(a) From the elemental analysis the empirical formula of
Compound O is C
6H
12O.
(b) The M
+ peak at
m/
z = 100 is evidence that the molar mass is 100 g mol
-1 and hence the molecular formula of
Compound O is the same its empirical formula, i.e. C
6H
12O. The fragment at
m/
z = 71 is due to loss of either –CHO or –CH
2CH
3 i.e. (M – CHO)
+ or (M – C
2H
5)
+ and the fragment at
m/
z = 29 is due to CHO
+ or C
2H
5+. (Although not on the IB syllabus the fragment at
m/
z = 72 is due to H
+ recombining after loss of an ethyl group to form C
2H
5CHCHOH
+ and the fragment at
m/
z = 43 is due to –CHCHOH
+ after the loss of the second ethyl group).
(c) The absorptions at just below 3000 cm
-1 are due to C–H and the sharp absorption at 1730 cm
-1 is due to the presence of a C=O. double bond.
(d) The
1H NMR spectrum shows shows that the hydrogen atoms are in four different chemical environments. The position of the single proton split into a doublet with a shift of 9.6 ppm is indicative of an aldehyde –CHO adjacent to a C atom containing one H atom. This means that
Compound O is an isomer of hexanal. The triplet at 0.9 ppm (due to –CH
3) and the complex signal at 1.6 ppm (due to –CH
2–-) indicate two ethyl groups. These are bonded to a carbon atom bonded to one hydrogen atom (hence the complex splitting pattern) and the aldehyde functional group which gives the complex pattern for a single proton at 2.1 ppm.
All this information taken together confirms that
Compound O is
2-ethylbutanal, (C2H5)2CHCHO.