Presentation on theme: "Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not? Are these two hydrogens truly equivalent? Seemingly equivalent hydrogens."— Presentation transcript:
Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not? 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. Seem to be equivalent until we look at most stable conformation, the most utilized conformation.
Homotopic The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions.
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 This structure would be S Pro S hydrogen. Pro R hydrogen
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 This structure would be S Pro S hydrogen. (Making this a D causes the structure to be S.) Pro R hydrogen
Diastereotopic methyl groups (not equivalent), each split into a doublet by H c a and a’ Example of diastereotopic methyl groups.
13 C NMR 13 C has spin states similar to H. Natural occurrence is 1.1% making 13 C- 13 C spin spin splitting very rare. H atoms can spin-spin split a 13 C peak. ( 13 CH 4 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.
13 C NMR spectrum 4 peaks 4 types of carbons.
13 C chemical shift table
Hydrogen NMR: Analysis: Example 1 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 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 CH 2. Fragments: (CH 3 ) 3 C-, -CH 2 -, CH Have accounted for all atoms but one C and one O. Conclude: Carbonyl group! -(C=O)- 6. Absence of splitting between CH 2 and CH 3. Conclude: they are not adjacent.
Example 2, C 3 H 6 O 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.
Figure 13.8, p.505 Chemical shift table… Observed peaks were 2.5 – 3.1 vinylic ethers Observed peaks were 2.5 – 3.1. Ether!
NMR example Formula tells us two pi bonds/rings Three kinds of hydrogens with no spin/spin splitting. What can we tell by preliminary inspection….
Now look at chemical shifts 2. From chemical shift conclude geminal CH 2 =CR 2. Thus one pi/ring left. 3. Conclude there are no single C=CH- vinyl hydrogens. Have CH 2 =C-R 2. This rules out a second pi bond as it would have to be fully substituted, CH 2 =C(CH 3 )C(CH 3 )=C(CH 3 ) 2, to avoid additional vinyl hydrogens which is C 8 H 14. X In CH 2 =CR 2 are there allylic hydrogens: CH 2 =C(CH 2 -) 2 ? 1. Formula told us that there are two pi bonds/rings in the compound.
Do the R groups have allylic hydrogens, C=C-CH? 1.Four allylic hydrogens. Unsplit. Equivalent! 2.Conclude CH 2 =C(CH 2 -) 2 3.Subtract known structure from formula of unknown… C 7 H 12 - CH 2 =C(CH 2 -) C 3 H 6 left to identify Remaining hydrogens produced the 6H singlet. Likely structure of this fragment is –C(CH 3 ) 2 -. But note text book identified the compound as
Infrared Spectroscopy Chapter 12
Table 12.1, p.472 Energy
Final Exam Schedule, Thursday, May 22, 10:30 AM Fang, MD10A Kunjappu, MD10B Kunjappu, MD10C 320A Metlitsky, MD10D1127N Zamadar2143N
Infrared spectroscopy causes molecules to vibrate
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.) 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) Consider C=C bond stretch… ethylene 1,1 difluoro ethylene What about 1,2 difluoro ethylene?
Table 12.4, p.478 Different bonds have different resistances to stretching, different frequencies of vibration
Figure 12.2, p.475 Frequency, measured in “reciprocal centimeters”, the number of waves in 1 cm distance. Energy. wavelength Typical Infra-red spectrum.
Figure 12.2, p.475 C=O C-H “fingerprint region”, complex vibrations of the entire molecule. Vibrations characteristic of individual groups.
Table 12.5, p.480 BDE of C-H
Table 12.5, p.480 BDE and CC stretch
Figure 12.4, p.480 Alkane bands
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
Recognition of Groups: Amines Easiest to recognize is N-H bond stretch: 3300 – 3500 cm -1. Same area as alcohols. Note tertiary amines, NR 3, 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 cm -1