# Why does the weaker C–H bond have a higher wavenumber than the C=O bond?

My understanding is that a stronger bond has a higher wavenumber in IR spectrum. But why does the C–H vibration have a higher wavenumber than the C=O vibration? The latter is a double bond, so I think it should be a stronger bond than the C–H single bond.

A property of the harmonic oscillator is that the oscillation frequency, $$\omega$$, is dependent not only on $$k$$ (the spring constant) but also on the mass $$m$$ of the object:

$$\omega = \sqrt{\frac{k}{m}}$$

We can crudely model a chemical bond as a two-body harmonic oscillator, which largely obeys the same rule, except that the mass must be replaced with the reduced mass $$\mu$$, defined by:

$$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}$$

where $$m_1$$ and $$m_2$$ are the masses of the two bodies. So, for a C–H bond, we have

$$\frac{1}{\mu_\ce{CH}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{1 u}} \implies \mu_\ce{CH} = \pu{0.923 u}$$

and for a C=O bond, we have

$$\frac{1}{\mu_\ce{CO}} = \frac{1}{\pu{12 u}} + \frac{1}{\pu{16 u}} \implies \mu_\ce{CO} = \pu{6.86 u}$$

From J. Chem. Phys. 1946, 14 (5), 305–320 the force constant $$k$$ for the C=O bond in formaldehyde is approximately $$2.46$$ times that for the C–H bond in methane (the units are in the old cgs system, so I am lazy to quote the actual values).

So, we can come up with a very rough theoretical estimate. The wavenumber used in IR spectroscopy, denoted by $$\bar{\nu}$$, is directly proportional to the (angular) frequency $$\omega$$.

\begin{align} \frac{\bar{\nu}_\ce{CH}}{\bar{\nu}_\ce{CO}} &= \frac{\omega_\ce{CH} / 2\pi c}{\omega_\ce{CO} / 2\pi c } \\ &= \frac{\omega_\ce{CH}}{\omega_\ce{CO}} \\ &= \sqrt{\frac{k_\ce{CH}}{k_\ce{CO}} \cdot \frac{\mu_\ce{CO}}{\mu_\ce{CH}}} \\ &= \sqrt{\frac{1}{2.46} \cdot \frac{\pu{6.86 u}}{\pu{0.923 u}}} \\ &= 1.74 \end{align}

This is not too far from the experimental value of $$\pu{2900 cm-1}/\pu{1730 cm-1} = 1.68$$ (using values in the middle of the typical ranges for alkanes and aldehydes).

According to Skoog, Analytical Chemistry:

Using classical mechanics, assuming a diatomic molecule, the frequency of vibration $$\nu$$ may be described by $$\nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}$$

where $$k$$ represents the force constant of this bond, and $$\mu$$ the reduced mass of the particles bond together, defined as

$$\mu = \frac{m_1 \cdot m_2} {m_1 + m_2}$$

An equivalent quantum chemical / mechanial description includes $$c$$, the speed of light, yielding a result expressing the radiation in wavenumbers: $$\bar{\nu} = \frac{1}{2\pi \cdot c} \sqrt{\frac{k}{\mu}}$$

So (1), with the same reduced mass $$\mu$$, the observed frequency increases with increasing force constant $$k$$. And (2) $$\nu$$ equally increases while lowering the reduced mass $$\mu$$.

Skoog further mentions as typical force constant of single bonds the range of $$3 \times 10^2 \ldots 8 \times 10^2\,\pu{N m}^{-1}$$ with an average of $$5 \times 10^2\,\pu{N m}^{-1}$$; while stating $$1 \times 10^3\,\pu{N m}^{-1}$$ and $$1.5 \times 10^3\,\pu{N m}^{-1}$$ for typical double, and triple bonds, respecitively.

Consequently, for a carbonyl $$\ce{C=O}$$ double bond, assuming as the mass $$m_1$$ of a carbon atom of

$$m_1 = \frac{12 \times 10^{-3}\,\pu{kg / mol} } {6.0 \times 10^{23}\,\pu{atoms / mol}} \times 1\,\pu{atom} = 2.0 \times 10^{-26}\,\pu{kg}$$

and for oxygen a mass $$m_2 = 2.7 \times 10^{-26}\,\pu{kg / atom}$$, yields a reduced mass $$\mu = 1.1 \times 10^{-26}\,\pu{kg}$$. Using the above mentioned equation determines as vibration frequency (in wavenumbers) a value of

$$\bar{\nu} = 5.3 \times 10^{-12}\,\pu{s \cdot cm}^{-1} \sqrt{\frac{1 \times 10^3\,\pu{N \cdot m^{-1}}}{1.1 \times 10^{-26}\,\pu{kg}}} = 1.6 \times 10^3\pu{cm}^{-1}$$.