Spin-Spin Splitting in Alkenes
Transcription
Spin-Spin Splitting in Alkenes
Spin-Spin Splitting in Alkenes Alkenes are not free to rotate around the C-C bonds; the two protons attached to the same carbon may or may not be equivalent Can have two-bond as well as three-bond couplings in alkenes Three-bond couplings are referred to as vicinal couplings Typical values are 6-15 Hz for cis Typical values are 11-18 Hz for trans Two-bond couplings are referred to as geminal couplings Typical values are 0-1 Hz In order to split one another, the protons must be nonequivalent In cis-difluoroethylene both protons are equivalent and do not split on another In trans-difluoroethylene both protons are equivalent and do not split one another In monofluoroethylene the three protons are not equivalent and will split one another Trans-1,2Cis-1,2difuoroethylene difuoroethylene Both protons equivalent Both protons equivalent Monofluoroethylene Protons Nonequivalent Example: Vinyl Acetate (from text) Each of the three protons (a, b, and c) is split by the others: Proton a is split by proton c (cis proton) into a doublet and again by proton b (geminal proton – small splitting) into a doublet of doublets Proton b is split by proton by proton c (trans proton) into a doublet and again by proton a (geminal proton – small splitting) into a doublet of doublets Proton c is split by proton by proton b (trans proton) into a doublet and again by proton a (cis proton) into a doublet of doublets; both splittings relatively large The assignments are based on the appearance of the multiplets Analysis of multiplet a Small splitting is the interaction with proton b (geminal splitting) 1347.76 – 1373.29 = 1.47 Hz 1368.51 – 1367.04 = 1.47 HZ Large splitting is the interaction with proton c (cis splitting) This splitting is the difference between the centers of the two pairs of peaks (1347.76 + 1373.29) / 2 = 1374.03 (1468.51 + 1367.04) / 2 = 1367.78 1374.03 – 1367.78 = 6.25 Hz Analysis of multiplet b Small splitting is the interaction with proton a (geminal splitting) 1472.57 – 1471.01 = 1.48 Hz 1458.59 – 1456.75 = 1.84 Hz (should be 1.48) Large splitting is the interaction with proton c (trans splitting) This splitting is the difference between the centers of the two pairs of peaks (1472.57 + 1471.09) / 2 = 1471.83 (1458.59 + 1456.72) / 2 = 1457.67 1471.83 – 1457.67 = 14.16 Hz Analysis of multiplet c In this case both splittings are relatively large; only vicinal couplings are involved The smaller of the two is due to coupling between proton c and proton a (a cis splitting) The larger of the two is due to coupling between proton c and proton b (a trans splitting) Again, the splitting is the difference between the centers of the peaks (2192.85 + 2186.60) / 2 = 2189.73 (2178.88 + 2172.63) /2 = 2175.76 2189.73 – 2175.76 = 13.97 Hz Note that if two or more protons are coupled, then the coupling constant between protons a and b is the same as the coupling constant between protons b and a. For protons a and c: In multiplet a the coupling constant between protons a and c was measured to be 6.25 Hz In multiplet c the coupling constant between protons c and a was measured to be 6.25 Hz For protons a and b: In multiplet a the coupling constant between protons a and b was measured to be 1.47 Hz In multiplet b the coupling constant between protons b and a was measured to be 1.48 Hz For protons b and c: In multiplet b the coupling constant between protons b and c was measured to be 14.16 Hz In multiplet c the coupling constant between protons c and b was measured to be 13.97 Hz (well within margin of error) Can say that two protons are coupled if they have the same coupling constant; can be useful in assigning peaks. Another Alkene Example: Crotonic Acid Multiplet a is a pair of doublets and must arise as a result of one large and one small splitting. This can be assigned to the methyl group. Multiplets b and c arise as a result of two splittings; a splitting by a single proton into a doublet and a second splitting into quartets by three equivalent protons. The split into a doublet will be the same for both protons, but for one the coupling constant with the methyl group is much stronger than the other. The proton cis to the methyl group will experience a much larger coupling constant with the methyl group; multiplet c is assigned to this proton on this basis The proton trans to the methyl group will experience a much smaller coupling constant with the methyl group; this is a long-range interaction. Analysis of Multiplet a (methyl group) The methyl protons are split into a doublet of doublets by the two single protons The large splitting is due to interaction with the proton cis to the methyl group (a normal three-bond coupling) This coupling constant is calculated to be 6.07 Hz The small splitting is due to interaction with the proton trans to the methyl group (a long-range coupling, will be small) Two values calculated from spectra – 1.46 Hz and 1.80 Hz Ideally these should be the same Analysis of Multiplet b Multiplet b arises from the hydrogen atom trans to the methyl group The signal from proton b is split into a doublet by proton c (trans proton – large coupling constant) Each doublet is split into quartet by the methyl group (proton a) (long range coupling – small coupling constant) The small coupling constant is 1.64 Hz The large coupling constant is 15.51 Hz Analysis of multiplet c Multiplet c arises from the proton cis to the methyl group The signal from proton c is split into a doublet by proton b (trans proton –large coupling constant) Each doublet is split into a quartet by the methyl proton This coupling is much larger than in the preceding case (6.89 Hz versus 1.64 Hz) As a result there is partial overlap of these multiplets The trans splitting is calculated by taking the difference between the centers of the quartets Works out to be 14.91 Hz Close to 15.51 Hz – well within a margin of error Spin-Spin Splitting in Aromatics Benzene – all protons equivalent, give rise to a singlet Monosubstituted benzene – protons divided into three sets those ortho to the substituent group those para to the substituent group those meta to the substituent group Typical values for coupling constants are: 7-10 Hz for ortho protons 2-3 Hz for meta protons (may not be observed) 0-1 Hz for para protons (may not be observed) Often do not get the predicted number of peaks for symmetrically substituted benzene Toluene (methyl benzene) gives a single line for all the aromatic protons on a low-field instrument Can also get more than the predicted number of peaks due to a breakdown of first-order splitting Only asymmetrically substituted compounds follow expected splitting patterns Aromatic Example 1: 3-nitrobenzaldehyde 300-MHz spectrum of 3-nitrobenzaldehyde In this case the ortho couplings appear to be the only ones observed Ha experiences only meta and para couplings; appears as a singlet Hb and Hd experience one ortho coupling and appear as doublets Not possible to assign based on the above spectrum alone Some of the meta couplings show up if spectrum is expanded Hc experiences two ortho couplings and appears as a triplet The aldehyde proton appears downfield at 10.147 ppm Aromatic Example 2: 1,10-phenanthroline 360-MHz Spectrum of 1,10-Phenanthroline Appears that only ortho couplings are observed Ha and Hc should appear as doublets; the signal at 9.18 ppm is assigned to Ha because it is closer to the electronegative nitrogen atoms and should be shifted furthest downfield The signal at 7.62 ppm is assigned to Hb since this is spit by both Ha and Hc; should appear as triplet or doublet of doublets The signal at 7.76 ppm is assigned to the Hd protons since this signal appears as a singlet; no other protons in ortho positions Aromatic Example 3: 2-nitrophenol (from text) Both ortho and meta couplings are present in this case The b and d protons experience two couplings, one large and one small, and appear as a doublet of doublets The large coupling is an ortho coupling The small coupling is a meta coupling The a and c protons experience two large couplings (causing a split into doublet of doublets, which appear as a triplet) and one small coupling (causing each component of the triplet to be split) The large coupling is an ortho coupling The small coupling is a meta coupling