COOC₂H₅COOC₂H₅
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II. C₆H₃—SO₂Cl + C₂H₅OH = C₆H₃—SO₂OC₂H₅ + HCl
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NO₂NO₂

COOC₂H₅COOC₂H₅
//
III. C₆H₃—SO₂OC₂H₅ + C₂H₅OH = C₆H₃—SO₂OH + (C₂H₅)₂O
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NO₂NO₂

Kastle, it will be observed, gave the symmetrical formula to this mixture of chlorides. Several acid etherial salts were made, and a series of the neutral salts of various metals described by him.

The action of pure symmetrical chloride was studied in the same general manner to see if the resulting products would be the same as those formed from the mixed chlorides.

1. Action of Methyl Alcohol upon the Symmetrical chloride.

A portion of the chloride was dissolved in methyl alcohol, and the solution boiled until a drop added to cold water gave no precipitate, of unchanged chloride. The alcohol was then distilled off, and the thick syrup remaining, diluted with water. This solution was neutralized with barium carbonate and filtered. On cooling, the barium salt crystallized in shining mica-like plates, or in yellowish needles corresponding accurately with those described by Kastle. They gave the following analytical results.

0.2664 gram lost 0.0211 gram at 150°,
and gave 0.0870 gram BaSO₄.
Cal. for[COOCH₃
/
C₆H₃—SO₂O
\
NO₂		]2Ba + 3H₂OFound.
H₂O = 7.797.88
[anhydrous salt] Ba = 20.8520.85

2. In like manner the barium ethyl salt was made. It also agreed perfectly with Kastle’s description, crystallizing in fine, colorless needles, forming in tufts from a not too concentrated solution. In case it is necessary to concentrate these solutions, it is of advantage to add a small quantity of alcohol to the solution as this prevents any great amount of saponification, which otherwise takes place to a noticeable extent.

Analysis.

I. 0.2824 gram lost 0.0276 gram at 180°,
and gave 0.0860 gram BaSO₄.
II. 0.2655 gram lost 0.0262 gram at 190°,
and gave 0.0815 gram BaSO₄.
Cal. for[COOC₂H₃
/
C₆H₃—SO₂O
\
NO₂		]2Ba + 4H₂O.Found.
I II
H₂O = 9.519.779.86
Ba = 20.0019.8420.02