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I have been trying to solve this NMR question for about four hours and I feel like I'm always really close but never manage to find the perfect combination.

Anyway here's the info in the question

Formula is $\ce{C11H14O2}$

1.triplet at 2.4 ppm (2H)

2.singlet at 3.3 ppm (3H)

3.triplet at 3.5 ppm (2H)

4.doublet at 7.4 ppm (2H)

5.doublet at 7.5 ppm (2H)

6.singlet at 2.0 ppm integrates for 3H

I have tried a number of structures and the one I feel i got very close to was this one, each number on the side corresponds to the signal that I think it is. The X is just the area where I know i am missing a bond, I just don't know how to seal it up.

enter image description here

Edit: After the advice given by you all, I came up with this structure which seems to be pretty sound

enter image description here

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  • $\begingroup$ Do the doublet and triplet come with coupling constants, or signals that can be picked to be translated in such? $\endgroup$ – Martin - マーチン Apr 22 '15 at 17:44
  • $\begingroup$ unfortunately the information I have written is the only info available in the question. I just assumed that the triplet at 3.5 and the singlet at 3.3 were very close in proximity due to the similar peak values $\endgroup$ – Matthieu Kratz Apr 22 '15 at 17:47
  • $\begingroup$ I just wanted to make that sure. If it is not given in the question itself, then it is usually also not necessary for the problem to be solved. That assumption you make is very reasonable. We have a couple of experts on the network, I am pretty sure, this question won't be without answer long. $\endgroup$ – Martin - マーチン Apr 22 '15 at 17:52
  • $\begingroup$ I do hope so, the real problem I am having with this question is managing to satisfy the atom count, I just can't find a combo of proper peak strucutre and atom that works.... $\endgroup$ – Matthieu Kratz Apr 22 '15 at 17:55
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    $\begingroup$ Are you using table with NMR values - you really should. $\endgroup$ – Mithoron Apr 22 '15 at 18:03
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  1. Calculate your DBEs (double bond equivalents):

\[\mathrm{DBE} = \frac{2c-h+2}{2} = \frac{2\cdot 11 - 14 +2}{2} = 5\] Note that a benzene has 4 DBEs!

  1. Shifts at 7.5 ands 7.4 are definitely not alkine protons. Aromatic protons are more likely, aren't they?
  2. Largest couplings between benzene protons are typically ortho couplings. How large are they in your case?
  3. Two sets of doublets at 7.5 and 7.4 ppm might indicate a 1,4-disubstituted benzene. You might want to play with some increment rules here.
  4. You have two triplets at 3.5 and 2.4 ppm with each 2 H. These might be two neighbouring methylene groups in $\ce{C_{aryl}-CH2-CH2-O-R}$.
  5. R in (5) might be $\ce{CH3}$.
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    $\begingroup$ Thank you very much, I was way too focusd on getting a linear structure that I totally forgot about trying an aromatic structure. $\endgroup$ – Matthieu Kratz Apr 22 '15 at 18:28
  • $\begingroup$ Your DBE's are the same as what I call Index of hydrogen deficiency (IHD), right? $\endgroup$ – gannex Nov 20 '16 at 2:18

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