# Determination of structure with HNMR

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.

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

• Do the doublet and triplet come with coupling constants, or signals that can be picked to be translated in such? Apr 22 '15 at 17:44
• 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 Apr 22 '15 at 17:47
• 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. Apr 22 '15 at 17:52
• 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.... Apr 22 '15 at 17:55
• Are you using table with NMR values - you really should. Apr 22 '15 at 18:03

$\mathrm{DBE} = \frac{2c-h+2}{2} = \frac{2\cdot 11 - 14 +2}{2} = 5$ Note that a benzene has 4 DBEs!
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 $\ce{CH3}$.