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Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not? C(CH3)3 H2 C C(CH3)3 H3C H CH3 Seem to be equivalent until we look at most stable conformation, the most utilized conformation. H3C H Are these two hydrogens truly equivalent? H H CH3 C(CH3)3 H H Seemingly equivalent hydrogens may be homotopic, enantiotopic, diastereotopic. H3C H 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. CH3 Homotopic H Achiral H replace one H with D H D Achiral The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions. Enantiotopic H H replace one H with D Achiral H D Chiral, have two enantiomers. 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 R hydrogen H D This structure would be S Pro S hydrogen. Diastereotopic H D H H D H H3C H3C H3C replace H with D Cl Cl H H Cl H produced diastereomers 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 R hydrogen H D H3C Cl H This structure would be S Pro S hydrogen. (Making this a D causes the structure to be S.) Example of diastereotopic methyl groups. H3C c a H d OH H3C a' a and a’ H CH3 b H Diastereotopic methyl groups (not equivalent), each split into a doublet by Hc 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. 13C NMR spectrum 4 peaks 4 types of carbons. 13C chemical shift table Hydrogen NMR: Analysis: Example 1 Fragments: (CH3)3C-, -CH2-, CH3- -(C=O)- O 1. 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 nonequivalent 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. 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. Possible structures OH O HO OCH3 O O 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 Possible structures OH O HO OCH3 O O 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. Now look at chemical shifts 1. Formula told us that there are two pi bonds/rings in the compound. 2. From chemical shift conclude geminal CH2=CR2. Thus one pi/ring left. 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? X Do the R groups have allylic hydrogens, C=C-CH? 1. Four allylic hydrogens. Unsplit. Equivalent! 2. Conclude CH2=C(CH2-)2 3. Subtract known structure from formula of unknown… C7H12 - CH2=C(CH2-)2 ------------------------------------------ C3H6 left to identify Remaining hydrogens produced the 6H singlet. Likely structure of this fragment is –C(CH3)2-. But note text book identified the compound as Infrared Spectroscopy Chapter 12 Energy Table 12.1, p.472 Final Exam Schedule, Thursday, May 22, 10:30 AM Fang, MD10A Kunjappu, MD10B Kunjappu, MD10C Metlitsky, MD10D 1127N Zamadar 2143N 320A Infrared spectroscopy causes molecules to vibrate 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… H H F H H H F H ethylene 1,1 difluoro ethylene What about 1,2 difluoro ethylene? Different bonds have different resistances to stretching, different frequencies of vibration Table 12.4, p.478 Typical Infra-red spectrum. wavelength Frequency, measured in “reciprocal centimeters”, the number of waves in 1 cm distance. Energy. Figure 12.2, p.475 C-H C=O “fingerprint region”, complex vibrations of the entire molecule. Vibrations characteristic of individual groups. Figure 12.2, p.475 BDE of C-H 414 464 556 472 Table 12.5, p.480 BDE and CC stretch 376 727 966 Table 12.5, p.480 Alkane bands Figure 12.4, p.480 Recognition of Groups: Alkenes (cyclohexene). Compare these two C-H stretches Sometimes weak if symmetric Recognition of Groups: Alkynes (oct-1-yne) This is a terminal alkyne and we expect to see 1. Alkyne C-H 2. Alkyne triple bond stretch (asymmetric) Recognition of Groups: Arenes. (methylbenzene, toluene) Out-of-plane bend; strong 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 Recognition of Groups: Alcohols Recognition of Groups: Ethers No O-H bond stretch present but have C-O in same area as for alcohol. C-O stretch in assymetric ethers sp3 CH3 O sp2 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 Esters One C=O stretch and two C-O stretches. 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 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