196 Miss Homer, The Resolution of Salts of Asymmetric 
The Resolution of Salts of Asymmetric Nitrogen Compounds 
and Weak Organic Acids. By Miss Annie Homer, Bathurst 
Student of Nevvnham College, Cambridge. (Communicated by 
Mr H. O. Jones.) 
[Read 11 March 1907.] 
It was thought advisable to ascertain whether some of the 
optically active nitrogen compounds could be used for the reso- 
lution of weak organic acids. At present the only method 
available for the resolution of organic acids is the crystallization 
of their salts with the bases brucine, strychnine, morphine and 
some synthetic bases. These bases are all very similar and offer 
little scope for variation, and on account of the weak basic 
properties of these substances their salts are readily hydrolysed, 
and especially so those with weak organic acids. The optically 
active nitrogen compounds are theoretically better suited for the 
purpose as they are salts of strong bases. 
With this object in view an optically inactive iodide was taken, 
viz. methylphenylbenzylisopropyl ammonium iodide (for prepara- 
tion see paper by Miss Thomas and Mr Jones, Carnh. Phil. Proc. 
1905). The iodide was dissolved in fifty per cent, alcohol, and 
treated with slight excess of moist silver oxide. Silver iodide was 
formed and the substituted ammonium hydroxide left in solution. 
The liquid was filtered ; to the filtrate was added the calculated 
quantity of (d) tartaric acid to form the acid tartrate. The 
solution was evaporated on a water bath and finally allowed to 
crystallize in a desiccator over sulphuric acid. The crystals 
separating out were hygroscopic. After the third crystallization 
from a mixture of alcohol and petroleum ether, the rotation of the 
salt was taken and the mother liquors set to crystallize in order to 
yield enough tartrate for an analysis. 
On analysis : 
•2315 grs. yielded ’5334 grs. CO., and T445 grs. H.,0. 
C = 62-8°7 0 , H = 6'99 %• 
T247 grs. „ ‘2872 grs. C0 2 and ’0789 grs. H.,0. 
0 = 62-8%, H = 7 08. 
Theory for (CH 3 C 7 H 7 C 6 H 5 C 3 H 7 N) 2 H 4 C 4 0 6 requires C = 72-8, H = 764. 
„ (CH 3 C 7 H 7 C 6 H 5 C 3 H 7 N)H 5 C 4 0 6 „ C=65T,H=6-98. 
CH 3 C 7 H 7 C 6 H 5 C 3 H 7 NH 6 C 4 0 6 .H a 0 „ C=61'9,H=7T3. 
The substance is evidently CH 3 C 7 H 7 C (j H 5 C 3 H 7 NH 3 C 4 0 B . H 2 0. The 
high percentage of carbon may be due to the presence of some 
