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I am currently a Chemistry undergrad and I love to fact check everything my O Chem professor says against my research advisor, a physical/computational chemist. Well, my O Chem professor, when discussing IR absorptions, said something which was a simplification for teaching purposes but which seemed rather unusual to me. He said that high intensity peaks are due to a large change in dipole (clealy true) which often corresponds to more electronegative bonds and then gave the example of carbonyl absorptions which are intense, or nitrile absorptions, or $\ce{CO}$ single bond absorptions, etc.

What he then said, and says he has no good explanation for, is the following:

The absorption of an $sp^3$ hybridized $\ce{CH}$ stretch is high-medium intensity.

The absorption of an $sp^2$ hybridized $\ce{CH}$ stretch is medium intensity.

The absorption of an $sp$ hybridized $\ce{CH}$ stretch is medium-low intensity.

Of course the hybridizations given refer to the hybridization state of the carbon to which the hydrogen is attached.

The reason he has no good explanation for this phenomenon is that in most other situations, an $sp$ hybridized carbon will behave as if it is more electronegative than an $sp^2$ or $sp^3$ hybridized carbon.

An example of this would be that $pK_a$ of an $sp < sp^2 < sp^3$. That is a characteristic of how electronegative they are in an acid-base sense.

Here, however, we see the opposite trend. That is, when the more electronegative stretch should have a higher intensity peak, it is weaker, and the less electronegative stretch has a high intensity peak.

For the record, the arbitrary assignment of more and less electronegative based on behavior isn't super satisfactory to me, and is probably why I'm digging deeper on this.

So, my advisor essentially said all that matters is this equation: $$f=\left\langle\psi_0\right| \mu\left| \psi_v\right\rangle^2$$ Basically, he said I can ignore the wavefunction details of the equation because the wavefunctions of the three stretches are not going to differ by much because they all have identical reduced masses and some other stuff I can't remember.

So, all that matters is the dipole moment change as the bond is displaced from its equilibrium position.

His physical explanation of the intensities described above is that there is a smaller dipole moment change for $sp^2$ and $sp$ hybrid carbons because the electron density is trapped in the double and triple bonds more so than in the single bond. That seems reasonable to me.

The Organic Chemist fires back, however. He says that he would expect intensity to increase with the double and triple bonds because the electron in pi bonds are in a higher energy state than the sigma bond electrons and are thus more easily displaced which ought to result in a larger dipole moment change. That also makes a lot of sense I think.

So, is my advisor correct or is there a more complicated explanation?

Sorry for being so long-winded, I just wanted to get past the simple explanations so that we can have a deeper, better answer.

Edit: A very good answer would also explain why $sp$ hybridized C-H stretches are sometimes very high intensity. Even as high as $sp^3$ stretches at times.

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    $\begingroup$ I have no idea what the correct answer is, but when your organic prof talks about the electrons in pi bonds being higher in energy, how is that relevant to the C-H bond itself (which of course isn't a pi bond)? $\endgroup$
    – orthocresol
    Commented Oct 29, 2015 at 0:20
  • $\begingroup$ The idea is that the change in dipole has to do with the electron in that carbon-carbon bond, so having pi bonds ought to allow the electrons to be displaced easier as the hydrogen vibrates, so you might expect a larger change in dipole and thus greater intensity, but that's not what happens. $\endgroup$
    – jheindel
    Commented Oct 29, 2015 at 0:57
  • $\begingroup$ @jheindel - Your adviser is correct. The IR spectrum intensity goes as the magnitude of the transition dipole moment, which in turn is reflected by the (induced) electronic dipole moment (your formula gives the intensity for the expectation value of the linear approximation of the transition dipole moment operator). I don't see how the hybridization argument holds water, especially when you're in the ground electronic state and are looking at roto-vibrational states. $\endgroup$
    – Todd Minehardt
    Commented Dec 3, 2015 at 20:27
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    $\begingroup$ Right I agree that he's right simply because that's what actually happens but I'm looking for a good explanation as to where the reasoning of my O chem Prof is wrong. Cause it seems to make sense to me when though it gives the wrong result. $\endgroup$
    – jheindel
    Commented Dec 4, 2015 at 6:40
  • $\begingroup$ The intensity is proportional to the transition moment but not quite as in your question. $\mu$ should be expanded as a series in x (displacement about equilibrium position) then $f=\mu\int \psi_0 \psi_vdx+(d\mu/ dx)_e\int\psi_0~x~\psi_vdx+ \cdot$. Due to wavefunction orthogonality the first term is zero so the intensity depends on the slope the dipole has at the equilibrium geometry as the bond vibrates, not the absolute value. So this may cause the effect you are describing. $\endgroup$
    – porphyrin
    Commented Oct 24, 2016 at 21:38

1 Answer 1

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In considering the C-H stretch, let's consider that the C atom is attached to something else via an "immobile" bond(s).

For $sp^3$ hybridized the C would have 3 other bonds besides the immobile one.

For $sp^2$ hybridized the C would have 2 other bonds besides the immobile one.

For $sp^1$ hybridized the C would have 1 other bonds besides the immobile one.

For a non linear molecule there should be 3N-6 modes of vibration, but a linear on has 3N-5. 

   Hybridization  # Atoms  Linear   Vibration Modes
       sp3            4       No         6
       sp2            3       No         3
       sp1            2       Yes        1

The more vibration nodes, the more microstates for the molecule in both its current state and its excited state. Thus having more microstates increases the probability that there are more transitions from the current state to some excited state.

For an analog on the microstate overlap, think about atomic spectra vs UV-VIS. Atomic spectra are line spectra, UV-VIS is band spectra.

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  • $\begingroup$ Isn't it true, however, that many of those modes are vibrational modes involving the other atoms bonded to carbon and not just the hydrogen? Are you saying those vibrations increase the intensity of the C-H vibration because there is a change in dipole moment regardless of whether or not the C-H is the thing vibrating? I think I just need you to be a little more explicit in explaining the trend of intensities. $\endgroup$
    – jheindel
    Commented Nov 2, 2015 at 23:46
  • $\begingroup$ I'm not thinking dipole at all that explains the energy shift. // Think of a $\ce{=CH_2}$ group bonded to a wall. The various CH motions are shown here. en.wikipedia.org/wiki/… But this is resonance stretching as if the molecule were in a continuous wave. A microstate would be the 3 atoms having some motion which would quickly damp out. In a real matrix the 3 atoms just aren't rock steady until a photon comes along. Does that make more sense? $\endgroup$
    – MaxW
    Commented Nov 3, 2015 at 0:16
  • $\begingroup$ Uhh... There's no damping out in molecular vibrations. The ground state of all vibrations has an energy greater than zero. Also, change in dipole most definitely has something to do with the absorption since that's what determines the intensity of a transition which is what IR shows... The microstate argument might have something to do with it, but I'm not convinced. $\endgroup$
    – jheindel
    Commented Nov 3, 2015 at 5:03
  • $\begingroup$ Also, the modes which you're referencing (bending, scissoring, etc.) are absorbed at a much lower energy than straight up C-H stretch. Bending modes are down below 1000 wavenumbers usually. $\endgroup$
    – jheindel
    Commented Nov 3, 2015 at 5:05
  • $\begingroup$ So if ethylene is in carbon tetrachloride then the ONLY movement of the atoms in the molecule has to be the "pure" modes like the ethylene is in a rarefied gas? // The point is that if the molecule was in a non-pure vibration mode then that motion damps out in a very short time. The absorption of a photon is must be fast. The photon isn't really a standing wave in which the atoms move in resonance. It is really more like a particle going 3E8 meters/sec. The gist is that in a solution the "allowed" vibrations last a time at least comparable to the time between intermolecular collisions. $\endgroup$
    – MaxW
    Commented Nov 3, 2015 at 5:27

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