Designer
Abdulkder Salad
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Organic Practical
O
rganic practical encompasses different methods that extend from purification to identification to quantitation of natural or synthetic substance. Each practical topic carries its own importance inside this wide application of the discipline .
We already mentioned , in the introduction , the role of the organic chemistry in our modern life ; an example can be taken for the preparation of synthetic products for medical purpose . In this case level of purity and concentration has to be maintained seriously. In environmental chemistry , toxicity of the hazardous substance must be precisely known by frequently determining the level of contamination . Nowadays fast and reproducible methods of determination has been developed as a result of the modern technology . we will briefly discuss certain practical methods beginning with purification systems .
1. Purification of solid substance
We rarely obtain a pure chemical product at the end of the chemical reaction during the preparation of a specific substance . Aspirin is a solid substance that can easily be prepared in the laboratory ; the product is liable to different sources of contaminations like inappropriate side reaction , occurrence of reverse reaction , application of impure reagents etc. . There must be a way to purify the product
A. Recrystallization: In this process, the solid substance is dissolved in a least possible amount of a suitable solvent to make saturated solution and heat to boiling until the solid material disappears , the solution is then gently cooled ; pure crystals of the sought for substance will be formed .The crystals are filtered , washed , and dried in a desiccator .
B. Sublimation: To apply this method , the desired substance must be volatile one ; that is a substance that sublimes ( changes directly from solid state to gaseous state ) . The mixture sample is heated and the vapor of the sublimed substance is collected on a cool surface as pure crystals .
Simple distillation unit
2. Purification of liquid substance.
Separation and purification of liquids is commonly achieved by distillation methods besides other advanced techniques like chromatographic procedures . Every pure solvent has its own unique physical properties like boiling point , melting point , vapor pressure etc. Water boils at 100oC. No other pure solvent boils at that temperature .
Purification of liquids by distillation employs this distinct physical property. The process involves heating the liquid substance to its gaseous state and condensing back to liquid state by cooling. Different types of distillation sets are used depending on the nature of the solvent to be purified or separated.
A. Simple distillation: This process involves direct heating of the components in which solvent evaporates according to its boiling point and cools and condenses inside the condenser as pure liquid drops into receiving flask . This type of distillation is applied if there is wide difference between the boiling points of the constituents .
Fractional distillation unit
B. Fractional distillation: The only difference between the simple distillation and the fractional distillation is the presence of fractionating column . This column is glass tube packed with glass beads or metal pieces to increase the surface area and to make efficient separation . This type of distillation is applied if the boiling points of the constituents are relatively closer together. In this method the separation takes place inside the fractionating column .
C. Steam distillation: This type of distillation is used for the purification or separation of volatile organic solvents ; in this process insoluble steam is introduced into the mixture and as result the sensitive volatile compound vaporizes and condenses into the receiving flask mixed with water vapor. The water can be eliminated by simple distillation or other drying methods.
Steam distillation unit
D. Vacuum distillation: This kind of distillation is used when the boiling point of the liquids to be distilled is very high at ordinary atmospheric pressure . These liquids may dissociate or fractionate when subjected to higher temperature at normal atmospheric pressure . In vacuum distillation the vapor pressure of the liquid mixture is reduced causing easy evaporation of the volatile liquid without damaging the substance . This type of distillation has wide commercial application and available in various set-ups .
Vacuum distillation unit
Qualitative Organic Analysis
1. Identification of elements, C, H, O
We know that all organic compounds contain carbon besides hydrogen and oxygen in most cases ; it not so important to test for these elements . The only way to test for these elements (C,H,O) is to heat a sample of the organic compound in the presence of dried copperic oxide (CuO) , and check for CO2 and water vapor evolved .
2. Identification of elements S, N, X, P
In classical wet chemical analysis , there are two ways to test for the presence of these elements.
· Middleto
Take about 0.2g of the compound and transfer into a test tube . Add nearly 1.5g of a mixture of zinc powder with anhydrous sodium carbonate (2 : 1 respectively) . Heat the mixture strongly in a hot Bunsen flame for about three minutes . Immerse the red-hot tube into a small beaker containing about 30ml distilled water . Boil the contents to dissolve it . Filter the solution and perform the following tests.
i. The sulfur can be tested from the residue on filter paper by adding dilute HCl (4M) , evolution of hydrogen sulfide gas is detected by its unpleasant odor and by exposing piece of paper wetted with lead acetate solution , it turns black .
ZnS + HCl ZnCl2 + H2S
S2- + Pb2+ PbS
Another test for element sulfur:
Add freshly prepared sodium nitroprusside solution to about 3ml of the filtered solution , purple-red color confirms the presence of sulfur .
1) Test for nitrogen:
Since nitrogen is changed into cyanide (CN-) in this test , it will be tested as cyanide. Add few crystals of iron(II) sulfate to 3ml of the filtrate solution and a few drops of 2M NaOH solution . Boil the mixture for one minute , cool it and add 2-3ml ferric chloride solution and enough amount of dilute H2SO4 or concentrated HCl to dissolve the precipitate . Blue solution or precipitate indicates the presence of nitrogen .
Fe(OH)2 + 6NaCN Na4[Fe(CN)6] + 2NaOH
Na4 [Fe(CN)6] + 2Fe2(SO4)3 Fe4[Fe(CN)6] + 6Na2SO4
(Blue)
2) Test for halides:
If nitrogen or sulfur is already detected , add 6M HNO3 to the solution and boil it for three minutes to expel the hydrogen sulfide and hydrogen cyanide gases . Cool the solution and add very dilute silver nitrate solution and observe the result :
1. White precipitate soluble in ammonia solution indicates the presence of Cl - .
AgNO3 + NaCl AgCl + NaNO3
2. Light yellow precipitate soluble in excess ammonia solution indicates the presence of Br - ,
AgNO3 + NaBr AgBr + NaNO3
3. Yellow precipitate insoluble in ammonia solution indicates I -.
AgNO3 + NaI AgI + NaNO3
The other method for the identification of these elements is :
a) Lassaigne’s test:
This test is more or less similar to that of Middleton’s except little modifications has to be made . In this method small piece of sodium (0.2 – 0.4g) is put into an ignition tube containing about 0.2g of the sample.
The mixture is gently heated at the beginning so that the reaction begins smoothly , and then strongly heated to red-hot ; continue heating for another one or two minutes. Plunge the red-hot tube into small amount of distilled water . Heat the mixture and filter it ; follow the steps as Middleton’s except that there is no test on the residue .
· Detection of phosphorus:
Take about 3ml of Lassaigne’s or Middleton’s solution in a test tube and add to it 5ml of concentrated nitric acid . Boil the mixture to oxidize the phosphorus into phosphate . Add 1ml of ammonium molybdate solution after cooling to around 40oC.
Formation of yellow solution or precipitate indicates the presence of phosphorus .
· Chemical tests on organic compounds:
1. Hydrocarbons:
A. Unsaturated aliphatic hydrocarbons:
We have seen that hydrocarbons are generally classified into saturated hydrocarbons which undergo substitution reaction and unsaturated hydrocarbons which undergo addition reaction.
Procedure: Take 3ml of oleic acid or any other cooking oil and add few drops of bromine in CCL4 , note the immediate disappearance of the color of the bromine .
Repeat the same experiment with dilute aqueous potassium permanganate and observe the disappearance of the violet color of the permanganate .
B. Aromatic hydrocarbon:
Aromatic hydrocarbons undergo both substitution and addition reactions , the reason is already explained in the theoretical part.
Procedure: Take about 150 mg of ACl3 in a test tube (Pyrex) and heat in a Bunsen burner in a nearly horizontal position. Continue to heat until the solid vaporizes (sublimes) above the bottom of the test tube for about 2-3cm ; cool it to around 40o C . Add 25mg of the aromatic hydrocarbon if it is solid or 2drops if it is liquid followed by few drops of chloroform . The formation of red to blue color where the sample contacts with aluminum chloride indicates the presence of aromatic hydrocarbon .
Functional Group Identification
i. Alcohols
Most of the chemical properties of alcohols was discussed in the theoretical part at chapter 10. Since there are three structural forms of alcohols as primary , secondary and tertiary , the chemical test of the hydroxyl group not only identifies the functional group but also indicates the position of the group .
A. With chromic acid
In this test the sample must be aldehyde free.
Procedure: Take about 2ml of potassium dichromate (K2Cr2O4) in a test tube , add 1ml of conc. sulfuric acid , cool the mixture in a tap water , add 1ml of methyl alcohol and heat gently ; notice the emergence of formaldehyde smell and the formation of a green solution .
B. Reaction with carboxylic acids (esterification).
Take 2ml of ethanol and 2ml of acetic acid in a dry test tube . Add 1ml of conc. H2SO4 to the mixture . Heat the contents in a water bath for about 2 minutes . Transfer the contents to a beaker containing sodium bicarbonate solution . Notice the pleasant smell of the ester .
C. Lucas test
Take three test tubes containing ethanol , 2-propanol (isopropyl alcohol) , and tertiary butyl alcohol respectively . To each one add 10ml of Lucas reagent (anhydrous zinc chloride in conc. HCl) . Stopper each tube and shake . Leave the tubes to stand at room temperature and record the time taken by the alcohol to form second layer or an emulsion .
D. Reaction with sodium
There is a reaction between the ordinary liquid alcohols and sodium similar to that of water , except that the reaction is comparatively more gentle in alcohols. In both cases , hydrogen gas is liberated and alkoxide solution is formed.
CH3CH2OH + Na CH3CH2ONa + H2
Ethanol Na-ethoxide
ii. Phenols
We defined the phenols as the compounds that contain hydroxyl group connected directly to the benzene ring . Phenols are soluble in approximately 0.5M NaOH but insoluble in dilute sodium bicarbonate solution .
1. With bromine (in water & CCl4)
Add bromine water to about 1% aqueous solution of phenol and notice the decolorization of the bromine solution . Perform the same test with bromine in CCl4 and compare the rate of color disappearance .
2. With ferric chloride
Take one drop of phenol into a test tube and 3ml of distilled water , shake well then add three drops of ferric chloride solution . Note the violet color formed which disappears by adding dilute HCl .
Blue Violet
A. Liebermann test
Phenols react with sodium nitrite (NaNO2) and conc. H2SO4 forming nitro-compounds which give different colors in acidic and basic solutions . Procedure: Transfer few crystals of NaNO2 in dry test tube. Add 3ml of conc. H2SO4 . Heat the mixture gently until the crystals dissolve. Add one or two drops of phenol and notice the brown color formation that changes into blue color . transfer the contents into small beaker containing 10-15ml distilled water and notice the red color formed . Add to it sodium hydroxide solution until it becomes basic notice the blue color again .
B. Sodium hypochlorite
Add few drops of ammonium hydroxide to a phenol solution ; then add few drops of sodium hypochlorite , freshly prepared . Notice the formation of blue color . Add dilute HCl to this blue solution and notice the red color formed. The solution changes into blue again by adding excess NaOH .
C. Phthalic test
In a dry test tube , put 0.5g of phenol ; add 1g of phthalic anhydride and few drops of conc. H2SO4 , heat the contents gently until the materials fuse together , then add 1ml of water and excess of sodium hydroxide . Note the red color due to formation of phenolphthalein .
Colorless Colorless Blue
Phenol Phthalic anhydride Phenolphthalein
1. Carbonyl compounds (aldehydes & ketones)
Aldehydes and ketones both contain carbonyl group (-CO-) . In the case of aldehydes , hydrogen atom is directly connected with the carbonyl carbon while in the case of ketones , alkyl or aryl group is connected to the carbonyl carbon and for this reason , aldehydes easily oxidize while ketones do not easily oxidize. But they both share the same test .
A. Brady’s test:
Take 3ml of the carbonyl compound , add 1ml of Brady’s reagent (2,4-dinitrophenyl hydrazine) , yellow or red precipitate indicates the presence of an aldehyde or ketone .
B. Tollens test (distinguish between aldehyde & ketone)
To 3 drops of the substance add 0.5ml of Tollens reagent ([Ag(NH3)2]NO3) . Formation of black precipitate or silver mirror on the walls of the test tube indicates aldehyde .
Ag(NH3)2+ + RCHO Ag + RCOOH + 2NH3
Silver mirror
C. Fehling test ( for another differentiation )
Add 1ml of the sample to Fehling solution ( prepared by mixing equal volumes of Fehling A , and Fehling B ) . Heat the contents to boiling . Formation of red precipitate of Cu2O indicate the presence of an aldehyde .
Cu2+ + 2OH - + RCHO Cu2O + RCOOH + H2O
Blue red precipitate
i. Schiff ‘s reagent
Add 1ml of Schiff ‘s reagent to the substance under investigation (several drops) . Formation of violet red color confirms the presence of aldehydes.
ii. Aldehydes oxidize in the presence of reducing agent such as potassium permanganate KMnO4 or potassium dichromate K2Cr2O7 , in acidic solution producing green solution .
D. Carboxylic acids
We have defined the organic carboxylic acid as any compound that contains carboxylic functional group. These organic acids can be classified into aliphatic carboxylic acids which are the fatty acids and aromatic carboxylic acids in which the carboxylic group is directly attached to benzene ring. We also divided the aliphatic carboxylic into monocarboxylic acids, dicarboxylic and tricarboxylic.
1) Aliphatic monocarboxylic acids
A. With sodium bicarbonate:
Carboxylic acids react with carbonates and bicarbonates resulting evolution of carbon dioxide gas and the corresponding salt of the acid .
RCOOH + NaHCO3 CO2 + RCOONa + H2O
To 3ml of formic acid (HCOOH) , add 1ml of sodium bicarbonate solution. Notice the evolution of CO2 which can be tested with clear lime water making it turbid .
HCOOH + NaHCO3 CO2 + HCOONa + H2O
CO2 + Ca(OH)2 CaCO3 + H2O
lime water
B. Esterification:
Carboxylic acids react with alcohols in the presence conc. H2SO4 giving ester and water. Take 3ml of ethyl alcohol (ethanol) add to it 3ml of acetic acid and 1.5ml of conc. Sulfuric acid . Heat the mixture in a water bath for five minutes . Cool the solution and transfer to a small beaker containing sodium bicarbonate solution . Notice the liberation of a gas with a fruity smell .
C. With ferric chloride:
Take 3ml of formic or acetic acid and neutralize with sodium hydroxide ; then add 1ml of dilute ferric chloride solution . Notice the formation of red solution that changes into brown precipitate by heating .
3RCOO - + Fe3+ (RCOO -)3Fe
red solution
(RCOO -)3Fe + H2O heat (RCOO-)2Fe(OH) + RCOOH
Brown precipitate
2) Polycarboxylic acids
There are certain aliphatic carboxylic acids that contain more than one carboxylic group . These carboxylic acids can give different salts , esters and amides according to the number of carboxylic groups they contain .
3) Oxalic acid
Oxalic acid is white crystalline substance soluble in water . It has many household and industrial uses . It is also used as cleaning agent like removing rust . Oxalic acid is important reducing agent.
1. With sodium bicarbonate
Add 1ml of sodium bicarbonate solution to 2ml of oxalic acid solution , note the evolution of colorless gas (CO2) . Test it with clear lime water .
H2C2O4 + NaHCO3 CO2 + Na2C2O4 + 2H2O
2. Dehydration of oxalic acid
Heating oxalic acid with conc. H2SO4 decomposes by loosing water . Transfer few mg. of solid oxalic acid into test tube . Add to it 1ml of conc. H2SO4 , heat if necessary . Notice the evolution of gases (CO , CO2) , without charring .
H2C2O4 + H2SO4 Heating CO + CO2 + H2O
3. Oxalic acid as reducing agent
Oxalic acid oxidizes to CO2 and CO thus acting as reducing agent . Take 3ml of oxalic acid solution into a test tube . Add 1ml of dilute sulfuric acid ; add 0.5ml of KMnO4 solution . Heat gently and note the immediate disappearance of the permanganate color.
2KMnO4 + 5H2C2O4 + 3H2SO4 2MnSO4 + K2SO4 + 10CO2 + 8H2O
Violet colorless
4. With silver nitrate
Add 2ml of dilute silver nitrate to 2ml neutral oxalic acid solution and notice the formation of white precipitate of silver oxalate soluble in both ammonia solution and dilute nitric acid
AgNO3 + H2C2O4 Ag2C2O4 + 2NaNO3
A. Tartaric acid
Tartaric acid is white crystalline substance that naturally occurs in many plants like grapes. Tartaric acid is used in domestic cooking as constituent of baking soda . It has also many industrial and analytical applications.
1. With sodium bicarbonate
To 3ml of aqueous solution of tartaric acid add 1ml of sodium bicarbonate solution and observe the CO2 evolved .
C4H6O6 + 2NaHCO3 Na2C4H4O6 + 2CO2 + 2H2O
2. Oxidation of tartaric acid
A. Add 1ml of dilute H2SO4 to 3ml of aqueous solution of tartaric acid then add about 1.5ml of potassium dichromate . Heat the mixture and observe the evolution of CO2 and formation of green colored solution.
B. Add 1ml of dilute H2SO4 to 3ml of aqueous solution of tartaric acid , then add few drops of potassium permanganate solution . Note the disappearance of the permanganate color.
3. Tartaric acid as reducing agent
A. Add 3ml Tollens reagent (ammonia silver nitrate solution) to 2ml of tartaric acid solution . Notice the formation of silver mirror on the walls of the test tube .
B. Add few drops of ferrous sulphate solution to 3ml of tartaric acid solution , then add hydrogen peroxide drop by drop until the color of the solution becomes green . Add excess of sodium hydroxide solution and note the formation of violet color (Fenton’s test).
4. Citric acid
Citric acid contain three carboxylic groups and for that reason give three different salts and esters . It is white crystalline solid that dissolves in water. Citric acid is naturally occurring substance which is found in various fruits and vegetables like citrus fruits such as lemon . It is used as food additive and in certain cosmetic products .
A. With conc H2SO4
Citric acid reacts with conc. Sulfuric acid in the same way as oxalic acid . There is no charring in this reaction but brownish-yellow color is formed .
Add 1ml of conc. H2SO4 to few milligrams of the solid citric acid . Then heat the mixture and note the evolution of gases CO , CO2 , SO2 and the yellow color of the solution .
B. Oxidation
Add 1ml of HgSO4 solution to 3ml of citric acid solution . Heat the mixture to boiling . Add two drops of 2% KMnO4 solution . Note the disappearance of the violet color and formation of white precipitate .
C. Reduction
Citric acid can act as reducing agent even though it has less reducing effect in comparison to the tartaric acid .
Add 2ml of ammoniacal silver nitrate solution (Tollens reagent) to 1ml of neutral citric acid solution ; warm the mixture in hot water bath . Note that there is no any change . Heat the mixture to boiling and note the silvery deposits on the walls of the test tube.
D. With calcium chloride
Add 1ml of calcium chloride solution to 1ml of neutral citric acid solution. Notice that there is no any precipitate formed . Heat the mixture to boiling and observe the formation of precipitate (test to distinguish between citric acid and oxalic acid ) .
Aromatic acids
1. Benzoic acid
Benzoic acid is white crystalline solid substance that easily sublimes on heating.
a) With NaHCO3
Take few milligrams of benzoic acid into test tube containing 3ml of ethanol diluted with water (1:1) . Add 2ml of sodium bicarbonate solution . Notice the CO2 evolved .
C6H5COOH + NaHCO3 C6H5COONa + CO2 + H2O
b) Removal of the carboxylic group.
Mix together equal quantities of benzoic acid and soda lime in a test tube . Heat the contents strongly and observe the liberation of benzene fumes .
C6H5COOH + Soda lime heating C6H6 + CO
Aromatic carboxylic group can easily be removed relative to that of aliphatic acids.
c) With conc Sulfuric acid:
Benzoic acid reacts with concentrated sulfuric acid and in this case, the benzoic acid sublimes instead of charring.
d) Esterification
Dissolve few milligrams of benzoic acid in 3ml ethanol . Add several drops of concentrated sulfuric acid . Heat the mixture in a water bath for five minutes. Then transfer the contents into a beaker containing saturated sodium carbonate solution . Notice the fragrant smell of ethyl benzoate
2. Salicylic acid
Salicylic acid is white crystalline substance insoluble in water . It sublimes on heating like benzoic acid .
a) Removal of carboxylic group
Same procedure as that of benzoic acid is taken but phenolic fumes is evolved instead of benzene fumes .
C6H4(OH)COOH soda lime C6H5OH + CO
b) With ferric chloride
Take few milligrams of salicylic acid and dissolve in ethyl alcohol . Then add several drops of ferric chloride . Notice the violet color formed which disappears with the addition of dilute HCl .
c) Esterification
Dissolve few milligrams of the acid in 3ml of methanol .Add three drops of conc. Sulfuric acid and heat mixture in a water bath for five minutes . Transfer the contents to a beaker containing sodium carbonate solution . Note the pungent smell of methyl salicylate
Amines and amides
We have seen that amines contain amine functional group (-NH2) while amides contain carbonyl carbon attached to the amine group . Amides are formed by carboxylic acid reaction with ammonia or amine . Tests of amides will be within the amino acids .
We will make tests on aniline as a representative for amines . Aniline is colorless liquid that easily changes to black when exposed to air or light with fishy odor .
1. Salt formation
Add 1ml of aniline to 2ml of water ; notice the formation of two layers due to insolubility of aniline in water . Add 1ml of dilute HCl to the solution and notice the disappearance of the aniline as a result of formation of soluble salt aniline hydrochloride .
Add small amount of sodium hydroxide solution and notice the formation of two layers again
2. With bromine water
Add 1ml of aniline to 2ml of water and shake ; add excess bromine water . Notice the immediate formation of yellow precipitate .
3. With sodium hypochlorite: Add few drops of aniline to 2ml of water , then add 1ml of sodium hypochlorite solution , notice the violet color of the solution .
4. Formation of azo compounds: Azo compounds are organic compounds that contain functional group, RN = NR’, R being alkyl or aryl group . These compounds are derivatives of diazene (HN = NH) in which both the two hydrogens are substituted for an alkyl or aryl groups .These compounds have bright colors due to electronic delocalization.
Take several drops of the aniline into a test tube containing 3ml dilute HCl , insert the test tube in an ice bath for few minutes , then add few crystals of sodium nitrite and put aside for seconds . Add several drops of beta – naphthol in sodium hydroxide solution ; notice the formation of crimson (purple) color. Heat the liquid that remained to boiling , notice the evolution of nitrogen gas and the phenolic smell.
Chemical Tests Relating on Biochemistry
We have already introduced brief idea of the subject “ Biochemistry “ in the theoretical part of this book and there is no need to repeat the discussion again . We will introduce certain qualitative and quantitative tests on topics pertaining biochemistry .
1. Carbohydrates We defined carbohydrates as polyhydroxy-aldehydes or ketones based on their primary functional group ; here are certain qualitative tests for carbohydrates .
a) Molisch Test: In this test , the carbohydrate is hydrolysed with conc. sulfuric acid to corresponding monosaccharides which combines with sulfonated alpha – naphthol giving purple color .
Procedure: Add 1ml of alpha – naphthol in ethanol (Molisch reagent) to 1ml of the carbohydrate aqueous solution , shake the mixture very well, then add carefully 1ml and half of conc. sulfuric acid through the wall of the test tube , notice the purple ring formed at the contact of the two solutions .
b) Barford's Test: The significance of this test is that it can differentiate between the monosaccharides and the disaccharides since Barford's reagent readily reduces the monosaccharides unlike the case of disaccharides which takes far longer time to reduce .
Procedure: Add 1ml of Barford's reagent (1liter solution containing 66.5g of copper acetate and 9.0ml glacial acetic acid) to 1ml of the carbohydrate sample heat the solution in water bath for two minutes ; the formation of red precipitate (Cu2O) shows the presence of a monosaccharide like glucose and fructose . If the precipitate forms after long time of boiling , the carbohydrate is disaccharide like sucrose .
RCHO + 2Cu2+ + 2H2O Cu2O + RCOOH + 4H+
c) Anthrone Test: The carbohydrate is dehydrated with conc. H2SO4 to form furfurals which reacts with anthrone forming blue – green compound .
Procedure: Add 2ml of anthrone solution (2g/L in conc.H2SO4) to 2ml of the carbohydrate solution . Heat the solution in boiling water , cool the solution and observe the blue-green color formed . This test can be applied to estimate quantitatively the amount of the carbohydrate by using colorimetry method .
|
|
+ |
|
= |
blue- green complex |
Furfural anthrone
d) Bial’s test: This test is mainly specific for pentoses where the substance is dehydrated with conc. HCl to form furfural which reacts with orcinol in the presence of ferric ions to generate colored product . The test has both qualitative and quantitative importance by using colorimetric method
Procedure: Add 1ml of the sample solution to 2.5ml of the orcinol solution (prepared by dissolving 1.5g of the orcinol in 500ml of conc. HCl and 20 drops of 100g/L ferric chloride ) . Heat the contents in a boiling water bath until the color develops ; cool the solution and add three drops of amyl alcohol . Formation of bluish-green color shows the presence of a pentose and brown color for hexoses .
e) Osazone test: Reducing sugars react with excess phenylhydrazine forming osazone crystals . Sucrose being non-reducing sugar do not form osazone crystals , therefore osazone test can be general test to distinguish between different sugars ; each sugar forms characteristic osazone crystal shapes with the phenylhydrazine at boiling temperature . The time of crystal formation and melting point further contributes to the identification test .
Procedure: Dissolve a mixture of 0.2g of the sugar sample and 0.4g of phenylhydrazine hydrochloride and 0.6g of sodium acetate in 4ml water in a clean test tube ; immerse the test tube in a boiling water for about 15-20 minutes remove the test tube and let it cool gradually . Notice the formation of the osazone crystals.
f) Iodine test for polysaccharides:
We have already seen that polysaccharides consist of different groups like starch , glycogen etc. the nature of the polysaccharide is detected with iodine test in which starch gives blue color , glycogen gives red color while monosaccharides and disaccharides do not give color change .
Procedure: take 5ml of the sample solution , then add 2-3 drops of iodine solution (prepared from 33.7g of KI in 118ml water, shake until it dissolves then add 16.9g of sublimed iodine crystals , shake well to dissolve ) . Starch gives blue color glycogen gives reddish-brown; brown-yellow color shows negative test .
2. Amino acids
a) Ninhydrin test ,
To detect the presence of amino acids , ninhydrin is the main reagent ; this reagent has qualitative and quantitative application in amino acids analysis . Ninhydrin being a powerful oxidizing agent , reacts with alpha-amino acids at certain range of pH values forming purple colored compounds and liberating CO2 gas . In the following tests , we are only concerned with qualitative tests since quantitative determination necessitates spectrometric measurements which is beyond the scope of this practical .
Procedure: Take 1ml of the unknown amino acid solution into a
clean test tube . Add 5 drops of ninhydrin solution ( 2g/L) , boil the
solution n in a boiling water bath . Notice the purple color formed.
Red color
b) Millon’s Test :
This test is specific for amino acids that contain phenyl group like phenylalanine and tyrosine .
Procedure: Take 1ml of the amino acid solution (phenylalanine or tyrosine) into a test tube ; add about 5 drops of the Millon’s reagent (prepared by dissolving 150 g of mercuric sulphate in 1liter of 15% (v/v) of H2SO4 solution) . Heat the contents in a boiling water bath for 10 minutes ; cool the solution to room temperature , add 5 drops of sodium nitrite solution (10g/L) . Notice the formation of red color or precipitate .
c) Nitroprusside test: This is a qualitative test to detect the presence of free thiols . It is specific for sulphur containing amino acids like cysteine . Sodium nitroprusside reacts with the thiol groups in basic solution giving red colored compounds .
Procedure: Take 2ml of the amino acid solution , add 0.5ml freshly prepared sodium nitroprusside solution (20/L) , then add 0.5ml of concentrated ammonium hydroxide solution , mix very well . Notice the appearance of red color .
3. Proteins:
There are different methods to test the presence of proteins or to estimate its quantities in a sample . But before any analytical treatments , the protein must be purified and isolated for certain analytical purpose . There is no common method to be followed to purify all proteins ; that is the procedure for one protein may denaturate another one ; but certain basic principles are available to purify proteins depending on solubility , molecular size and ionic nature .
a) Salt precipitation: The most common method to precipitate proteins from aqueous solution is by salting out , this is carried out by adding ionic salt to the protein solution . Ions furnished by the salt will attract the water molecules reducing the number of water molecules around the charged sites of the protein molecule and decreasing its solubility .
Procedure: Take one or two eggs and separate the white portion ; add 200ml of ammonium sulfate solution (saturated) . Stir the mixture very well and transfer to centrifuge test tubes ; centrifuge them for 20 minutes . Collect the precipitate and transfer to another test tube. Decant the supernatant to a beaker containing 100ml saturated ammonium sulfate solution and mix well . Centrifuge the mixture same as before and collect the precipitate in another test tube . store the samples for analysis .
b) Biuret test: This is important in protein analysis since it detects the presence of peptide bond and can be used for both qualitative and quantitative application . In this test , alkaline Cu2+ solution is treated with the sample containing peptide bond in which violet- colored co-ordinated complex is formed .
Procedure: Heat small amount of the sample (100- 200mg) in a dry test tube until it melts ; notice the liberation of ammonia gas.
Dissolve the remained solid in sodium hydroxide solution followed by 1drop of very dilute copper sulfate solution . Violet color will be formed if protein or peptide bond is present otherwise the color remains blue .
c) Kjeldahl’s method
This method is used to determine the amount of nitrogen in a sample which leads to estimation of the protein in that sample by introducing a conversion factor . This procedure goes through several steps .
1. Digestion of the sample in which the following reaction takes: Sample + conc. H2SO4 Heating (NH4)2SO4 + CO2 + SO2 + H2O
2. Liberation of nitrogen in the form of NH3 , the following reaction takes place :
(NH4)2SO4 + 2NaOH Na2SO4 + 2H2O + 2NH3
3. Trapping the ammonia in an excess of dilute HCl ,
2NH3 + H2O + 2HCl 2NH4+ + H2O + 2Cl-
Procedure: weigh accurately 1g of dry ground sample into a digestion tube or Kjeldahl’s flask ; add 12ml of conc. H2SO4 and 2 tablets4 of catalyst (CuSO4) , if not available use hydrogen peroxide appropriately and carefully . Heat the contents at 1500C for 20 minutes , then raise the temperature to 4200C and continue heating for another 20 minutes until clear solution is formed .
Cool the solution to 50oC . Add 50ml of 20% NaOH solution followed by 50ml distilled water ; set up steam distillation unit leading to a conical flask containing 50ml of 0.1N HCl . Continue distilling until at least 100ml of the distillate is obtained . Back titrate the excess HCl with 0.1N NaOH . Calculate the amount of nitrogen and convert into % protein by multiplying the factor 6.2 , the idea of this factor is clearly explained in the theoretical section .
Kjeldahl’s set for nitrogen determination (the process is automated nowadays).
Calculations : %N =
%protein = above result x the appropriate factor for the sample.
3.Lipids:
We have defined the lipids in theoretical section that lipids are natural organic compound composing of different classes ; but we can give more simpler definition to fit our practical purpose ; that fats are esters of large fatty acids which are insoluble in water but easily dissolve in organic solvents like benzene and acetone . Our experiments will be confined within that simplified definition as fatty acid esters .
a) Solubility test:
Lipids differ in their solubility which is characteristic physical property that can be employed to extract and isolate them from complex organic materials .
Procedure: Prepare six dry test tubes each one containing 5ml of one of the following solvents ( chloroform , ether , acetone , hexane , ethanol and water). To each test tube add 3drops of oil or small piece of solid fat ; shake well to mix . Observe the rate of the solubility of the oil or the fat in various solvents .
b) Saponification value:
The process of saponification is already explained in the theoretical part of the biochemistry section , and was defined as the number of milligrams of potassium hydroxide required to neutralize 1g of the fatty acid .
Procedure: Weigh accurately 1g of fat and transfer it to a small beaker ; add 3ml of 95% ethanol/ether mixture (1:1) , and dissolve the fat thoroughly . Transfer the mixture into quick fit boiling flask ; rinse the beaker with further millilitres of the solvent . Set up the refluxing condenser for the sample and another one containing everything except fat as blank . Heat the two sets in boiling water bath for 30 minutes . Cool to room temperature and titrate with 0.5N HCl using phenolphthalein as an indicator . Record the result .
Calculation:
1ml of 0.5N HCl neutralize 28.055 mg of KOH
Therefore the saponification value =
c) Iodine number of fats
Procedure: Transfer 10ml of the fat solution (20g/L in chloroform) into stoppered bottle ; add 25ml of ICl solution (32.5g/L) , shake well and keep it in the dark for 1hr . Prepare a blank (that contain all solutions except fat) in another bottle . Add 10ml of KI solution (99.6g/L) to each bottle followed by 1ml of starch solution (10g/L) pale straw color is formed . Titrate the solutions with standard 0.1N Na2S2O3 solution until the color disappears ; the titration process must take place as quick as possible . Report the result and calculate the iodine number in mg/L .
Calculations:
Iodine No. of fat
CH
3
(CH
2
)
7
CH
CH(CH
2
)
7
COOH + Br
2
CH
3
(CH
2
)
7
CH - CH(CH
2
)
7
COOH
BrBr
OrangeColorless
CCCC
KMnO
4
O
OHOH
CH
3
- OH
H
+
K
2
Cr
2
O
4
HCHO
HCOOH
O
MethanolMethanal
Methanoic acid
CH
3
- CH
2
- OH + OH - C
O
CH
3
H
+
CH
3
- CH
2
- O - C - CH
3
O
Ethanol
Acetic acid
Ethylacetate (ester)
R - CH
2
OH
Conc. HCl
ZnCl
2
(anhyd.)
no turbidity at room temp. (no reaction)
1
o
alcohol
R
2
CH - OH
Conc. HCl
ZnCl
2
(anhyd.)
turrbidity appears in less than 10 min.
R
2
CHCL + H
2
O
R
3
COH
2
o
alcohol
Conc. HCl
ZnCl
2
(anhyd.)
R
3
CCl + H
2
O
turbidity appears immediately
3
o
alcohol
OH + Fe
3+
O
3
Fe
NH - NH
2
NO
2
+ O
C
R
R
NO
2
NH - N
NO
2
NO
2
C
R
R
+ H
2
O
H
+
2,4 - Diphenylhydrazine
2,4 - diphenylhydrazone
Bradys reagentyellow precipitate
carbonyl
compound
R - CHO
RCOOH
K
2
Cr
2
O
7
H
+
CH
3
- C
O
OH
CH
3
- C
O
O - CH
2
-CH
3
+ H
2
OHO- CH
2
- CH
3
+
Acetic acid
Ethanol
Ethylacetate (ester)
C
C
OHO
O
HO
HC
HC
COOH
COOH
OH
OH
COOH
CH
2
- C - COOH
CH
2
COOH
H
Oxalic acidTartaric acid
Citric acid
COOH
COOH
OH
COOH
COOH
Benzoic acid
Salicylic acid
Phthalic acid
(Benzene carboxylic acid
(2-hydroxybenzoic acid
Benzene-1-2-dicarboxylic
acid
COOH
+ CH
3
OH
COOCH
3
+ H
2
O
Methylbenzoate
H
+
COOH
OH
+ FeCl
3
O
C
OH
o
-
3
Fe + HCl
3
Violet color
COOH
OH
+ CH
3
- OH
COOCH
3
OH
+ H
2
O
Methylsalicylate
NH
2
+ HCl
NH
4
+
+ Cl
-
Aniline
aniline hydrochloride (salt)
NH
2
Aniline
+ 3Br
3
NH
2
Br
Br
Br
+ 3HBr
2,4,6-tribromoaniline
