therefore won’t show up on the spectrum R – OH ⇆ R – OD - If a peak appears that you suspect could be for an exchangeable proton, you can prove it by adding a couple drops of D2O to the NMR tube, give it a shake and when the spectrum is run again, the proton will exchange with the deuterium in D 2O and the peak will disappear ➢ Symmetry - Symmetry is important in identifying environments – highly symmetrical molecules have fewer environments ➢ Integration - Integration is the number of protons that the signal represents - It is a ratio, not an absolute number. We measure the value of all the integrals then set the smallest value equal to 1 proton – this gives us an integer ration of the protons in all the other environments - Sometimes the ratio may not be a whole number → e.g. 1.5:1 - In this case we need to multiply all the integrals in order to get a whole number as we cannot have half a proton → 1.5:1 because 3:2 ➢ Chemical Shift - Expressed in ppm. Often a universal reference added to spectra (TMS 𝛿 = 0 ppm) - Chemical shift is affected by the atom to which the proton is attached and the electronic environment (substituents) surrounding the proton Electron Withdrawing Group Electron Donating Group Ester or Nitro groups ➢ Typical 1H Chemical Shifts Ether, Amines or Halogens ➢ Multiplicity (AKA Splitting Pattern or Coupling) - Any proton will couple with protons on adjacent atoms – we call this 3JHH coupling – it means that protons are coupling through 3 bonds: - Coupling leads to a splitting pattern in the signal → result: a doublet for each H - The exception is that protons on heteroatoms do not couple: - This looks like the same 3 bond coupling, however we will not see any coupling here and both signals will appear as singlets - The n+1 rule o The splitting pattern for a proton will have n+1 lines where n is the number of adjacent protons that it couples with o Line intensities determined by Pascal’s triangle ➢ Coupling in Aromatic Compounds - Aromatic protons also couple with their nearest neighbours, but we must remember the symmetry planes to identify different environments Ortho-substitution Meta-substitution Para-substitution - Consider the structure of aspirin and what you would expect the aromatic protons to look like on an NMR spectrum - In reality, more complicated than expected as aromatic peaks tend to display “fine coupling”, but for our purposes, in a spectrum like aspirin’s, we are not concerned with it as we can still discern the doublets and triplets - Sometimes the fine splitting becomes very complicated, however, and the peak appears as a multiplet. When this happens, we can no longer discern any information from the multiplicity of the aromatic peaks and can only use the integration and chemical shift to give us structural details ➢ 13 C NMR - The most abundant isotope of carbon (12C) cannot be observed by NMR spectroscopy. Instead, we look at the 13C isotope → only has a 1.1% natural abundance, so you need to make up your NMR tubes a bit more concentrated than for 1H in order to get a good quality spectrum - As the 13C isotope is so rare, it is extremely unlikely that we would get two neighbouring carbon atoms to both be 13C, therefore we don’t see 13C-13C coupling/splitting. 13C does couple with neighbouring 1H atoms, but we tend to remove this splitting from the spectra by ‘decoupling’ them to produce a better quality spectrum - As a result, 13C NMR spectra usually appear as a series of singlets - It is also usually not possible to integrate 13C NMR spectra ➢ Typical 13C Chemical Shifts ➢ DEPT - DEPT (Distortionless Enhancement by Polarisation Transfer) gives us some extra information about each carbon environment o CH – peak points upwards o CH2 – peak points downwards o CH4 – peak points upwards o CR4 – no peak ➢ Use 13C & DEPT in Combination 4 3 2 1 4/5 5 2 ➢ Elemental Analysis - Knowing the molecular formula before attempting to deduce the structure from spectra is invaluable. You can determine the molecular formula by combining two techniques: mass spectrometry and elemental analysis - Most common technique to get this data is combustion analysis - Essentially the sample is burned and the resulting combustion products are quantified - Usually reported as a percentage of each element - Halogens hard to detect in this manner, so conspicuous by their absence in any reported compositions that do not add up to 100% ➢ Calculating the Molecular Formula - Check the mass spectrum to find the molecular ion (and therefore molecular mass of the molecule) – sometimes denoted on the spectrum by a triangle - Sum the elemental composition data → does it add up to 100%? If not, the remainder is a heteroatom (usually a halogen) - If the molecule does contain a halogen, check the MS fragments to look for isotopes: o Fluorine: ~100% 19F = 1 peak o Chlorine: ~75% 35Cl & ~25% 37Cl = 2 peaks, 3:1 ratio o Bromine: ~51% 79Br & ~49% 81Br = 2 peaks, 1:1 ratio o Iodine: ~100% 127I = 1 peak - Worked example: o Work out the molecular formula for compound A which has elemental composition C 40.6%, H 5.35%, O 15.45%, with a molecular ion of 206/208 (1:1) 40.6 + 5.35 + 15.45 = 61.4% o Does not sum to 100% → remaining 38.6% is a halogen o Mass spectrum shows this must be bromine as there is a 1:1 ratio of isotopes, therefore we know that the elemental composition of 𝐵𝑟 = 38.6% Divide by the RAM Divide by the smallest Carbon Hydrogen Oxygen Bromine (40.6%) (5.35%) (15.45%) (38.6%) 40.6 = 3.38 12 5.35 = 5.35 1 15.45 = 0.97 16 38.6 = 0.48 80 3.38 = 7.04 0.48 5.35 = 11.15 0.48 0.97 = 2.02 0.48 0.48 =1 0.48 7 11 2 1 C7 H11 O2 Br Nearest integer (may need to multiply) Empirical formula o Check if empirical formula is the molecular by looking at the molecular ion C7H11O2Br Mr = (7 x 12) + (11 x 1) + (2 x 16) + 80 = 207 g/mol o Molecular ion peaks at 206/207 (if we had used 79 and 81 for Br isotopes rather than 80, we would have got 206 and 208). Therefore, empirical formula = molecular formula o If they are different, need to multiply empirical formula so they match ➢ Key Atomic Masses Hydrogen – 1.01 g/mol Fluorine – 19.00 g/mol Carbon – 12.01 g/mol Chlorine – 35.45 g/mol Nitrogen – 14.01 g/mol Bromine – 79.90 g/mol Oxygen – 16.00 g/mol Iodine – 126.90 g/mol