The shifting carbonyl stretching frequency

Which of the following structures would you expect to have the strongest (higest wavenumber) streching frequency for the carbonyl stretch?
The key to this type of problem is realizing that there are two contributing forms to any carbonyl compound resonance hybrind. The one shown above, but also the polar charged form below.
You would expect that any group which could cause elctron donation to the carbonyl carbon would favor the polar form (right) and decrease the strecthing frequency. Similarly, any group that causes electron withdraw would enhance the double bond character (left) which increases the stretching frequency.
In this example you are looking at the effects through the conjugation of a phenyl ring. The stongest electron withdrawing group listed above is the nitro group, followed by chlorine, and the methyl group is slightly donating.
You also need to draw the resonance structures. Illustrating the withdraw of the nitro group on both the para and meta substituted phenol rings brings up a good point. You need to be able to draw a resonance strucure where the positive charge build up is next to the carbonyl. Think about what this would look like if you had the dipolar carbonyl-two adjacent positive charges = not a heavily contributing structure, therefore the double bond character is favored by this charge build up. The para substitution allows this but the meta does not (below). The correct answer for the problem above, methyl 4-nitrobenzoate, has a carbonyl strecthing frequency of 1700 cm-1.

Distinguishing Aldehydes and Ketones using IR

The following two spectra are simple carbonyl compounds. Formaldehyde, the simplest aldehyde, and acetone, the simplest ketone. Students often come to me frustrated because they can not tell one carbonyl compound from the next, or the peak will be right between the two literature values. The aldehyde or ketone question is simple. In both you will see a very prominent C-O stretch around 1700cm-1 area.

Ketone

Aldehyde

But in the aldehyde you should also see see a peaks around 2820 and 2720cm-1. They often look like a doublet and are sometimes referred to as a Fermi doublet. These are the C-H stretches between the aldehydic proton and the carbonyl carbon. The presence of these peaks along with a carbonyl peak is a good indication that you have an aldehyde.

Beginning IR Problems

You have successfully isolated a compound you know to be one of three narcotic analgesics. How could you use IR to determine which compound you have isolated? What specific bands would you be looking to be present or absent?

The key to this problems is simply recognizing the fuctional groups. While the structures may be pretty similar overall there are key differences. For example heroine should display a carbonyl peak, while this would be true of oxycodone as well, in oxycodone you would also have an OH peak. Morphine should have no peaks in the carbonyl region.