jwhitesell.ucsd.edujwhitesell.ucsd.edu/documents/2015spKeyFinalBlue.pdf · 2015-06-10 · Provide a complete mechanism including curved arrows showing the flow of electrons for all
The solution to the NMR question of both exams is quite simple:the formula indicates the presence of one ring or one double bond:all 10 hydrogens can be seen as two CH3 and three CH2 groups;these five groups account for five of the six carbons and all of the hydrogens leaving CO;can not be an alcohol as no OH; can not be an ether as there are no H on carbon bearing O;thus, the O must be a ketone (C=O) and thus there is a double bond but no ring.thus, you must put together CH3 CH3 CH2 CH2 CH2 C=O so as to account for the splitting;as there is not methyne carbon, there is not branching and all six carbons form a straight chain;the O can not be on the end of the chain (that would be an aldehyde which would show a single proton;thus, the carbonyl group must be on the 2nd or 3rd carbon of the chain; on the 3rd carbon there are two CH3CH2 groups; on the 2nd carbon there is only one.
