Amrah:

What  you are attempting to synthesize is  NH2Cl. the mono chlorinated adduct, without too much of the tri chlorinated amines. To accomplish this you need to perform the reaction with an excess of ammonia to minimize the chances that you will form any of the multii-chlorinated adducts. This is the opposite to the stoichiometric molar ratio shown above 4 : 1 Cl to ammonia molar ratio.

In 250 ml solution of 2N sodium hydroxide with good cooling, chlorine is passed with amount necessary for the preparation of sodium hypochlorite (but no more!). The solution is cooled to 0 ° C and few ice cubes together with 250 ml of pre-cooled 1N ammonium hydroxide solution were poured. The reaction mixture is transferred to a distillation flask connected with a condenser cooled with ice-cold water. Immediately a vacuum is applied and distillation is performed at 30-40° C collecting about 50-60 ml. The solution thus obtained is stable chloramine at 0 ° C for some time, but the hydrogen chloride formed during its decomposition, further accelerates the decomposition. This process could be slow down if  a slight excess of ammonia solution is used. The isolated chloramine is poured into to a distillation flask connected to a drying tube filled with anhydrous potassium carbonate. This drying tube is additionally connected with a trap which is placed in liquid air. After evaporation of the gas from the flask with the solution it is subjected to a slight heating. Chloramine passes over anhydrous potassium carbonate and is collected in the trap. Then, at -60 ° ammonia is removed by evaporation.

It should be noted that by this method small amount nitrogen trichloride could be obtained. Nitrogen trichloride is unstable and can cause a violent explosion.

Hi Mrah,

you can use one of the published methods of reaction between ammonia and sodium hypochlorite

NH3 + OCl− ⟶ NH2Cl + HO−

Good Luck

I prepared monochloramines from a reported method but DPD titration method showed no monochloramines, while free chlorine was very high. Same happened when I tested results with colorimeter, Free chlorine value was high and combined chlorine value was not significant at all. What could be the possible reason? Moreover, when I mean to set chloramine dose to 2 mg/l, is it supposed to be combined chlorine value only(i.e obtained by total chlorine - free chlorine)?