3
$\begingroup$

I have the following two spectra (IR and 1H-NMR) see below.

The chemical formula is: C16H16O2

Thus DBE = 9

My approach:

From the IR, I can see that there's no broad peak around 2500-3300, thus I can rule out any carboxylic acids and alcohols.

Then I'll look at the sharp peak at 1734 which makes me suspect an aldehyde.

From the H-NMR I see the following:

enter image description here

This is where I run into some issues.

I see peak a and c as -CH2- groups next to another R-CH2-, but in two different chemical environments.

And I suspect that peak d might be caused by two benzene rings (2x5H=10H) in a symmetric molecule, like; Ar-R-Ar but I can't stitch these informations together to something useful.

I hope that someone can help me out. enter image description here

|improve this question
$\endgroup$
  • $\begingroup$ There's no aldehyde, you do not have the NMR peak for CHO $\endgroup$ – Waylander Oct 31 at 11:50
  • $\begingroup$ Do you have an idea of what it could be then? Or at least what could cause the 10H peak? $\endgroup$ – Tino Petersson Oct 31 at 11:51
  • $\begingroup$ Your signal at 4.4 could be ArOCH2R $\endgroup$ – Waylander Oct 31 at 11:53
  • $\begingroup$ Well shouldn't that R-group then be another CH2 for it to have a multiplicity of a triplet? And do you have any idea of the peak at ~7.35? $\endgroup$ – Tino Petersson Oct 31 at 11:56
  • 2
    $\begingroup$ I'd suppose it's probably Ph-CH2-CH2-O-C(O)-CH2-Ph? $\endgroup$ – PCK Oct 31 at 12:48
0
$\begingroup$

This is not a hard question when compared to other questions of same kind. The first easy clue is the molecular formula, which is already given. That make your life much easier. The given molecular formula for the compound is $\ce{C16H16O2}$, hence you can determine the degree of unsaturation of the molecule ($\ce{C_nH_nO_x}$):

$$u= \frac{n \times 2 + 2 - m}{2}=\frac{16 \times 2 + 2 - 16}{2}=9$$

Based on $\ce{^1H}$-$\mathrm{NMR}$, you correctly proposed that there are two mono-substituted aromatic nuclei in the molecule, since there is a 10-proton multiplet ($2 \times 5\ce{H} = 10 \ce{H}$) in the aromatic region ($\pu{7-8 ppm}$). However, instead of your suggestion of two symmetric aromatic rings like $\ce{C6H5-R-C6H5}$, these two phenyl groups are not probably identical rings. That because, you can see a clear 2-proton doublet (not 4-proton) at around $\pu{7.2 ppm}$, separating from the rest of 8-proton multiplet. Regardless, these two aromatic nuclei would count for 8 degrees of 9 possible unsaturations and $\ce{C12H10}$ portion of the molecule leaving only $\ce{C4H6O2}$ to be accounted for.

The presence of strong peak at $\pu{1734 cm-1}$ in given FT-IR spectrum suggests that there is a carbonyl group in the molecule. As you suspected, it is neither a carboxylic acid group (absence of broad peak at $\pu{2500-3300 cm-1}$ in the given FT-IR), nor an aldehyde group (absence of 1-proton resonance at $\pu{9-10 ppm}$ in the given $\ce{^1H}$-$\mathrm{NMR}$). Nevertheless, it has counted for the last degree of unsaturation. Furthermore, that also counted for one oxygen atom and a carbon atom of the molecule as well. Now, we need count for only $\ce{C3H6O}$.

Now, let's analyze $\ce{^1H}$-$\mathrm{NMR}$ spectrum. There is a 2-proton singlet at around $\pu{3.6 ppm}$ for isolated methylene group (without a neighboring hydrogen), indicating it is most probably next to a oxygen atom in the molecule or in between one of phenyl group and the carbonyl group, based of its deshielding nature. Also, there are two 2-proton triplets present in the $\ce{^1H}$-$\mathrm{NMR}$: One is present at around $\pu{2.9 ppm}$ and the other at around $\pu{4.3 ppm}$, suggesting the presence of isolated (no other protons at vicinity) -$\ce{CH2CH2}$- group in the molecule. These 3 -$\ce{CH2}$- groups are accounted for the rest of $\ce{C3H6}$ portion of the molecule.

Furthermore, the highly deshielding nature of the resonance at around $\pu{4.3 ppm}$ strongly suggest that corresponding -$\ce{CH2}$- group is next to the final oxygen atom to be counted. Also, the other -$\ce{CH2}$- group corresponds to the resonance at around $\pu{2.9 ppm}$, which is only moderately deshielded, clearly suggests that it is neighboring the second phenyl group.

Based on these assignments, one can deduce the structure as 2-diphenylethyl 2-phenylacetate as depicted in the diagram below (inserted blue box) with suggested $\ce{^1H}$-$\mathrm{NMR}$ spectrum by ChemDraw, which gives a close resemblance to the actual one:

2-diphenylethyl 2-phenylacetate

|improve this answer
$\endgroup$
2
$\begingroup$

I will ignore the IR as I always do and work only with the NMR.

  • The ten aryl protons are probably two phenyl groups ($\ce{C6H5-{}}$). That leaves $\ce{C4H6O2}$ and takes care of eight of the nine double-bond equivalents.

  • The two triplets are very likely (examining the coupling constant would confirm that) part of a $\ce{-CH2-CH2-{}}$ fragment. On either side of these is either an oxygen or a quarternary carbon as there are no additional couplings.

  • The final singlet points to a $\ce{-CH2-{}}$ group with the same constraints as above.

  • We still have $\ce{CO2}$ and one double-bond equivalent to take care of. One functional group particularly suggests itself. Then, we have to play connect the fragments.

  • $\ce{CO2}$ is likely an ester functionality since we do not have a carboxylic hydrogen to spare.

  • One of the two triplet $\ce{CH2}$ is shifted downfield rather strongly. This suggests that it is directly bound to the ester group oxygen, giving $\ce{-CH2-CH2-OC(=O)-{}}$.

  • Both of the remaining $\ce{CH2}$ groups will be benzylic but the singlet is more deshielded than the triplet. That suggests that the singlet $\ce{CH2}$ is not only benzylic but also α to the ester.

  • Therefore, the final result is $\ce{C6H5-CH2-CH2-OC(=O)-CH2-C6H5}$.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.