Part 6: STRUCTURE DETERMINATION

EA:
Since the data doesn't adds up to 100% there must be another element. IT CAN'T BE A HYDROCARBON !
Prerequisite knowledge should indicate that this is probably oxygen. But we don't need to know yet because we have the molecular weight :

%C = 66.63  :   66.63 % of  72.107 = 48.044 which means there are 48.044 / 12.011 = 4 carbons
%H = 11.18  :   11.18 % of  72.107 = 8.061 which means there are 8.061 / 1.008  = 8 hydrogens

If we use that we get that C : H ratio is 1 : 2 and that the % account for C4H8.  That leaves 72.107 - ( 4 x 12.011 + 8 x 1.008) = 15.999 which exactly matches  the atomic weight of oxygen.

Remember NEVER TO ROUND DATA during EA calculations (it will invariably mean you get the wrong answer) AND if you add up the MW using accurate atomic weights it should match exactly.

The molecular formula = C4H8O.

So what functional groups could we have ?  The formula is unsaturated, so we need to remember double bonds too.
We can have the following functional groups :  alkene, alcohol, aldehyde, cycloalkane, epoxide, ether, ketone  (for examples, see diagram below).
Too 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. There are lots of possibilities. Only a  few are shown. Biggest error was students not checking the molecular formula and not even drawing a pair of isomers !

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 are only a few possibilities for C4H8O.  The biggest error was drawing that lacked a chirality center or didn't name them. The most common answer was the butenol system. In this case you need to make sure you number it for the higher priority -OH as an alcohol and remember for name to be complete you must assign as R or S.