COH, is the simplest tertiary alcohol, and was obtained by A.
4,as mannose is produced by reversing the sign of the asymmetric system adjoining the terminal COH group.
If the asymmetric system adjoining the COH group, which is that introduced in synthesizing the hexose from the pentose, be eliminated, the formulae at disposal for the two pentoses are CH 2 (OH) - - - COH CH 2 (OH)+-- COH.
As a matter of fact, only arabinose gives an active product on oxidation; it is therefore to be supposed that arabinose is the - - - compound, and consequently CH 2 (OH) - - - + COH = /-glucose CH 2 (OH) + - - - COH = l-gulose.
It can be shown that d-galactose is CH 2 (OH) + - + - COH, and hence d-talose is CH 2 (OH) + - + + COH.
Now in oxidizing, or introducing more oxygen, for instance, by means of a mixture of sulphuric acid and potassium bichromate, and admitting that oxygen acts on both compounds in analogous ways, the two alcohols may give (as they lose two atoms of hydrogen) CH 3 CH 2 COH and CH 3 C0 CH 3.
The first compound, containing a group COH, or more explicitly 0 = CH, is an aldehyde, having a pronounced reducing power, producing silver from the oxide, and is therefore called propylaldehyde; the second compound containing the group - C CO C - behaves differently but just as characteristically, and is a ketone, it is therefore denominated propylketone (also acetone or dimethyl ketone).