The method at first employed for preparing the hydrazine compound consisted in treating the amido acid, suspended in HCl, with potassium nitrite and then with stannous chloride. The tin was then removed from the solution by the addition of sodium carbonate and the hydrazine compound thrown down with HCl. This method however gave poor results the yield being only about 50% of the theoretical.
Another method was accordingly substituted for the above, namely that of Strecker and Römer (Ber. IV. s 784.) By this the diazo compound is made first and isolated. This is done by suspending the finely powdered acid in absolute alcohol, cooling and passing a current of the oxides of nitrogen through in the ordinary way. The acid changes in appearance, becoming more crystalline and slightly darker and settles quickly on being shaken. The reaction here may be expressed thus—
| ⎧ | CH3 | ⎧ | CH3 | ||
| C6H3 | ⎨ | SO2OH + HNO2 = | C6H3 | ⎨ | SO3 + 2H2O |
| ⎩ | NH2 | | | ╲ | ||
| ⎩ | N=N |
When the reaction is completed as shown by the appearance of the suspended powder it is filtered and while still fresh is added to a solution of acid sodium sulphite as long as it continues to dissolve readily.
To this solution there is added a quantity of solution of acid sodium sulphite equivalent to that already used and the solution is then boiled. It has at first a deep red color but in a few moments becomes light reddish yellow. The reaction of HNaSO3 on the diazo compound may be represented in two stages, the first portion forming an addition product and the second acting as a reducing agent. Thus,
| ⎧ CH3 | ⎧ CH3 | |||
| 1. C6H3 | ⎨ SO3 + HNaSO3 = | C6H3 | ⎨ SO2ONa | |
| │ ╲ | ⎩ N=NSO3H | |||
| ⎩ N=N |
| ⎧ CH3 | ⎧ CH3 | ||
| 2. C6H3 | ⎨ SO2ONa + HNaSO3 + H2O = | C6H3 | ⎨ SO2ONa |
| ⎩ N=NSO3H | ⎩ NH—NHSO3 | ||
To the hot solution an excess of conc. HCl is added when the hydrazine compound separates in a few moments in lustrous yellow scales which completely fill the solution. On the addition of the HCl a large amount of SO2 is given off from the excess of HNaSO3 and the solution becomes deep red. When the hydrazine has separated the mother liquor is again yellow.
The reaction is represented as follows:
| ⎧ CH3 | ⎧ CH3 | ||
| C6H3 | ⎨ SO2ONa + HCl + H2O = | C6H3 | ⎨ SO2OH + H2SO4 + NaCl |
| ⎩ NH—NHSO3H | ⎩ NH—NH2 |