Example Expect the signal for this hydrogen to be split into seven by the six equivalent neighbors. Expect the peak for the methyl hydrogens to be split into two peaks by the single neighbor. Overall: small peak split into seven (downfield due to the Cl). larger peak (six times larger) split into two (further upfield). 1

Attempt to anticipate the splitting patterns in each molecule.

p. 491

Spin-spin splitting. Coupling constant, J. The actual distance, J, between the peaks is the same within the quartet and the doublet. Figure 13.12 1H-NMR spectrum of 1,1-dichloroethane. Split into a group of 4 Split into a group of 2 4

In preparation for discussion of origin of Spin-Spin recall earlier slide More Shielding due to electrons at nucleus being excited. Due to shielding, less of the magnetic field experienced by nucleus, Lower energy needed to excite. Peaks on right are “upfield”. Reduced shielding, more of the magnetic field experienced, higher energy of excitation. Peaks are “downfield”. 5

Origin of spin-spin splitting In the presence of a external magnetic field each nuclear spin must be aligned with or against the external field. Approximately 50% aligned each way. Non-equivalent hydrogen nuclei separated by two or three bonds can “spin – spin split” each other. What does that mean? Consider excitation of a hydrogen H1. Energy separation of ground and excited states depends on total magnetic field experienced by H1. Now consider a neighbor hydrogen H2 (passive, not being excited) which can increase or decrease the magnetic field experienced by H1. About 50% of the neighboring hydrogens will augment the applied magnetic field and about 50% will decrement it. Get two peaks, a double The original single peak of H1 has been split into two peaks by the effect of the neighbor H2. The energy difference is J Energy H1, being excited Here H2 augments external field, peak moved downfield. Here H2 decreases external field, peak moved upfield. 6

Coupling constant, J, in Hz 3-pentanone Same as gap here. Coupling constant, J, in Hz 3-pentanone Figure 13.14 The quartet-triplet 1H-NMR signals of 3-pentanone showing the original trace and a scale expansion to show the signal splitting pattern more clearly. The left side of molecule unaffected by right side. Peak identification… Figure 13.14, p.511 7

Magnitude of Coupling Constant, J The magnitude of the coupling constant, J, can vary from 0 to about 20 Hz. This represents an energy gap (E = hn) due to the interaction of the nuclei within the molecule. It does not depend on the strength of the external field. J is related to the dihedral angle between bonds. J largest for 0 (eclipsed) or 180 (anti), smallest for 90, intermediate for gauche. 8

gauche anti vinyl systems Table 13.4, p.511 9

Hb is augmenting external field causing a larger energy gap. Spin-Spin Splitting Now look at some simple examples. Examine the size of the peaks in the splitting. Hb is augmenting external field causing a larger energy gap. Hb decrementing external field causing a smaller energy gap. Ha is being excited. Hb is causing spin-spin splitting by slightly increasing or decreasing the magnetic field experienced by Ha. Figure 13.15 The origins of signal splitting patterns. Each arrow represents an Hb nuclear spin orientation. 10

One neighbor assists, one hinders. No effect. Two neighboring atoms assist external field. More energy needed to excite. Peak is “downfield”. Again Ha is flipping, resonating. The two Hb are causing spin-spin splitting by slightly changing the magnetic field experienced by Ha. One neighbor assists, one hinders. No effect. Both neighbors oppose. Less energy needed to excite, “upfield”. Figure 13.15 The origins of signal splitting patterns. Each arrow represents an Hb nuclear spin orientation. Recall that for the two Hb atoms the two states (helping and hindering the external field) are almost equally likely. This give us the 1 : 2 : 1 ratio. Figure 13.15b, p.512 11

Three neighboring Hb’s causing splitting when Ha is excited. All Hb augment Two augment, one decrement. One augment, two decrement. All decrement. Ha being excited. Three equivalent Hb causing spin spin splitting. Figure 13.15 The origins of signal splitting patterns. Each arrow represents an Hb nuclear spin orientation. Figure 13.15c, p.512 12

Naturally if there are two non-equivalent nuclei they split each other. Figure 13.17 Vicinal coupling between two nonequivalent H atoms. Figure 13.17, p.513 14

Technique: use a tree diagram and consider splittings sequentially. Three nonequivalent nuclei. Ha and Hb split each other. Also Hb and Hc split each other. Technique: use a tree diagram and consider splittings sequentially. Figure 13.19 Coupling that arises when Hb is split by two different nonequivalent H atoms Ha and Hc. This analysis assumes no other coupling in the molecule and that Jab _ Jbc. Figure 13.19, p.513 15

More complicated system Figure 13.20 Complex coupling that arises when Hb is split by Ha and two equivalent atoms Hc. Again, this analysis assumes no other coupling in the molecule and that Jab _ Jbc. Figure 13.20, p.514 16

Return to Vinyl Systems Not equivalent (R1 is not same as R2) because there is no rotation about the C=C bond. Figure 13.21 Geminal coupling that occurs when two H atoms on the same carbon atom are not equivalent. This is most common in unsymmetrical alkenes and cyclic molecules. Figure 13.21, p.514 17

Example of alkenyl system Figure 13.22 300 MHz 1H-NMR spectrum of ethyl propenoate. We will perform analysis of the vinyl system and ignore the ethyl group. Three different kinds of H in the vinyl group. We can anticipate the magnitude of the coupling constants. 18

Each of these patterns is different from the others. Analysis Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE Figure 13.23 Tree diagrams for the complex coupling seen for the alkenyl H atoms in the 1H-NMR spectrum of ethyl propenoate. Now examine the left most signal…. 19

Ha being excited. Both Hb and Hc are coupled and causing splitting. Hb causes splitting into two peaks (big splitting, JAB) Hc causes further splitting into a total of four peaks (smallest splitting, JAC) JAB = 11-18 Hz JAC = 0 - 5 JBC = 5 - 10 Figure 13.23 Tree diagrams for the complex coupling seen for the alkenyl H atoms in the 1H-NMR spectrum of ethyl propenoate. 20

Each of these patterns is different from the others. Analysis in greater depth based on knowing the relative magnitude of the splitting constants. Aim is to associate each signal with a particular vinyl hydrogen. Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE Figure 13.23 Tree diagrams for the complex coupling seen for the alkenyl H atoms in the 1H-NMR spectrum of ethyl propenoate. Look at it this way... This signal appears to have big (caused by trans H-C=C-H) and small (caused by geminal HHC=) splittings. The H being excited must have both a trans and geminal H. The H must be Ha. 21

Analysis in greater depth - 2. Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE Figure 13.23 Tree diagrams for the complex coupling seen for the alkenyl H atoms in the 1H-NMR spectrum of ethyl propenoate. And the middle signal. This signal appears to have big (caused by trans H-C=C-H) and middle (cis H-C=C-H) splittings. The H being excited must have both a trans and cis H. The H must be Hb. 22

Analysis in greater depth - 3. Each of these patterns is different from the others. JAB = 11-18 Hz, BIG JAC = 0 – 5 Hz, SMALL JBC = 5 – 10 Hz, MIDDLE Figure 13.23 Tree diagrams for the complex coupling seen for the alkenyl H atoms in the 1H-NMR spectrum of ethyl propenoate. And the right signal. This signal appears to have small (caused by geminal HHC=) and middle (cis H-C=C-H) splittings. The H being excited must have both a geminal and cis H. The H must be Hc. 23

As with pi bonds, cyclic structures also prevent rotation about bonds Approximately the same vinyl system as before. Non equivalent geminal hydrogens. Analyze this. No spin spin splitting of these hydrogens. Nothing close enough 24

Note the “roof effect”. For similar hydrogens the inner peaks can be larger. Figure 13.25 Tree diagrams that indicate the complex coupling seen in 1H-NMR signals for the vinyl group and the oxirane ring H atoms of 2-methy-2-vinyloxirane. Figure 13.25, p.516 25

Ha will be a triplet (two Hb); Likewise for Hc. We analyze Hb. Coincidental Overlap: Non-equivalent nuclei have same coupling constant. Ha will be a triplet (two Hb); Likewise for Hc. We analyze Hb. A triplet of triplets Here Ha and Hc have same coupling with Hb (Jab = Jbc), ,, coincidental overlap: splits to 5, four equivalent neighbors. 26

Analyze what happens as Jab becomes equal to Jbc. First get peak heights when Jab does not equal Jbc. Recall heights in a triplet are 1 : 2 : 1 2 1 1 2 1 1 First split the Hb by Ha in ratio of 1:2:1. 2 x 2 Each component is split by Hc in ratio of 1:2:1. Result for each final peak is product of probabilities 1 x 1 1 x 2 2 x 1 27

Peak heights shown when Jab does not equal Jbc. Examine middle peak. Let Jbc become larger until it equals Jab and add overlapping peaks together. 1 2 1 2 4 2 1 2 1 1+4+1 Peak heights shown when Jab does not equal Jbc. 28

Now adjacent peak. 1 2 1 2 4 2 1 2 1 2+2 1 4 6 4 1 29

Fast Exchange Expect coupling between these hydrogens. Three bond separation. There is no coupling observed especially in acid or base. Reason: exchange of weakly acidic hydrogen with solvent. The spectrometer sees an “averaged hydrogen”. No coupling and broad peak. ethanol 30

Return to Question of Equivalent hydrogens Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not? Seem to be equivalent until we look at most stable conformation, the most utilized conformation. Are these two hydrogens truly equivalent? Seemingly equivalent hydrogens may be homotopic, enantiotopic, diastereotopic. How to tell: replace one of the hydrogens with a D. If produce an achiral molecule then hydrogens are homotopic, if enantiomers then hydrogens are enantiotopic, if diastereomers then diastereotopic. We look at each of these cases. 31

Homotopic The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions. 32

Enantiotopic The hydrogens are enantiotopic and equivalent in the NMR unless the molecule is placed in a chiral environment such as a chiral solvent.. The hydrogens are designated as Pro R or Pro S Pro S hydrogen. Pro R hydrogen This structure would be S 33

Diastereotopic If diastereormers are produced from the substitution then the hydrogens are not equivalent in the NMR. Diastereotopic hydrogens. The hydrogens are designated as Pro R or Pro S Pro S hydrogen. (Making this a D causes the structure to be S.) Pro R hydrogen This structure would be S 34

Example of diastereotopic methyl groups. a and a’ Diastereotopic methyl groups (not equivalent), each split into a doublet by Hc 35

13C NMR 13C has spin states similar to H. Natural occurrence is 1.1% making 13C-13C spin spin splitting very rare. H atoms can spin-spin split a 13C peak. (13CH4 would yield a quintet). This would yield complicated spectra. H splitting eliminated by irradiating with an additional frequency chosen to rapidly flip (decouple) the H’s averaging their magnetic field to zero. A decoupled spectrum consists of a single peak for each kind of carbon present. The magnitude of the peak is not important. 36

13C NMR spectrum 4 peaks  4 types of carbons. 37

13C chemical shift table 38

Hydrogen NMR: Analysis: Example 1 Fragments: (CH3)3C-, -CH2-, CH3- -(C=O)- Molecular formula given. Conclude: One pi bond or ring. 2. Number of hydrogens given for each peak, integration curve not needed. Verify that they add to 14! 3. Three kinds of hydrogens. No spin-spin splitting. Conclude: Do not have non-equivalent H on adjacent carbons. 4. The 9 equivalent hydrogens likely to be tert butyl group (no spin-spin splitting). The 3 equivalent hydrogens likely to be methyl group. The two hydrogens a CH2. 5. Have accounted for all atoms but one C and one O. Conclude: Carbonyl group! 6. Absence of splitting between CH2 and CH3. Conclude: they are not adjacent. 39

Example 2, C3H6O 1. Molecular formula  One pi bond or ring 2. Four different kinds of hydrogen: 1,1,1,3 (probably have a methyl group). 3. Components of the 1H signals are about equal height, not triplets or quartets 4. Consider possible structures. 40

Possible structures 41

Chemical shift table… Observed peaks were 2.5 – 3.1 Figure 13.8 Average values of chemical shifts of representative types of hydrogens. These values are approximate. Other atoms in the molecules may cause signals to appear outside of these ranges. ethers Observed peaks were 2.5 – 3.1. Ether! vinylic Figure 13.8, p.505 42

Possible structures 43

What can we tell by preliminary inspection…. NMR example What can we tell by preliminary inspection…. Formula tells us two pi bonds/rings Three kinds of hydrogens with no spin/spin splitting. 44

X Now look at chemical shifts 2. From chemical shift conclude geminal CH2=CR2. Thus one pi/ring left. 1. Formula told us that there are two pi bonds/rings in the compound. X 3. Conclude there are no single C=CH- vinyl hydrogens. Have CH2=C-R2. This rules out a second pi bond as it would have to be fully substituted, CH2=C(CH3)C(CH3)=C(CH3)2 , to avoid additional vinyl hydrogens which is C8H14. In CH2=CR2 are there allylic hydrogens: CH2=C(CH2-)2? 45

Do the R groups have allylic hydrogens, C=C-CH? Four allylic hydrogens. Unsplit. Equivalent! Conclude CH2=C(CH2-)2 Subtract known structure from formula of unknown… C7H12 - CH2=C(CH2-)2 ------------------------------------------ C3H6 left to identify But note text book identified the compound as Remaining hydrogens produced the 6H singlet. Likely structure of this fragment is –C(CH3)2-. 46