TBR O Chem Section 3 Example 3.10 and 3.16

[OverlyNeurotic]

---

Members don't see this ad.

Hey there y'all!

So I was finishing up reviewing spectroscopy and analysis today and ran into a few example questions that left me feeling a bit confused.

Here's a picture of the first example:

The solution says "the septet and doublet in a 1:6 ratio are a dead give-away for the isopropyl group." How does the author know it's a definite 1:6 ratio from the NMR graph? To me, the ratio of integrals looks closer to a 1:4 ratio than a 1:6 ratio. Is there any way to assess the ratio without having to estimate from the graph? Also, why isn't a singlet peak displayed or did they just casually leave it out?

Here's the second example:

Shouldn't the singlet peak slightly upfield of 8 ppm be a 4H peak considering there are 4 protons directly associated with the benzene ring? I get the 3H singlet peak slightly upfield of 4 ppm which represents the methoxy hydrogens but I am having trouble understanding why the other peak was incorrectly labeled (at least to my understanding).

Anyways, thanks y'all for any help you can provide. Greatly appreciated!

Steven

 

[aldol16]

The solution says "the septet and doublet in a 1:6 ratio are a dead give-away for the isopropyl group." How does the author know it's a definite 1:6 ratio from the NMR graph? To me, the ratio of integrals looks closer to a 1:4 ratio than a 1:6 ratio. Is there any way to assess the ratio without having to estimate from the graph? Also, why isn't a singlet peak displayed or did they just casually leave it out?

Unless they give you the integrations, it's very difficult to figure out the ratios just by eyeballing it. So it's not a particularly good question but at least it gives you NMR practice. Now, you should realize though, that just the doublet and septet pattern is a dead giveaway for the isopropyl group. The tertiary proton will be split by six methyl protons and the six methyl protons will, in turn, be split by the tertiary proton (one can confirm this easily by COSY, or two-dimensional NMR). You don't need the integrations to figure this out. Six equivalent protons must be present adjacent to the tertiary proton in order for it to be split into a clean septet at all.

What singlet are you referring to? If the molecule is 2,4-dimethyl-penta-3-one, then you don't have singlets at all. It's basically a ketone with two isopropyl substituents. So the only protons are two equivalent tertiary ones and twelve equivalent methyl ones.

Shouldn't the singlet peak slightly upfield of 8 ppm be a 4H peak considering there are 4 protons directly associated with the benzene ring? I get the 3H singlet peak slightly upfield of 4 ppm which represents the methoxy hydrogens but I am having trouble understanding why the other peak was incorrectly labeled (at least to my understanding).

That signal does correspond to four protons. You should realize that the integrations are useful only for giving ratios of protons and not absolute numbers - especially when molecules have symmetry. So in this case, you have p-methoxy anisole, which is a benzene ring with two methoxy groups para to each other. The axis of symmetry runs straight through the middle of the ring. So the methoxy protons are equivalent and will give the same signal and the pair of benzene protons are equivalent to each other and will give the same signal. So the integration will be in a ratio of 4:6 benzene:methoxy or any multiple of this.

 

[OverlyNeurotic]

Unless they give you the integrations, it's very difficult to figure out the ratios just by eyeballing it. So it's not a particularly good question but at least it gives you NMR practice. Now, you should realize though, that just the doublet and septet pattern is a dead giveaway for the isopropyl group. The tertiary proton will be split by six methyl protons and the six methyl protons will, in turn, be split by the tertiary proton (one can confirm this easily by COSY, or two-dimensional NMR). You don't need the integrations to figure this out. Six equivalent protons must be present adjacent to the tertiary proton in order for it to be split into a clean septet at all.

What singlet are you referring to? If the molecule is 2,4-dimethyl-penta-3-one, then you don't have singlets at all. It's basically a ketone with two isopropyl substituents. So the only protons are two equivalent tertiary ones and twelve equivalent methyl ones.



That signal does correspond to four protons. You should realize that the integrations are useful only for giving ratios of protons and not absolute numbers - especially when molecules have symmetry. So in this case, you have p-methoxy anisole, which is a benzene ring with two methoxy groups para to each other. The axis of symmetry runs straight through the middle of the ring. So the methoxy protons are equivalent and will give the same signal and the pair of benzene protons are equivalent to each other and will give the same signal. So the integration will be in a ratio of 4:6 benzene:methoxy or any multiple of this.

Wow, thank you for such a rapid response! That makes a lot of sense - the two equivalent tertiary protons would display a 2H septet peak and the twelve equivalent methyl protons would display a 12H doublet peak. For some reason, I thought the two tertiary protons would also display a 2H singlet peak to represent the lack of protons on the carbonyl carbon but I'm starting to realize now that that sort of reasoning most likely stemmed from trying to cram a chapter at 2 in the morning haha.

And that's a great point regarding the integrations. TBR hadn't listed that the integrals represent ratios and not absolute numbers so that's a major help!

Thanks once again for all of the help bud!