Part 6: STRUCTURE DETERMINATION
We have to identify two compounds A
and B. An analysis
of the data from the question is presented below:
A and B are isomers
=> they have the same molecular
formula
The molecular formula can be
obtained from the combustion analysis
data as: C3H6O2 ( HOW
? )
This gives an
IHD
of 1
which implies 1 ring or 1 pi bond.
Using this and an analysis of the
NMR
provides us with:
A:
-
We have two signals, implying 2 types of H.
-
3.7ppm, singlet , 1H
-
2.1ppm, singlet, 1H
-
Since we have 1:1 integrals, this implies they are both 3H
-
As they are 3H signals, this suggests two different Me groups.
-
Both are deshielded (compared to a hydrocarbon) but one much more that
the other.
-
Chemical shift at 2.1ppm could be CH3C=O
-
Chemical shift at 3.7ppm could be CH3O-
This accounts for all of the formula,
so the structure of A has
to be the ester, methyl ethanoate (methyl acetate) CH3CO2CH3
B:
-
We have three signals, so 3 types of H.
-
8.0ppm, singlet 1H
-
4.2ppm, quartet, 2H
-
1.3ppm, triplet, 3H
-
The quartet and the triplet suggest an ethyl group: -CH2CH3
-
The downfield position of the -CH2- indicates it is attached
to -O- rather than -C=O, so we have -OCH2CH3
-
The very downfield peak at 8.0ppm is similar to that of an aldehyde
(9-10ppm),
but not downfield enough for a carboxylic acid (12ppm)
This leaves H, C, and O from the
formula and the IHD = 1 to account for
which must be as a C=O. Then the only way to fit this together is that
B must be an ester (hence the similar boiling points) in this case
ethyl
methanoate (ethyl formate), HCO2CH2CH3