We have one band in this region and it must obviously be due, in a molecule like chloroform to a C-H stretching mode. V1 is the only CH stretch so we must assign the band at 3019cm-1 to this mode. Note that the frequency is very high for a saturated molecule. The value is more appropriate to an aromatic CH. why do you think this is so? It's hard to see that the mass of the chlorines has much to do with it, so it suggests strongly that the bond is stiffer than normal [i.e. the C-H force constant is anomolously large]. I suspect the reason is the very high electronegativity of chlorine atoms. The electron withdrawing effect of each chlorine increases the strength and rigidity of the CH bond. 1200cm-1
region A band in this region could
well be due to a CH deformation. Its frequency would be a little less than half of the CH
stretch. We have one appropriate mode V2 so we assign the band at 1216cm-1
to this mode. Now let's turn to the C-Cl stretching and deformation modes. The stretches should be higher in frequency than the deformations so the two bands above 400cm-1 are the stretches and the remaining two must be the deformations. The band at 757 is weak, that at 668 strong so we assign the former to the asymmetric mode and the latter to the symmetric one i.e. 757 weak V4 668 strong V3 This leaves us with two modes to assign and two bands remain but both are pretty strong, so we could chose either way. In Spectrum II, I show the low
frequency part of the spectrum recorded at 1cm-1 resolution.
35Cl:37Cl approx. 3:1- thus the atomic weight is close to 35.5 In the umbrella mode V5,
the three chlorines move together but the frequency will be mass sensitive. A vibration of
chlorine 35 will exceed that of mass 37. The 3 chlorines are a random mix of the isotopes.
Try and work out the importance of species and hence the pattern you expect to see in the
spectrum. STOP HERE |
