V. 0.1505 gram gave 0.0820 gram BaSO₄.

OH ]

/ ]

C[C₆H₃ ]

/ \ \ ]

/ O OH ]2

/ /

Cal. for C₆H₃——SO₂

\

NO₂

Found

I II III IV V

C = 52.66 52.18 52.42 52.67 —— ——

H = 3.46 3.76 3.27 3.66 —— ——

S = 7.39 —— —— —— 7.57 7.48

An effort to obtain the anhydride was unsuccessful. Some loss of weight

was observed, but the compound underwent decomposition before this loss

amounted to much.

6. The Action of Pyrogallol upon the Symmetrical Chloride of

Paranitroorthosulphobenzoic Acid.

The product of this action dissolves readily in dilute sodium hydroxide

without residue, producing a very deep purple-black color when

concentrated, passing to grayish-violet as the solution is diluted. On

adding hydrochloric acid, precipitation occurs, as in most of these

reactions. On attempting to filter off this precipitate, it forms a

sticky, black mass on the filter with which little can be done. It is

best to evaporate to dryness before filtration and powder the residue.

This powder can then be washed fairly clean from alkali salts and

acid.

Nothing to suggest the formation of an etherial salt was observed.

Analysis of this product for sulphur showed that in this galleïn, as

in the case of the hydroquinone phthaleïn more than two pyrogallol

residues had entered the acid residue. The indications were that

six had entered into one of the chloride. This also agrees with the

observation of Lyman[23], who describes a hexapyrogallol galleïn of

orthosulphoparatoluic acid.

Probably a mixture of varying composition was obtained, and little

importance was attached to the results save as they showed that no

etherial salt is formed in the reaction.

[23] Am. Ch. Journ. XVI-527.

7. The Action of β-Naphthol upon the Symmetrical Chloride of

Paranitroorthosulphobenzoic Acid.

It was hoped that here, as in the case of the monohydroxy phenols an

etherial salt would be obtained. It was found, however, that very

little action occurred, save such as was indicated by the development

of a bright carmine color in the melted mixture, until a temperature of

about 160° was reached. At this point hydrochloric acid was evolved,

but the chloride itself undergoes decomposition. Nothing definite could

be isolated among the reaction products, save unchanged β-Naphthol.

VIII. The Action of Aniline upon the Symmetrical Chloride of

Paranitroorthosulphobenzoic Acid.

As has been pointed out in the Introduction, it was in connection

with the aniline derivatives of orthosulphobenzoic acid, that the

isomerism of the chlorides was first noticed, two anilids being

obtained. Accordingly, when Gray began his study of the chlorides

of paranitroorthosulphobenzoic acid, his first effort was to obtain

evidence of the existence of two anilids. These were not obtained,

however, until after the chlorides themselves had been isolated, as

their properties made their isolation and preparation a matter of

difficulty.

Some points still remained in doubt after Gray’s study, and a further

investigation was thought to be desirable to clear these up.

Some time was spent in an endeavor to obtain a method by which a good

yield of fusible, or symmetrical, anilid could be obtained. The yield

in all cases tried, is not a good one. The presence of the nitro group

appears to complicate the reaction, leading to secondary reactions

whose course could not be followed. Upon bringing aniline and the

chloride together, a very vivid red color was always observed, and the

same was true when it was necessary to employ alkali. The fact that

such colors develop when nitro compounds are treated with alkali has

been noticed in many instances and some progress has been made in the

study of these compounds. Jackson and Ittner[24] have lately reviewed

this subject.

If a solution of the symmetrical chloride in ether is slowly added to a

similar solution of aniline, no appreciable amount of heat is evolved.

If the resulting solution is allowed to stand at ordinary temperatures,

action proceeds very slowly, aniline hydrochloride being precipitated

as the reaction proceeds. This can be filtered off from time to time

and the rate of action so observed. In such a way it was found that

five grams of chloride required about fifty hours time to react

completely with an excess of aniline. Similar results were obtained

with chloroform as the solvent. By boiling the solution for an hour or

more the reaction is complete.

[24] Am. Chem. Journ. XIX-199

The method employed was to bring the chloride and an excess of

aniline—somewhat more than four molecules—together in chloroform

solution. The flask was then boiled for about an hour, when the

chloroform was distilled off. During the boiling as well as the

distillation more or less bumping occurs in consequence of the aniline

hydrochloride which separates, and constant shaking of the flask is

sometimes necessary. The residue which is in a thick, gummy condition

in consequence of the presence of an excess of aniline, was digested

with water acidulated with hydrochloric acid. The excess of aniline

is thus removed, and the reaction product obtained as a reddish-brown

solid substance. This was treated with dilute sodium hydroxide, all

lumps being broken up with a stirring rod. The undissolved substance

is largely anil, which was filtered off. The anilid was then regained

by acidifying the alkaline solution, in which it was dissolved. It

separates immediately as a curdy colorless precipitate, though it is

frequently colored pink by impurity. It was found that this color could

be removed, in case not much was present, by redissolving the anilid in

alkali, and slowly pouring the solution into an excess of dilute acid.

In all cases a considerable amount of anil was obtained, even when the

substances were employed in the molecular ratios of 1:10. The reactions

involved, so far as the formation of anilid and anil are concerned are,

COCl CO.NH.C₆H₅

/ /

C₆H₃——SO₂Cl + 4C₆H₅NH₂ = C₆H₃——SO₂.NH.C₆H₅ + C₆H₅NH₃Cl

\ \

NO₂ NO₂

CO

/ \

COCl / N.C₆H₅

/ / /

C₆H₃——SO₂Cl + 3C₆H₅NH₂ = C₆H₃——SO₂ + 2C₆H₅NH₃Cl

\ \

NO₂ NO₂

On the whole the reaction seemed to be the most satisfactory in

chloroform solution, the main objection being, that, owing to the

simultaneous presence of chloroform, alkali, an a trace of aniline,

phenyl isocyanide is always formed, and renders the work more or less

unpleasant.

A number of experiments were also made to see if the yield could

be increased be employing a modification of the “Schotten-Baumann

Reaction”[25] for the formation of anilids. For this purpose an etherial

solution of the chloride was added to a like solution of aniline in

which was suspended finely powdered anhydrous potassium carbonate. The

proportions of the substances were those demanded by the equation

COCl CO.NH.C₆H₅

/ /

C₆H₃——SO₂Cl + 2C₆H₅NH₂ + 2K₂Cl₃ = C₆H₃——SO₂NH.C₆H₅ + 2KCl + 2KHCO₃

\ \

NO₂ NO₂

Very little anilid was, however obtained, but in its place a

substance soluble in water, of acid reaction capable of forming salts

and yielding several well characterized derivatives. I hope to

investigate this reaction more fully at some future time.

[25] Ber. XVII-2545; XXIII, 3430.

* * * * *

The anilid is rather sparingly soluble in alcohol, from which it is

deposited on cooling in very small needles. These melt, as stated by

Gray, at 222°. It is also soluble in chloroform and glacial acetic

acid, but does not form well defined crystals from any solvent. It

dissolves in dilute alkali from which solution acids precipitate it

unchanged.

* * * * *

The anil is also soluble in alcohol, glacial acetic acid etc. It

crystallizes in much better-formed crystals than does the anilid. These

melt at 188°.

On boiling the anil with aniline for a time, it is converted into the

anilid

CO

/ \

/ N.C₆H₅ CO.NH.C₆H₅

/ / /

C₆H₃——SO₂ + C₆H₅NH₂ = C₆H₃——SO₂NH.C₆H₅

\ \

NO₂ NO₂

In none of these reactions was any infusible anilid observed.

A Further Investigation of the Symmetrical Chloride of Paranitroorthosulphobenzoic Acid · 第 15 / 29 题