2.2ppm: a CH3 group (since integrates to 3H) adjacent to a single hydrogen (since a doublet).
6.4ppm: a CH group in which the carbon atom is part of an alkene (CH since integrates to 1H and the chemical shift indicates that the hydrogen attached to an alkene). The multiplicity of this signal (a doublet of quartets) indicates that there are three identical hydrogens and one other hydrogen adjacent to this group. The three identical hydrogens will be the CH3 group at 2.2ppm since they share a common coupling constant (7.0Hz).
6.8ppm: a CH group in which the carbon atom is part of an alkene (CH since integrates to 1H and the chemical shift indicates that the hydrogen attached to an alkene). The multiplicity of this signal (a doublet) indicates that it is adjacent to a single hydrogen which will be the the =CH at 6.4ppm since they share a common coupling constant (15.5Hz)
7.2-7.3ppm: A monosubstituted benzene ring since the peak integrates to 5H and the chemical shift is characteristic of aromatic compounds.
Combining this information gives the basic structure Ph-CH=CH-CH3.
The remaining question is the stereochemistry of the double bond since there is both a cis- and a trans-isomer of this compound. The coupling constant between the two hydrogens attached to the alkene (15.5Hz) is too large to be a coupling constant between two hydrogens which are cis to one another (6-12Hz), but is consistent with two hydrogens which are trans to one another (14-18Hz). Thus, the compound must be E- (or trans-) 1-phenylpropene as shown below.
back to simple answer to question 4 back to CHAPTER 2 answers
back to answers to problems back to STEREOCHEMISTRY home page