# Chemistry - Why does the C–C bond have extremely weak absorptions?

## Solution 1:

For vibrational spectra the primary transition under investigation is the $v = 1 \leftarrow 0$ excitation (because $\hbar\omega >> k_\mathrm{B}T$, so excited states have negligible thermal population). The intensity of the transition depends on the transition dipole moment:

$$R_{10} = \langle 1 | \hat{\mu} | 0 \rangle$$

in that the intensity $I \propto |R_{10}|^2$. Taking a Taylor expansion of the operator $\hat{\mu}$ with respect to $q$, the vibrational coordinate, at $q = 0$, we have:

$$\hat{\mu} = \mu_{q = 0} + \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} + \cdots$$

and plugging this in we get

\begin{align} R_{10} &= \langle 1 | \mu_{q = 0} | 0\rangle + \left< 1 \middle| \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} \middle| 0 \right> + \cdots \\ &= \mu_{q=0}\langle 1 | 0\rangle + \left( \frac{\mathrm d \mu}{\mathrm d q}\right)_{q=0} \langle 1 | \hat{q} | 0 \rangle + \cdots \end{align}

where we took out the values of $\mu$ and $\mathrm d\mu/\mathrm d q$ evaluated at $q = 0$ because those are just ordinary numbers. The first term in the Taylor expansion vanishes because $|1\rangle$ and $|0\rangle$ are orthogonal (a key property of the harmonic oscillator, which is a good model for the low-lying vibrational states at the bottom of the well), and if we neglect the second- and higher-order terms, we have†

$$R_{10} \propto \left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0}$$

The RHS represents how much the dipole moment $\mu$ changes upon vibration of the bond (which is measured by $q$). As such:

• for something like a C–C bond, where vibration leads to nearly no change in the dipole moment, this quantity is small, the transition dipole moment is small, and the peak is weak.
• On the other hand, the stretching of the C=O unit leads to a large change in dipole moment and the corresponding peak is therefore stronger.
• The C=C stretch of alkenes tends to be weak as well, unless it is conjugated to an electron-withdrawing group like a carbonyl group. This can also clearly be linked to the magnitude of $(\mathrm d \mu/\mathrm dq)_{q=0}$.
• And incidentally, this is also why molecules like $\ce{H2}$ or $\ce{O2}$ cannot be studied by IR spectroscopy (no matter how much you vibrate, there is no dipole moment), and also why the symmetric stretch of $\ce{CO2}$ is not IR active. In all of these cases $(\mathrm d \mu/\mathrm dq)_{q=0}$ is identically zero.

Unfortunately most of what you've written is kind of off the mark.

1. It is indeed "reluctant" to absorb the photon, but that is not because of the stiffness of the bond. If the bond is stiffer ($k$ is larger), then it simply needs a photon of shorter wavelength (as the vibrational frequency $\omega = \sqrt{k/\mu}$ will be larger). Instead it is related to the poor interaction of the C–C vibrational mode with the oscillating electric field of the photon. This is going into stuff that I don't know about, so can't explain much more (search "Fermi golden rule" in Atkins MQM or some other quantum book).

2. $D_\mathrm{e} = kr^2/2$ is not a valid equality here as it simply makes no physical sense. If the energy of the oscillator is at $D_\mathrm{e}$, then physically speaking, the corresponding bond length $r$ should be infinity, as the bond is dissociated. Furthermore, the equilibrium bond length which you used in the calculation ($r_\mathrm{e} = \pu{154 pm}$) refers to the middle of the potential well. If you really wanted to model the C–C bond as a classical oscillator, then your equation should read $$U = \frac{1}{2}k(r - r_\mathrm{e})^2$$ because the thing that gets squared is the displacement from equilibrium.

3. In that calculation you used the dissociation energy for one mole of carbon–carbon bonds, and that doesn't make sense if you're trying to calculate the force constant of one carbon–carbon bond.

4. Even changing no. 3 wouldn't make any difference, because the harmonic oscillator model simply does not work near dissociation (in fact if a bond were to be a harmonic oscillator, it would never dissociate). Therefore bond dissociation energies cannot be treated within a harmonic model; you have to use something else such as a Morse potential.

5. A typical value of a C–C force constant is on the order of $\pu{400 N m-1}$. A better way of calculating the force constant is with $$\bar{\nu} = \frac{\omega}{2\pi c} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \implies k = \mu(2\pi c\bar{\nu})^2,$$ where $\bar{\nu}$ is the IR wavenumber and $\mu$ is the reduced mass of the oscillator. A very rough calculation is as follows. Plugging in $\mu = \pu{10^-26 kg}$ (for a $\ce{C-C}$ bond), $\bar{\nu} = \pu{1000 cm^-1} = \pu{10^5 m^-1}$, and $c = \pu{3 x 10^8 m s^-1}$, we have $k = \pu{355.3 N m^-1}$. Correct order of magnitude, at least.

6. It's also nearly impossible that $k_\ce{C=O} < k_\ce{C-C}$, as force constants tend to be correlated with bond order. A quick Google search shows that $k_\ce{C=O}$ is on the order of $\pu{1000-2000 N m^-1}$. (If you go searching for this, you may wish to know that $\pu{1 mdyne/Å} = \pu{100 N m^-1}$, as CGS units are widely used.)

† Of course, there's an additional point that must be made, which is that $\langle 1 | \hat{q} | 0 \rangle \neq 0$, but you'll have to take my word for it as I'm not keen on proving it here. It is most readily shown using ladder operators, and it should be covered in any QM text under the harmonic oscillator.

## Solution 2:

If we assume the simplyfied description, i.e. atoms being tethered together by springs, the spring constant allows describe where (energy wise, along the scale of wavenumbers, the abscissa in the specta plot) the absorption becomes observable. This however does not say a iota if the absorption band is strong, leading to a low remnant transmittance (the ordinate in the spectra plot).

From a practioner's perspective, C-C bonds often are referred as skeletal vibrations, because they represent vibrations of the backbone of the molecules characterized. As such, overall the span where to find these absorptions is rather wide ($\pu{1500 - 600 cm^{-1}}$), which contasts with an easy-to-spot absorption band like the one of $\ce{C#N}$ around $\pu{2260 cm^{-1}}$, regardless if you expand the spectral representation, or not. The better known example are the ones to attribute substitution patterns of arenes (source)

but as these hints are no longer taught as regularly as in the past, often need an expansion (who routinely looks at the bands at $\pu{900 cm^{-1}}$ and below as well?), and most importantly, since NMR spectroscopy provides this information more rapidly, the focus of attention while inspecting an IR spectrum often shifts to more prominent pattern.

## Solution 3:

Also in your analysis, you should consider Raman spectroscopy as a complementary method. Raman is also vibrational, however what show up weakly in the IR tends to show up in the Raman. In your example c-c bonds are very prominate in the Raman despite being weak in the IR. Because in the course of a vibration, IR spectroscopy is sensitive to the change of dipolar moments, and Raman spectroscopy is susceptible to the change of polarizabilities of the "electron cloud" involved.