Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Organic Spectroscopy => Topic started by: MaximumCupcakes on April 03, 2012, 03:59:05 PM
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Hello all - thanks for taking the time to read. I'm really having problems with spectroscopy in general, so any help would be very much appreciated!
My compound's formula is C10H11O2Br. The IR spectrum clearly shows an -OH peak and a carbonyl peak (1689 cm-1)
H-NMR Data: http://img36.imageshack.us/img36/5506/hnmr.jpg
Summary:
-Two peaks at 1.3 ppm. I'm unsure whether these are singlets or a doublet. (6H)
-Quintet at 3 ppm (1H)
-Triplet at 7.6 ppm (1H)
-Triplet at 7.9 ppm (1H)
-Triplet/Multiplet at 8 ppm (1H)
-There's also an odd peak at 11 ppm labeled "exchanges with D2O. I haven't ever seen this before. (1H)
I have C-NMR and MS data, but I'm trying to work through these first.
Attempt at a solution:
Honestly, I'm not very good at this. My best guess is that this compound contains a monosubstituted benzene ring, with the 5 hydrogens around the ring. The other 6 hydrogens must be part of two methyl groups. Am I on the right track here?
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Hello all - thanks for taking the time to read. I'm really having problems with spectroscopy in general, so any help would be very much appreciated!
My compound's formula is C10H11O2Br. The IR spectrum clearly shows an -OH peak and a carbonyl peak (1689 cm-1)
H-NMR Data: http://img36.imageshack.us/img36/5506/hnmr.jpg
Summary:
-Two peaks at 1.3 ppm. I'm unsure whether these are singlets or a doublet. (6H)
-Quintet at 3 ppm (1H)
-Triplet at 7.6 ppm (1H)
-Triplet at 7.9 ppm (1H)
-Triplet/Multiplet at 8 ppm (1H)
-There's also an odd peak at 11 ppm labeled "exchanges with D2O. I haven't ever seen this before. (1H)
I have C-NMR and MS data, but I'm trying to work through these first.
Attempt at a solution:
Honestly, I'm not very good at this. My best guess is that this compound contains a monosubstituted benzene ring, with the 5 hydrogens around the ring. The other 6 hydrogens must be part of two methyl groups. Am I on the right track here?
Are you sure the peak around 3 is a quintet? Where are the 5 aromatic hydrogens you've assigned? I agree to an extent that the 6 hydrogens are two methyl groups, but I think there is something a bit subtle here. What are your J-values? What does your carbon NMR and mass spec show?
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Did you calculate the degrees of unsaturation?http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Degree_of_Unsaturation (http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Alkenes/Degree_of_Unsaturation)
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Thank you for the input. The quintet at 3 ppm is pretty clearly resolved - you can see it in the picture on the original post. I'll get back to you with the J values - will those help in the identification?
IR, MS: http://i40.tinypic.com/ebdf8x.jpg
The degree of unsaturation is 5.
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If you're interested, could I get some ideas on interpreting the MS? I know how to figure out the fragmentation patterns after the compound has been identified, but I don't know how to work backwards and actually identify it.
My proton decoupled CNMR shows peaks at 25, 35, 122, 128, 131, 135, 152, and 170.
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Well one thing should certainly pop-out at you on the mass spec. Hint: (M+2) :)
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Does the M+2 peak have to do with the Bromine atom?
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Does the M+2 peak have to do with the Bromine atom?
Think about isotopes!!
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Thank you for the input. The quintet at 3 ppm is pretty clearly resolved - you can see it in the picture on the original post. I'll get back to you with the J values - will those help in the identification?
IR, MS: http://i40.tinypic.com/ebdf8x.jpg
The degree of unsaturation is 5.
What specific pattern of heights would you expect for a quintet?
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Well one thing should certainly pop-out at you on the mass spec. Hint: (M+2) :)
If you are referring to bromine, you should read the question. Bromine presence is a given, so you don't have to guess it from MS.
Sorry if you mean something else.
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You've got 8 different peaks in the 13C-NMR. That tells you there's a degree of symmetry in the molecule.
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That there are 5 degrees of unsaturation tells you about the existence of another functional group, and it is easy to infer what it is using the downfield peak in the H-1 NMR.
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The large peak at 1689 in the IR should give a hint to the same functional group as that gives the peak that exchanges with D2O if i'm on the right track of course. It's an interesting problem.. I haven't yet figured out the coupling of the 1H NMR spectrum although I'm convinced the 2 peaks at 1.3 ppm are singlets.
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The 1H NMR is really irritating. The triplets want to be hydrogens on a benzene ring but as far as I can make out that would imply there have to be some doublets nearby!
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Without determining exactly what the structure is, you should still be able to determine a limited number of possibilities. I suggest using the Tinker Toy method. You have C10H11BrO2. From the integral, there is one or two signals at 1.3 ppm with 6H's. There is a signal at 3 ppm with 1H, this signal has splitting. There are 3H's in the aromatic region and 1H that can be exchanged with D2O. Hydrogens come in 1s, 2s, and 3s. What could the peaks at 1.3 ppm be? If you have a benzene ring, how many carbons are left, 10-6 = 4. How many of them have hydrogens attached, 3 or 4? If it is only three, what must the other carbon have attached to it? What atoms are left? I don't foresee that many possibilities. (I think this is only a question of regiochemistry.) Start by drawing at least one of them. Does that structure suggest any other possible structures, especially if you switch the order of the groups? You should be able to come up with some structures that use all of the atoms from the formula. For there, you should begin to recognize some structures are better than others. You will need to review the adjacency rule, especially for aromatic compounds. I cannot say exactly where the Br and OH will be except from reversing how it would likely to have been synthesized. For this part of the problem, you may need to look at values from a chemical shift table to help.
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The 1H NMR is really irritating. The triplets want to be hydrogens on a benzene ring but as far as I can make out that would imply there have to be some doublets nearby!
Why?
Try to add up what you have so far:
benzene compound --> substitution pattern? number of protons on the benzene ring?
bromine somewhere
exchangeable proton --> ?
Singlet or doublet --> ?
Your quintet is actually a septet!!! (plz have a closer look) --> only one functional group shows a clean septet