30 PROCEEDINGS OF THE AMERICAN ACADEMY 



determine the products of the decomposition of methyhiric acid under 

 these conditions. 



Two tubes, each containing 1.3 grm. methyluric acid, and an excess 

 of hydrochloric acid saturated at 0°, were heated four or five hours at 

 170'^. The gas wliich escaped on opening the tubes was found to con- 

 tain no methyl chloride. The excess of acid was driven off on the 

 water bath, and the residue distilled with plumbic hydrate until the 

 distillate was no longer alkaline. The ammoniacal distillate was caught 

 in hydrochloric acid, and evaporated to dryness on the water bath. 

 The residue was treated with a small (juantity of absolute alcohol, and 

 the filtered solution again evaporated to dryness. There was then left 

 a white saline residue, wliich gave with great readiness Hofmaiin's iso- 

 cyanide reaction, showing the presence of a monamine. The chloride 

 was converted into the platinum salt, and this was analyzed after re- 

 crystallization from hot water. 



0.4760 grm. gave on ignition 0.1991 grm. platinum. 



Calculated for Found. 



. (CH3NH,).PtClg 

 Pt 41.61 41.82 



Methylamine is, therefore, one of the products of the reaction. 



From the residue left on distillation, it was easy to isolate glycocoll 

 in the ordinary way. The licjuid was filtered from the basic [dumbic 

 chloride, the lead removed from the solution by hydric sulphide, and 

 the filtrated evaporated. On standing, glycocoll crystallized out with 

 its characteristic properties. For its identification, it was converted 

 into the copper salt by boiling with freshly precipitated cupric oxide, 

 and precipitation of the blue solution by alcohol. Of this salt, — 



0.4400 grm. lost at 130° 0.0388 grm. 



Calculated for Found. 



(C2H,N0,)_,Cu . li.fi 

 Bfi 7.85 7.68 



A determination of copper in the dry salt gave, — 



0.4068 grm. left on ignition 0.1523 grm. CuO. 



Calculated for Found. 



(C,H,NO,),Cu 

 CuO 37.00 37.43 



The reaction in this case may therefore be writtten, — 



C,H3(CH3) N,03 + 5 H,0 = 3C0, + 2NH3 + CH.NH, + CJI.NO,. 



