Stannic salts, reactions Stannous chloride

Stannic and stannous chloride are best prepared by the reaction of chlorine with tin metal. Stannous salts are generally prepared by double decomposition reactions of stannous chloride, stannous oxide, or stannous hydroxide with the appropriate reagents. MetaUic stannates are prepared either by direct double decomposition or by fusion of stannic oxide with the desired metal hydroxide or carbonate. Approximately 80% of inorganic tin chemicals consumption is accounted for by tin chlorides and tin oxides. 

Anhydrous stannous chloride, a water-soluble white soHd, is the most economical source of stannous tin and is especially important in redox and plating reactions. Preparation of the anhydrous salt may be by direct reaction of chlorine and molten tin, heating tin in hydrogen chloride gas, or reducing stannic chloride solution with tin metal, followed by dehydration. It is soluble in a number of organic solvents (g/100 g solvent at 23°C) acetone 42.7, ethyl alcohol 54.4, methyl isobutyl carbinol 10.45, isopropyl alcohol 9.61, methyl ethyl ketone 9.43 isoamyl acetate 3.76, diethyl ether 0.49, and mineral spirits 0.03 it is insoluble in petroleum naphtha and xylene (2). 

There are two series of compounds of tin and the halogens those related to stannous oxide, SnO, in which the metal has a valence of 2 (stannous salts), and those related to stannic oxide, Sn02, in which it has a valence of 4 (stannic salts). The compounds with lower valence are usually prepared by the action of the halogen acids on the metal. Stannous chloride, for example, is formed when hydrochloric acid reacts with tin. Stannic salts, on the other hand, are prepared by the reaction of tin and the free halogen. In this preparation stannic bromide results from the reaction of bromine and tin. The reaction between the two ele-... 

Nitriles may be converted to their imino chloride salts by the action of dry hydrogen chloride in ether. These intermediates ate reduced by anhydrous stannous chloride to stannic aldimonium chlorides, which on hydrolysis yield aldehydes. Chloroform may be added to facilitate the solution of the nitrile. The quality of the stannous chloride catalyst is important the preparation of an active and dependable form has been described. The yields are usually high for many aromatic nitriles, as in the preparation of /3-naphthaldehyde (95%). The reaction has also been employed in the heterocyclic series, as in the synthesis of 4-methyl-thiazole-5-aldehyde (40%). The reduction of the cyano group in the... 

To 1 molecule of a mononitro-compound, i atoms of tin, or 3 molecules of stannous chloride, are therefore used. In calculating tire amount of the latter necessary for a reaction, it is to be remembered that the salt crystallises with two molecules of water (SnCl,+ 2H,0). If the reduction is to be effected by metallic tin, double the above quantity is frequently used, i.e. to 1 nitro-group, 3 atoms of tin. In this case, the tin is not converted into stannic chloride, but into stannous chloride ... 

Discussion As would be predicted from the position of tin in the electromotive series, non-oxidizing acids will react with this metal to form salts and evolve hydrogen. If hydrochloric acid is used, mercuric chloride can be used to determine whether stannous chloride, or stannic chloride is produced. This is the same reaction which was employed in the reduction of mercuric chloride to mercurous chloride, studied in Experiment 128 a. 

The product of the reaction is a mixture of the double-salts of aniline hydrochloride with stannous chloride and stannic chloride. If the product is allowed to cool when reaction is complete, the double-salts are obtained in crystalline condition. Aniline forms many double-salts with the halides of the less active metals. The tin salts have the composition expressed by the formulas... 

Tin hydride has been made by the reaction of hydrochloric acid with magnesium-tin alloy (1) or metallic tin (2) by reduction of stannous salts electrolytically (3), by metals (4, 5), or sodium borohydride (6) and by reduction of stannic chloride by atomic hydrogen (7) or metals (S). Perhaps the most practical method involves the reduction of stannic chloride with lithium aluminum hydride under nitrogen (9, 10, 11) containii 0 1% oxygen 12) at — 70° C. The presence of oxygen inhibits the decomposition, which ordinarily occurs at room temperature. This inhibition is at least partially due to the fact that metallic tin is a catalyst for the decomposition, but in the presence of oxygen becomes coated with a film of oxide, which is ineffective as a catalyst 12). 

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