C[NH.C₆H₅]₂
/ \
/ O
/ /
Cal. for C₆H₃——SO₂
\
NO₂
Found.
I. II. III.
N = 10.58 10.85 —— ——
S = 8.06 —— 8.00 8.16
By this series of transformations it is possible to pass from one
anilid to the other, the steps being:
CO.NH.C₆H₅ C=N.C₆H₅ C[NH.C₆H₅]₂
/ / \ / \
/ / N.C₆H₅ / O
/ / / / /
C₆H₃——SO₂NH.C₆H₅ ➡ C₆H₃——SO₂ ➡ C₆H₃——SO₂
\ \ \
NO₂ NO₂ NO₂
This is of special interest as affording a means of passing from a
derivative of one of the chlorides, to a substance derived from the
other, by steps that can be clearly followed.
Conclusions.
In the course of this investigation several facts have been established.
1. By the methods described, the symmetrical chloride of
paranitroorthosulphobenzoic acid can be obtained in fine crystalline
form, perfectly free from its isomer, with an average yield of forty
percent.
2. By treatment of the chloride with benzene and aluminium chloride,
only one chlorine atom can be replaced by a phenyl group.
3. The barium salt of paranitroorthobenzoyl benzenesulphonic acid, when
perfectly pure, crystallizes constantly with seven molecules of water
of crystallization.
4. With alcohols, the symmetrical chloride yields directly the acid
etherial salt of paranitroorthosulphobenzoic acid, no evidence
having been obtained of an intermediate chloro-etherial salt. The
unsymmetrical chloride on the other hand yields the intermediate
product.