Essential practical nmr for organic chemistry - S. A. Richards

t in a fairly spectacular fashion, when it is present in a molecule. Once again, I = 1/2, so we only have two allowed states to worry about. Unlike 13C however, f uorine has only one isotope, 19F, and as this of course, has 100 % natural abundance, we see the whole proton signal split, instead of a couple of tiny satellites on either side of our signals! This point is well illustrated with a spectrum of 3-fluor propanol (Spectrum 6.10), which shows a fairly dramatic example of fluorin coupling. The F-CH2- coupling is about 47 Hz, and the F-CH2-CH2- coupling, is 27 Hz. The coupling to the third methylene group is non-existent in this example but can be seen sometimes (0–3 Hz). Another example of 19F coupling, this time in an aromatic system, (4-fluor benzoic acid) is shown in Spectrum 6.11. Note how the 19F couplings to the aromatic protons give the AA′BB′ system an 2.02.53.03.54.04.5 ppm 16 .0 0. 2 26 .5 0. 7 26 .3 26 .4 Spectrum 6.10 3-Fluoro propanol. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 85 7.307.357.407.457.507.557.607.657.707.757.807.857.907.958.008.05 ppm 15 .0 15 .0 Spectrum 6.11 4-Fluoro benzoic acid. asymmetric appearance. The actual values in this case are 9.0 Hz (ortho) and 5.6 Hz (meta) which are fairly typical. More useful 19F coupling data is given in Table 6.1. Fluorine can sometimes throw up some unexpected couplings in certain situations and spectra need to be handled with care! Sometimes, fluorin can be seen to couple over an unfeasible number of bonds (we have seen a seven bond coupling in the past). This is because fluorin is so electron hungry that it can couple through space as well as through-bond! We have also noted some strange behaviour with fluorinate pyridines, for example, 3-fluor nicotinic acid (Structure 6.19 and Spectrum 6.12). The signal for Hc (approx. 8.1 ppm) clearly shows couplings of 9.1, 2.9 and 1.7 Hz. The 9.1 Hz coupling must be from the fluorin as it does not appear anywhere else in the spectrum and its chemical shift distinguishes it from either of the other two protons. Of the other two protons, the signal at 8.82 ppm, (Hb) shows only a 2.9 Hz coupling which is also found in Hc, whilst Ha exhibits two small couplings (2.0 and 1.7 Hz), the smallest of these also appearing in Hc. These observations lead to the conclusion that the fluorine–proto couplings in this molecule are as given in Table 6.2. N F O OH HaHb Hc Structure 6.19 3-Fluoro nicotinic acid. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 86 Essential Practical NMR for Organic Chemistry Table 6.1 Some typical 19F–proton couplings. Structure 19F-1H position Typical 19F-1H coupling (Hz) FCH3 F-CH2-F-CH2-CH2- F-CH2-CH2-CH2- 45 24 0–3 F HH H F-H (geminal) F-H (cis) F-H (trans) 85 20 50 CH3~C=C H F F-CH3 2–4 H~C=C~CF3 F3C-H 0-1 CH3 F F F F3C-CH2- 8–10 F H F-H (ortho) F-H (meta) F-H (para) 6.2–10.3 3.7–8.3 0–2.5 F CH3 F-CH3 (ortho) F-CH3 (meta) F-CH3 (para) 2.5 0 1.5 F F H H Faxial–Haxial Faxial–Hequatorial Fequatorial–Hequatorial 34 11.5 5–8 F-Hc coupling did not surprise and neither did F-Ha coupling. But the F-Hb coupling of less than a single Hz is totally bafflin and defie obvious logic! Having learnt the lessons from this simple little compound, it would seem reasonable to expect similarly surprising couplings in other fluorinate heterocycles. Tread carefully! P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 87 8.058.108.158.208.258.308.358.408.458.508.558.608.658.708.758.808.858.908.95 ppm 8. 0 7. 7 7. 7 Spectrum 6.12 3-Fluoro nicotinic acid. Table 6.2 Fluorine–proton couplings in 3-fluoro nicotinic acid. Position Coupling (Hz) F-Ha 2.0 F-Hb very small, <1.0! F-Hc 9.1 6.6.3 Coupling between Protons and 31P Phosphorus is the other heteroatom of major coupling importance to the organic chemist. Like 19F, 31P has a spin of 1/2 and a 100 % natural abundance, so you know what to expect! The actual size of the couplings observed with 31P can vary considerably, depending on the oxidation state of the 31P atom. You’ll fin some useful examples in Table 6.3. 31P shows one particularly interesting feature. The size of couplings normally decreases dramatically with the number of intervening bonds, but this is not always the case with 31P (Table 6.3). A proton directly bonded to a 31P atom can be split by an enormous coupling of as much as 700 Hz (depending on the oxidation state of the phosphorus)! That means that the two parts of such a signal would be separated by almost 3 ppm at 250 MHz! So huge is this coupling that you could easily fail to recognise or accept it as a coupling at all, if you came across it. Structure 6.20 and Spectrum 6.13 show an example of 31P-1H coupling. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 88 Essential Practical NMR for Organic Chemistry Table 6.3 Some typical 31P–proton couplings. Structure 31P-1H relative position Typical 31P-1H coupling (Hz) (CH3)3P P-CH3 2.7 (CH3)3P=O P-CH3 13.4 (CH3)4P+I− P-CH3 14.4 (CH3-CH2)3P P-CH2-CH3 P-CH2-CH3 0.5! 13.7! (CH3-CH2)3P=O P-CH2- P-CH3 11.9 16.3 R P H H P-H 180–200 P O HRO OR P-H 630–710 CH3 P OR R R R P-CR2-CH3 10.5–18.0 CH3 P OR R R R P-CR2-CH3 10.5–18.0 CH3 P OR OR R R O P-CR2-CH3 10.5–18.0 P O O CH3 P-CH2- 6 P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 89 P+ BrX- Structure 6.20 Compound showing 31P-1H coupling. 1.81.92.02.12.22.32.42.52.62.72.82.93.03.13.23.33.43.53.63.73.83.94.04.1 ppm 7. 1 7. 4 7. 6 7. 1 Spectrum 6.13 31P-1H coupling. The complex multiplet at 4.02 ppm shows a 13 Hz 2-bond 31P coupling to the firs -CH2 in the chain and spin decoupling enables the 3-bond 31P coupling to the next -CH2 in the chain (1.86 ppm) to be measured (8 Hz). 6.6.4 Coupling between 1H and other Heteroatoms If you ever run a sample which is contaminated with an ammonium salt, in DMSO, you will see 14N–proton coupling, as shown in Spectrum 6.14. Note that the three lines of the multiplet are of equal intensity (the middle line is a little bit taller than the outer ones, but this is because of the width of the peaks at their bases. The central signal is reinforced because it stands on the tails of the outer two). This is because 14N has a spin of I = 1, and the allowed states are therefore –1, 0 and +1. This three line pattern with its 51 Hz splitting is highly characteristic and once seen, should never be forgotten. 14N coupling is only observed when the nitrogen atom is quaternary. In all other cases, any cou- pling is lost by exchange broadening, or quadrupolar broadening, both of which we’ve discussed before. Two-bond couplings, [e.g., 14N+-(CH2)4] are not observed, even when the nitrogen is qua- ternary, in ‘quat salts’ such as (n-butyl)4N+− Br−, presumably because the coupling is very small. So the phenomenon is only ever observed in the +NH4 ion! Note: The -CH2- attached to the qua- ternary nitrogen in compounds like tetra N-butyl ammonium chloride does present as a distorted P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 90 Essential Practical NMR for Organic Chemistry 6.856.906.957.007.057.107.157.207.257.307.357.407.457.507.557.607.657.707.75 ppm Spectrum 6.14 Typical appearance of +NH4 ion in DMSO. triplet with its central line split into a narrow triplet but this has nothing to do with 14N coupling as the same distortion can sometimes be seen in -CH2- groups next to certain other moieties, e.g., -SO2R. It is a non-firs order phenomenon caused by slight non-equivalence of the two protons in question. Boron has two isotopes, both of which have spin! 10B has a natural abundance of 18.8 %, and a spin of I = 3 (allowed spin states –3, –2, –1, 0, +1, +2, +3; i.e., one signal will be split into seven lines of equal intensity), whilst 11B has a natural abundance of 81.2%, and a spin of I = 3/2 (allowed spin states –3/2, –l/2, +1/2, and +3/2; i.e., one signal will be split into four lines of equal intensity). This gives rise to amazing effects in the borohydride, BH4− ion (Spectrum 6.15), which can some- times be formed accidentally during borohydride reductions. Note that the 10B-H couplings are of a different size to the 11B-H couplings. All 11 lines of the BH4− ion are to be found between 0 and –0.7 ppm. Note that, like 14N, 11B has a quadrupolar nucleus, but once again the symmetrical envi- ronment of the borohydride ion negates the relaxation pathway that would otherwise cause significan line broadening. Boron coupling is not generally seen in asymmetric environments or over multiple bonds. -0.65-0.60-0.55-0.50-0.45-0.40-0.35-0.30-0.25-0.20-0.15-0.10-0.050.00 ppm 6. 2 8. 4 0. 8 7. 5 6. 7 Spectrum 6.15 Boron–proton coupling in the borohydride ion. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 91 -0.10.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 ppm 0. 5 4. 5 0. 4 1. 0 9. 9 1. 0 Spectrum 6.16 Mixture of two organotin compounds. One other heteroatom worth mentioning is tin as organotin compounds are significan in organic synthesis. Tin has no fewer than ten naturally occurring isotopes, but fortunately, only three of them have nuclear spin. 115Sn has a natural abundance of a mere 0.32 %, which makes it spectroscopically insignificant of course. The only isotopes of tin that need concern us, are 117Sn (natural abundance 7.67 % and I=1/2), and 119Sn (natural abundance 8.68 %, and also, I=1/2). These two isotopes are both capable of two-bond and three-bond couplings in alkyl organotin com- pounds. This is demonstrated in Spectrum 6.16 which shows a mixture of two organotin compounds. The compound with a strong central peak at 0.5 ppm is thought to be (CH3)3-Sn-OH. The inner satel- lites result from a 117Sn-CH3 coupling of 69 Hz and the outer satellites to a 119Sn-CH3 coupling of 72 Hz. The second compound with the major signal centred at 0.13 ppm is (CH3)3-Sn-Sn-(CH3)3. In this case, we see once again, satellites resulting from two-bond couplings but also a second set of inner satellites resulting from smaller three-bond couplings of about 16 Hz for both 117Sn and 119Sn (i.e., Sn-Sn-CH3). Note too from the chemical shifts of these methyl groups that tin has quite a strong shielding effect. Finally, 29Si is an isotope that you should be aware of – every time you acquire a well prepared sample using TMS as a standard! 29Si satellites (accounting for about 4.7 % of the total signal, J 29Si – -CH3, 6.6 Hz) should be visible at the base of your TMS peak. The small coupling provides a good test of shimming quality (Spectrum 6.17). 6.6.5 Cyclic Compounds and the Karplus Curve As we have already mentioned, chemical shifts and couplings are heavily influence by molecular constraint and for this reason, some guidance in dealing with cyclic (saturated) compounds might well prove useful. We have already seen that in straightforward open-chain alkyl systems, the size of proton-proton couplings is governed by the electronegativity of neighbouring atoms. But the most P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 92 Essential Practical NMR for Organic Chemistry -0.04-0.03-0.02-0.010.04 0.03 0.02 0.01 0.00 ppm Spectrum 6.17 TMS showing 29Si satellites. important factor which governs the size of couplings between vicinal protons is the dihedral angle between them. In open-chain systems, this angle is usually averaged by rotation about the C-C single bond, and so is not normally of significance But in carbocyclic systems, dihedral angles are usually fi ed, since the structures are generally rigid. It is therefore vital that we understand how the size of vicinal couplings varies with dihedral angle. This data can be obtained by using the Karplus equation but the information derived from this equation (or equations as there are various versions of it) is more usefully portrayed graphically. A family of curves thus constructed makes additional allowance for factors other than dihedral angles which influenc vicinal proton couplings, e.g., localised electronegativities (Figure 6.2) but we have opted for a simplifie graph showing only three curves. Selection of the best curve for a given situation is perhaps rather a matter of trial and error, but is best approached by positively identifying an axial–axial coupling, since this arrangement ensures (in six-membered rings at least) a dihedral angle of 180◦ between the protons. Choose the curve that best fit the value that you observe for an axial–axial coupling in your molecule. Note that in the absence of any extreme electronic effects, this should give rise to a coupling of about 12 Hz. Similarly, a dihedral angle of 90◦ gives rise to a coupling of approximately 0 Hz, and where the angle is 0◦, we may expect a coupling of about 10 Hz. Making a model of the molecule becomes very important in the case of carbocyclic compounds, as it is important to be able to make fairly accurate estimates of dihedral angles. Now let us consider Structure 6.21 and Spectrum 6.18 and see how the Karplus curve can be used to aid assignment of the spectrum. (This compound will be referred to from now on as the morpholine compound as we will use it to demonstrate several different techniques) Note that the aromatic region has been omitted as it contains little of interest and we wish to concentrate on the carbocyclic region of the spectrum. It was acquired in CDCl3. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 93 θ 0 2 4 6 8 10 12 14 16 18 1801701601501401301201101009080706050403020100 Angle θ (degrees) Co up lin g co ns ta nt (H z) H H Figure 6.2 The Karplus curve – for relating the observed splitting between vicinal protons to their dihedral angle, θ . To derive maximum benefi from this exercise, we recommend that you make a model of this molecule, and refer to it as we go through the spectrum. Note that the morpholine ring falls naturally into a ‘chair’ conformation. Note also that in this example, the -CH2-Cl function will seek to minimise the morpholine ring energy by occupying an equatorial environment as this minimises steric interactions between it, and protons, and other substituents on the ring. All groups do this. The benzyl function will do likewise by inversion of the nitrogen stereochemistry. It is also worth noting that nine times out of ten, equatorial protons absorb at somewhat lower f eld than the corresponding axial protons. This can be reversed in certain cases where the specifi anisotropies of the substituents predominate over the anisotropies of the rings themselves but this is relatively rare. The difference is typically 0.5–1.0 ppm, but may be more. The structure is depicted as a Newman projection below (Figure 6.3). Aromatic protons aside, (they give the expected fi e-proton multiplet centred at around 7.3–7.4 ppm) the firs signal we encounter as O N Cl Structure 6.21 The morpholine compound. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 94 Essential Practical NMR for Organic Chemistry 2.02.53.03.54.04.55.05.56.06.57.07.5 ppm 25 .4 43 .8 30 .2 2.12.22.32.42.52.62.72.82.93.03.13.23.33.43.53.63.73.83.94.0 ppm Spectrum 6.18 The morpholine compound in CDCl3 with expansion. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 95 N O CH2Ph CH2 Cl Hc Hg Hj Hi Hb Hh Hd a e Figure 6.3 The morpholine compound shown as a Newman projection. we work from left to right is a complex multiplet – which is actually, a doublet of doublet of doublets [ddd] – at 3.95 ppm. Careful measurement of the couplings reveals them to be 11.4, 3.4 and 2.0 Hz. Since the multiplet is dominated by one large coupling, we can be safe in the knowledge that it must be an equatorial proton. This is because the dihedral angles between equatorial protons and both their equatorial and axial vicinal partners are always such that they give rise to relatively small couplings (check model and the Karplus curves). The only large coupling (i.e., 10 Hz or more) an equatorial proton can have will always be to its geminal partner – if it has one. So in this case, the 11.4 Hz coupling is clearly a geminal coupling. If we now make the entirely reasonable deductions that the proton giving rise to this signal is likely to be alpha to oxygen rather than nitrogen (on the basis of chemical shift) and that as the -CH2-Cl will be equatorial (as explained earlier), then this multiplet can only be assigned to the equatorial proton ‘b’ since there are no other equatorial protons that are alpha to oxygen in the molecule. The other two couplings can be rationalised in terms of the equatorial–axial coupling (3.4 Hz which is reciprocated in the ddd at 2.27 ppm) and the equatorial–equatorial coupling (2.0 Hz which is reciprocated in the dddd at 2.71 ppm). Note: methods of unpicking couplings will be discussed at length in later sections. Such methods are very useful when dealing with more complex spin systems like this one. The degree of roofin of ‘b’ indicates that its geminal partner must be fairly close to it in terms of chemical shift and sure enough, the six-line multiplet (another ddd) centred at 3.76 ppm satisfie the requirements for this proton (‘d’). Note that the second large coupling to this signal is also 11.4 Hz, the axial–axial coupling being the same size as the geminal coupling in this instance. The small remaining coupling (approx. 2 Hz) is reciprocated in the dddd at 2.71 ppm and is an axial–equatorial coupling. Proton ‘c’ can be define by the fact that it is not equatorial and it is highly coupled. The multiplet at 3.82 ppm satisfie these requirements. It is in the right ball park for chemical shift and is highly complex in that this proton is already the X part of an ABX system coupled to both protons alpha to the chlorine (the AB part). It is then further coupled with a 10 Hz, axial–axial coupling (reciprocated in the dd at 2.07 ppm) and with a 2 Hz axial–equatorial coupling which is reciprocated in the ddd at 2.90 ppm. Note that ‘c’ and ‘d’ are not fully resolved from each other. Such overlap inevitably complicates the issue. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 96 Essential Practical NMR for Organic Chemistry The N-benzyl protons are accidentally equivalent, presenting as a singlet at 3.59 ppm and overlap with the two protons alpha to the chlorine atom which present as the heavily roofed AB part of an ABX system (i.e., eight lines) centred at 3.55 ppm. Without slavishly dissecting the remaining four signals (2.90, 2.71, 2.27, 2.07 ppm), we hope that the principles of carbocyclic analysis have now been established. You should see at a glance that the 2.90 and 2.71 ppm signals must belong to equatorial protons because they are each dominated by only one large coupling and the remaining two must correspond to their axial partners. You should now be able to verify which equatorial proton belongs to which axial proton just by inspection. There is one last coupling which we have not yet mentioned and that is the apparent extra small coupling that can be seen on the equatorial protons alpha to the nitrogen (2.90 and 2.71 ppm). These two signals are in fact coupled to each other by what is known as a W path coupling. These are 4-bond couplings (unusual in saturated systems) which can be seen in situations where all the intervening proton–carbon and carbon–carbon bonds lie in the same plane. You can see from the model which you have next to you (?) that by definition such protons can only be equatorial. Note that whilst all the assignments in this section have been made purely on the basis of observations of couplings and multiplet appearance, this type of assignment is often simplifie by having definit ve knowledge of coupling pathways. We will discuss the options available for acquiring this type of data in a later chapter. Whilst six-membered rings may often give rise to quite complex spectra, they are at least generally rigid and based on the ‘chair’ conformation. As we have seen, this means that dihedral angles can be relied on and the Karplus curve used with reasonable confidence Unfortunately however, the same approach will end in tears if applied to other ring systems. Five-membered rings for example, are notoriously difficul to deal with as they have no automatic conformational preference. They are inherently fl xible, their conformations driven by steric factors. Cis protons on fi e-membered rings can have dihedral angles ranging from approximately –30◦ to 0◦ to +30◦ and exhibit a range of couplings to match. Trans protons on the other hand can range from +90◦ to +150◦. Deductions that can be made on the basis of observed vicinal couplings are therefore limited. If the observed coupling is very small, the two protons can only be trans to each other but if it is not, then they may be either cis or trans. We council against reliance on molecular modelling packages to produce a valid conformation of such structures. The energy difference between potential conformers is often small and could change in different solvents. Modelling packages consider molecules in isolation and thus make no allowances for solvent effects. Stereochemical assignments of such ring systems can only be confidentl made on the basis of NOE experiments which we will cover in detail in Section 8.5. 6.6.6 Salts, Free Bases and Zwitterions Sometimes, misunderstandings can arise when dealing with compounds containing protonatable centres. Hopefully, in this section, we will be able to clarify a few key issues that are relevant to such compounds. As we have already mentioned, CDCl3 should be avoided as a solvent for salts for two reasons. Firstly, salts are unlikely to be particularly soluble in this relatively nonpolar solvent but more importantly, spectral line shape is likely to be poor on account of relatively slow proton exchange at the protonatable centre. The remedy is simple enough – avoid using CDCl3 and opt for one of the more polar options instead, e.g., deuterated DMSO or MeOH and you should obtain spectra every bit as sharp as those of free bases. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 97 In practical terms, it is invariably a nitrogen atom that is protonated in salt formation. This always leads to a downfiel shift for protons on carbons both alpha and beta to the nitrogen concerned. In alkyl amines, the expected shifts would be about 0.7 and 0.3 ppm respectively. Remember that some heterocyclic compounds (e.g., pyridine) contain nitrogen atoms that are basic enough to protonate and comparable downfiel shifts can be expected (Spectrum 5.9). A misconception that we commonly encounter is that a spectrum can be a ‘mixture of the salt and the free base.’ This is an excuse that is often used by chemists to explain an inconveniently messy looking spectrum! Don’t be tempted by this idea – proton transfer is fast on the NMR timescale (or at least, it is when you use a polar solvent!) and because of this, if you have a sample of a compound that contains only half a mole-equivalent of an acid, you will observe chemical shifts which reflec partial protonation and not two sets of signals for protonated and free-base forms. It doesn’t happen – ever! Of course, whether a compound forms a salt or not depends on the degree of availability of the lone pair of electrons on any nitrogen atoms in the compound (i.e., their pKb values) and on how strong the acids involved in the salt formation (pKas). As a rough rule of thumb, alkyl and aryl amines do form salts whilst amides, ureas, most nitrogen-containing heterocycles and compounds containing quaternary nitrogen atoms do not. It should always be remembered of course, that the NMR spectrum reflect a compound’s behaviour in solution. It is quite possible for a compound and a weak acid to crystallise out as a stoichiometric salt and yet in solution, for the compound to give the appearance of a free base. For this reason, care should be taken in attempting to use NMR as a guide to the extent of protonation. If the acid has other protons that can be integrated reliably, e.g., the alkene protons in fumaric or maleic acid, then there should be no problem but if this is not the case, e.g., oxalic acid, then we would council caution! Do not be tempted to give an estimate of acid content based on chemical shift. With weak acids, protonation may not occur in a pro rata fashion though it is likely to in the case of strong acids. Sometimes, you may encounter compounds which have more than one protonatable centre. It is often possible to work out if either one or more than one are protonated in solution. A good working knowledge of pKbs is useful to help estimate the likely order of protonation with increasing acidity. Assume that the most basic centre will protonate firs and assess the chemical shifts of the protons alpha to each of the potentially protonatable nitrogen atoms. Zwitterionic compounds are worthy of special mention: H2N−R − COOH ↔ H3N+−R − COO− By their very nature, their partial charge separation can make them fairly insoluble and the degree of this charge separation (and hence resultant NMR spectra) tends to be highly dependant on concentration and pH. For these reasons, we recommend dealing with such compounds by ‘pushing’ them one way or the other, i.e., by adjusting the pH of your NMR solution so that the compound in question is either fully protonated (addition of a drop of DCl) or de-protonated (addition of a drop of saturated sodium carbonate in D2O): H3N+−R − COOH H2N − R − COO− Fully protonated Fully de-protonated P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 98 Essential Practical NMR for Organic Chemistry NH+ N N O A A = Aryl Cl- Structure 6.22 A protonation example. Whilst dealing with protonation issues, it is well worth considering the time dependence of the process in the context of the NMR timescale. A compound of the type shown in Structure 6.22 provides an interesting example. As a free base, the Ar-CH2-N protons would present themselves as a simple singlet. The lone pair of electrons on the nitrogen invert very rapidly on the NMR timescale and so the environment of the two protons is averaged and is therefore identical. However, on forming a salt, the whole process of stereochemical inversion at the nitrogen is slowed down dramatically because the sequence of events 3.33.43.53.63.73.83.94.04.14.24.34.44.54.64.74.84.95.05.1 ppm 3. 2 1. 1 0. 9 1. 0 Spectrum 6.19 Slow inversion of a protonated tertiary amine nitrogen. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come Delving Deeper 99 A HN+ A H A = Aryl Structure 6.23 Compound showing ‘pseudo enantiomeric’ behaviour. would be; de-protonation, inversion and re-protonation. Although as we said earlier, proton transfer is in itself a very fast process on the NMR timescale, it is the time taken for the entire process to occur that determines the nature of the spectrum that we observe. What we actually observe for the Ar-CH2-N protons of the salt in this molecule is a pair of broad, featureless signals at 4.6 and 5.0 ppm. The explanation for this is simple enough once the concept of time dependency for the inversion sequence has been appreciated. The protons in question fin themselves in different environments (within the context of the NMR timescale) and therefore have distinct chemical shifts. The signals are broad because the dynamic exchange process is taking place with a time period comparable to the NMR timescale, the broadening masking the geminal coupling between them (see Spectrum 6.19). A logical extension of these ideas will lead you to a recognition of the fact that a phenomenon of this type could yield species in solution which appear to behave as if they contain a chiral centre – even when they don’t. We have seen ‘pseudo enantiomeric’ behaviour in compounds of the type shown in Structure 6.23 (when protonated). All the protons of the CH2s in a molecule of this type may be non-equivalent (i.e., you observe essentially three AB systems). Note that the coupling from the alkene CH is would be small to both of the cyclic CH2s when the spectrum is acquired in the presence of HCl (see Spectrum 6.20). When the free 3.954.004.054.104.154.204.254.304.354.404.454.504.554.604.654.704.754.80 ppm 1. 2 1. 2 2. 2 1. 1 1. 0 Spectrum 6.20 Protonated nitrogen of a tertiary amine acting as a ‘chiral centre’. P1: JYS c06 JWST025-Richards October 7, 2010 10:32 Printer: Yet to come 100 Essential Practical NMR for Organic Chemistry base is liberated, all the AB systems collapse to give singlets. The explanation follows on logically from a consideration of the previous example. Protonation of the tertiary amine generates a chiral centre at the nitrogen atom, forcing all the geminal pairs of protons into different environments – hence the three AB systems. But this does not in any way imply that it would be possible to separate out enantiomers of the compound in salt form. These ‘pseudo enantiomers’ can only be differentiated within the context of a technique which has a timescale of a couple of seconds. Attempting to separate them on an HPLC column for example, would be unsuccessful as this technique has a timescale of several minutes (define by how long compounds take to travel down the column and enter the detector). During this time, proton exchange and consequent ‘enantiomer’ interconversion would have occurred many times in the course of the analysis. The only manifestation of this might be a slightly broader than normal (single) peak. This whole area of spectroscopy touches on many different topics and can only be approached confidentl with a reasonable working knowledge of basic NMR, stereochemistry and certain aspects of physical chemistry. P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come 7 Further Elucidation Techniques – Part 1 If a spectrum does not yield the definit ve information that you require on inspection, there are many other ‘tools of the trade’ that we can use to further elucidate structures. Broadly speaking, these fall into two categories – chemical techniques and instrumental techniques. 7.1 Chemical Techniques We will take a brief look at chemical techniques first It is true to say that the development of more and more sophisticated instrumental techniques has to a considerable extent, rendered these less important in recent years but they still have their place and are worthy of consideration in certain circumstances. Before embarking on any of these chemical techniques, however, be advised that they are not without a measure of risk as far as your sample is concerned! One of the great strengths of NMR is that it is a nondestructive technique but that can change quite rapidly if you start subjecting your compounds to large changes in pH or to potentially aggressive reagents like trifluoroaceti anhydride (TFAA)! Be sure that you can afford to sacrific the sample as recovery may not be possible. In the case of precious samples, chemical techniques should be regarded as a last option rather than a firs choice. 7.2 Deuteration Deuteration is the most elementary of the chemical techniques available to us, but it is still useful for assigning exchangeable protons which are not obviously exchangeable, and for locating exchangeables masked by other signals in the spectrum. There are of course, other ways of identifying exchangeables. The signal can be scrutinised closely to see if it has any 13C satellites associated with it, though this is not viable in the case of broad signals. Alternatively, irradiation of the water peak in an NOE experiment can be used as we’ll see later. Nonetheless, deuteration does provide a quick and easy method of identificatio which is still perfectly valid. Just to recap on the procedure, add a couple of drops of D2O to your solution, and shake vigorously for a few seconds. Note that with CDCl3 solutions, the best results are obtained by passing the resultant solution through an anhydrous sodium sulfate filte to remove as much emulsifie D2O as possible. (Note Essential Practical NMR for Organic Chemistry S. A. Richards and J. C. Hollerton © 2011 John Wiley & Sons, Ltd. ISBN: 978-0-470-71092-0 P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come 102 Essential Practical NMR for Organic Chemistry 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 1. 5 1. 5 1. 0 1. 0 Spectrum 7.1 n-Butanol in CDCl3 with -OH obscured by multiplet at 1.39 ppm. also that CDCl3 and D2O are not miscible, the CDCl3 forming the bottom layer as it is more dense than D2O.) You then re-run the spectrum and check for the disappearance of any signals. Careful comparison of integrations before and after addition of D2O should reveal the presence of any exchangeable protons buried beneath other signals in the spectrum. If they are slow to exchange, like amides, a solution of sodium carbonate in D2O, or NaOD may be used. The technique is demonstrated using n-butanol in Spectra 7.1 and 7.2. Note the reduction in integration of the multiplet centred at 1.39 ppm. 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 1. 5 1. 0 1. 0 1. 0 Spectrum 7.2 n-Butanol in CDCl3 after shaking with two drops of D2O P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come Further Elucidation Techniques – Part 1 103 R O CH3 R OH CH3 Structure 7.1 Keto-enol exchange. Remember that any proton which is acidic enough is prone to undergo deuterium exchange. Methylene protons alpha to a carbonyl for example, may exchange if left standing with D2O for any length of time, as they can exchange via the keto-enol route (i.e., Structure 7.1). Note that deuterium exchange of the -OH leads to incorporation of deuterium alpha to the carbonyl in the ketone form. This may happen, even if there is no evidence of any enol signals in the spectrum initially, i.e., it can occur even when the equilibrium is heavily in favour of the ketone. Aromatic protons of rings which bear two or more -OH groups are also prone to undergo slow exchange, as are nitrovinyl protons. 7.3 Basification and Acidification This topic has been dealt with quite extensively in Section 6.6.6 so we won’t go over the material again but there is perhaps one other type of problem that may be worth looking at with a view to solving by a change of pH. Consider the two structures in Structure 7.2. Whilst the preferred method of differentiating these structures would be by an NOE experiment, it would be possible to accomplish this by running them in DMSO and then adding a drop of base to each solution and re-running. (Note: DMSO is the preferred solvent for this experiment as both the neutral and the charged species would be soluble in it.) In both cases, the phenoxide ion (Ar-O−) would be formed and the extra electron density generated on the oxygen would feed into the ring and cause a significan upfiel shift of about 0.3–0.4 ppm in any protons ortho- or para- to the hydroxyl group. In the example above, the compound on the left would show such an upfiel shift for only a single doublet (a), whilst the compound on the right would show an analogous upfiel shift for both a narrow doublet (b) and a doublet of doublets (c). Caution should be exercised if attempting any determination of this type as it is not the preferred method and it is always safest if both compounds to be distinguished are available for study in this way. OH O CH3 CH3 O OH CH3 CH3 bc a Structure 7.2 Compounds which show one or two upfield shifts. P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come 104 Essential Practical NMR for Organic Chemistry 7.4 Changing Solvents If a signal of particular interest to you, is obscured by other signals in the spectrum, it is often worth changing solvent – you might be lucky, and fin that your signal (or the obscuring signals) move sufficientl to allow you to observe it clearly. You might equally well be unlucky of course, but it’s worth a try. Running a sample in an anisotropic solvent like D6-benzene or D5-pyridine, can bring about some even more dramatic changes in chemical shifts. We tend to use benzene in a fairly arbitrary fashion, but in some cases, there is a certain empirical basis for the upfiel and downfiel shifts we observe. For example, benzene forms collision-complexes with carbonyl groups, ‘sitting’ above and below the group, sandwich-style. When the carbonyl is held rigidly within the molecule, either because it forms part of a rigid system, or because of conjugation, we can generally expect protons on the oxygen side of a line drawn through the carbon of the carbonyl, and at right angles to the carbonyl bond to be deshielded. Conversely, those on the other side of the line are shielded. 7.5 Trifluoroacetylation This is quite a useful technique which can give a rapid, positive identificatio of -OH, -NH2, and -NHR groups in cases where deuteration would be of little value. Even though the technique can be a little time-consuming and labour-intensive in terms of sample preparation, it can nonetheless yield results in less time than it would take to acquire definit ve 13C data – particularly if your material is limited. Consider Spectrum 7.3. The bottom trace shows the ordinary spectrum of cyclohexanol, run in CDCl3. Distinguishing it from chlorocyclohexane is not easy (without the use of 13C NMR) – the chemical shift of the proton alpha to the functional group would be similar in both compounds, and in the case of the alcohol, the -OH need not show coupling to it. Furthermore, in problems of this type, the -OH proton itself may well be obscured by the rest of the alkyl signals or combined with the solvent water peak. Integration of the alkyl multiplet before and after deuteration will not necessarily be very reliable, since looking for 1 proton in a multiplet of 10 or 11, will give only a relatively small change in integral intensity (and let us not forget that water in the CDCl3 which will absorb in this region, along with any water that may be residual in the compound). The top trace shows what happens when the sample is shaken for a few seconds with a few drops of TFAA. The reaction shown in Structure 7.3 occurs. The resultant spectrum is clearly very different from the alcohol, as the trifluoroaceti ester function is far more deshielding with respect to the alpha proton than is the -OH group. A downfiel shift of >1 ppm can be seen. This clearly distinguishes the alcohol from the analogous chloro compound which would of course give no reaction. This is a relatively quick and convenient technique, the reagent reacting quite readily (assuming no great steric hindrance of course) with alcohols, primary and secondary amines. (Note the possibility of complication if you react a secondary amine with TFAA – it will yield a tertiary amide!) If the reaction is a little slow, as is often the case with phenolic -OH groups, you can ‘speed it up a little’ by gentle warming, more shaking, and even adding a drop of D5-pyridine to base-catalyse the reaction. Use of this reagent is however, somewhat limited. You can only use it in solvents which don’t react with it (D4-methanol, and D2O are obviously out of the question), or contain a lot of water, i.e., D6-DMSO. Another slight drawback is that the cleaving of the anhydride liberates trifluoroaceti acid, P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come Further Elucidation Techniques – Part 1 105 0.51.01.52.02.53.03.54.04.55.05.5 ppm 91 .27. 8 0.51.01.52.02.53.03.54.04.55.05.5 ppm 80 .77. 5 Spectrum 7.3 The use of TFAA to identify an -OH group. which has nuisance value if your compound is very acid-sensitive, and will also protonate any unreacted amine function present. If this salt-formation is a problem, it is worth adding sodium bicarbonate in D2O, dropwise, to neutralise the acid. You’ll know when the excess TFA has been neutralised, as further additions of bicarbonate fail to produce any further effervescence (shake thoroughly, and don’t forget to release built up pressure by releasing carbon dioxide!) Dry your solution through an anhydrous sodium sulfate filte before rerunning. OH O F F F O F F F O + O O F F F + O OH F F F Structure 7.3 Using TFAA to identify an -OH group. P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come 106 Essential Practical NMR for Organic Chemistry 7.6 Lanthanide Shift Reagents Unfortunately, the use of lanthanide shift reagents such as the europium compound, Eu(fod)3 is a practise that has been largely consigned to the dustbin of history so we will say very little about them. The problem with trying to use them in high-fiel spectrometers is that the fast relaxation times of the collision complexes brought about by paramagnetic relaxation (courtesy of the europium or other lanthanide atom), leads to severe line broadening. This paramagnetic broadening is very much worse in high fiel spectrometers so if you are using a 250 MHz or above, don’t bother trying. If, however, you are still soldiering on with a spectrometer of 100 MHz or less, then by all means try using them to ‘stretch’ a spectrum out – if your compound is suitable. They work by coordinating with an atom that has a lone pair of electrons available for donation. The more available the lone pair, the greater will be the affinit (-NH2/NHR > -OH > >C=O > -O- > -COOR > -CN). Note that they will only work in dry solvents that don’t contain available lone pairs. Good luck! 7.7 Chiral Resolving Agents We have seen that the spectra of enantiomers, acquired under normal conditions, are identical. The NMR spectrometer does not differentiate between optically pure samples, and racemic ones. The wording is carefully chosen, particularly ‘normal conditions’, because it is often possible to distinguish enantiomers, by running their spectra in abnormal conditions – in the presence of a chiral resolving agent. Perhaps the best known of these is (–)2,2,2,trifluoro-l-(9-anthryl ethanol, abbreviated understandably to TFAE. (W.H. Pirkle and D.J. Hoover, Top. Stereochem., 1982, 13, 263). Structure 7.4 shows its structure. This reagent may form weak collision-complexes with both enantiomers in solution. As the reagent is itself optically pure, these collision-complexes become ‘diastereomeric.’ That is, if we use (–)TFAE (note that the (+) form can be used equally well), the complexes formed will be: (–)reagent – (+)substrate, and (–)reagent – (–)substrate. These complexes often yield spectra sufficientl different to allow both discrimination and quantificatio of enantiomers. This difference will be engendered largely by the differing orientations of the highly anisotropic anthracene moieties in the two collision complexes. You won’t be able to tell which is which by NMR, of course – that’s a job for polarimetry or circular dichroism, but if you know which enantiomer is in excess, you can get a ratio, even in crude samples which would certainly give a false reading in an optical rotation determination! The use of TFAE is demonstrated in Spectrum 7.4, which shows the appearance of proton ‘b’, before and after the addition of 30 mg of (+)TFAE to the solution. (This is the region of interest – it is usually protons nearest the chiral centre, which show the greatest difference in chemical shifts in the pair of F F F H OH Structure 7.4 Structure of TFAE. P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come Further Elucidation Techniques – Part 1 107 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 Chemical Shift (ppm) 4.552.001.016.17 4.95 4.90 4.85 4.80 4.75 4.70 4.65 4.60 4.55 4.50 4.45 4.40 4.35 4.30 4.25 4.20 4.15 4.10 4.05 4.00 Chemical Shift (ppm) 1.01 4.95 4.90 4.85 4.80 4.75 4.70 4.65 4.60 4.55 4.50 4.45 4.40 4.35 4.30 4.25 4.20 4.15 4.10 4.05 4.00 Chemical Shift (ppm) 2mg substrate + 30mg +TFAE NH2 OH 'b' Spectrum 7.4 The use of TFAE as a chiral resolving agent. complexes formed). The middle trace shows the expansion of proton ‘b’ prior to the addition of TFAE and the top trace, its appearance after the addition of the reagent. It is clear from this trace that proton ‘b’ is no longer a simple X-part of an ABX system. Apart from shifting upfield it has broadened, and eight lines are apparent. This is because the ‘b’ protons in the two collision complexes have slightly different chemical shifts and we can now see them resolved from one another. (Clearly, this sample was a racemate as the ratio of the resolved ‘X’ multiplets is 50/50.) This difference is often quite small, and so as to exploit it to the full, experiments of this type are best performed at high fiel (e.g., 400 MHz or more). TFAE is a very useful reagent for this type of work, as it is very soluble in CDCl3, which is just as well, as a considerable quantity of it is often needed to produce useful separations – even at high field Its other advantage is that its own proton signals are generally well out of the way of the sort of substrate signals you are likely to be looking at. It should be noted that the compound under investigation should have at least one, and preferably two potentially lone pair donating atoms to facilitate interaction with the reagent. Another useful reagent of this type is ‘chiral binaphthol’ (see Structure 7.5). P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come 108 Essential Practical NMR for Organic Chemistry OH OH Structure 7.5 (R)-(+)-1,1’-bi-2-naphthol This is a member of an interesting class of compounds which are chiral, without actually containing a define chiral centre. They are chiral because their mirror images are non-superimposable. In the case of this molecule, there is no rotation about the bond between the two naphthol rings because of the steric interaction between the two hydroxyl groups. ‘d’ and ‘l’ forms can be isolated and are perfectly stable (Optical purity determination by ‘H NMR, D. P. Reynolds, J. C. Hollerton and S. A. Richards, in Analytical Applications of Spectroscopy, edited by C. S. Creaser and A. M. C. Davies, 1988, p346). Optically pure mandelic acid (see Structure 7.6) can be a useful chiral resolving agent where the compound you are looking at has a basic centre, as it can form an acid-base pair with it, which is a stronger form of association. This compound is of sparing solubility in CDCl3 however and can precipitate out your compound if, as is often the case, its protonated form is of low solubility in CDCl3. The technique of using resolving agents is obviously a useful one in following the synthesis of a compound of specifie chirality. To summarise, we take our compound, which has a chiral centre of unknown rotation and form some sort of complex by introducing it to a reagent of known optical purity. The complexes we form have diastereoisoromeric character, which can give rise to a difference in the chemical shifts of one or more of the substrate signals. This enables us to determine the enantiomer ratio, either visually, or by integration, if we have sufficien signal separation. In practise, if using one of these reagents to follow the course of a chiral separation, it is essential to determine whether resolution is possible, by performing a test experiment either on a sample of racemate, or at least a sample known to contain significan quantities of both enantiomers. Once useable resolution has been established, the technique can be used to monitor solutions of unknown enantiomer ratios with reasonable accuracy, down to normal NMR detection limits. It is a good idea to keep the ratio of reagent to sample as high as possible. We recommend starting with about 1–2 mg of compound in solution with about 10 mg of reagent. In this way, you can minimise both the quantity of your sample and the amount of (expensive) reagent used. Keeping the initial sample small has another advantage – it avoids line broadening associated with increased viscosity of O OH OH Structure 7.6 Mandelic acid. P1: JYS c07 JWST025-Richards October 2, 2010 18:41 Printer: Yet to come Further Elucidation Techniques – Part 1 109 very concentrated solutions, whilst at the same time leaving the option open for further increasing the concentration of reagent, if needed. One f nal point – the use of chiral resolving agents is restricted to nonpolar solvents, i.e., CDCl3 and C6D6, though combining these can sometimes augment separation. That just about concludes the ‘Chemical techniques’ section. As we’ve seen, some important types of problem can be tackled using them, and if your sample is scarce, all is not lost – it can often be recovered, though this might take some effort on your part. Deuterated samples can be back-exchanged by shaking with an excess of water, trifluoroacetylate samples can be de-acetylated by base hydrolysis, and shift reagents can be removed by chromatography. Now, we shall have a look at some of the most important ‘Instrumental techniques’. . .