B. V. Tronov.

On the Reactions of Cleavage and Formation of Double Bonds.

When a student, in studying chemistry successively, passes from inorganic compounds to organic ones, he sometimes has occasion to encounter the necessity of abandoning certain ideas that have already become customary. The most ordinary reagents, for some reason, behave in organic chemistry quite differently than in inorganic chemistry. Thus acids, especially strong ones such as nitric and sulfuric acids, always dissociate with the splitting off of a hydrogen ion; alkalis—KOH, NaOH—split off a hydroxyl ion. But these same acids interact with organic compounds according to a directly opposite type:

\[ \mathrm{C_6H_6 + HO.NO_2 \longrightarrow C_6H_5.NO_2 + HOH} \]

\[ \mathrm{C_6H_6 + HO.SO_2OH \longrightarrow C_6H_5SO_2OH + HOH,} \]

and, according to the electronic scheme, even:

\[ \mathrm{(C_6H_5)'H^{\cdot} + (HO)'(NO_2)^{\cdot} \longrightarrow (C_6H_5)'(NO_2)^{\cdot} + (H_2O),} \]

i.e., one of the strongest acids dissociates like an alkali.

The reaction of the action of alkalis on amides of organic acids is written as follows:

\[ \mathrm{CH_3CO|NH_2 + H|OK \longrightarrow CH_3COOK + NH_3} \]

or

\[ \mathrm{(CH_3CO)'(NH_2)^{\cdot} + H^{\cdot}(OK)' \longrightarrow CH_3COOK + NH_3.} \]

Here the alkali has given up its hydrogen, i.e. it has reacted as an acid; and yet the reaction proceeds very well in dilute aqueous solution, i.e. under conditions in which almost complete dissociation into \(\mathrm{K^{\cdot}(OH)'}\) takes place, and not at all into \(\mathrm{(KO)'}\) and \(\mathrm{H^{\cdot}}\).

Thus one must somehow redraw concepts that have already been firmly established, and the thought involuntarily arises whether it is not possible to explain the processes encountered in some other way, without these unnatural assumptions that an acid becomes an alkali and an alkali an acid.

Let us take the same reactions of nitration and sulfonation, but let us direct our attention not to the hydroxyl group, but to the other parts of the molecules of sulfuric and nitric acids:

\[ \begin{array}{cc} \mathrm{HO} & \mathrm{O} \\ \ \ \backslash & \ \ \| \\ \mathrm{S} & \\ /\ \ \| & \\ \mathrm{HO} & \mathrm{O} \end{array} \qquad \mathrm{HO{-}N\!\left(=O\right)=O} \]

If instead of sulfuric acid we take its anhydride,

\[ \mathrm{O{=}S\!\left(=O\right)} \]

and act with it on benzene, the same sulfonation will occur—only here it proceeds considerably more readily. Meanwhile, in the anhydride there is no hydroxyl at all, and the reaction can be represented only as follows:

\[ \mathrm{C_6H_6 + O{=}S\!\left(=O\right) \longrightarrow C_6H_5{-}S\!\left(=O\right)\!\left(=O\right){-}OH} \]

It is very probable, therefore, that sulfuric acid also acts on the hydrocarbon with its hydroxyl group, and on the part of the molecule common with the anhydride, i.e.

\[ \ce{ C6H6 + O=S(OH)2(=O) -> C6H5-S(=O)(OH)2(OH) } \]

The further course of the process is understandable if one recalls that compounds with several hydroxyls at one atom of \(\ce{C}\), \(\ce{N}\), \(\ce{S}\), etc., are often unstable and readily decompose, giving off water:

\[ \ce{ C6H5-S(=O)(OH)2(OH) -> C6H5-S(=O)(OH)(O) + H2O } \]

The same explanation is applicable also to nitric acid. It can act not only as an acid, but also as an oxidizing agent, for example, with the evolution of hydrogen:

\[ \ce{ HO-N(=O)(=O) + 2H -> HO-N(H)(OH)(=O) -> HN(=O)(=O) + H2O } \]

With a hydrocarbon having a sufficiently mobile hydrogen atom, an entirely analogous addition may occur, followed by elimination of water:

\[ \ce{ C6H6 + O=N(OH)(=O) -> C6H5-N(=O)(OH)(OH) -> C6H5-N(=O)(=O) + H2O } \]

With such an interpretation, the acid does not react at all according to the type of a salt, and even from the point of view of the ionic theory it is quite understandable that the positive hydrogen will go to the negative oxygen, and the hydrocarbon residue to nitrogen. If, however, one takes into account the possibility of neutral decomposition, then even a side reaction is explained, which is always observed to a greater or lesser degree during nitration—namely, the appearance of brown fumes, i.e., reduction of nitric acid. A certain fraction of the acid and benzene molecules may react in this direction:

\[ \ce{ C6H5H + N(=O)(=O)OH -> HN(OC6H5)(=O)OH -> HN(=O)(=O) + C6H5OH } \]

further, \(\ce{HNO2}\) will decompose into \(\ce{H2O}\), \(\ce{NO}\), and \(\ce{NO2}\), while the phenol under the conditions of nitration of benzene will immediately be nitrated and oxidized further.

If, instead of phenyl, we introduce some electropositive radical, which will more readily attach itself to oxygen, then the oxidation reaction predominates. Therefore fatty hydrocarbons cannot be nitrated with strong nitric acid; it is necessary to use dilute acid (with strong heating). It probably reacts in the hydrate form: \(\ce{NO(OH)3}\) or \(\ce{N(OH)5}\);

these hydroxyls, indeed, are held weakly; they are easily split off with the liberation of water, but one of them can be detached together with hydrogen; the hydrocarbon residue is sufficiently mobile:

\[ \mathrm{N(OH)_3 + HC(CH_3)_2 \;\longrightarrow\; H_2O + HO\!-\!N(=O)\!-\!CH(CH_3)_2 \;\longrightarrow\; O{=}N(=O)\!-\!CH(CH_3)_2} \]

Considering now the hydrolysis of amides by alkalis, we note that in the amide molecule, besides the group \(\mathrm{NH_2}\), there is a double bond between \(\mathrm{C}\) and \(\mathrm{O}\). A double bond is in general considerably more active than a single one. In the majority of cases it is very difficult to break a single bond between atoms, whereas a double bond is easily cleaved under the influence of the most varied reagents. In addition, it is known that a simple bond of carbon with oxygen is, in general, broken more readily than the bonds between the links of the carbon skeleton.

Hence it would naturally be expected that the carbonyl bond, i.e. the double bond between carbon and oxygen, will be still more mobile. Indeed, aldehydes and ketones undergo addition reactions even more readily than ethylenic hydrocarbons, adding even many such molecules which do not act upon hydrocarbons, such as \(\mathrm{NaHSO_3}\), \(\mathrm{NH_3}\), \(\mathrm{NH_2OH}\), \(\mathrm{HCN}\), [[unclear: obscured formulae/text]], and so on. They do not act on [[unclear: obscured word]], or they combine with difficulty with oxygen [[unclear: obscured text]] molecules of the type \(\mathrm{HX}\), where \(\mathrm{X}\) is a halogen, [[unclear: obscured text]] addition is extraordinarily easy. Meanwhile, for example, \(\mathrm{NH_3}\) does not add to ethylene precisely for this reason, because the compound \(\mathrm{C_2H_5NH_2}\) is sufficiently stable; \(\mathrm{CH_3—CH_2—OSOONa}\) also should not very readily give up \(\mathrm{HOSOONa}\); simply, in \(\mathrm{CH_2{=}CH_2}\) the double bond is cleaved with greater difficulty.

Thus, the carbonyl bond, where it is present by itself, proves to be extraordinarily active. However, in considering compounds in which the carbonyl carbon is also joined to a halogen, a hydroxyl, an amino group, or an organic radical, it is almost always assumed that it is precisely this atom or radical that reacts first, and that the double bond is not cleaved. Thus they write the already mentioned reaction of saponification of amides. In the action of ammonia on acid chlorides they assume simply the exchange of halogen for the amino group:

\[ \mathrm{CH_3—C(=O)Cl + HNH_2 \;\longrightarrow\; CH_3—C(=O)NH_2 + HCl.} \]

In doing so, they completely forget that if we take molecules with a halogen or a carbonyl separately, i.e., for example, \(\mathrm{CH_3—CH_2Cl}\) and \(\mathrm{CH_3—CH{=}O}\), then the first, for interaction with ammonia, requires (at ordinary temperature) several days, whereas the second reacts instantaneously, almost in the same way as an acid chloride:

\[ \mathrm{CH_3—C(=O)H + HNH_2 \;\longrightarrow\; CH_3—C(H)(NH_2)(OH) \;\longrightarrow\; CH_3—C(H){=}NH + H_2O.} \]

Precisely likewise, water also interacts very rapidly with acetyl chloride, giving, according to the equation:

\[ \mathrm{CH_3—C(=O)Cl + HOH \;\longrightarrow\; CH_3—C(=O)OH + HCl.} \]

Meanwhile, \( \mathrm{CH_3-CH_2Cl} \) exchanges chlorine for hydroxyl with water extremely slowly, whereas the addition of water to carbonyl where the product is more or less stable (as chloral hydrate, hydrates of oxalic acid, mesoxalic acid, etc.) proceeds very rapidly.

All this compels one to think that in acid chlorides, amides, esters, etc., it is often precisely the carbonyl double bond that first enters into reaction, and addition products are first formed:

\[ \mathrm{ CH_3-C(=O)Cl + HNH_2 \longrightarrow CH_3-C(OH)(Cl)(NH_2) } \]

\[ \mathrm{ CH_3-C(=O)Cl + HOH \longrightarrow CH_3-C(OH)(OH)(Cl) } \]

In amides with alkali, addition of the ions \( \mathrm{K^\cdot} \) and \( \mathrm{(OH)'} \) first occurs:

\[ \mathrm{ CH_3-C(=O)NH_2 + KOH \longrightarrow CH_3-C(OK)(OH)(NH_2) } \]

Such substances cannot be isolated, but this contradicts nothing: they are simply extremely unstable and at once decompose with liberation of \( \mathrm{H_2O} \), \( \mathrm{NH_3} \):

\[ \mathrm{ CH_3-C(OH)(Cl)(NH_2) \longrightarrow CH_3-C(=O)NH_2 + HCl } \]

\[ \mathrm{ CH_3-C(OH)(OH)(Cl) \longrightarrow CH_3-C(=O)OH + HCl } \]

\[ \mathrm{ CH_3-C(OK)(OH)(NH_2) \longrightarrow CH_3-C(=O)OK + NH_3 } \]

quite analogously to the first product of the chlorination of alcohol:

\[ \mathrm{ CH_3-CH_2OH + Cl_2 \longrightarrow CH_3-C(H)(OH)(Cl) + HCl \longrightarrow CH_3-C(=O)H + 2HCl } \]

or to the product of the oxidation of alcohol:

\[ \mathrm{ CH_3-C(H)(H)(OH) + O \longrightarrow CH_3-C(H)(OH)(OH) \longrightarrow CH_3-C(=O)H + H_2O } \]

A very clear indication of the participation in the reaction first of all of the carbonyl, and not of chlorine, etc., I find in the following comparison. Different acid chlorides differ greatly from one another in their capacity for reactions. Acetyl chloride fumes in air and with water boils up instantaneously. Benzoyl chloride, in order to be completely decomposed, must be shaken for a long time with alkali. In structure, \( \mathrm{CH_3COCl} \) and \( \mathrm{C_6H_5COCl} \) differ in that in one the acid-chloride group is bound to methyl, in the other to phenyl. Let us compare the combination of these same radicals with the separate parts of this group: with \( \mathrm{C-Cl} \) and with \( \mathrm{C=O} \).

\[ \begin{array}{c} \mathrm{C{-}Cl\ and\ C_6H_5{-}C{-}Cl,}\\[-2mm] \diagdown\!\!\!\diagup\quad\quad\quad\diagdown\!\!\!\diagup\\ \mathrm{H}\quad \mathrm{H}\quad\quad\mathrm{H}\quad \mathrm{H} \end{array} \]

then the former is more mobile.

In carbonyl compounds, acetone more readily than all other ketones gives reactions characteristic of the carbonyl bond: it combines with bisulfite, adds hydrocyanic acid, forms oxime, phenylhydrazone, semicarbazone, condenses with pyrrole, etc. Benzophenone either does not enter into these reactions at all, or only with difficulty. Consequently, the mobility of chlorine, owing to the replacement of methyl by phenyl, instead of increasing, decreased very strongly; the mobility of the carbonyl group likewise decreased, as did its ability to react with the chlorine of the acid chloride. All this becomes clear if one assumes that in acid chlorides the reaction is begun not by the chlorine, but by the carbonyl.

Some experimental confirmation of such an interpretation may be seen in an experiment made by me:

In preparing methyl-α-pyridyl ketone from bromo-magnesium-pyrrole with acetyl chloride, a solid mass is formed. By shaking it five times with portions of ether, I was able to extract from it only 20% of the ketone. The rest was retained and was separated only after decomposition with water. This fact is again readily explained if one assumes a preliminary addition at the carbonyl bond¹:

\[ \mathrm{R{-}MgBr + C\!\left(\begin{array}{c} \mathrm{CH_3}\\ \mathrm{Cl}\\ \Vert\\[-1mm] \mathrm{O} \end{array}\right) \longrightarrow BrMg{-}O{-}C\!\left(\begin{array}{c} \mathrm{R}\\ \mathrm{CH_3}\\ \mathrm{Cl} \end{array}\right)} \]

This product can, of course, also react thus:

\[ \mathrm{ \begin{array}{c} \mathrm{R}\\[-1mm] \mathrm{CH_3} \end{array} >C< \begin{array}{c} \mathrm{OMgBr}\\[-1mm] \mathrm{Cl} \end{array} \longrightarrow \begin{array}{c} \mathrm{R}\\[-1mm] \mathrm{CH_3} \end{array} >C{=}O + ClMgBr } \]

nevertheless it was extracted with ether, but rapid decomposition of the whole mass occurs only with water:

\[ \mathrm{ \begin{array}{c} \mathrm{R}\\[-1mm] \mathrm{CH_3} \end{array} >C< \begin{array}{c} \mathrm{OMgBr}\\[-1mm] \mathrm{Cl} \end{array} + HOH \longrightarrow BrMgOH + \begin{array}{c} \mathrm{R}\\[-1mm] \mathrm{CH_3} \end{array} >C< \begin{array}{c} \mathrm{OH}\\[-1mm] \mathrm{Cl} \end{array} \longrightarrow \begin{array}{c} \mathrm{R}\\[-1mm] \mathrm{CH_3} \end{array} >C{=}O + HCl } \]

In exactly the same way one can interpret many other reactions, for example the formation of ethers in the interaction of acid chlorides with alcohols:

\[ \mathrm{ CH_3{-}C\!\left(\begin{array}{c} \Vert O\\ Cl \end{array}\right) + HOCH_3 \longrightarrow CH_3{-}C\!\left(\begin{array}{c} OCH_3\\ OH\\ Cl \end{array}\right) \longrightarrow CH_3{-}C\!\left(\begin{array}{c} OCH_3\\ \Vert O \end{array}\right) + HCl } \]

the formation of acid anhydrides on heating acid chlorides with salts:

\[ \mathrm{ CH_3{-}C\!\left(\begin{array}{c} \Vert O\\ Cl \end{array}\right) + Na{-}O{-}C\!\left(\begin{array}{c} \Vert O\\ CH_3 \end{array}\right) \longrightarrow CH_3{-}C\!\left(\begin{array}{c} ONa\\ O{-}C\!\left(\begin{array}{c}\Vert O\\ CH_3\end{array}\right)\\ Cl \end{array}\right) \longrightarrow CH_3{-}C\!\left(\begin{array}{c} \Vert O\\ O{-}C\!\left(\begin{array}{c}\Vert O\\ CH_3\end{array}\right) \end{array}\right) + NaCl } \]

decomposition of anhydrides by water:

\[ \ce{CH3-C(=O)-O-C(=O)-CH3 + HOH -> CH3-C(OH)2-O-C(=O)-CH3 -> CH3-C(=O)OH + HO-C(=O)-CH3} \]

reactions of esters, e.g.:

\[ \ce{CH3-C(=O)-OC2H5 + HNH2 -> CH3-C(NH2)(OH)-OC2H5 -> CH3-C(=O)NH2 + HOC2H5} \]

Even the acids themselves are sometimes cleaved as though not according to the acid type (i.e., into hydrogen and a negative acid ion), but interact according to the ketone type. Let us take the most typical, it would seem, reaction of an acid: esterification. It is always represented as an ordinary exchange decomposition, carried out in the same way as reactions between mineral salts (therefore complex esters are even called¹ “ester salts”). However, I shall point to one fact that is difficult to reconcile with such an understanding.

It is known that if, in benzoic acid, both ortho hydrogens are replaced by any atoms or radicals (methyl, nitro group, halogens², etc.), then the carboxyl becomes astonishingly immobile³. Such acids scarcely or not at all esterify, for example with absolute methyl alcohol in the presence of HCl. If, however, the prepared ester is taken, it is just as difficult to saponify. This is explained by steric hindrance, i.e., by the fact that the substituents in the ortho positions shield the carboxyl and prevent molecules of various reagents from approaching it. All this is well enough, but the same steric hindrance for some reason proves ineffective if we prepare the ester not directly from the acid with alcohol, but from the silver salt with methyl iodide. According to the generally accepted understanding these two reactions are completely analogous:

\[ \ce{ \left(\chemfig{*6(--(-R)-(-C(=[:30]O)-O^{-})-(-R)---)}\right) Ag+ + J(CH3) -> \chemfig{*6(--(-R)-(-C(=[:30]O)-OCH3)-(-R)---)} + AgJ } \]

\[ \ce{ \left(\chemfig{*6(--(-R)-(-C(=[:30]O)-O^{-})-(-R)---)}\right) H+ + HOCH3 -> \chemfig{*6(--(-R)-(-C(=[:30]O)-OCH3)-(-R)---)} + H2O } \]

and it is incomprehensible why steric hindrance interferes with one of them, while the other is insensitive to it.

In my understanding, the action of methyl iodide on the salt remains an ionic reaction, whereas the interaction of the acid with the alcohol must be represented as follows:

\[ \ce{CH3-C(=O)OH + H:OCH3 <=> CH3-C(OCH3)(OH)OH <=> CH3-C(=O)OCH3 + HOH} \]

\[ \begin{gathered} \text{[[structural scheme: substituted cyclopropanecarboxylic acid]]+\mathrm{HOCH_3} \rightleftarrows \text{[[structural scheme: cyclohexane derivative with two }R\text{ substituents and }C(OH)_2(OCH_3)\text{ group]]}\\ \rightleftarrows \text{[[structural scheme: methyl ester of a substituted cyclohexanecarboxylic acid]]+\mathrm{HOH}. \end{gathered} \]

But if in the case under consideration the second carboxyl [[unclear: group]] is transformed only intermediately, then its swelling, and ultimately its [[unclear: cleavage]], may be hindered by the neighboring parts of the molecule.

From this point of view many reactions become comprehensible which until now have had no explanation. I shall cite the following: the formation of hydrocarbons from acids on heating with alkalis:

\[ \mathrm{ CH_3{-}C\!\left(\begin{array}{c} OH\\[-2mm] ONa\\[-2mm] ONa \end{array}\right) \longrightarrow CH_4 + C\!\left(\begin{array}{c} O\\[-1mm] ONa\\[-1mm] ONa \end{array}\right) } \]

\[ \mathrm{ C_6H_5{-}C\!\left(\begin{array}{c} O\\[-1mm] OH \end{array}\right) + Ca{-}O \longrightarrow C_6H_5{-}C\!\left(\begin{array}{c} O\\[-1mm] O{-}Ca \end{array}\right) \cdots OH } \]

\[ \mathrm{ \longrightarrow C_6H_5{-}C\!\left(\begin{array}{c} OH\\[-1mm] O\\[-1mm] O \end{array}\right)Ca \longrightarrow C_6H_6+ C\!\left(\begin{array}{c} O\\[-1mm] O\\[-1mm] O \end{array}\right)Ca } \]

the conversion of formic acid into oxalic acid:

\[ \mathrm{ H{-}C\!\left(\begin{array}{c} ONa\\[-1mm] O \end{array}\right) + H{-}C\!\left(\begin{array}{c} ONa\\[-1mm] O \end{array}\right) \longrightarrow H{-}C\!\left(\begin{array}{c} ONa\\[-1mm] OH \end{array}\right) {-} C\!\left(\begin{array}{c} ONa\\[-1mm] O \end{array}\right) \longrightarrow } \]

\[ \mathrm{ \longrightarrow H_2+ C\!\left(\begin{array}{c} ONa\\[-1mm] O \end{array}\right) {-} C\!\left(\begin{array}{c} ONa\\[-1mm] O \end{array}\right) } \]

Acid cleavage of acetoacetic ester:

\[ \begin{array}{c} \mathrm{CH_3}\\ |\\ \mathrm{C{=}O}\\ |\\ \mathrm{CH_2}\\ |\\ \mathrm{C{=}O}\\ |\\ \mathrm{OC_2H_5} \end{array} + 2\mathrm{KOH} \;\Longrightarrow\; \begin{array}{c} \mathrm{CH_3}\\ |\\ \mathrm{C}\langle^{\mathrm{OK}}_{\mathrm{OH}}\\ |\\ \mathrm{CH_2}\\ |\\ \mathrm{C}\langle^{\mathrm{OK}}_{\mathrm{OH}}\\ |\\ \mathrm{OC_2H_5} \end{array} \;\Longrightarrow\; \mathrm{CH_3COOK} + \mathrm{CH_3COOK} + \mathrm{C_2H_5OH}. \]

Even the Cannizzaro reaction (conversion of benzaldehyde under the action of alkalis into an acid and an alcohol):

\[ \mathrm{C_6H_5{-}C}\langle^{\mathrm H}_{\mathrm O} + \mathrm{C}\langle^{\mathrm O}_{\mathrm H}\mathrm{C_6H_5} \;\Longrightarrow\; \mathrm{C_6H_5{-}C(=O){-}C}\langle^{\mathrm{OH}}_{\mathrm H}\mathrm{C_6H_5}, \]

\[ \mathrm{C_6H_5{-}C(=O){-}C}\langle^{\mathrm{OH}}_{\mathrm H}\mathrm{C_6H_5} + \mathrm{KOH} \;\Longrightarrow\; \mathrm{C_6H_5{-}C}\langle^{\mathrm{OK}}_{\mathrm{OH}} {-} \mathrm{C}\langle^{\mathrm{OH}}_{\mathrm H}\mathrm{C_6H_5} \;\Longrightarrow \]

\[ \Longrightarrow\; \mathrm{C_6H_5{-}C(=O)OK} + \mathrm{H{-}C}\langle^{\mathrm{OH}}_{\mathrm H}\mathrm{C_6H_5}. \]

Thus, very many reactions of organic and inorganic compounds find a complete explanation if it is assumed that, first of all, the double bond is broken and addition takes place. True, in the majority of cases there immediately follows a new elimination, with the formation again of the same kind of double bond, i.e. the final result is obtained the same as if the double bond had remained unchanged throughout the entire process. But in this way it is possible to clarify something that was incomprehensible from the former point of view, to avoid certain inconsistencies, and to introduce a generalization over a very wide field.

The replacement of one reaction by two presents nothing inconvenient here. It is in any case known that such elimination reactions proceed very easily; that two hydroxyls, or chlorine and hydroxyl at one carbon, three hydroxyls at one sulfur atom, and the like, are not held firmly and quickly give off a molecule of \(\mathrm{H_2O}\) or \(\mathrm{HCl}\).

Some doubt may be caused by the circumstance that, if in the molecule there are present simultaneously, for example, \(\mathrm{OH}\) and \(\mathrm{ONa}\), then for some reason hydrogen departs, and not sodium, which in general is more mobile; for example, in

\[ \mathrm{CH_3{-}C}\langle^{\mathrm{OH}}_{\mathrm{ONa}}\mathrm{ONa} \]

\(\mathrm{CH_4}\) is split off, and not \(\mathrm{CH_3Na}\). But the eliminating groups must nevertheless, with greater or lesser force, attract one another; they must nevertheless form a comparatively strong system, and if \(\mathrm{CH_3Na}\) (which in general can hardly be prepared) were formed and separated, then it would immediately react with

\[ \mathrm{C}\langle^{\mathrm{OH}}_{\mathrm{ONa}}{=}\mathrm O, \]

giving \(\mathrm{CH_4}\) and

\[ \mathrm{C}\langle^{\mathrm{ONa}}_{\mathrm{ONa}}{=}\mathrm O. \]

In general, sodium does not pass from oxygen to carbon, but hydrogen can pass. It is another matter when it is not a hydrocarbon radical that is split off simultaneously, but chlorine: \(\mathrm{OH}\) can take sodium with it (as in the formation of anhydrides from acid chlorides with salts).

1. Ya. I. Mikhailenko, Carbon Compounds. ↩↩

2. In addition, it seems, fluorine. ↩

3. Works of V. Meyer. ↩