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Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: CrimpJiggler on December 20, 2011, 05:54:17 PM

Title: Two distinct HNMR splitting patterns
Post by: CrimpJiggler on December 20, 2011, 05:54:17 PM
I noticed these on my NMR:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg38.imageshack.us%2Fimg38%2F9320%2Fscreenshotyv.png&hash=2d41fd9df9b8529ace2270caf944825c964bfd15)
they both have a small integration ratio indicating that they're produced by a single proton. Is the one on the left a triplet of doublets? If so, how is that splitting pattern produced? As for the one on the right, I don't know what to make of that.

Heres the compound:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.rdchemicals.com%2Fmolimg%2Fbig%2F7960.gif&hash=a8debbff9a2d431db05b39aeee83a63afc2ac0ca)
I know that bicyclic compounds like this are fairly rigid so there is probably a lot of geminal coupling going on there. I'm still a beginner when it comes to HNMR interpretation so trying to interpret this compounds spectrum has me bewildered. Besides looking for a singlet produced by the hydroxyl proton, I don't even know where to start.
Title: Re: Two distinct HNMR splitting patterns
Post by: ATMyller on December 21, 2011, 03:12:04 AM
The first one is 1:1:2:2:1:1 so it is truly a triplet of doublets and most likely belong to the 3rd carbon (counting from CH next to the OH group) with a CH, CH2 and C-(CH3)2 groups as neighbors.

The other one is a 1:1:1:1:1:1:1:1 octet so actually a doublet of doublet of doublets. I can't find any one proton group with three different single proton neighbors so atleast one of the splitting is caused by something other that mere J3(H-H) coupling.
Title: Re: Two distinct HNMR splitting patterns
Post by: sjb on December 21, 2011, 05:50:27 AM
The first one is 1:1:2:2:1:1 so it is truly a triplet of doublets and most likely belong to the 3rd carbon (counting from CH next to the OH group) with a CH, CH2 and C-(CH3)2 groups as neighbors.

The other one is a 1:1:1:1:1:1:1:1 octet so actually a doublet of doublet of doublets. I can't find any one proton group with three different single proton neighbors so atleast one of the splitting is caused by something other that mere J3(H-H) coupling.

Counting clockwise from the carbon with hydroxy group as 1, protons on carbon 2 seems to fit, perhaps, especially with one relatively large coupling and two smaller ones. Note that these protons are not equivalent.
Title: Re: Two distinct HNMR splitting patterns
Post by: discodermolide on December 21, 2011, 07:01:36 AM
Don't forget the "W" coupling between the bridgehead protons.
Title: Re: Two distinct HNMR splitting patterns
Post by: CrimpJiggler on December 21, 2011, 10:41:22 AM
The first one is 1:1:2:2:1:1 so it is truly a triplet of doublets and most likely belong to the 3rd carbon (counting from CH next to the OH group) with a CH, CH2 and C-(CH3)2 groups as neighbors.
Wouldn't the two -CH3 groups split that into a septet? If I had to guess what causes a triplet of doublets, I would have said a single proton which splits it into a doublet, then another pair protons which splits the doublet into a triplet. Is that correct? As for a doublet of doublet of doublets, thats what I suspected but I couldn't find any proton to fit the criteria for that either.

Don't forget the "W" coupling between the bridgehead protons.
Whats W coupling?
Title: Re: Two distinct HNMR splitting patterns
Post by: discodermolide on December 21, 2011, 10:47:52 AM
Look at the shape of the lines between the two bridgehead protons, it is in the form of a "W" lying in a plane. It is around 2-2.8Hz.
Any good NMR book should give you an exact definition.
Title: Re: Two distinct HNMR splitting patterns
Post by: ATMyller on December 23, 2011, 05:06:17 AM
Wouldn't the two -CH3 groups split that into a septet?
No they are 5 bonds away so they cause no splitting. It is the plain carbon in C-(CH3)2 group that is the neighborg.
Title: Re: Two distinct HNMR splitting patterns
Post by: key435 on December 28, 2011, 11:53:01 AM
Counting clockwise from C(CH3) is 1
The first one is truly a triplet of doublets so it is split by 2 H (of CH2 => triplet) and then by a H (of CH => triplet of doublets)  so the best position is 4 (--CH-OH).
And the next one is a douplets of doublets of doublets so it may be split by a H  and a H (of 2 CH neighbors) and a H of itself (of CH2). Because CH2 is near a asymmetric carbon it has different shifts. And the best position is 5 ( a H of CH2-)
That's what I think. ^^