Lipids
Inroduction
Lipids are non-polar organic biomolecules that are totally or nearly insoluble in water but are quite soluble in non-polar organic solvents like ether, chloroform or benzene. They are rather heterogeneous group of compounds differing in their structure and perform diverse biological functions. They serve as major structural components of the membranes and also form a protective coating on many organisms. Some of the vitamins and hormones are lipids. Here are several different classes of lipids but all of them derive their distinct properties due to the hydrocarbon nature of their structure. Lipids have been classified in several ways but the most acceptable classification is the one based on the structure of their backbone. Based on this, they are divided into two groups: complex lipids and simple lipids. The complex lipids are esters of fatty acids. In various forms of complex lipids like acylglycerides, phospholipids, sphingolipids and waxes, fatty acids are covalently joined via an ester linkage to a trihydroxy alcohol, glycerol, or its derivative. Since the complex lipids yield soap on alkaline hydrolysis, they are also called saponifiable lipids. The simple lipids, on the other hand, do not contain fatty acids and are therefore called non-saponifiable lipids. They include compounds like terpenes, sterols etc.
A) DETERMINATION OF ACID VALUE OF FATS AND OILS
Principle
Different fat samples may contain varying amount of fatty acids. In addition, the fats often become rancid during storage and this rancidity is caused by chemical or enzymatic hydrolysis of fats into free acids and glycerol. The amount of free fatty acids can be determined volumetrically by titrating the sample with potassium hydroxide. The acidity of fats and oils is expressed as its acid value or number, which is defined as mg KOH required to neutralize the free fatty acids present in 1g of fat and oil. The amount of free fatty acids present or acid value of fat is a useful parameter, which gives an indication about the age and extent of its deterioration.
Materials and Reagents
1. Burette.
2. Conical flasks.
3. Test compounds (olive oil, butter, margarine etc.; fresh and the samples that have been stored at room temperature for several days may be used for comparison).
4. 1 per cent phenolphthalein solution in 95% alcohol.
5. 0.1N KOH: Weigh 5.6g of KOH and dissolve it in distilled water and make the final volume to 1 L. Standardize this solution by titrating a known volume of 0.1N oxalic acid (Prepare by taking 630 mg oxalic acid in 100 ml water) using phenolphthalein as an indicator till a permanent pink color appears. Calculate the actual normality (N2) of KOH solution from equation N1V = N2V2, where N1 and V1 are normality and volume of oxalic acid taken for titration and V2 is the volume of KOH solution used.
6. Fat solvent (95% ethanol: ether 1:1, v/v).
Procedure
1. Take 5g of fat sample in a conical flask and add 25 ml of fat solvent (Reagent No. 6) to it. Shake well and add a few drops of phenolphthalein solution and again mix the contents thoroughly.
2. Titrate the above solution with 0.1N KOH until a faint pink color persists for 20-30 seconds.
3. Note the volume of KOH used.
4. Repeat the steps 1-3 with a blank, which does not contain any fat sample.
Calculations
0.1 N KOH solution used for blank = x ml
0.1 N KOH solution used for sample = y ml
Titre value for sample = (y – x)
titre value x Normality of KOH x 56.1
Acid value (mg KOH/g fat) = __________________________________________________
weight of sample (g)1 ml of 1N KOH contains 56.1 mg of KOH. Hence a factor of 56.1 is incorporated in the numerator in the above equation to obtained weight of KOH from the volume of 0.1N KOH solution used during this titration.
Titrate the liberated iodine with standard sodium thiosulphate solution till the content of the flask become pale yellow in color. Add a few drops of starch solution and continue to titrate it further with sodium thiosulphate solution till the blue color disappears.
Calculations
The difference between the blank and test reading gives amount of 0.1 sodium thiosulphate required to react with an equivalent volume of iodine. One L of 0.1N iodine solution contains 12.7 g of iodine. The iodine number can thus be calculated as follows:
Volume of 0.1 N sodium thiosulfate used for blank = x ml
Volume of 0.1 N sodium thiosulfate used for sample= y ml
(x – y) x 12.7 100
Iodine number = ___________________ x ____________________
1000 wt. of sample (g)
Precautions
The bottles must be shaken thoroughly the titration to ensure that all the iodine is expelled from the chloroform layer.