THE DETERMINATION OF ACETYL NUMBER. 73 



be run with every series of tests under preciselj'' similar conditions as to tinae and 

 treatment, except that the ceresine may be omitted. However, every lot of cere- 

 sine must be tested, should be free from soluble matter and not assimilate any alkali 

 on saponification. The difference between the titration of the blank and that of 

 the excess alkali in the test is the acid equivalent of the fat after acetylation, which 

 should be calculated to milligrams of potassium hydroxide for 1 gram of fat. 



One c.c. of N/2 acid is equivalent to 28.054 milligrams of potassium hydroxide. 



The difference between the saponification number pf the fat before and after 

 acetylation is the acetyl number. In case the original fat contains free soluble 

 acids, their titer should be determined and proper correction made for the same. 



Limit of error, 0.50 acetyl number. 



Synopsis of Reaction. 

 A better conception of the method may be obtained by a summary of 

 the reactions: — 



Acetylation of glycerides of monohydroxy and dihydroxy acids, monoglycerides 



and diglycerides and free alcohols. (See formulas.) 

 Saponification of the acetylated product. (See formulas.) 

 Saponification of the original or unacetylated product. 

 Titration of excess alkali. 

 Acetyl number by difference. 



Glycerides of Monohydroxy and Dihydroxy Acids. 



Acetylation: — 



(R . OH . C00).C3H.. +3(CH3CO) .0 = (R . OCH3CO . COO)3G3H5 +3 CH3COOH 

 triglyceride of acetic acetylated acetic 



monohydroxy acid anhydride glyceride acid 



Example: Ricinolein (CitHk . OH . COO)3C3H3 



Saponification : — 



(R . OCH3CO . COO)3C3H, +6 KOH 



acetylated glyceride alkali 



= 3 R . OH . COOK +3 CH3COOK +C3Hg(OH)3 



potassium potassium glycerol 



salt of hydroxy acid acetate 



[R(OH)2COO]3C3H.-, + (CH3CO)20 = [R(OCH3CO)2COO]3C3H5 +H2O 



triglyceride of acetylated glyceride 



dihydroxy acid 



Example: Dihydroxystearin [Ci7H33(OH)2COO]3C3H5 



[R(OCH3CO)2COO]3C3H5+9 K0H = 3 R(OH)2COOK+6 CH3COOK+C3H,(OH)r 



Monoglycerides and Diglycerides. 



3)2 + (CH3CO)20 = (RCOO)(CH,C 

 diaceto-gly( 



(RCOO)(CH3COO)2C3H.+3 K0H = RC00K+2 CH3COOK+C3H.-,(OH)3 



(RCOO)C3H..(OH)2 + (CH3CO)20 = (RCOO)(CH,COO)2C3H5+H20 

 monoglyceride diaceto-glyceride 



(RC00)2C,>Hr,(0H) +(CH3CO)20 = (RCOO)2(CH3COO)C3H5+CH3COOH 

 diglyceride monaceto-glyceride 



(RCOO)2(CH3COO)C3H,+3 KOH = 2 RCOOK4-CH3COOK+C3H,-,(OH)3 



