Determination of structure with HNMR
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 $\ce{CH3}$.