130 ANIMAL CHEMISTRY LECTURE VI. 



Dehyd-uric? Water Mesoxalic Urea 



C 5 N 4 H 2 3 + 4 H 2 = C 3 H 2 5 + 2CN 2 H 4 0. 



In the above equations I have represented both atoms of urea to 

 be simultaneously separated from the mesoxalic acid ; but in 

 reality their separation is usually effected at two successive stages, 

 the first of which results in the formation of alloxan, and the 

 second in its decomposition, thus : 



Dehyd-uric? Water Alloxnn Urea 



C 5 N 4 H 2 3 + 2H 2 = C 4 N 2 H 2 4 + CN 2 H 4 0. 



Alloxan Water Mesoxalic Urea 



C 4 N 2 H 2 4 + 2H 2 = C 3 H 2 5 + CN 2 H 4 0. 



We have, you perceive, three mesoxalic compounds, first the 

 non-nitrogenous acid, then the compound of the acid with one 

 atom of urea minus 2H 2 0, and lastly, the compound of the acid 

 with two atoms of urea minus 4.H 2 0, thus : 



Mesoxalic Alloxan Dehyd-uric ? 



C 3 II 2 5 C 4 N a H 2 4 C 5 N 4 H 2 3 



Now, by hydrogenising mesoxalic acid, we obtain tartronic 

 acid C 3 H 4 O 5 , and by hydrogenising alloxan we obtain dialuric 

 acid C 4 N 2 H 4 O 4 , which two bodies accordingly bear to uric acid 

 the same relation that mesoxalic acid and alloxan bear to dehvd- 

 uric acid, thus, 



Tartronic Dialuric Uric acid 



C 3 H 4 5 C 4 N a H 4 4 C 5 N 4 H 4 3 ; 



and, just as our hypothetical dehyd-uric acid yields mesoxalic 

 acid and alloxan, so should actual uric acid yield us the tartronic 

 and dialuric acids. 



In reality, however, these bodies have not been obtained by 

 the mere breaking up of uric acid, but only by rehydrogenising 

 the mesoxalic acid and alloxan which result from the breaking 

 up of its dehydrogenised product. Despite, however, this flaw in 



