Proton NMR - Unknown structure

[pkthunder]

Can someone please explain to me these questions: http://imgur.com/a/uzHNQ

 

 

I thought a 3.7ppm indicated ether, however it is apparently an aldehyde proton?

 

 

 

[hypervalent_iodine]

3.7ppm is most definitely not an aldehyde proton. You would expect such a proton to occur at around 9ppm, give or take. Also, I am skeptical of foreign links, so could you post the picture in the thread or have it as a pdf attachment?

[pkthunder]

3.7ppm is most definitely not an aldehyde proton. You would expect such a proton to occur at around 9ppm, give or take. Also, I am skeptical of foreign links, so could you post the picture in the thread or have it as a pdf attachment?

 

 

That's what I thought as well. My friend was trying to convince me otherwise.

Sorry about the link though, I'm new.

[hypervalent_iodine]

Had a look at the pictures. You should have been able to figure out it wasn't an aldehyde given the structures you're provided with in the second question (none of them are aldehydes, but one of them has to be your molecule). Have you managed to work out the second question yet?

Edited May 28, 2011 by hypervalent_iodine

[pkthunder]

I changed it to A a few hours ago. Would that be correct?

 

 

If you have the time, could you please have a look at this problem?

 

 

 

 

 

I'm thinking its A, but that's only through elimination of the other 3.

 

My reasoning:

 

B - there should be a singlet associated with the -OH group

 

C - there should be a singlet for the -OH

 

D - there should be a singlet for -CH3

 

 

But for A (I'm not sure how to name it), I'm finding more H environments than given.

 

 

 

Any help would be appreciated.

[hypervalent_iodine]

It wouldn't be A. 270nm is typical of an aromatic group, so A is eliminated. Also, your ¹H NMR shows aromatic peaks at ~ 7.3ppm. You don't necessarily see and -OH peaks in an NMR spectrum if you haven't done a D₂O shake. So don't worry that you can't see it. Also, if that reasoning is good enough for B-D, then how come you chose A, which also has an OH group on it? I would look at a few things for this: firstly, are any of the shifts typical of methyl group and secondly, what would give rise to the splitting pattern observed on the peaks at 2.8ppm and 3.6ppm? From that you should be able to work it out. The other way you could look at it is by looking at how the aromatic protons are split. Different substitution patterns on an aromatic ring (para, meta, ortho) will give rise to different splitting in the aromatic region. It sometimes takes a trained eye to see it though.

[pkthunder]

It wouldn't be A. 270nm is typical of an aromatic group, so A is eliminated. Also, your ¹H NMR shows aromatic peaks at ~ 7.3ppm. You don't necessarily see and -OH peaks in an NMR spectrum if you haven't done a D₂O shake. So don't worry that you can't see it. Also, if that reasoning is good enough for B-D, then how come you chose A, which also has an OH group on it? I would look at a few things for this: firstly, are any of the shifts typical of methyl group and secondly, what would give rise to the splitting pattern observed on the peaks at 2.8ppm and 3.6ppm? From that you should be able to work it out. The other way you could look at it is by looking at how the aromatic protons are split. Different substitution patterns on an aromatic ring (para, meta, ortho) will give rise to different splitting in the aromatic region. It sometimes takes a trained eye to see it though.

 

Yea, I had looked around and found about the aromatic multiplet (not taught to me by my lecturer) and I had thought it was C (phenethyl alcohol) since the peaks matched, but the NMR I found of it had the singlet for -OH, so I eliminated it, since I didn't know that -OH peaks may not show up without a D2O shake. So I'm choosing B.

 

 

Thanks for your help. You've taught me more than my lecturer has

[hypervalent_iodine]

For your first question, I can't remember if Isaid A or B. It was definitely one of the esters though.

 

Also, I think you need to have anther quick look at your spectra for this question. The alkyl peaks you have are a little too high for a methyl group. The splitting for both looks to be two triplets, which means that both are methylene protons (CH2 groups). As well, you have a pretty strong MS peak with an m/z of 65, which corresponds to a loss of CH2CH2OH.

 

And no problems! NMR can be a bit tricky sometimes. The best way to learn it is to keep practicing with examples like this.

 

Oh and you should look up information on how para, ortho and meta disubstitution on aromatic rings changes splitting patterns. It's very useful information to have under you belt in these questions.

[pkthunder]

For your first question, I can't remember if Isaid A or B. It was definitely one of the esters though.

 

Also, I think you need to have anther quick look at your spectra for this question. The alkyl peaks you have are a little too high for a methyl group. The splitting for both looks to be two triplets, which means that both are methylene protons (CH2 groups). As well, you have a pretty strong MS peak with an m/z of 65, which corresponds to a loss of CH2CH2OH.

 

And no problems! NMR can be a bit tricky sometimes. The best way to learn it is to keep practicing with examples like this.

 

Oh and you should look up information on how para, ortho and meta disubstitution on aromatic rings changes splitting patterns. It's very useful information to have under you belt in these questions.

 

Sorry, I meant choosing C.

 

 

 

Thanks again

[hypervalent_iodine]

No problems. Feel free to come back with any other questions