RDR 12

Results and Discussion

The experiment was comprised of two parts — the the determination of the saponification number and iodine number of lipid extracts from an oil sample (lard). The saponification saponification number indicates the amount (mg) of KOH needed to saponify a gram of fat or in other words, the amount of KOH required to neutralize the fatty acids  brought about about by the complete complete hydrolysis of of 1 gram of the lipid lipid extract, and to obtain obtain its molecular molecular weight. weight. Initially, 1.0 g of the oil sample was weighed in a previously prepared clean and dry round bottom flask. Then, exactly 15.00 mL of 5% alcoholic KOH was delivered into the flask, the reaction upon the addition of KOH is  presented below.

(1) Boiling chips were added to the flask and the sample was refluxed in alkali for an hour to hydrolyze triacylglycerols to give glycerol and potassium salts of fatty acids which is the soap. Afterwards, the sample was cooled to room temperature. temperature. Five milliliters of the resulting solution was transferred into a 125-mL Erlenmeyer flask, and then 3 drops of 1% phenolphthalein indicator (colorless in acidic solutions and magenta in basic solutions) was added to the solution. Afterwards, the solution was titrated with 0.5 M HCl until the originally colorless solution turned faint pink, which signaled the end of titration. Note that there will be existing unreacted KOH in the solution, which was titrated as previously stated, thus saponification number is determined through the back titration of excess KOH shown below. HCl + KOH  KCl + H2O

(2)

This required ____ mL of the titrant to reach the end point. Another trial performing the same procedure was done which needed ___ mL of the titrant to observe the color change. The saponification number and the molecular weight of the extracted triacylglycerol were found out to be _____________, respectively, using the formula below.

 

  



( ) (  )

 

     

()

()

The obtained saponification numbers, ______ were relatively __ smaller __ smaller or greater   __ as to its literature literature value which ranges from 190-202. Having solved the saponification number and molecular weight hence, will determine the nature of the fatty acids of the lipid extract; the greater the molecular weight implies longer chains of fatty acids existing in the sample, thus the smaller the number of fatty acids which are liberated per gram of fat, which means having a smaller saponification number, and vice versa. From there, it can be implied that molecular weight and saponification value are inversely proportional. The determination of the iodine number indicates the degree of unsaturation of the lipid extract by finding out how much iodine is taken up by 100 g of the sample. First, 0.1 g triacylglycerol was weighed in each of the two previously prepared clean and dry 250-mL Erlenmeyer flasks. Afterwards, 5 mL ether was added to the flasks and then the samples were dissolved. The addition of 10 mL Hanus reagent (a solution of iodine

monobromide in glacial acetic acid) was added and then the flasks were covered with aluminum foil. In this reaction, iodine monobromide reacts with the unsaturated bonds (double bonds) to produce a di-halogenated single bond, as depicted by the equation below. As can be noticed, in each double bond existing in the hydrocarbon chain, there corresponds one atom of iodine bonded to one carbon.

(5) The flasks were allowed to stand in the dark for 30 minutes, with occasional shaking, this is necessary to avoid side reactions that may occur before titration. Afterwards, the flasks were brought out and then 5 mL of 15% KI was added to the solution. The mixtures were shook thoroughly in order to extract any remaining free iodine into the KI solution. Since the Hanus reaction already took place upon executing this step, it can be inferred that all iodine atoms needed to convert all unsaturated bonds into single bonds were already used up, thus the remaining unreacted ICl will determine how much iodine was used up. The addition of KI solution then will react to the unreacted ICl hence, liberating iodine atoms in the solution, as shown in the equation below.

(6) Then, 12.5 mL distilled water was added into each flask to dilute the solutions. The resulting solutions were titrated with 0.1 Na 2S2O3 until the previously betadine-red colored solution turned pale yellow, these required ____ mL and ____ mL (V1) of titrant to reach the end point, respectively. The reaction that took upon titration is shown below.

(7) Afterwards, 1 mL of freshly prepared 1% starch solution was added into each flask and then shook vigorously in order to extract any iodine existing in ether. The solution turned blue-violet in color, brought about  by the iodine/starch complex. These solutions were again titrated with 0.1 Na 2S2O3 until the blue-violet color disappeared indicating the liberation of iodine atoms from starch, which required ____ mL and ____ mL (V 2) of the titrant to reach the end point, respectively. This step aimed to use starch as an indicator and have an identical endpoint as to V 1. Afterwards, the iodine number was calculated to be ___ and ____, respectively using the formula below.

 

   



  (     ) 

()

The obtained iodine numbers were _____, which were relatively __ smaller or greater  ___ compared to its literature value which ranges from 45-70. References

Slade, P. E. (1998). Handbook of fiber finish technology . New York: Marcel Dekker. Bailey, A. V., Harris, J. A., & S kau, E. L. (1967). Relationship between the calculated glyceride composition and the iodine value of cottonseed and pecan oils.  Journal of The American Oil Chemists Society . doi:10.1007/BF02558167

ANALYSIS OF LIPIDS. (n.d.). Retrieved from http://people.umass.edu/~mcclemen/581Lipids.html