francis
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Identify the structure
I was given a sheet containing NMR, IR, MS and UV-Vis spectral data.
I determined the structure to be
Is this correct? It appears to fit with the data - though I'm not sure about the UV vis data, which has two wavelengths of maximum absorption (215 and
274nm) with two corresponding molar absorptivities (6000 and 1500).
Also, if my proposed structure is correct, I can't find much information at all about this structure? ChemDraw names it to be
1-(2-ethylphenyl)ethanol, is there a more widely used name?
Lastly...
ChemDraw predicts the -CH- carbon (attached to the hydroxyl group) to have a much larger shift in the 13CNMR than my spectra has: the proposed
structure has 69ppm or so for that carbon in ChemDraw.
No signal is in that region on my spectra (as you can see). The highest shifted non-aromatic carbon is 34ppm or so.
Is my proposed structure still viable?
Thanks alot
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DDTea
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Your structure is viable. Here's how I worked through this:
I started with the mass spectrum. The mass ion is at m/z = 150. From that, I calculated the molecular formula C10H14O (I glanced at the FTIR to
check for obvious functional groups). By the formula DU = C + 1 - 0.5(X - N), where C = # carbon atoms, X = # monovalent atoms, N = # nitrogen atoms
, I calculated 4 degrees of unsaturation.
4 degrees of unsaturation led me to suspect the presence of a benzene ring. Looking at the IR spectrum, this is confirmed by C-sp2 absorptions >
3000 1/cm as well as the ring breathing modes at ~1500 and ~1600 1/cm. Continuing to the 1H NMR now, between 6.9 and 7.5 ppm are 2 doublets and 2
triplets. That corresponds to ortho substitution pattern about the ring.
Returning to the IR, the strong, broad absorption centered at ~3400 1/cm is typical of O-H stretching. That's pretty much all you're going to get
from the IR spectrum--the fingerprint region is going to be tricky to interpret because the peaks are saturated.
So now let's pick apart the 1H NMR. At 1.1 ppm are three protons split into a triplet. That suggests a methyl group split
by a vicinal methine proton, or H3C-CH. That methine, if it has no other neighbors, would be split into a quartet. There is a quartet at 3.3 ppm.
This downfield shift could be caused by bonding of this proton's carbon to a heteroatom or some other electron-withdrawing group. So this must be our
alcoholic carbon. This spin-system is complete: we have H3C-CH(OH)- in addition to an ortho-substituted benzene ring. So, put the two together and
let's work on the benzene ring's other substituent.
Keeping in mind that, based on our molecular formula, we are now down to C2H5, there really aren't many options. It's an ethyl group. The methyl
group of this should appear as a doublet with integration = 3. This is seen at 1.5 ppm. The methylene protons are trickier. Not only are they split
by the methyl protons, but also by each other: they aren't magnetically equivalent. So the splitting pattern will be more complicated. These are the
multiplet at 1.9 ppm.
As far as the UV-Vis goes, you have a very strong absorption and a much weaker one. The strong one at 215 nm would be the pi-->pi* transition.
The weaker one would be an n-->pi* transition, which makes sense for a molecule with a conjugated pi system and also non-bonding electrons.
[Edited on 10-3-11 by DDTea]
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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