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NITRATION AGENTS AND METHODS MORE RARELY USED 103

2HNO 3 + N 2 O 3 <-> 2N 2 O 4 + H 2 O (d)

H 2 SO 4 .qH 2 O + H 2 O = H 2 SO 4 .(q+1)H 2 O (e)

Thus the overall equation would be:

RH + N 2 O 4 + nH 2 SO 4 .mH 2 O -> RNO 2 + NSO 5 H + (n-l)H 2 SO 4 .(m+l)H 2 O

The principal assumption was that nitration of aromatic compounds in the

medium under consideration depends on the saturation of the free valences of the

sulphuric acid with nitrous acid and water. To make use of the most N 2 O 4 the

ratio of free sulphuric acid to the amount of water (n-1) : (m+1), should by the

end of the reaction be not less than a certain minimum value, characteristic for

the compound being nitrated. For benzene the value is 4:1, for chlorobenzene

5:1, for toluene 1.8:1.
When using a solution of 1.1 mole N 2 O 4 in 45% oleum for the nitration of
1 mole of benzene, Titov obtained dinitrobenzene with a yield 97-89% of theory.

Similarly, by treating nitrotoluene with a solution of N 2 O 4 in oleum, 2,4-dinitro-

toluene was obtained with a yield 98% of theory. According to Titov, 2,4-dinitro-

toluene could be converted into 2,4,6-trinitrotoluene with a yield 85% of theory.

According to more recent views, nitrogen dioxide in sulphuric acid solution

gives the nitrosonium ion NO+ and a nitric acid molecule. The latter, treated with

an excess of sulphuric acid, gives a nitronium ion, which is the actual nitrating

agent :

N 2 O 4 + H 2 SO 4 + NO+ + HNO 3 + HSO 4 - (29)

HNO 3 + 2H 2 SO 4 + NO 2 + + 2HSO 4 - + H 3 O+ (30)

(Ingold, Gillespie, Graham, Hughes and Peeling [89]).

NITRATION WITH NITROGEN DIOXIDE IN THE PRESENCE

OF FRIEDEL-CRAFTS CATALYST

Schaarschmidt [38a] has examined the nitration of aromatic hydrocarbons

with nitrogen dioxide in the presence of aluminium or ferric chloride. Benzene

with N 2 O 4 and AlCl 3 forms a red complex at about 10°C, which decomposes on

adding water with the formation of a small quantity of chlorobenzene. The compo-

sition of the complex, according to Schaarschmidt, was 2AlC1 3 .3C 6 H 6 .3N 2 O 4.

When heated to 30-35°C however it underwent a transformation, followed by the

formation of nitrobenzene when diluted with water:

2AlCl 3 .3C 6 H 6 .3.N 2 O 4 + H 2 O -> 2AlC1 3 .H 2 O + 3C 6 H 6 .N 2 O 4 (31)

C 6 H 6 .N 2 O 4 + H 2 O -> C 6 H 5 .NO 2 + HNO 2 (31a)

Titov [34] assumed a different mechanism for this reaction, suggesting that

heating a hydrocarbon with nitrogen dioxide and AlCl 3 led to the reaction:

3RH + 3N 2 O 4 + 2AlC1 3 -> 3RNO 2 + 3NOCl + Al 2 Cl 3 (OH) 3 (32)