Mass spectrometry is used to identify structures of organic compounds. The sample to be analysed is vaporised, fed into the ioniser and fragmented by bombarding electrons. With knowledge on the relative masses and abundances of isotopes making up the compound or by referencing against the spectra of known substances, the structure of the compound can be determined. For example,
Question
With reference to the spectrum below, deduce the structure of an organic compound that forms a sweet smelling liquid with acetyl chloride.
Answer
The sweet smelling liquid could be an ester and hence the organic compound could possibly be an alcohol. Assuming that it is a simple alcohol, its general formula is CnH2n+1OH.
Look for the peak representing the molecular ion (M+), i.e. the unfragmented sample molecule with an electron knocked out. This peak must be one with one of the highest u/z and with a relatively high abundance.
The highest peak is at 33 u/z but has a negligible abundance. Hence, the peak at 32 u/z is most probably the molecular ion. If so,
Hence, the organic compound could be methanol, CH3OH. We can verify our guess by analyzing the other peaks:
|
u/z |
Molecule | Formula |
Logic |
|
33 |
M++1 | 13CH3OH |
13C has an abundance of only 1.07%, but much higher than that of 17O, 18O and 2H. |
|
32 |
M+ | 12CH3OH |
One of the highest u/z and high abundance |
|
31 |
M+-1 | 12CH3O+ |
Cleavage of polarised O-H bond; relatively stable and thus high abundance |
|
30 |
M+-2 | 12CH2O+ |
Loss of 1 proton attached to C and one to O |
|
29 |
M+-3 | 12CHO+ |
Loss of 2 protons attached to C and one to O |
|
28 |
M+-4 | 12CO+ |
Full deprotonation |
|
15 |
M+-7 | 12CH3+ |
Cleavage of polarised -OH bond |
With the above deductions, we can conclude that the spectrum is one for methanol.