The first spectrum is 1-hexene:
The key peaks are marked; later, we will recognize the strong absorption about 890 cm-1 as the C-H out of plane bend for a terminal alkene.
Next is trans-2-hexene:
The C=C absorption is much weaker, and would get weaker still if we put a third alkyl group on the double bond. The C=C-H stretch is hidden; don't count on always seeing it. Again, later, we will recognize the strong peak at about 970 cm-1 and the alkene CH out of plane for a trans alkene.
For comparison, here is the cis isomer:
The C=C stretch is again weak, but the C=C-H stretch is visible. The peak at around 700 cm-1 is diagnostic for a cis alkene.
Next, a typical ketone, 2-hexanone:
The carbonyl stretch is the only significant absorption. Reduction of the ketone leads to the alcohol, 2-hexanol:
The small absorption at about 1700 cm-1 suggests that a very small amount of the ketone may be a contaminant.
Here is a typical aldehyde, butanal:
In addition to the carbonyl stretch, aldehydes have the very cahracteristic doublet at 2720 cm-1 and 2820 cm-1 for the aldehyde H out of plane bending.
Finally, an ester:
Now let's try a few problems. You probably will not be able to get a single correct answer for most of these, but try to generate at least one valid possible answer.
Compound 1: C5H10O
Compound 2: C8H8O
In recitation on 14 October, a serious question arose about this problem. We had concluded that the compound was a monosubstituted benzene, but a student pointed out that it seemed to show the single absorption in the 735-770 cm-1 region that is characteristic of an ortho-disubstituted benzene (ortho = 1,2-). So I went looking for a spectrum with better resolution and found this:
With the much better resolution, we can see the requisite two peaks (730-770 and 690-710 cm-1) for the monosubstitution. Good point by the student, lousy selection of the original spectrum by me.
Compound 3: C7H14O
Compound 4: C6H14O
Compound 5: C5H10O2
We will work some of these in recitation and post answers before the exam.
