Question

: If the force constant is approximately the same for C−C, C−N, and C−O bonds, how can I predict the relative positions of their stretching vibrations in an IR spectrum?

Answer 1

Stretching frequency depends on 2 factors i.e., force constant of the bond and reduced mass of the 2 atoms. Since, the force constant is kept approximately the same; we will depend on other factors to answer this question.
Formula used: We use the following formula of stretching frequency:-
$\overline{\nu }=\dfrac{1}{2\pi c}\sqrt{\dfrac{K}{\mu }}$

Complete step-by-step answer: As we know that the amount of energy required to stretch a bond depends on the strength of the bond and the masses of the bonded atoms. The stronger the bond, the greater the energy required to stretch it. Also the frequency of the vibration is inversely proportional to the mass of the atoms, so heavier atoms vibrate at lower frequencies.
The approximate vibration frequency of a bond can be calculated by the following equation derived from Hooke’s law:-
$\overline{\nu }=\dfrac{1}{2\pi c}\sqrt{\dfrac{K}{\mu }}$
where,
$\mu$ = reduced mass of the atoms
C = velocity of light ($3\times {{10}^{8}}m/s$ )
K = force constant of the bond
-The force constant (K) is a measure of the strength of a bond. Thus, the value of stretching frequency of a bond increases with increasing the bond strength. But since the force constant is kept the same according to the question, then we have to solely depend on the other factor that is reduced mass.
As we can see that there are different atoms attached to carbon. Suppose a particular atom increases the overall mass, the reduced mass increases and hence the frequency (stretching wavenumber) of vibration decreases.
Atomic mass of O = 16amu
Atomic mass of N = 14amu
Atomic mass of C = 12amu
Since the other atom is the same in C-C, C-N, C-O bond, therefore we can directly compare the atomic masses of C, N, and O.
Atomic masses of $O\rangle N\rangle C$, therefore the reduced mass of $C-O\rangle C-N\rangle C-C$.
Therefore, by this we can conclude that C-C has a large stretching frequency and then comes C-N whereas C-O has the least stretching frequency. So accordingly, C-C stretches at a larger wave number followed by C-N and then C-O.
Relative positions of C-C, C-N and C-O stretching vibrations in an IR spectrum are as follows:-
$C-C\rangle C-N\rangle C-O$

Note: -If both the factors are not kept the same, then always answer by considering the force constant (K) factor as strength of the bond plays as a dominating factor in case of stretching frequency.