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With the solution of this question another possibility of the way A is to be used ( see step 120):
The compound contains a phenyl group of C6H5. Thus C4H5O must be looked up for univalent remainder, fitting the spectrum. The small number of H-atoms suggests that multiple bonds and thus possibly "long range" couplings.
  1. Predicates from the chemical shifts of the signals:
    d in ppm Possible groupings
    2,3 (triplet) -CH3 (aromatically) CH3-CO- CH3-CºC- -CH2-C=C
      -CH2-CO- -OH H-CºC  
    4,1(Dublett) CH3O- -CH2O- =CH-O- -OH
    4,6 =CH-O- -CH2O- -OH =C=CH2
  2. Analysis of intensities and couplings:
    1. The multiplicity of the signals with d = 2.3 ppm and d = 4.1 ppm shows that CH2- group with a CH-group couples: HC- · · · ·  -CH2-  .

      The intensities acknowledge this findings.
      From the structures possible for the first signal is applicable thus only CºC-. Altogether we maintain: HCºC-CH2-O-.
    2. The singlet signal with d = 4.6 ppm with same intensity as the Dublett can come then only from not coupling CH 2 - group.
  3. Check on the basis the sum formula:
    Since the univalent remainder contains only four C-atoms, the following structure corresponding with the spectrum comes into consideration: HCºC-CH2-O-CH2-

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