Return to Question of Equivalent hydrogens . Stereotopicity – Equivalent or Not?

1. 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.

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

3. 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

4. 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

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

6. 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.

7. 13C NMR spectrum 4 peaks  4 types of carbons.

8. 13C chemical shift table

9. Hydrogen NMR: Analysis: Example 1 -(C=O)- Fragments: (CH3)3C-, -CH2-, CH3- • 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.

10. 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.

11. Possible structures

12. Chemical shift table… Observed peaks were 2.5 – 3.1 ethers Observed peaks were 2.5 – 3.1. Ether! vinylic Figure 13.8, p.505

13. Possible structures

14. 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.

15. 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?

16. 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-.

17. Infrared Spectroscopy Chapter 12

18. Energy Table 12.1, p.472

20. Final Exam Schedule, Thursday, May 22, 10:30 AM

22. Infrared spectroscopy causes molecules to vibrate

23. A non-linear molecule having n atoms may have many different vibrations. Each atom can move in three directions: 3n. Need to subtract 3 for translational motion and 3 for rotations # vibrations = 3 n – 6 (n = number of atoms in non-linear molecule) Infrared radiation does not cause all possible vibrations to vibrate. For a vibration to be caused by infrared radiation (infrared active) requires that the vibration causes a change in the dipole moment of the molecule. (Does the moving of the atoms in the vibration causes the dipole to change. Yes: should appear in spectrum. No: should not appear.) Consider C=C bond stretch… ethylene 1,1 difluoro ethylene What about 1,2 difluoro ethylene?

24. Different bonds have different resistances to stretching, different frequencies of vibration Table 12.4, p.478

25. Typical Infra-red spectrum. wavelength Frequency, measured in “reciprocal centimeters”, the number of waves in 1 cm distance. Energy. Figure 12.2, p.475

26. “fingerprint region”, complex vibrations of the entire molecule. C=O C-H Vibrations characteristic of individual groups. Figure 12.2, p.475

27. BDE of C-H 414 464 556 472 Table 12.5, p.480

28. BDE and CC stretch 376 727 966 Table 12.5, p.480

29. Alkane bands Figure 12.4, p.480

30. Recognition of Groups: Alkenes (cyclohexene). Compare these two C-H stretches Sometimes weak if symmetric

31. Recognition of Groups: Alkynes (oct-1-yne) • This is a terminal alkyne and we expect to see • Alkyne C-H • Alkyne triple bond stretch (asymmetric)

32. Recognition of Groups: Arenes. (methylbenzene, toluene) Out-of-plane bend; strong

33. Recognition of Groups: Alcohols The O-H stretch depends on whether there is hydrogen bonding present Compare –O-H vs -O-H….O Hydrogen bonding makes it easier to move the H with H bonding as it is being pulled in both directions; lower frequency

34. Recognition of Groups: Alcohols

35. Recognition of Groups: Ethers No O-H bond stretch present but have C-O in same area as for alcohol.

36. C-O stretch in assymetric ethers

37. Recognition of Groups: Amines Easiest to recognize is N-H bond stretch: 3300 – 3500 cm-1. Same area as alcohols. Note tertiary amines, NR3, do not have hydrogen bonding. Hydrogen bonding can shift to lower frequency

38. Esters One C=O stretch and two C-O stretches.

39. Recognition of Groups: Carbonyl • C=O stretch can be recognized reliably in area of 1630 – 1820 cm-1 • Aldehydes will also have C(O)-H stretch • Esters will also have C-O stretch • carboxylic acid will have O-H stretch • Amide will frequently have N-H stretch • Ketones have nothing extra

40. What to check for in an IR spectrum C-H vibrations about 3000 cm-1 can detect vinyl and terminal alkyne hydrogens. O-H vibrations about 3500 cm-1 C=O vibrations about 1630 – 1820 cm-1 C-O vibrations about 1000-1250 cm-1