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Anilinium chloride is the organic compound with the formula C 6 H 5 NH + 3 Cl −. A white solid, it is the chloride salt of anilinium, which is the conjugate acid of aniline, C 6 H 5 NH 2. Anilinium chloride is produced by treatment of aniline with hydrochloric acid. The cation consists of a phenyl ring attached to a tetrahedral ammonium center.
For example, reaction of aniline with sulfuric acid at 180 °C produces sulfanilic acid, H 2 NC 6 H 4 SO 3 H. If bromine water is added to aniline, the bromine water is decolourised and a white precipitate of 2,4,6-tribromoaniline is formed. To generate the mono-substituted product, a protection with acetyl chloride is required:
2,4,6-Tribromoaniline can be prepared by treating bromine water with aniline in a solution of acetic acid or dilute hydrochloric acid: [1] By reacting bromine with aniline in water, a white precipitate immediately forms and that is 2,4,6-tribromoaniline
2,4,6-Trichloroaniline can be prepared by reaction of dry aniline with chlorine gas while in an anhydrous solution of carbon tetrachloride. 2,4,6-Trichloroaniline precipitates from solution as a white solid. In the presence of water in the solution the white material will be contaminated with aniline black. [3]
The sodium fusion extract is made alkaline by adding NaOH. To this mixture, freshly prepared FeSO 4 solution is added and boiled for some time and then cooled. A few drops of FeCl 3 are added and Prussian blue (bluish green) color forms due to formation of ferric ferrocyanide along with NaCl. This shows the presence of nitrogen in the organic ...
The same equation relating the concentrations of acid and base applies. The concept of neutralization is not limited to reactions in solution. For example, the reaction of limestone with acid such as sulfuric acid is also a neutralization reaction. [Ca,Mg]CO 3 (s) + H 2 SO 4 (aq) → (Ca 2+, Mg 2+)(aq) + SO 2− 4 (aq) + CO 2 (g) + H 2 O
The hydrochloric acid and sodium hydroxide byproducts undergo a secondary reaction to release the byproducts of water and sodium chloride. The overall reaction is thus NaOCl + 2NH 3 → N 2 H 4 + NaCl + H 2 O. Excess ammonia and sodium chloride are removed by distillation, followed by azeotropic distillation with aniline to remove water.
Therefore, the side of the equation with water will be formed preferentially. If, for example, water, instead of hydroxide, was used to deprotonate the carboxylic acid, the equilibrium would not favor the formation of the carboxylate salt. This is because the conjugate acid, hydronium, has a pK a of -1.74, which is lower than the carboxylic ...