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.

EA:
The data adds up to 100% so we know the elements involved : C, H and O.

%C = 62.04 divide by atomic weight  62.04 / 12.011 = 5.165
%H = 10.41 divide by atomic weight  10.41 / 1.008 = 10.327
%O = 27.55 divide by atomic weight  27.55 / 15.999 = 1.722

Now divide by the smallest to get the simplest ratio :  C = 5.165/1.722 = 2.999, H = 10.327/1.722 = 5.997, O = 1.722/1.722 = 1.000

Remember to be very careful when rounding during during EA calculations (it will invariably mean you get the wrong answer if you round too far) AND if you add up the MW using accurate atomic weights it should match exactly.

So we have the empirical (molecular) formula = C3H6O.

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

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  (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. There is only one possibilities for C3H6O.  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.  Here the groups are in priority order O > CH2O > CH3 > H.

Note that cyclopropanol is not chiral as it has a mirror plane or where is the C with 4 different groups attached ?