would be facilitated by the high concentration of hydrogen-ion.
Summary.
Oxidation of Organic Compounds.
Formaldehyde, like other aldehydes, is readily oxidized. A favorite reagent, used in oxidizing it, is an ammoniacal solution of silver nitrate (exp.), the separation of silver from such a solution being a characteristic reaction of aldehydes. The reagent is rendered still more sensitive by the addition of sodium or potassium hydroxide.[579] We may ask how we would interpret, from the point of view of the electric theory of oxidation and reduction, the oxidation of formaldehyde and the reduction of silver nitrate to silver, under these conditions. According to the theory, the oxidizing agent in silver nitrate is the silver-ion, the discharge of which gives positive electricity, which the oxidized substance, the formaldehyde, must absorb. But in a silver nitrate solution there is a far larger concentration of the silver-ion than in an ammoniacal solution (p. [220]), containing the same total concentration of silver. The complex silver-ammonium-ion Ag(NH3)2+, it may be recalled, is a rather stable one,[580] and, consequently, the addition of ammonia to silver nitrate should decidedly weaken its oxidizing power. Still, the practical use of ammonia, especially in combination with sodium hydroxide, is found to be most effective. We are led to suspect that, in spite of the untoward effect of ammonia on the oxidizing power of the silver compound, an alkaline solution is desirable for the sake of the effect of the alkali on formaldehyde, the reducing substance involved. To follow up this conclusion, we must next consider, in some detail, the nature of formaldehyde; we shall presently find that the conclusion, which we have just reached, as to the probably favorable effect of alkali on the reducing power of formaldehyde, will be verified by experiments, which the consideration of formaldehyde will suggest.
The oxidation of formaldehyde may most clearly be formulated on the basis of views, developed by Nef, on the formation of methylene[581] derivatives, containing bivalent carbon atoms. A solution [p291] of formalin contains formaldehyde in a variety of forms, in a very complex condition of equilibrium. Of these compounds, the aldehyde, CH2O, probably exists in two forms, which have the same composition and molecular weight, but which differ in the arrangement of the atoms in the molecules (in the structure of the molecules); we probably have CH2═O ⥃ CH(OH), the former of which (CH2═O) is, most likely, by far the more stable and the chief one of these two substances, present under ordinary conditions. The second compound CH(OH) may be present in traces only. One difference, we note, lies in the position of one of the hydrogen atoms in the respective molecules; the second form contains a hydroxide group (OH), which gives it the properties of an acid and renders it capable of forming salts CH(OMe) with bases. But the molecule of this second form also would contain a carbon atom, only two of whose valences are satisfied (by H and OH), two of the ordinary four valences of a carbon atom being thus left free or unsaturated. We may indicate the two free carbon valences in the formula ═CH(OH). Such an unsaturated, bivalent carbon atom ═C would be particularly sensitive to oxidation.[582]
Besides these two forms, a formalin solution also contains a polymerized form, probably (CH2O)2, which in dilute solution, or under the influence of heat, slowly breaks down into formaldehyde, (CH2O)2 ⇄ 2 CH2O.
The addition of alkali to the mixture probably leads to the formation of the salt ═CH(OMe), thus disturbing all the conditions of equilibrium and leading to the transformation of a very much larger part of the aldehyde into a compound containing the characteristic unsaturated (bivalent) carbon, than was originally present. The aldehyde will thus become more susceptible to oxidation as a result of the enormous increase in the concentration of the oxidizable component. We may assume this to be either the salt, ═CH(OMe), or its negative ion, ═CH(O−), or both, or some analogous derivative. Further, the two free valences of a bivalent carbon atom may be considered to consist of a positive and a negative charge of electricity, either actual or potential,[583] and the oxidation will consist [p292] primarily in the absorption of two positive charges, from the oxidizing agent, to convert the negative charge on the carbon atom, say in ±CH(ONa), into a positive charge.[584] If the oxidizing agent is alkaline silver nitrate solution, we may formulate the successive actions as follows:
(NaO)HC± + 2 Ag+ → (NaO)HC2+ + 2 Ag ↓.
The two positive silver ions correspond to two negative ions, e.g. hydroxide ions HO−, which are set free by the discharge of the silver ions, and which, in turn, will combine with the oxidized carbon atom holding the two positive charges:
(NaO)HC2+ + 2 HO− → (NaO)HC(OH)2 → (NaO)HC:O + H2O.