It really depends a lot on how much calculating your teacher expects you to do. There are a few ways of looking at substituent effects on the NMR of protons attached to a benzene ring, depending on how deep you get.
The first is exactly what you did, and it is an excellent start and usually as far as you need to look. Recognize splitting patterns among the protons, and use that and the integration to determine which carbons of the benzene ring are substituted. It is very easy to recognize a 1,4; 1,3; or 1;2 substituted benzene ring based on splitting patterns, and in this case you recognized a 1,2,4 substituted benzene ring. You just can't tell what groups are where. You do know, however, that the semi-singlet will be the proton by itself in between two substituents (4 position), that the semi-doublet will be para to the semi-singlet (1 position), and that there will be a slightly more complicated peak, sort of a semi-doublet of doublets that is next to it (2 position)
Now you have to look at the groups themselves, and what effect they will have on protons at various positions around the ring. These are, fortunately, basically additive. If you pick a proton on a benzene ring, there is a Z value for substituents at each point on the ring that will give you a fairly good prediction of where the proton will be. If there are no substituents at all, the proton will be at 7.36 ppm, which is the chemical shift for benzene.
Electron withdrawing substituents, like carbonyls, will pull electrons away from the benzene ring, reducing the effect of the ring current. Due to resonance effects, this is stronger in the ortho and para positions. Protons on a benzene ring which contains a carbonyl group will have a higher chemical shift than 7.36, and the ortho and para protons will be higher than the meta protons. Electron donating groups, like hydroxyls and amines, have the opposite effect, strengthening the ring current and reducing the chemical shift of protons on the ring, but it is still most pronounced in the ortho and para positions. The halogens fit somewhere between the two extremes, having little effect on chemical shift, and due to back bonding, the effects on ortho, para, and meta positions aren't always as clear cut. For chlorine, for example, the effect is weak in the para position, weaker in the meta position, and even weaker in the ortho position.
So let's look at the two possibilities. For the 3,4-dichloroacetophenone, you would expect the 2 proton (semi-singlet) to be raised significantly by the carbonyl in the ortho position, lowered slightly by the 4 chloro in the meta position, and lowered just a trace by the 3 chloro in the ortho position. The 5 proton (semi-doublet) would be raised slightly by the meta ketone, and lowered slightly by an ortho chloro and a meta chloro. The 6 proton (more complicated semi-dd) would be raised significantly by the ortho ketone, and lowered by the para and meta chloro. So you would expect the 6 proton (dd) to be highest, then the 2 proton (s), then the 5 proton (d).
For the 2,4-dichloroacetophenone, you would expect the 3 proton (semi s) to be raised slightly by the meta ketone, and lowered a tiny bit by the two ortho chlorines. The 5 proton (semi-dd) is raised slightly by the meta ketone, lowered a tiny bit by the ortho chloring, and lowered more by the para chloring. The 6 proton (semi-d) is raised quite a bit by the meta ketone, and lowered somewhat by the two meta chlorines. That puts the 6 proton highest (d), then the 3 proton (s), then the 5 proton (dd).
In your spectrum, you show the semi-d at around 7.5 ppm, the semi-s around 7.4, and the semi-dd around 7.3. This is the order found in the second example (6-3-5), so my guess would be that this is the 2,4- rather than the 3,4-dichloroacetophenone.
Of course, this whole analysis is easier if you can actually attach some numbers to it. I like this site -
http://www.chem.wisc.edu/areas/reich/handouts/nmr-h/hdata.htm - scroll down on the left side to "benzene - substituent effects". Using this table, you can calculate (for 2,4-dichloroacetophenone) the 6 proton to be about 7.82, the 3 proton to be about 7.50, and the 5 proton to be 7.31.
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Topic: NMR spectra problem (Read 14165 times)