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 HNaSO_{3} 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, ┌─ ┌─ │ CH_{3} │ CH_{3} 1. C_{6}H_{3} ┤ SO_{3} + HNaSO_{3} = C_{6}H_{3}┤ SO_{2}ONa │ \ │ │ N═N │ N═NSO_{3}H └─ └─ ┌─ │ CH_{3} 2. C_{6}H_{3} ┤ SO_{2}ONa + HNaSO_{3} + H_{2}O │ N═NSO_{3}H └─ ┌─ │ CH_{3} = C_{6}H_{3} ┤ SO_{2}ONa + HNaSO_{4}. │ NH-NHSO_{3}H └─ 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 SO_{2} is given off from the excess of HNaSO_{3} and the solution becomes deep red. When the hydrazine has separated the mother liquor is again yellow. The reaction is represented as follows: ┌─ │CH_{3} C_{6}H_{6}┤SO_{2}ONa + HCl + H_{2}O │NH-NHSO_{3}H └─ ┌─ │CH_{3} = C_{6}H_{3}┤SO_{2}OH + H_{2}SO_{4} + NaCl │NH-NH_{2} └─ The yield of hydrazine when both the diazo and the NaHSO_{3} are freshly prepared is practically quantitative. The hydrazine thus prepared was treated with a hot 10% solution of copper sulphate till a permanent blue color was obtained in the solution. Nitrogen is evolved and the copper sulphate is reduced to cuprous oxide which is precipitated as a red powder. The reaction is as follows. ┌─ │CH_{3} C_{6}H_{3}┤SO_{2}OH + 2CuSO_{4} + H_{2}O │NH-NH_{2} └─ ┌─ = C_{6}H_{4}┤CH_{3} + Cu_{2}O + N_{2} + 2H_{2}SO_{4} │SO_{2}OH └─