Combining analytical techniques

In a nutshell

Various different types of spectroscopic techniques can be used in tandem to work out the structure of an unknown organic compound.

Combining spectroscopic techniques

Lots of different spectroscopic techniques can be used to identify an unknown molecule. However, certain techniques can be more useful than others in helping to identify unknown organic compounds.

Mass spectrometry

Mass spectrometry tells you the molecular mass of different fragments and from this, the molecular mass can be determined.

Infrared spectrometry

Infrared spectrometry can help identify functional groups as certain functional groups absorb at a particular wavenumber range.

Proton NMR

Proton NMR can show all the different hydrogen environments as well as the number of hydrogens in an environment via integration traces. The adjacent number of hydrogens can also be found by high-resolution proton NMR and it can be determined by the peak splitting.

Carbon-$$13$$​ NMR

Carbon NMR can show all the different carbon environments. The number of carbons cannot be calculated since carbon NMR only works on the isotope $$^{13}C$$ , which makes up only $$1\%$$​ of carbon isotopes.​

Elemental analysis

In elemental analysis, experiments are conducted to determine the percentage compositions of different elements in a compound. This can give either the empirical or the molecular formulae.

Example

From this we are able to deduce that we have at peak with an $$m/z\ =\ 15$$, this is likely due to a $$CH_3^{+}$$. Peak at $$m/z\ =\ 43$$, this is likely to be a peak due to a $$H_3C-C^+=O$$ fragment. The peak at $$m/z\ =\ 45$$ is likely to be a peak due to a $$^+COOH$$ fragment. Lastly the peak at $$m/z\ =\ 60$$ indicates a molecular mass of $$60$$.​

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The IR spectrum of ethanoic acid shows a sharp peak at $$\approx\ 1760cm^{-1}$$​ is a $$C=O$$ band. The broad peak at $$\approx\ 3100\ cm^{-1}$$ indicates an $$O-H$$ stretch from a carboxylic acid.​

The carbon NMR spectra shows that there are two carbon environments. The first peak occurs at $$\approx\ 180\ ppm$$. This indicates at carbonyl environment $$(C=O)$$. Lastly, the peak at $$\approx\ 20\ ppm$$ indicates an alkyl group.​

Knowing that the molecular mass of the unknown compound is $$60$$ as well as knowing that the molecule most likely contains a carboxylic acid, this can be removed from the molecular mass to calculate the length and molecular mass of the alkyl chain. 

Carboxylic acid contains $$(COOH)$$ has an atomic mass of: $$12(C)\ +\ 16(O)\ +\ 16(O)\ + 1(H)\ =\ 45$$. Removing this from the total mass gives the remaining $$60M_r\ -\ 45M_r\ =\ 15M_r$$.  The alkyl chain which is equal to this mass is $$(CH_3)$$. 

The resultant molecule is:

This can be cross-referenced with the information above; we have two carbon environments, a molecular mass of $$60$$ and a carboxyl functional group.