Question

What qualitative assay would you use to distinguish vanillin from 5-bromovanillin and what would you expect to be the major difference between the GC/MS of vanillin and 5- bromovanillin?

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Answer 1

1) Qualitative analysis: To distinguish vanillin and 5-bromovanillin the element detection method (Lassaigne’s test) can be used.

5-bromo vanillin will give a confirmative test for the bromine. Whereas, vanillin gives a negative test for bromine.

In the element detection test, the substance is fused with sodium metal to give different salts of sodium. e.g. 5-bromo vanillin will give NaBr in the extract whereas vanillin will not give any of such salt.

Thus, the extract upon reaction with AgNO3 gives the yellowish-white ppt of AgBr.  

AgNO3(aq) + Br-(aq) -------> AgBr(s)

However, vanilline sodium fussion extract will not give any such ppt as it does not contain any Br atom.

2) The melting point of vanillin is 81-83 and that of 5-bromo vanillin is 352^oC, hence in GC vanillin will have lower retention time than the 5-bromo vanillin.

In the mass spectrum, the M+ peak for vanillin will be 152 with no prominent M+2 peak and that of 5-bromo vanillin will give M+ at 232 with M+2 at almost of the same intensity.

Because bromine-containing compounds give characteristic M+2 peak with equal intensity of M+ ion peak.

And this peak is due to the natural abundance of bromine isotopes with the almost the same amount.