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Topic: Two distinct HNMR splitting patterns  (Read 9773 times)

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Offline CrimpJiggler

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Two distinct HNMR splitting patterns
« on: December 20, 2011, 05:54:17 PM »
I noticed these on my NMR:

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:

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.

Offline ATMyller

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Re: Two distinct HNMR splitting patterns
« Reply #1 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.
Chemists do it periodically on table.

Offline sjb

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Re: Two distinct HNMR splitting patterns
« Reply #2 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.

Offline discodermolide

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Re: Two distinct HNMR splitting patterns
« Reply #3 on: December 21, 2011, 07:01:36 AM »
Don't forget the "W" coupling between the bridgehead protons.
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Offline CrimpJiggler

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Re: Two distinct HNMR splitting patterns
« Reply #4 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?

Offline discodermolide

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Re: Two distinct HNMR splitting patterns
« Reply #5 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.
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Offline ATMyller

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Re: Two distinct HNMR splitting patterns
« Reply #6 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.
Chemists do it periodically on table.

Offline key435

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Re: Two distinct HNMR splitting patterns
« Reply #7 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. ^^

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