Equivalent H

[deleted647690]

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Are the Hydrogens at carbons 2 and 5, and then the H at carbons 3 and 4, considered equivalent? I'm looking at the H NMR for this compound in my BR Review book, and I'm wondering why the peaks for those two sets of hydrogens are split into what looks like "triplet of triplets"? Wouldn't they appear as one full peak if they are equivalent?

 

[aldol16]

Are the Hydrogens at carbons 2 and 5, and then the H at carbons 3 and 4, considered equivalent? I'm looking at the H NMR for this compound in my BR Review book, and I'm wondering why the peaks for those two sets of hydrogens are split into what looks like "triplet of triplets"? Wouldn't they appear as one full peak if they are equivalent?

I think you're confusing two concepts as well as two signals. I can't see it too clearly but the signal corresponding to the protons in question is the set of signals at 1.75 ppm. These protons are deshielded more so than the protons at the bottom of the ring due to proximity to the carbonyl. Second, equivalence has nothing to do with splitting pattern. To deduce splitting pattern, just look at one set of equivalent protons in question. Here, let's look at carbon 2. There are two protons attached to that carbon. It's going to couple with the proton at the tip of the ring and the two protons at the bottom of the ring, each with some coupling constant. So you're going to get either a triplet of doublets or doublet of triplets, depending which pair has the higher coupling constant. Here, you get a doublet of triplets, which is correct.

The number of equivalent carbons only gives you the integration. Here, carbons 2 and 5 are equivalent and each have two protons attached to them, so you would expect an integration of 4.

 

[deleted647690]

I think you're confusing two concepts as well as two signals. I can't see it too clearly but the signal corresponding to the protons in question is the set of signals at 1.75 ppm. These protons are deshielded more so than the protons at the bottom of the ring due to proximity to the carbonyl. Second, equivalence has nothing to do with splitting pattern. To deduce splitting pattern, just look at one set of equivalent protons in question. Here, let's look at carbon 2. There are two protons attached to that carbon. It's going to couple with the proton at the tip of the ring and the two protons at the bottom of the ring, each with some coupling constant. So you're going to get either a triplet of doublets or doublet of triplets, depending which pair has the higher coupling constant. Here, you get a doublet of triplets, which is correct.

The number of equivalent carbons only gives you the integration. Here, carbons 2 and 5 are equivalent and each have two protons attached to them, so you would expect an integration of 4.

I confused integration with coupling. Isn't carbon 2 neighbored by a CH and then a CH2, meaning it should have a quartet coupling pattern because of the n+1 rule?

 

[aldol16]

I confused integration with coupling. Isn't carbon 2 neighbored by a CH and then a CH2, meaning it should have a quartet coupling pattern because of the n+1 rule?

You should go over NMR theory again - I think you're a bit rusty then. Those two groups exert different fields because they're just different and so will split the signal in a non-equivalent manner. You're right that if you had 3 otherwise equivalent hydrogens vicinal to the proton in question, you would get a quartet. But a CH and CH2 group are not equivalent. A case where you would expect a quarter would be something like H3C-CH-(X)2 or something like that. The signal corresponding to the CH carbon here would be split into a quartet due to the three methyl protons (which are equivalent to one another and thus will give the same coupling constant).

 

[deleted647690]

You should go over NMR theory again - I think you're a bit rusty then. Those two groups exert different fields because they're just different and so will split the signal in a non-equivalent manner. You're right that if you had 3 otherwise equivalent hydrogens vicinal to the proton in question, you would get a quartet. But a CH and CH2 group are not equivalent. A case where you would expect a quarter would be something like H3C-CH-(X)2 or something like that. The signal corresponding to the CH carbon here would be split into a quartet due to the three methyl protons (which are equivalent to one another and thus will give the same coupling constant).

...haha, I just finished the chapter on spec in BR too. I guess I'm very rusty overall........so the doublet of triplets means a doublet from the CH and a triplet from the CH2 then, whoops

 

[aldol16]

...haha, I just finished the chapter on spec in BR too. I guess I'm very rusty overall........so the doublet of triplets means a doublet from the CH and a triplet from the CH2 then, whoops

Yes, basically. Based only on the structure, it also could have been a triplet of doublets but in this case, the CH coupling had a bigger coupling constant.