A functional group is the atom or group of atoms that gives an organic compound its characteristic reactions. Identifying the functional group is the first step in working out the structure of an unknown compound.At this stage you need to recognise these groups:
alkene: C=C
alcohol: -OH
aldehyde: -CHO
ketone: C=O within the chain (RCOR′)
carboxylic acid: -COOH
haloalkane: C-X, where X is a halogenChemists use three types of evidence together.1. Chemical (test tube) tests. A reagent gives a visible change with one functional group but not others. For example, bromine water is decolourised by alkenes, and sodium carbonate fizzes with carboxylic acids.2. Mass spectrometry gives the relative molecular mass, Mr, from the molecular ion peak. High resolution mass spectrometry can give the molecular formula.3. Infrared spectroscopy shows which bonds are present, for example a broad O-H absorption or a sharp C=O absorption.Worked example: a compound has molecular formula C3H6O. It does not decolourise bromine water and is not oxidised by acidified potassium dichromate(VI). Its IR spectrum shows a strong absorption at 1715 cm−1.
No reaction with bromine water: no C=C.
Not oxidised: not a primary or secondary alcohol, and not an aldehyde.
Absorption at 1715 cm−1: C=O present.
So it is a ketone: propanone, CH3COCH3.Combining several pieces of evidence is essential, because a single test rarely identifies a compound on its own.
alkene: C=C
alcohol: -OH
aldehyde: -CHO
ketone: C=O within the chain (RCOR′)
carboxylic acid: -COOH
haloalkane: C-X, where X is a halogenChemists use three types of evidence together.1. Chemical (test tube) tests. A reagent gives a visible change with one functional group but not others. For example, bromine water is decolourised by alkenes, and sodium carbonate fizzes with carboxylic acids.2. Mass spectrometry gives the relative molecular mass, Mr, from the molecular ion peak. High resolution mass spectrometry can give the molecular formula.3. Infrared spectroscopy shows which bonds are present, for example a broad O-H absorption or a sharp C=O absorption.Worked example: a compound has molecular formula C3H6O. It does not decolourise bromine water and is not oxidised by acidified potassium dichromate(VI). Its IR spectrum shows a strong absorption at 1715 cm−1.
No reaction with bromine water: no C=C.
Not oxidised: not a primary or secondary alcohol, and not an aldehyde.
Absorption at 1715 cm−1: C=O present.
So it is a ketone: propanone, CH3COCH3.Combining several pieces of evidence is essential, because a single test rarely identifies a compound on its own.