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
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!


Shifts at 7.5 ands 7.4 are definitely not alkine protons. Aromatic protons are more likely, aren't they?
Largest couplings between benzene protons are typically ortho couplings. How large are they in your case?
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. 
You have two triplets at 3.5 and 2.4 ppm with each 2 H. These might be two neighbouring methylene groups in $\ce{C_{aryl}-CH2-CH2-O-R}$.
R in (5) might be $\ce{CH3}$.