Class Summary and Notes from 11/16/2000 - CHEM3312

Topic: Amines (Spectroscopy)

Related Reading

Related Problems

Summary

• Mass Spectroscopy
  1. The Nitrogen Rule: Compounds with an odd number of nitrogen atoms will have an odd molecular weight, compounds with an even number of nitrogen atoms will have an even molecular weight
  2. Most common cleavage: a-cleavage
• Infrared Spectroscopy
  1. N-H bonds vibrate at ~3300-3500 cm^-1. Note that this is similar to the position of an O-H stretching vibration. The shapes of the peaks, however, are distinctly different. Use the link from the class home page to the AIRS (Animated Infra-Red Spectroscopy) website if you want to see examples of IR spectra for amines and alcohols.
  2. N⁺-H bonds appear at ~2200 cm^-1. This is often useful for identifying tertiary amines (which will not have the N-H peak mentioned previously). A small amount of acid can be added prior to recorded the IR spectrum. The appearance of a peak in this range only when acid is added indicates that an amine is present.
• Nuclear Magnetic Resonance Spectroscopy
  1. N-H signals are often broad with no coupling to neighbors
  2. N-H + D₂O gives N-D + HOD
     The N-H signal will therefore not appear when D₂O is used as the solvent
  3. H-C-N signals are deshielded (~2.5 ppm) due to the electronegative nitrogen
  4. Carbons near N are deshielded (in the ¹³C NMR) by about 20 ppm

Problems Discussed In Class

Results of Class Discussion

• The IR peak in compound A indicates the presence of a carbonyl group
• The reaction involving a carbonyl, an amine, and a reducing agent is called reductive amination. The net result is replacing the C=O with CH-NR₂ (R in this case is H).
• The IR for compound B has a peak at 3200 cm^-1 that represents the N-H group
• The IR for compound B also has a peak at 1900 cm^-1 that represents the C-H bonds in the molecule
• The proton NMR for compound B shows peaks with relative areas of 9:3:1 (and since N-H peaks are often broad, none of these peaks are likely to be for the N-H)
• The relative areas give us the relative number of hydrogens contributing to the peaks. So the peak at 0.9 is likely to be made by 3 CH₃ groups, the peak at 1.0 is likely to be from a CH₃ group, and the peak at 2.6 is likely to be from a CH group
• The splitting of the peaks in the NMR tells us about neighboring hydrogens. The peak at 0.9 is a singlet, and so the three methyl groups do not have any neighboring hydrogens. The peak at 1.0 is a doublet, so the CH₃ making this peak must be next to a CH. The peak at 2.6 is a quartet, so the CH making this peak must be next to a CH₃.

Special Notes

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