[arover]

Hey all. I'm trying to map out the hydrogens in an H-NMR of Benzophenone [http://en.wikipedia.org/wiki/Benzophenone] for an experiment I did in the lab. I couldn't find a H-NMR to post on here (at least from the glimpse I took on google), but I have the chart in front of me. Theres 3 sets of peaks: A triplet peak, a triple triplet, and a double doublet. These peaks were confirmed by my professor. I've found the hydrogens that contribute to the triplet and the double doublet, but I can't for the life of me find how you can have a triple triplet. I know you can travel through one pi-bond to talk to another hydrogen, but no matter how I try and work it out, I don't know how one could possibly have two sets of 2 matching hydrogens, all talking to the same C-H, on each ring.

Am I missing something here?

[nj_bartel]

Is this a real nmr spectrum?  If it is, I believe aryl protons can undergo something called W-coupling, but that's not typically covered in intro organic classes.  This is a type of j4 coupling.

[arover]

Yes, this is a real nmr spectrum. I think I could scan it real quick...I'll have an image in a few.

[arover]

http://img255.imageshack.us/img255/6047/hnmrbenzo.jpg

[macman104]

I was also going to ask if this would might be w-coupling...

http://www.ch.ic.ac.uk/local/organic/nmr_1H.html

Search the page for "W" coupling

[arover]

Hmmm. I'm still not sure I understand what W-coupling is exactly? I can't remember if we've gone over it, but I do know my professor for lab lecture (separate from regular Ochem lecture) has gone into more detail than normal for what intro Ochem covers as far as NMR goes. I did get a little lost at times tho....

[nj_bartel]

Picture a benzene with the hydrogens in positions 1 and 3 drawn explicitly.  Do you see how the bonds from the hydrogen to the carbon, to the carbon, to the hydrogen form a W?  Those two hydrogens can couple with each other.

And wtf @ nmr with no integrations -_-

From my PoV, it seems like, if you're to take W-coupling into account, you'd have a doublet of doublets, another doublet of doublets, and a triplet of triplets.  If you're not to take W-coupling into account, it seems to me you'd have a doublet, another doublet, and a doublet of doublets.  So I don't know 

[azmanam]

sorry... pet peeve...  1H spectra show proton signals, not hydrogens.

[nj_bartel]

No problem 

[AWK]

You can find 90 MHZ 1H NMR spectrum of benzphenone on
http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

[McCoy]

Mmm the proton on C-3 has a chemical shift of 7.6(or whatever that is) ppm.  just by looking at it might appear to be a triplet, which is what you would expect since there are two protons on adjacent carbons, one on C-2 and one on C-4.  But if you expand the aromatic region, you can see that the signal for H-3 is actually a doublet of doublets.  it's coupled to H-2 and H-4.  Since H-2 and H-4 are not identical, their interactions with H-3 are not identical,(in other word the value of 3JH-3,H-2 is not equal to the value of  3^JH-3,H-4)..the n+1 rule seems not to apply here.

If the value of 3^JH-3,H-2 were equal to that of 3^JH-3,H-4, then the peak for H-3 would be a triplet. And i think your 1H NMR has just done that!