19
$\begingroup$

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.

$\endgroup$
27
$\begingroup$

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).

| improve this answer | |
$\endgroup$
10
$\begingroup$

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

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

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.