[Bendzo]

Hello!

This is actually a problem that puzzles my friend at his research, but I could not resist asking for your opinion, because I too find the problem must intriguing.
So here's the thing: My friend has a compound which has a Si-H bond, that is very well visible in the IR spectrum. This same compound must in the next step be attached on a solid support (carrier). It does so by attaching its self via the phenolic hydroxy functional group. This is a very important step for his further work, so he must assess the efficacy of the attachment and confirm the newly attached product. So he goes and records the IR spectrum, where the Si-H bond signal is missing. Because I do not work on the same field as he does, I will not write his whole experimental and indentification process. I will only conclude that the Si-H bond is still present in his sample, fact.
Question: How can an IR signal vanish, although the bond is present in the sample. We were thinking in the direction  sort of: If it is attached to a carrier, maybe "the vibrations and rotations" are limited, and so the peak is not seen (because it is very broad and weak). If anyone has any other guess, or knowledge, I would be delighted to learn your opinion!

Thank you,
Ben

[Corribus]

In IR spectroscopy peak height is (among other things) related to the dipole moment change during the vibrational mode.  If the symmetry of the molecule changes such that the dipole moment change becomes smaller, the corresponding peak will also become smaller.  Effects like lifetime broadening can also appear to diminish the intensity of a peak, and this occurs in cases where, for example, an H in an XH stretch is rapidly exchanged with other protons in the molecule. As in the case of an acidic proton, or protons involved in hydrogen bonds.  Finally, your peak could be vanishing because Si-H is no longer in your molecule. You say it is, but... how do you know?

[Bendzo]

OK, for better understanding of the problem, I attached p-dimethylsilylphenol to a polystyrene resin. The reaction is as simple as it gets, the resin has benzyl chloride functional groups that react with the hydroxyl group of the phenol. The reaction is done in DMF with CsCO3 acting as a base. When the reaction was completed, there were no organic compounds dissolved in DMF (GC/MS analysis).
The "invisible" peak is probably not affected by the change of the symmetry, as I can see the absorptions of Si-CH3 bonds that were present before i attached the compound to the polystyrene resin. They look intact.  As for the question, how do I know that the bond is still present, I tested for the Si-H bond via titration with bromine (Si-H bond is easily substituted with Si-Br). Of course, I also performed the titration on the starting resin and the reaction with bromine did not occur. I also linked pure phenol with the resin, and it also gave no reaction. I am baffled with this, because I have no idea what could have gone wrong in the linking phase, why the titration is positive, and why there is no Si-H peak, which should be very intense. Thank you for your time.

[Corribus]

There are often two Si-H stretches (on around 2200 wavenumbers, the other below 1000 wavenumbers). Do both of them disappear? Have you tried to look for an Si-Br stretch to confirm this reaction (should be really low frequency, maybe at 400 wavenumbers or so.

[Bendzo]

The problem is, that the peak near 1000 is among many other peaks, thus a correct assignment of the exact peak is impossible. As for the Si-Br bond, we have not checked there. The thing is, that the most intense Si-H bond peak dissapears...

Thank you,
Ben