explain why the C-H and C-C stretching frequency increases in the order of alkane alkene alkyne
Answers
Explanation:
Vibrational modes within a molecule can be described using the anharmonic oscillator model. This model assumes that the two masses (with known weight) are connected with a spring (with known strength). With the help of quantum mechanical calculations (Schroedinger equation) you can find the frequencies of basic stretching and bending modes:
,
masses of involved atoms
F=force constant
From this equation, one can deduce some basic trends can be deducted:
a. If the force constant F (= bond strength) increases, the stretching frequency will increase as well (in cm-1)
CC bond triple double single
(in cm-1) 2100 1650 1130
functional group alkyne alkene alkane
b. If the masses of the involved atoms increase, the peak will shift to lower wavenumbers e.g. H/D-exchange in labeling experiments although the bond strength remains the same.
C-X bond C-H C-D
(in cm-1) ~3000 ~2200
Based on the equation, one would always expect a sharp lines at a well defined wavenumber. Unfortunately, the change in vibration modes is always accompanied with change in rotational mode (Stokes and Anti-Stokes). The required energy for this process is much smaller (1-5 cm-1) and causes together with some other effects the broading of the 'lines'.
The number of basic stretching and bending modes expected for a molecule increases with the number of atoms in the molecule. For non-linear molecules 3N-6 (2N-5 bending, N-1 stretching) vibrations are observed (e.g. CH2Cl2). For linear molecules e.g. CO2 one expect to find 3N-5 (3*3-5=4) modes. If there is no symmetry in the molecule most of them will be observed the IR spectrum; the remaining modes will be observed in a Raman spectrum. The more complicated the molecule is (the more atoms it possesses and the lower the symmetry), the more peaks can be observed in the IR spectrum