I don't understand spectroscopy or spectrometry analysis.
May 2, 2011 4:58 AM Subscribe
I need help understanding analysis of mass spectrometry and UV, IR, NMR spectroscopy readouts. I'm looking up tutorials on websites and re-reading my textbook, and while I think I understand the theory behind each method I'm drawing a blank when it comes to actually interpreting results to complete problems.
Somewhere between understanding why these methods work and how to connect their printouts to actual molecular structures my brain is breaking. I'm looking for the simplest tutorials I can find and it's not connecting.
For example, IR spectroscopy. How do I know a certain peak corresponds to a certain type of bond when there may be three or four different types of bonds that take up that range? Consider this problem here. What am I supposed to be learning from those peaks?
And mass spec--how do I know if the base peak is the molecular ion or not? And if it's not, how do I know what the molecular ion is?
I'm similarly confused on NMR and UV. Can anyone help?
Somewhere between understanding why these methods work and how to connect their printouts to actual molecular structures my brain is breaking. I'm looking for the simplest tutorials I can find and it's not connecting.
For example, IR spectroscopy. How do I know a certain peak corresponds to a certain type of bond when there may be three or four different types of bonds that take up that range? Consider this problem here. What am I supposed to be learning from those peaks?
And mass spec--how do I know if the base peak is the molecular ion or not? And if it's not, how do I know what the molecular ion is?
I'm similarly confused on NMR and UV. Can anyone help?
What level are you studying these things at? My undergrad organic chemistry professor recommended an excellent book on NMR spectroscopy that I probably still have sitting around; I will look for it when I get home.
posted by TedW at 5:53 AM on May 2, 2011
posted by TedW at 5:53 AM on May 2, 2011
Also, it just so happens I am sitting in the control room of an MRI scanner right now; if I see any of the physicists (who do spectroscopy on these machines) I will ask them if they have any resources.
posted by TedW at 5:56 AM on May 2, 2011
posted by TedW at 5:56 AM on May 2, 2011
Response by poster: I'm an undergrad, Orgo I. I understand this stuff is like a puzzle, but I can't seem to even identify the pieces!
posted by Anonymous at 6:21 AM on May 2, 2011
posted by Anonymous at 6:21 AM on May 2, 2011
There's a reason that these spectroscopic techniques are referred to as "sporting methods", (as opposed to the non-sporting X-ray diffraction, say, which does not give your quarry a chance). No single method will usually give you the answer, but it will rule in or out various possibilities, and you usually have to then cross-reference. For organics you should find that 1H NMR, and 13C if available, will identify the components of your target molecule. Understanding 1H coupling patterns is obviously crucial, and getting a feel for where differently-shielded protons will appear in the spectrum. In mass spec, there's no reason that the base peak has to be the molecular ion, you're generally looking for the heaviest real signal, from which you may be able to deduce (or at least narrow down) the empirical formula, and then seeing what fragments off it. UV or IR may be less definitive, but will show the presence or absence of certain functional groups.
posted by wilko at 6:39 AM on May 2, 2011 [1 favorite]
posted by wilko at 6:39 AM on May 2, 2011 [1 favorite]
To solve these problems I first write down individual clues without any regard to the solution, so for instance the first thing we know (from the molecular ion) is that the total mass of the molecule is 116. The fact that this is an even number also tells you that the number of nitrogens is even (or zero). The next peak below the molecular ion is a small peak of 15, so I would guess that this is a methyl group. Below that is a peak at 85 which is due to the loss of something with mass of 31, the standard group of that mass is methoxide.
The base peak at 57 corresponds to C4H9, according to this table
For example, IR spectroscopy. How do I know a certain peak corresponds to a certain type of bond when there may be three or four different types of bonds that take up that range?
It's true that multiple bond types can cause peaks very close to each other, they do sometimes differ in strength and width. In your example the peaks at 1166 and 1194 are strong. C-O stretches are the only sources of strong peaks at those wavelengths, so now you know that you have two C-O bonds.
The peak at 1730 falls in the C=O (carbonyl) range, there isn't much else around there.
There are no C=C bonds in the IR spectrum.
So in this case we know:
Total mass = 116
At least one methyl group
A methoxide
A C4H9 fragment
Two C-O bonds
One C=O bond
No C=C bonds
posted by atrazine at 6:51 AM on May 2, 2011
The base peak at 57 corresponds to C4H9, according to this table
For example, IR spectroscopy. How do I know a certain peak corresponds to a certain type of bond when there may be three or four different types of bonds that take up that range?
It's true that multiple bond types can cause peaks very close to each other, they do sometimes differ in strength and width. In your example the peaks at 1166 and 1194 are strong. C-O stretches are the only sources of strong peaks at those wavelengths, so now you know that you have two C-O bonds.
The peak at 1730 falls in the C=O (carbonyl) range, there isn't much else around there.
There are no C=C bonds in the IR spectrum.
So in this case we know:
Total mass = 116
At least one methyl group
A methoxide
A C4H9 fragment
Two C-O bonds
One C=O bond
No C=C bonds
posted by atrazine at 6:51 AM on May 2, 2011
This was my favorite part of organic chem, though I don't do tons of it now. The tricky part is yeah, as you say, there's not necessarily a 1:1 relationship. If it helps, you may not want to think of it as a puzzle, but as an argument. To me, it's a serial process of finding the lowest-hanging fruit and then seeing if the guess that comes from that fits any other data. Atrazine covered the first problem so I'll look at the second one just so you can see a different thought process.
I start with IR because I like it better and because it can narrow things down faster. I look for stuff that couldn't be a lot of different things. In the IR spectrum, that peak at 2230 stands out to me because it's in an odd spot--it's something that *can't* be a lot of other things, so it'll narrow down the field a bunch. Could be an alkyne, could be a nitrile, couldn't be much else. The other funky guy is the peak at 750--probably a benzene ring without much attached, maybe a C-Cl. Interestingly, there's zip in the alkane region which means there won't be a lot of junk to sort through. Maybe it's a small molecule? The other peaks could be a bunch of different things, so I'll go back to them later.
Mass spec gives a big old peak at 103--so, yikes, not a small molecule, but a sturdy one! Maybe when I said "small" before, what I really should have said was "simple", meaning there's not a lot of groups hanging out every which way that can bust apart, as evidenced by the mass spec. Medium sized, very sturdy molecule? That yells "aromatic ring". And hey look, one of my IR guesses was "minimally substituted benzene ring". Sweet. A benzene with an alkyne or two or a nitrile or two hanging off. There's no peak from an H on an alkyne in the spectrum, so if there WAS an alkyne, under the circumstances, it'd need to be, what, a 1,2 phenyl ethyne? That won't work; the mass spec says the molecule's smaller than that. How about benzonitrile? Ooh, the math works on that one. That makes me pretty sure but I double check that I can attribute all the peaks to that, and maybe I look up the melting point just to make sure (rather than checking my work though I just typed that into the box at this point because I knew I was close and it's not my homework).
NMR is similar except I usually start by figuring out how many hydrogens and how many neighbors are involved, along with paying attention to the peaks out in the boonies.
posted by tchemgrrl at 9:42 AM on May 2, 2011
I start with IR because I like it better and because it can narrow things down faster. I look for stuff that couldn't be a lot of different things. In the IR spectrum, that peak at 2230 stands out to me because it's in an odd spot--it's something that *can't* be a lot of other things, so it'll narrow down the field a bunch. Could be an alkyne, could be a nitrile, couldn't be much else. The other funky guy is the peak at 750--probably a benzene ring without much attached, maybe a C-Cl. Interestingly, there's zip in the alkane region which means there won't be a lot of junk to sort through. Maybe it's a small molecule? The other peaks could be a bunch of different things, so I'll go back to them later.
Mass spec gives a big old peak at 103--so, yikes, not a small molecule, but a sturdy one! Maybe when I said "small" before, what I really should have said was "simple", meaning there's not a lot of groups hanging out every which way that can bust apart, as evidenced by the mass spec. Medium sized, very sturdy molecule? That yells "aromatic ring". And hey look, one of my IR guesses was "minimally substituted benzene ring". Sweet. A benzene with an alkyne or two or a nitrile or two hanging off. There's no peak from an H on an alkyne in the spectrum, so if there WAS an alkyne, under the circumstances, it'd need to be, what, a 1,2 phenyl ethyne? That won't work; the mass spec says the molecule's smaller than that. How about benzonitrile? Ooh, the math works on that one. That makes me pretty sure but I double check that I can attribute all the peaks to that, and maybe I look up the melting point just to make sure (rather than checking my work though I just typed that into the box at this point because I knew I was close and it's not my homework).
NMR is similar except I usually start by figuring out how many hydrogens and how many neighbors are involved, along with paying attention to the peaks out in the boonies.
posted by tchemgrrl at 9:42 AM on May 2, 2011
Try to tackle one type of spectroscopy at a time.
NMR:
I know you said you understand background, but make sure you're solid on key concepts:
- chemical shift - how much the electron cloud ('shield') is being pulled away from the atom (atom = H if you are doing H-NMR), so if you're closer to an electron withdrawing group (EWG), you will be farther left (downfield) on the spectra
- peak regions - how many peak areas there are tells you how many unique H's (for H-NMR) you have. This can give you big clues about symmetry in your molecule, because H's that are symmetrical will be in identical chemical environments, and appear at the same place on the spectra.
- multiplicity - how many times the peak is split tells you how many neighbors that H has (doublet means 1 neighbor, triplet means 2, etc.). Neighbors are H's on an adjacent C’s.
- integration - how many H's are represented by that peak, i.e. how many H's have that same electronic environment.
If you are given a spectra and trying to solve for the structure, make yourself a table for every peak. Make columns labels at the top of the table for (a) chemical shift, (b) multiplicity - like singlet, doublet, triplet, etc., (c) neighbors - determined from multiplicity, (d) integration, and (e) guess. For the guess, you should put down possible things this peak could represent. For instance, if it is a doublet with integration 6, it is likely to represent 2 CH3's next to a CH, but may also be 3 CH2's next to a CH, etc. When I was taking undergrad organic, my professor taught us how to do this table, and it really helped me to convert the overwhelming amount of info conveyed from the spectra into usable pieces of information.
Example: C4H8O2
By calculating the index of hydrogen deficiency, you know there is one double bond
Chemical Shift | Multiplicity | Neighbors | Integration | Guess
........ 4.0............quartet............3...............2 ........ - CH2 – CH3 (CH – CH2 – CH2 can be ruled out b/c of integration of other peaks); near EWG so most likely R - O - CH2 – CH3
........ 2.0............singlet.............0...............3 ........ - CH3 (no H neighbors)
........ 1.2............triplet..............2...............3 ........ CH3 - CH2 - ; farthest from EWG
Because you have four carbons and only three signals, one of your carbons doesn't have any H’s attached to it or there is symmetry in your molecule. Based on your molecular formula, start by assuming that this carbon is involved in a C=O bond. There is no singlet peak that integrates to 1, so you can rule out that the carbonyl is an aldehyde. You do have a singlet peak that integrates to 3, so it’s possible that this comes from a methyl attached to the carbonyl carbon, CH3-C(=O)-R. If you look at the first and third peaks, you can see that they are probably neighboring each other in a fragment that looks like R – O – CH2 – CH3. So now you have two fragments: one is a carbonyl, and the other is an oxygen attached to an alkyl chain. If you look at the molecular formula, you’ll see that you’ve already used all of your atoms between the two fragments, so you just need to put them together. Thus, you end up with an ester: CH3 – C(=O) – O – CH2 – CH3.
Another example:
C4H10O, which has an IHD of 0, so no double bonds or rings
Chemical Shift | Multiplicity | Neighbors | Integration | Guess
........ 4.0............singlet............0...............1 .......... OH (Can’t be CH because you already use 4 carbons for the below peak assignments)
........ 3.5............doublet...........1...............2 .......... CH – CH2 – R, where R is electron-withdrawing (b/c 3.5 ppm)
........ 1.8..........multiplet........many............1 .......... CH next to many H’s
........ 0.9...........doublet............1...............6 .......... 2 CH3’s next to CH, or 3 CH2’s next to CH (unlikely)
Based on the fact that there are only 4 carbons, you can assign the peak at 4.0 ppm to OH. The peak at 3.5 is then the CH2 that is attached to the OH (farthest downfield), the peak at 1.8 ppm is attached next (next farthest downfield), and the peak at 0.9 ppm is the farthest away from the OH. Putting this all together, you get (CH3)2 – CH – CH2 – OH as your solution.
IR:
Tells about functional groups, and is more of a qualitative type of spectra at this level. At the very least, you should be able to identify:
- NH : peak(s) around 3500. R-NH2 has two N-H bonds, so you typically see two peaks in this area. R – NH – R has one N-H bond, so expect only one peak.
- OH : broad peak somewhere around 2500-3300
- C=O : sharp peak around 1700. Depending on the group you have (amide, ester, aldehyde, etc.) you can predict better where it will appear.
When NMR and IR are used in the same problem, the IR helps to decide on which functional groups you may have. For instance, if from the molecular formula you have an oxygen, the IR should be able to tell you whether that oxygen is from an alcohol, ether, or carbonyl.
Feel free to MeMail me if you’d like more help. I’ve been an organic TA for several years, so I have plenty of practice problems with NMR, IR, and combined NMR/IR and can answer questions if you still have them. The key, really, is to practice and get familiar with solving this style of question. As for practice, this website has tons of spectral problems (with solutions) that are organized by the level of difficulty. Good luck!
posted by genekelly'srollerskates at 12:08 PM on May 2, 2011 [2 favorites]
NMR:
I know you said you understand background, but make sure you're solid on key concepts:
- chemical shift - how much the electron cloud ('shield') is being pulled away from the atom (atom = H if you are doing H-NMR), so if you're closer to an electron withdrawing group (EWG), you will be farther left (downfield) on the spectra
- peak regions - how many peak areas there are tells you how many unique H's (for H-NMR) you have. This can give you big clues about symmetry in your molecule, because H's that are symmetrical will be in identical chemical environments, and appear at the same place on the spectra.
- multiplicity - how many times the peak is split tells you how many neighbors that H has (doublet means 1 neighbor, triplet means 2, etc.). Neighbors are H's on an adjacent C’s.
- integration - how many H's are represented by that peak, i.e. how many H's have that same electronic environment.
If you are given a spectra and trying to solve for the structure, make yourself a table for every peak. Make columns labels at the top of the table for (a) chemical shift, (b) multiplicity - like singlet, doublet, triplet, etc., (c) neighbors - determined from multiplicity, (d) integration, and (e) guess. For the guess, you should put down possible things this peak could represent. For instance, if it is a doublet with integration 6, it is likely to represent 2 CH3's next to a CH, but may also be 3 CH2's next to a CH, etc. When I was taking undergrad organic, my professor taught us how to do this table, and it really helped me to convert the overwhelming amount of info conveyed from the spectra into usable pieces of information.
Example: C4H8O2
By calculating the index of hydrogen deficiency, you know there is one double bond
Chemical Shift | Multiplicity | Neighbors | Integration | Guess
........ 4.0............quartet............3...............2 ........ - CH2 – CH3 (CH – CH2 – CH2 can be ruled out b/c of integration of other peaks); near EWG so most likely R - O - CH2 – CH3
........ 2.0............singlet.............0...............3 ........ - CH3 (no H neighbors)
........ 1.2............triplet..............2...............3 ........ CH3 - CH2 - ; farthest from EWG
Because you have four carbons and only three signals, one of your carbons doesn't have any H’s attached to it or there is symmetry in your molecule. Based on your molecular formula, start by assuming that this carbon is involved in a C=O bond. There is no singlet peak that integrates to 1, so you can rule out that the carbonyl is an aldehyde. You do have a singlet peak that integrates to 3, so it’s possible that this comes from a methyl attached to the carbonyl carbon, CH3-C(=O)-R. If you look at the first and third peaks, you can see that they are probably neighboring each other in a fragment that looks like R – O – CH2 – CH3. So now you have two fragments: one is a carbonyl, and the other is an oxygen attached to an alkyl chain. If you look at the molecular formula, you’ll see that you’ve already used all of your atoms between the two fragments, so you just need to put them together. Thus, you end up with an ester: CH3 – C(=O) – O – CH2 – CH3.
Another example:
C4H10O, which has an IHD of 0, so no double bonds or rings
Chemical Shift | Multiplicity | Neighbors | Integration | Guess
........ 4.0............singlet............0...............1 .......... OH (Can’t be CH because you already use 4 carbons for the below peak assignments)
........ 3.5............doublet...........1...............2 .......... CH – CH2 – R, where R is electron-withdrawing (b/c 3.5 ppm)
........ 1.8..........multiplet........many............1 .......... CH next to many H’s
........ 0.9...........doublet............1...............6 .......... 2 CH3’s next to CH, or 3 CH2’s next to CH (unlikely)
Based on the fact that there are only 4 carbons, you can assign the peak at 4.0 ppm to OH. The peak at 3.5 is then the CH2 that is attached to the OH (farthest downfield), the peak at 1.8 ppm is attached next (next farthest downfield), and the peak at 0.9 ppm is the farthest away from the OH. Putting this all together, you get (CH3)2 – CH – CH2 – OH as your solution.
IR:
Tells about functional groups, and is more of a qualitative type of spectra at this level. At the very least, you should be able to identify:
- NH : peak(s) around 3500. R-NH2 has two N-H bonds, so you typically see two peaks in this area. R – NH – R has one N-H bond, so expect only one peak.
- OH : broad peak somewhere around 2500-3300
- C=O : sharp peak around 1700. Depending on the group you have (amide, ester, aldehyde, etc.) you can predict better where it will appear.
When NMR and IR are used in the same problem, the IR helps to decide on which functional groups you may have. For instance, if from the molecular formula you have an oxygen, the IR should be able to tell you whether that oxygen is from an alcohol, ether, or carbonyl.
Feel free to MeMail me if you’d like more help. I’ve been an organic TA for several years, so I have plenty of practice problems with NMR, IR, and combined NMR/IR and can answer questions if you still have them. The key, really, is to practice and get familiar with solving this style of question. As for practice, this website has tons of spectral problems (with solutions) that are organized by the level of difficulty. Good luck!
posted by genekelly'srollerskates at 12:08 PM on May 2, 2011 [2 favorites]
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posted by cholly at 5:40 AM on May 2, 2011