I recently did an identification of aldehydes and ketones lab at uni. The NMR spectrum attached is for my unknown. Now normally I can do these just
fine, but looking at the integrations for the peaks I am getting ~30 hydrogens in my unknown, I think this NMR is rather bad actually. The large peak
on the NMR at ~1.4ppm is seeming to indicate way too many hydrogens... The DNP derivative of the unknown melted at~95C although I have my doubts on
its purity, the recrystalization solvent they told us to use took very little to dissolve the hydrazone derivative, so washing it after filtering
would have caused too much loss.
I have it narrowed down to n-pentanal, n-heptanal or n-octanal. What does everyone think? I am leaning more towards the pentanal because of the
lower mp of the DNP derivitave (98C vs 107C)
Thanks all
(PS I'm on a bit of a deadline for this-monday)
Magpie - 11-3-2006 at 19:49
Its been a while since I did these but do have my notes so can review them.
You are suspicious of the 30 hydrogens and rightly so as this would likely be a solid right? Can you check the integration by cutting out the peaks,
weighing them, and then comparing?The_Davster - 11-3-2006 at 19:59
I had never heard of the cutting method before, but I just tried it and my scale does not even register the largest peak.
My unknown was a liquid, and it reeked something horrible, similar to the inside of a pumpkin but more pungent and bad.
So I was looking at it some more, the first peak, must be the hydrogen(1) on the same carbon as the carbonyl, the second must be the hydrogens on the
carbon next to the carbonyl(2) and the fourth peak must be the furthest hydrogens from the carbonyl(3). Just that damn multiplet is giving me hell...
[Edited on 12-3-2006 by rogue chemist]Magpie - 11-3-2006 at 20:08
I don't suppose you'll have a chance to do anymore testing like boiling point or refractive index.The_Davster - 11-3-2006 at 20:13
Unfotunatly not, I usually end up keeping my excess sample after the lab, but this stuff smelled so bad I did not want to keep it.
[Edited on 12-3-2006 by rogue chemist]turd - 12-3-2006 at 00:44
Whoa, what kind of stone-age NMR is that? Are you still using TMS as reference? Scary.
Anyway, it maybe not be what they want to hear, but you gave the only scientific correct answer: with the poor resolution you can't really tell.
Your interpretation of the aliphatic Hs seems correct. What I think where you are wrong is the 30 Hs. The signal of the single CHO-proton is way too
weak to be used as reference. What you should be doing instead is comparing the size of the integrals of the aliphatic Hs. How would the relations
look like for n-pentane or n-hexane? Could you have a tertiary or quarternary C?Nerro - 12-3-2006 at 03:42
Who cares what ref you use?
At what frequency was this spec taken?
Why is the resolution so bad on a <sup>1</sup>H-spectrum? I did a protonspectrum on a tiny fleck of product a while back and got a
perfectly defined sample. (200 MHz NMR apparatus).
Why do you base your integrals on a some tiny one if you know you have a liquid with a simple structure? 30 H's is pretty unlikely wouldnt you say?
Call that tiny peak 1 and then forget about it when comparing the others.The_Davster - 12-3-2006 at 10:24
Quote:
Originally posted by Nerro
Call that tiny peak 1 and then forget about it when comparing the others.
Yeah I just did that, I'm pretty sure it's pentanal now, although hexanal would make more sense, it is not a possible unknown.frogfot - 12-3-2006 at 11:05
Quote:
The signal of the single CHO-proton is way too weak to be used as reference
Why not, it seems to have a relatively big area. RC, although crappy resolution/shimming and impurities one can clearly see the four hydrogen types:
CHO, CH2CHO, n*CH2 and CH3 just as you said. Btw, doesn't the soft you're using have integration function..? It would be best to take the methyl group
as the reference..
[Edited on 12-3-2006 by frogfot]The_Davster - 12-3-2006 at 12:10
It's not software, its just a spectrum they handed out in the lab.
Thanks so much for that link, it appears to be heptanal now...just after I finished my report making it out to be pentanal...I guess this means my DNPH derivative was quite impure...damnturd - 12-3-2006 at 12:53
Quote:
Thanks so much for that link, it appears to be heptanal now...
In heptanal you have a CH2:CH3-proton ratio of 10:3 or more than 3:1. In your spectrum you don't have 3x as much CH2 as CH3, therefore this can't be
heptanal.
Edit: looking more closely this could actually be 3:1. Looks more like heptanal than pentanal after all. But the sample is dirty and the resolution is
poor, so it's hard to tell... :/
BTW: you don't see the CHO-proton on your spectrum at all - it's out there at 10ppm.
[Edited on 12-3-2006 by turd]Magpie - 12-3-2006 at 14:40
Assuming:
1. This HNMR is gospel as it was provided to you.
2. This is a normal aldehyde since you say it is
3. The 1st downfield peak is the reference at 0 ppm
4. Turd is correct that the -CHO peak is at 10 ppm and is therefore not visible to us.
Based on the Japanese library the 4th downfield peak is for the 2 protons of -CH2- adjacent to -CHO. If you accept this then by scaling the graph the
two remaining peaks together must have about 23 protons. Total protons would then be 1+2+23 = 26. This would then have to be C13H26O. But this is
not consistent with:
1. liquid state of the sample
2. low melting point of the derivative
So is it possible that this NMR is trash, or WTF? Fleaker - 12-3-2006 at 14:51
"My unknown was a liquid, and it reeked something horrible, similar to the inside of a pumpkin but more pungent and bad."
"Yeah I just did that, I'm pretty sure it's pentanal now, although hexanal would make more sense, it is not a possible unknown. "
From the descriptive side of it Rogue Chemist, I should mention hexanal has a different odor from which you described. Hexanal is very reminsicent of
freshly mown grass.Marvin - 13-3-2006 at 06:20
"BTW: you don't see the CHO-proton on your spectrum at all - it's out there at 10ppm."
There is an offset peak by 6ppm, add that to its position and get 9.75ppm, or nigh on exactly what is on the heptanal NMR, its a very good match.
This is a standard way of representing off scale peaks.turd - 13-3-2006 at 10:39
Quote:
There is an offset peak by 6ppm, add that to its position and get 9.75ppm, or nigh on exactly what is on the heptanal NMR, its a very good match.
This is a standard way of representing off scale peaks.
Interesting, I have not seen this before. But then neither have I seen TMS actually being used. You live to learn.Marvin - 13-3-2006 at 13:33
If fact, I should add thats the standard way for academia to distribute photocopies for students. I imagine the machines don't spit them out that way
normally.
An IR would probably have helped somewhat and purity should have been indicated by the nature and range of the derivitive melting. All very easy to
say this in retrospect.Quibbler - 14-3-2006 at 05:33
It's impossible to say because the integrations are so bad. Maybe there's some hydrocarbon impurity there. It's a long time since I've seen a CW
spectrum though.
BTW the reason old spectra are often used is
1. Laziness
2. At lower frequency the spin-spin interaction is more clearly seen.turd - 14-3-2006 at 12:38
How do you know it's CW? From the poor resolution?Xque - 14-3-2006 at 13:26
Quote:
Originally posted by turd
Whoa, what kind of stone-age NMR is that? Are you still using TMS as reference? Scary.
Could you please develop that point further? What is the reference superior to TMS? I've just finished going to lectures on NMR, and I had the
impression that TMS was the standard reference used in NMR.turd - 14-3-2006 at 13:37
Oh, there's nothing wrong with TMS, it's just that you don't use it anymore (or so I thought). Nowadays you just use the solvent peak.Nerro - 14-3-2006 at 23:52
You dont always have a solvent peak to go by. What if you're making a proton NMR of a liquid that you have plenty of? You don't dissolve that in
anythingso no solvent peak, so you use TMS.
And btw, even if you use the solvent peak when you can you still set that peak elative to the virtual TMS peak Quibbler - 15-3-2006 at 04:03
You can tell it is CW as sharp peaks show a beat pattern see the right side of the TMS peak.
You can also say this is probably a 60 MHz spectrum as the coupling in hydrocarbons is about 5 to 6 Hz.
Superconducting systems are so stable that a reference is often not needed. Any slight changes in field are compensated for by locking on to the
deuterium signal from the solvent (D is spin 1).
I suspect that the -CHO proton's chemical shift will vary (a lot) depending on the solvent and concentration so it cannot be used for identification.Sandmeyer - 28-4-2006 at 20:12
Kind of off topic but I always wondered why they buy expensive CDCl3 and not use CCl4 as solvent instead.turd - 28-4-2006 at 22:32
Quote:
Originally posted by Sandmeyer
Kind of off topic but I always wondered why they buy expensive CDCl3 and not use CCl4 as solvent instead.
I'm not an NMR-expert, but I can think of 3 reasons:
1) You lock on the D-signal
2) T1-relaxation is worse in CCl4
3) CCl4 sucks as a solvent
.
