14 The Philippine Journal of Science 



1913 



By solving algebraically the simultaneous equations : (1), (2), 

 (3), (4), and (7), we may express n, the free base in our ex- 

 periment, in terms of x as follows : 



v/X2— ax+0. 018824— (x+0. 0428) 

 n=- g . (8) 



in which a=0.0944. (Table II.) 



Now, when methyl salicylate is saponified, the rate of the 

 reaction will be proportional to the product of the free base 

 and the free ester; that is, 



dX= Ken ' (9) 



in which x 1 is the amount of ester hydrolysed and K is the 

 reaction velocity constant. 



However, upon consideration it will be apparent that this 

 reaction is accompanied by another saponification reaction which 

 results in the same products. For it is erroneous to suppose, as 



/NoNa 

 Goldschmidt does, that sodium methyl salicylate, 



k /COOCH3, 



can exist in the presence of free NaOH without undergoing 

 saponification, according to the equations 



C 6 H 4 ONaCOOCH 3 +NaOH ^C 6 H 4 -ONaCOONa+CH 3 OH 

 C 6 H 4 ONa-COONa+H 2 ^C 6 H 4 -OH-COONa+NaOH. 



The rate at which this saponification takes place will in all 

 probability be different from that at which the free ester is 

 saponified. That this rate is appreciable cannot be doubted, 

 since it is impossible to assign to the sodium in the phenol 

 position per se the peculiar property of entirely protecting the 

 adjacent — COOCH 3 group from attack. If x 2 is taken to rep- 

 resent the concentration of the substance transposed by this 

 reaction, we can express the velocity of this saponification by 

 the equation 



^=Kimn, (10) 



in which K 1 is the reaction velocity constant and m n is the prod- 

 uct of the concentrations of the reacting substances. But from 

 (1) and (7), m=5n. Substituting this value in (10), we obtain 



^ 2 = Ki5n2 (11). 



Now, if x x represents the amount of free ester transformed, 

 and x 2 the amount of sodium methyl ester transformed, their 



