NMR Spectroscopy. B = B o - B e ν o = γb/2π

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1 NMR Spectroscopy hem 345 Univ. Wisconsin, Madison hemical Shifts hemical shifts have their origin in the circulation of electrons induced by the magnetic field, which reduces the actual field at the nucleus. Thus a higher magnetic field has to be applied to achieve resonance. Different types of protons in a molecule are surrounded by different electron densities, and thus each one sees a slightly different magnetic field. B o e - A B = B o - B e ν o = γb/2π (magnetic field at nucleus) (Larmor precession frequency of A ) B e The Larmour precession frequency ν o depends on the magnetic field strength. Thus at a magnet strength of 1.41 Tesla protons resonate at a frequency of 60 Mz, at 2.35 Tesla at 0 Mz, and so on. Although z are the fundamental energy unit of NMR spectroscopy, the use of z has the disadvantage that the position of a peak is dependent on the magnetic field strength. This point is illustrated by the spectra of 2-methyl-2-butanol shown below at several different field strengths, plotted at a constant z scale. Effect of Spectrometer Magnetic Field Strength 2 Mz c d 3 b a 2 3 c 3 Me 4 Si c 0 Mz d b a Mz 0 0 For this reason, the distance between the reference signal (Me 4 Si) and the position of a specific peak in the spectrum (the chemical shift) is not usually reported in z, but rather in dimensionless units of δ, which is the same on all spectrometers. δ = (Frequency shift from Me 4 Si in z) (Spectrometer frequency, Mz) 1

2 1 hemical Shifts R X=,l, X X=N,S X Alkanes δ Downfield B o decreases B o increases Upfield Deshielded ν o increases ν o decreases Shielded igh frequency Low frequency The ranges above provide an estimate of the chemical shift for simple molecules, but don't help very much when there are multiple substituents. A simple scheme can be used to estimate chemical shifts of protons on sp 3 carbons. Use the base shift for methyl groups. 2 groups, and groups, and add to these the increments for each α substituent: Base Shift Increment (=)R 3.0 R l 2.4 Aryl 1.4 (=)R 1.0 = 1.0 Base shift : 1.5 α Ph: 1.4 α : 2.3 alculated: 5.3 bserved: 4.8 Ph l Base shift 2 : 1.2 α l: 2.4 alculated: 3.6 bserved:

3 Representative Proton hemical Shift Values (δ -4 to 6) Li (Me 3 Si) 3 Si-Te Ph-Te- (Me 3 Si) 3 Si-Se- Me-Te Se- 2 Ph 3 Li (Me 3 Si) 3 Si-S- ( 3 ) 2 Mg N ( 3 ) 3 l 3 - N Me ( 3 ) I ( 3 ) 3 ( 3 ) ( 3 2 ) 2 ( 3 ) 3 ( 3 ) 4 Sn ( 3 ) 4 Si Ph F S-Ph ( 3 ) 2 2 Ph ( 3 ) I 3 l N N ( 3 ) 2 N-Ph - -Ph (Me 2 N) 3 P= ( 3 ) 3 N 3 l 3 ( 3 2 ) 2 ( 3 ) 2 S 3 Ph Ph- 3 I 3 ( 3 2 ) 3 N l 2 2 l l 2 2 Me ( 3 ) 2 2 Ph N 2 -SnMe 3 2 l 2 Ph SiMe 3 (Et) 3 Ac Ph 2 l 2 SiPh Me 3 Si Ac Ph 2 3 N 2 3 F ( 3 ) 2 l Me

4 Representative Proton hemical Shift Values (δ 6 to 12) l 3 N Ac Me Me 2 3 Me Ph Et 2 2 Et l 3 l Ph Me 2 N N 2 N 2 N NMe Ph Me NMe 2 N t Bu S Me Me N Se g 16.7 S ipr 3 Si Me

5 Integration of NMR Spectra - Number of Protons hem 345 NMR is unique among common spectroscopic methods in that signal intensities are directly proportional to the number of nuclei causing the signal (provided certain conditions are met). In other words, all absorption coefficients for a given nucleus are identical. This is why proton NMR spectra are routinely integrated, whereas IR and UV spectra are not. A typical integrated spectrum is shown below, together with an analysis z 26.5 mm 11.8 mm 16.2 mm If given the molecular formula ( 9 ), there are in molecule Total area: = 54.5 mm Thus 5.5 mm per 26.5 / 5.5 = 4.86 i.e / 5.5 = 2.16 i.e / 5.5 = 2.97 i.e. 3 The vertical displacement of the integral gives the relative number of protons. It is not possible to determine the absolute numbers without additional information (such as a molecular formula). Sometimes a numeric value will be given, or sometimes, as in the example above, you have to measure the distance with a ruler. In this example, if we add up all of the integrals, we get 54.5; dividing by the number of hydrogens in the molecular formula gives 5.5 mm per. We can then directly estimate the number of protons corresponding to each multiplet by rounding to the nearest integer. It is generally possible to reliably distinguish signals with intensities of 1-8, but it becomes progressively harder to make a correct assignment as the number of protons in a multiplet increases beyond 8, because of the inherent inaccuracies in the method. The two parts of aromatic proton integral at δ can be separately measured as a 2:3 ratio of ortho to meta+para protons. 5

6 oupling - Splitting of NMR Signals If two protons have different chemical shifts and are within 3 bonds of each other (geminal or vicinal) then the protons will be coupled to each other: the signal will be split into a doublet (two lines separated by the coupling constant J) due to two magnetic orientations of the other proton. When there are two, three, or more neighbors, additional splittings can be observed α β J 1 J 2 E s d t s d dd s d t dd Two equal couplings. Two different couplings. When all of the couplings to a given proton are the same, then regular multiplets are formed, with the intensities shown below: # of Vicinal atoms Intensities Pascal's triangle) alled: Examples: 0 1 singlet X- 3 X X doublet X 2-3 X 2 -Y triplet X- 2-3 X 2-2 -X quartet X- 2-3 X- 2 --X pentet X X 3-2 -X sextet X 3 -X- 2 -R heptet X-( 3 ) 2 (X- 2 ) octet -( 3 ) nonet X 2 -( 3 ) 2 triplet n = 2 quartet n = 3 pentet n = 4 sextet n = 5 heptet n = 6 owever, when some of the coupling constants are different, then more complicated multiplets are seen. The simplest type is the doublet of doublets (dd) which arises from one proton coupled to two neighboring protons by different coupling constants. 6

7 oupling onstants oupling constants J vary widely in size, but the vicinal couplings in acyclic molecules that we are mostly going to be interested in are usually 7 z. The leading superscript ( 3 J) indicates the number of bonds between the coupled nuclei. 2 J = 2-15 z Typical: -12 z Typical: 7 z 3 J = 2- z geminal vicinal long-range 4 J = 0-3 z ne situation where the size of J provides important information is in the vicinal coupling across double bonds, where trans couplings are always substantially larger than cis couplings. J = z J = 8-12 z There are also a few situations where coupling across 4 bonds are observed in NMR spectra. This is rarely seen across single bonds, but small couplings (typically 1-3 z) are seen when there are intervening double or triple bonds. Meta 4 J = 2 to 3 z Allylic 4 J = 0 to 3 z Propargylic 4 J = 2 to 4 z Allenic 4 J = 6 to 7 z 7

8 NMR Spectra with no oupling hem Mz 1 NMR Spectrum Solv: Dl 3 Me Me Dimethyl malonate Mz 1 NMR Spectrum Solv: Dl 3 Source: Aldrich Spectral Viewer/ Me Methoxyacetone

9 Absence of Splitting between Equivalent Protons hem 345 Protons that have the same chemical shift do not show spin-spin splitting. Thus the 2 groups of both 1,2-dimethoxy- and 1,2-dibromoethane are singlets, whereas those of , where there is significant chemical shift between the 2 groups, are two triplets Mz 1 NMR Spectrum in Dl 3 Source: Aldrich Spectral Viewer/ 1,2-Dibromoethane Problem R-18U Mz 1 NMR spectrum in Dl 3 1-Methoxy-2-bromoethane Mz 1 NMR Spectrum Solv: Dl 3 Source: Aldrich Spectral Viewer/ Me Me 1,2-Dimethoxyethane

10 Simple oupling Patterns hem 345 Problem R-18N 2 3 l Mz 1 NMR spectrum in Dl 3 l l l 1,1,2-Trichloroethane Problem R l Mz 1 NMR spectrum in Dl 3 l l 1,1-Dichloroethane Problem R-18G Mz 1 NMR spectrum in Dl 3 omoethane

11 Simple oupling Patterns hem 345 Problem R-18F Mz 1 NMR spectrum in Dl 3 1-omopropane Problem R-18E Mz 1 NMR spectrum in Dl 3 2-omopropane

12 Practice Problems hem 345 Problem R-18Q: Mz 1 NMR Spectrum in Dl 3 Source: Aldrich Spectral Viewer/ 30 0 z Problem R Mz 1 NMR spectrum in Dl z

13 Size of oupling onstants hem 345 Vicinal coupling across double bonds shows a strong stereochemical dependence, with cis couplings (typically z) always being less than trans couplings (typically 15 z). Problem R-18P 3 3 l Mz 1 NMR spectrum in Dl l 30 0 z J = 13.6 z l 30 0 z Problem R-18Q 3 3 l Mz 1 NMR spectrum in Dl J = 8.1 z

14 and N Protons The chemical shifts of and N protons vary over a wide range depending on details of sample concentration and substrate structure. The shifts are very strongly affected by hydrogen bonding, with strong downfield shifts of -bonded groups compared to free or N groups. Thus signals tend to move downfield at higher substrate concentration because of increased hydrogen bonding (see the spectra of ethanol below). Pure ethanol proton % Et in l 4 5% Et in l 4 0.5% Et in l There is a general tendency for the more acidic and N protons to be shifted downfield. This effect is in part a consequence of the stronger -bonding propensity of acidic protons, and in part an inherent chemical shift effect. Thus carboxylic amides and sulfonamides N protons are shifted well downfield of related amines, and groups of phenols and carboxylic acids are downfield of alcohols. hemical Shift Ranges of, N and S Protons: Except for alcohols, the shifts are for dilute solutions in Dl 3 R-S 2 N 2 R=F3 R-N 2 R= 3 Ar-S Ar-N 2 R-S R-N 2 R-S 3 Ar- R- 2 R-N 3 + concentrated R- dilute δ

15 Second rder Effects hem 345 When two sets of protons that are coupled to each other are relatively close in chemical shift (i.e. when the chemical shift between them is similar in size to the coupling between them) simple multiplets are no longer formed. A commonly observed effect is that the intensities of the lines no longer follow the simple integer ratios expected - the multiplets "lean" towards each other. In other words, the lines of the multiplet away from the chemical shift of other proton (outer lines) become smaller and lines closer (inner lines) become larger. This can be seen in the marked triplets below (see next page for a simpler example). The leaning becomes more pronounced as the chemical shift difference between the coupled multiplets becomes smaller Mz 1 NMR spectrum in Dl 3 "leaning" "leaning" In addition there may be more lines than that predicted by the multiplet rules. A nice example is provided by the compound below. For the 2 2 group the two methylenes at δ 3.48 and δ 3.81 have a relatively large chemical shift separation, and they form recognizable triplets (although with a little leaning). For the Me 2 2 group the chemical shift between the 2 groups is small, and the signals are a complicated multiplet with only a vague resemblance to a triplet Mz 1 NMR spectrum in Dl

16 The "leaning" or "roof" effect Why equivalent protons do not show coupling ν AB = 50 z A B When the two protons are well separated in chemical shift, each one is a doublet due to coupling with the neighboring proton A B ν AB = 40 z ν AB = 30 z A B As the chemical shift becomes smaller, the the two peaks closest to each other (the inner peaks) become larger, and the outer peaks become smaller A B ν AB = z A B ν AB = z ν AB = 3 z A A B B Eventually the outer peaks disappear, and the inner peaks merge in to one - one sees only a singlet. So it is not that protons with the same shift don't couple, it is that the peaks that would show us the coupling (the outer peaks) have all disappeared z 16

17 oupling to Different Protons So far, we have seen only spectra where all of the couplings to a proton are the same, so that simple multiplets like triplets, quartets, etc are formed. owever, there are many circumstances where a proton may be coupled to two protons by different coupling constants, leading not to a triplet, but to a doublet of doublets. ne common situation of this type occurs in aromatic compounds, where both ortho and meta couplings are large enough to see, but the ortho coupling (8 z) is much larger than the meta (2 z). The para coupling is usually too small to see. This is thus one of the important exceptions to the rule that protons separated by more than 3 bonds do not show coupling. J 1 = J 2 J 1 J 2 t dd Problem R-23D 7 6 N Mz 1 NMR spectrum in Dl J ortho (coupling to 3 ) 30 0 z J meta (coupling to 6 ) N ther situations where protons separated by more than 3 bonds show coupling also involve intervening π bonds (double or triple bonds). Such couplings are typically smaller than the 7 z often seen for 3-bond couplings. See if you can assign the signals in the spectrum below, and identify the couplings. Problem R-27L Mz 1 NMR spectrum in Dl 3 Source: Adam Fiedler/ z

18 Diastereotopic Effects Diastereotopic protons are defined as two protons which have identical connectivity to the rest of the molecule, but have different chemical shifts because of some stereochemical feature of the molecule. The situation is simple with gem-alkene protons - it is easy to see how they are different. owever, it is more complicated for sp 3 carbons. l These two prrotons are diastereotopic These two prrotons are diastereotopic It turns out that 2 groups in any molecule that has a true asymmetic center (a center of chirality) anywhere in the molecule will be diastereotopic (see the substitution test in the text book). An typical example is 1,2-dibromopropane (NMR below). Rotation around the 1,2-- bond does not actually interchange the environment of the two hydrogens. To convince yourself of this, make two models of 1,2-dibromopropane, and put both in the same conformation. In one mark one of the hydrogens at 1, in the other mark the other one. Then see if you put the two marked hydrogens in exactly the same environment by rotating the bonds (this is the substitution test done with models). You can see that 1,2-dibromopropane has four sets of signals, with the two protons of the 2 group separated by about 0.3. Not only are the shifts of the two -1 protons different, but the coupling constant to the -2 proton is also different. The bond coupling (to the other proton at -1) is accidentally nearly the same as the bond coupling (to the proton at -2) for one of the protons at -1. This gives the triplet at δ 3.55 Some people call these "apparent triplets" because the two couplings are certainly different, but apparently not by much. For the other proton at -1 the --- coupling is much smaller, and so a dd is seen at δ The proton at -2 is pretty complicated - it is actually a doublet of doublets of quartets (ddq) from coupling to the two different protons at -1 and the methyl group at -3. Problem R Mz 1 NMR spectrum in Dl 3 Source: ASV/ 30 0 z

19 Effect of Electron Donating and Withdrawing Substituents on NMR hemical Shifts Problem R-19 ( 6 7 N) 300 Mz 1 NMR spectrum in Dl N Mz 1 NMR spectrum in Dl Me l 300 Mz 1 NMR spectrum in Dl l N Mz 1 NMR spectrum in Dl N

20 Isomeric Methoxynitrobenzenes hem 345 Problem R-19B ( 7 7 N 3 ) 300 Mz 1 NMR spectrum in Dl

21 Three Isomeric Trichlorobenzenes hem 345 Problem R-19A Three isomers of 6 3 l Mz 1 NMR spectrum in Dl z

22 arbon-13 NMR Spectroscopy hem 345 hemical Shift Ranges: N X arboxylic esters acids, amides = Alkyl halide alkyl amine Me 4 Si Ketones, aldehydes N R 3 -- Alkanes δ hemical shifts of some simple compounds: Me 3 Me 2 2 == Me Me Dl 3 N = Me 2 NMe 4 Mel 2 =N 2 Me Me-Me 4 Si N 4 MeS Me MeLi MeI δ == 2 Me + N Me Me 3 - N 2 2 =N-Me - N + -Me = 2 (Et) 3 2 = 2 ==NMe Li N-Me δ 0 22

23 arbon-13 NMR Spectroscopy hem 345 Shown below are the 13 NMR spectra of three isomers of 6 : Determine the expected numbers of carbons for each isomer. Determine which spectrum corresponds to which compound Identify the types of carbon signals, do a rough assignment Dl SiMe

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