Determination of Iodine Number of Fat/Oil
Halogenation of fats and oils (Triacylglycerols): The unsaturated fatty acids and their
esters absorb halogens like chlorine, iodine and bromine at their double bonds when treated
with reagent like iodine monochloride at room temperature in acetic acid and methanol
solution.
13 12 10 9 X2 (X2 = Cl2 , Br2 , I2)
CH3(CH2)4CH = CHCH2 CH = CH(CH2)7COOH
Linoleic acid
X X X X
| | | |
CH3(CH2)4 CH CH CH2 CH CH(CH2)7COOH
Stearic acid tetrahalide
(9,10,12,13 tetraiodostearic acid)
10 9 X2
CH3(CH2)7 HC = CH(CH2)7COOH
Oleic acid
X X
| |
CH3(CH2)7 HC-CH (CH2)7COOH
Stearic acid dihalide
Fig 2.6 Halogenation of fats and oils.
Iodine Number: This is defined as gms of iodine absorbed by 100gms of fat. As iodine is
absorbed only at the double-bonds of unsaturated fatty acid residues of triglycerides, iodine
number gives an estimate of the degree of unsaturation and so of the relative amounts of
unsaturated fatty acids in the triglyceride molecule of the fat. For example, coconut oil
(iodine Number 6-10) contains much less unsaturated fatty acids than olive oil (iodine No.80-
88) cottonseed oil (iodine No.103-110) contains much more unsaturated fatty acids than
butter (iodine No.26-28)
Iodine number is a useful characteristic for assessment of both purity and nutritive value of
the fat.
The iodine value is influenced by the percentage of each unsaturated fatty acid, the degree of
unsaturation of each fatty acid and the mean molecular weight of the fat.
The iodine number of some important fats and oils are given below:
Fats and Oils Iodine Number
Butter fat 26-28
Human fat 65-70
Peanut Oil 80-90
Corn Oil 110-115
Soya bean Oil 137-143
Linseed Oil 170-200
Beef fat or tallow 42
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Principle: The given amount of fat is treated with a measured excess of Hanus solution.
R- CH = CH – COOH + I – Br
Excess of
Hanus solution
I Br
| |
R – CH – CH – COOH + I- Br
Left over
To the left over Hanus solution is added potassium iodide solution. The iodine thus liberated
is titrated against standard solution of “hypo” (Na 2s2o3) using starch as an indicator. The color
change is from deep blue – black to white which mark the end point of the titration.
IBr + KI I2 + KBr
2Na2s2o3 + I2 Na2s4o6 + 2NaI
I2 + starch Blue colour
Materials:
1. Given oil sample.
2. Chloroform.
3. Hanus solution:
Preparation of Hanus soln: Iodine (13gms) was dissolved in glacial acetic acid,
Bromine (3ml) was added to it in the sol n was diluted with glacial acetic acid to 1
litre.
Iodine 13 gms
Bromine 3 ml
Glacial acetic acid Rest amount to prepare the soln of 1 litre.
4. Potassium iodide (10%): Dissolve 10g KI in 100ml dH2O.
5. 0.1 N sodium thiosulphate: Dissolve 24.0g of sodium thiosulphate in water and make
up to 1 L.
6. Starch solution (1%): Dissolve 1g starch in 100ml dH2O.
7. Stoppered flaks (250ml), burette, pipettes etc.
Procedure:
Test (With Oil):
In a 250ml conical flask, add 5ml of given oil sample (the oil sample is dissolved in CCl 4.
The concentration of the oil sample is 5g%), followed by 10ml of Hanes solution. Mix well,
cover the mouth of the flask with a paper and keep it for reaction take place. After 30 minutes
add 5ml of KI solution in to it. Mix well, followed by 25ml of distilled water. Add 4-5 drops
of starch indicator. The colour of the solution turn blue – black. Titrate the contents of the
flask with N/10 Na2s2o3 till the colour changes from blue black to white which marks the end
point of the titration. Note down the titre value which is xml.
Blank (Without Oil):
In 250ml conical flask add 5ml of CCl 4 only instead of oil sample and repeat the same
procedure as in the test. Note down the titre value which is Yml. The difference between the
two (i.e. blank-test) gives the amount of Na2s2o3 utilized in titrating the IBr which was used in
saturating the unsaturated fatty acid moiety I.e. the volume of IBr required to saturate the oil
= (Blank – test) value. In the test titration the excess of IBr, I.e. the left over IBr is titrated
against Na2s2o3. In blank titration the excess of IBr (as in the first case) and the actual volume
of IBr which would have been used up by oil to be saturated are together titrated against
Na2s2o3.
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Calculations:
Titre value obtained for test titration = Xml
Titre value obtained for blank titration = Yml
According to the Normality equation.
1ml of N/10 Na2s2o3 solution Ξ 1ml of N/10
IBr solution Ξ 1ml of N/10 iodine solution
Equivalent weight of iodine = 127
127 1
1ml of N/10 iodine = x = 0.0127gm
1000 10
(Two molecules of Na2s2o3 are equivalent to one molecule of iodine; thus one molecule of
Na2s2o3 is equivalent to one atom of iodine).
1ml of N/10 Na2s2o3 solution = 0.0127gm of iodine.
Amount of iodine absorbed by given amount of oil or fat = (Y-X) x 0.0127gm of iodine.
The concentration of oil sample is 5gm% I.e.5ml of oil Ξ 0.25gm of oil.
0.25gm of oil or fat consumes (Y-X) x 0.0127gm of iodine.
∴ 100 gm of oil or fat consumes gm of iodine
The iodine Number = gm of iodine.
Result: The desired iodine number of the given sample was................................
Note- Hanus iodine number:
An iodine number determined by the use of a solution of iodine in glacial acetic acid, with
iodine bromide serving as an accelerator of the reaction.
# Significance of Iodine Number:
1) The iodine number as a measure of the degree of unsaturation of the fatty acids in the
fat.
2) Indicates the relative amount of unsaturated fatty acids in the triglyceride molecule of
the fat.
3) It helps for assessment of both purity & nutritive value of the fat.
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# The iodine number gives no indication as to the number of double bonds present in the fatty
N
acid molecule. Amount of iodine in 1ml iodine is
10
1000 ml 1 N contain 127 gm iodine
1 ” ” ”
N
1 ” ” ”
10
= 0.0127 gm of iodine.
1ml of iodine contain 0.0127 gm of iodine.
1) Iodine number.
2) Significance.
3) Helogenation.
4) hanus soln 2 preparation.
5) Fat sample preparation.
6) Different iodine no.
7) Problems.
# Short Procedure :
5 ml oil sample
+
250 ml 10 ml hanus soln.
flask
[ 30 mins. gyL †X‡K AR K‡i †i‡L w`‡Z n‡e|
]
5 ml K I
+
25 ml of distilled water
+
4-5 drops starch
Gici titration Ki‡Z n‡e Na2S2O3 Gi mvnv‡h¨
hv burettee G _v‡K|
# Preparation of oil sample :
The oil sample is dissolved in ccl4. The concentration of the oil sample is 5g%. That is, 5 gm
is dissolved in 100 ml of ccl4.
Say, conc. of 5 ml of oil,
100 ml contain 5 gm oil
1 ” ” ”
5 ” ”
= 0.25 gm of oil.
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