14.3 Interpreting More IR Spectra | Organic Chemistry

[@JanetteWilson-xd2do]

Chad, just out of curiosity, is that a self-portrait in the top corner of the PowerPoint slides? 😁

[@ChadsPrep]

Of course - the same hairstyle!

[@iupacjr.4364]

Thanks 🙏

[@ChadsPrep]

You're welcome!

[@26d8]

great video!! Q: in dilute solution, the OH concentration would be lowered? how come the OH signal goes higher in energy when we have less concentration of OH in the 1ml of the solution? Q2: what if we compare HCN to a terminal alkyne? can we distinguish them from IR?

[@ChadsPrep]

In a more dilute solution of the alcohol you will have fewer other alcohol molecules to hydrogen bond with. The more hydrogen bonding the hydroxyl H is involved with the weaker the O-H bond becomes which is seen as a subtle shift to the right. So a more dilute solution with less hydrogen bonding would result in a small shift left instead. As far as distinguishing HCN to a terminal alkyne I have a couple of responses. First, you would never have to as HCN is not an organic compound, but inorganic instead. But still we could tell the difference in one way. The C-N bond is more polar than the C-C bond and therefore will give a stronger signal. If you line the corresponding spectra up side-by-side you can tell them apart, although I wouldn't expect an undergraduate to be able to tell which they have with only one of them in front of them as the difference is subtle. I thought there would be another difference in the C-H bond in HCN vs a terminal alkyne. HCN (pKa ~9-10) is significantly more acidic than a terminal alkyne (pKa ~ 26) and I suspected this would be reflected in a difference in bond strength and where the IR stretch appears. But I just looked it up and I'm wrong; it shows up in the same place (~3300cm-1). Hope this helps!

[@26d8]

@@ChadsPrep i appreciate your response, HCN will show up at the same peak but as you said its an inorganic compound, and its not likely to appear in IR spectra, I understand now, thank again for clarifying.

[@aaronkim8180]

The explanation for the difference in peaks between the primary and secondary amines is not completely accurate: "In fact, both N-H bonds of a single molecule will together produce only one signal. The reason for the appearance of two signals is more accurately explained by considering the two possible ways in which the entire NH2 group can vibrate. The N-H bonds can stretch in phase with each other, called symmetric stretching, or they can be stretching out of phase with each other, called asymmetric stretching. At any given moment in time, approximately half of the molecules will be vibrating symmetrically, while the other half will be vibrating asymmetrically. The molecules vibrating symmetrically will absorb a particular frequency of IR radiation to promote a vibrational excitation, while the molecules vibrating asymmetrically will absorb a different frequency. In other words, one of the signals is produced by half of the molecules. For a similar reason, the C-H bonds of a CH3 groups (appearing just below 3000 cm-1 in an IR spectrum) generally gives rise to a series of signal, rather than just one signal. These signals arise from various ways in which a CH3 group can be excited." -(Klein, David R. Organic Chemistry as a Second Language: Second Semester Topics. 3rd ed. Hoboken, NJ: Wiley, 2012.) Also i'm so thankful for professors like you! Love your videos! :^)