disturbed. It is thus possible to deduce the required con- stitution of a molecule from a knowledge of the constitu- tion of those chains of atoms into which it can be decom- posed or from which it can be built up by synthesis. Examples :
a. C 2 H 6 0. Two structures are possible ; viz.
CH 3 — CH 2 OH and CH 3 — 0— CH 3 .
Two isomeric compounds are known : ethyl alcohol and methyl ether.
The alcohol contains a typical hydrogen atom, and, as the formula CH 3 — CH 2 OH is the only one which satisfies this condition, it is assigned to the alcohol.
Thi$ is confirmed by the fact that ethyl chloride, which can only have the formula CH 3 .CH 2 C1, can be obtained from the alcohol by the following reactions :
and CH 3 CH 2 OH + HC1 = H 2 + CH 3 CH 2 C1.
The methyl ether must, therefore, have the other formula CH 3 — — CH 3 , which suits it perfectly. To begin with, this ether has no typical hydrogen atom, which fact is expressed by the formula, and further, it can be synthesised from substances which admit of only one formula, as for instance :
CH 3 ONa + CH 3 I = Nal + CH 3 — 0— CH 3 (Williamson).
b. Constitution of acetic acid, C 2 H 4 2 .
We have learned that the constitution of ethyl alcohol is CH 3 .CH 2 OH, and we know from experiment that by oxida- tion of it we can replace H 2 by 0, and obtain acetic acid. Now this acid contains a typical hydrogen atom, probably forming part of a hydroxyl group. Indeed, only one of the four hydrogen atoms is replaceable by metals or alcoholic radicals. And furthermore, the group (OH) can be replaced by CI in the following manner :
C 2 H 3 O.OH + PC1 6 = P0C1 3 + HC1 + 0,H,O-GL
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