Part 6: STRUCTURE DETERMINATION

Note that the question said "ALL the questions in this section should be answered based on the following data"....

Therefore, full marks were only possible if that was done.
Answers where functional groups were identified that did not match the molecular formula given by the student did not get credit.
Some students named substituents NOT the functional groups as the question required.
Enantiomers based on a different molecular formula did not get full credit since, in effect, a "different" question was being answered, not the one we asked ! In order to draw an enantiomer, a 3D drawing is needed, e.g. use wedge and hash or Fischer diagrams. Without the 3D it is impossible to assign R and S.

a.   Nice and easy to begin with.... we have the molecular formula = C4H8O2.
      Therefore the molecular mass = (4 x 12.011) + (8 x 1.008) + (2 x 15.999) = 88.106 g/mol

b.   Index of hydrogen deficiency ? Either draw out an example and count the pi bonds and rings, or use the formula.... IHD = 1

c.   So what functional groups could we have ?  The formula is unsaturated (IHD > 0) so we can have double bonds or rings too in the structures.....

      We could have the following functional groups : 

alkene, alcohol, aldehyde, cycloalkane, epoxide, ether, ketone, carboxylic acid and ester  (for examples, see the diagram below).

Many students failed to remember the basic rules of valence a drew O and C atoms that violated the octet rule by having too many bonds.

Constitutional isomers have different connectivities due to different functional groups or branching. The biggest errors were students not checking the molecular formula, and not even drawing a pair of isomers or poor nomenclature !


Enantiomers are non-superimposable mirror images, for that they need a chirality center, which is this case needs to be an sp3 C with 4 different substituents attached.  The biggest error was drawing a molecule that lacked a chirality center or didn't show it in 3D (they are stereoisomers after all).  Final task was to assign as R or S. 

f.    The only way to get a compound with the right molecular formula where there is only one type of C, one type of H and one type of O is shown below.  The image on the right shows the mirror planes to help you see the symmetry.

structure