Analysis of Organic Compounds

Qualitative Analysis of Organic Compounds

Detection of Elements

The most commonly occurring elements in organic compounds are carbon, hydrogen, oxygen, nitrogen, suphur and halogen elements. There is no direct method for the detection of oxygen. For detecting nitrogen, sulphur and halogens essential test is sodium fusion test (Lassaigne’s test).

Sodium Fusion Test

In order to detect nitrogen, sulphur and halogen in organic compounds, it is necessary to convert them into ionisable inorganic substances so that ionic tests of to fure the organic compounds with metallic sodium (Lassaigne’s test). In this way sodium cyanide, sodium sulphide and sodium halides, which are readily elements are present.

Test for Nitrogen

Cyanide ion and hence nitrogen ion in the sample, may be detected by the Prussian blue test. The filtered alkaline solution resulting from the action of water upon the sodium fusion is treated iron (II) sulphate and thus forms sodium hexayanoterrate (II). Upon boiling the alkaline iron (II) salt solution, some iron (III) ions are insensibly produced by the action of air. Now dilute sulphuric acid is added which dissolves the iron (II) and (III) hydroxides. The hexacyanoferrate (II) reacts with the iron (III) salt producing iron (III) hexacyano ferrate (II), Prussian blue. A Prussian blue precipitate or colouration indicates that nitrogen is present.

FeSO₄ + 6NaCN → Na₄[Fe(CN)₆] + Na₂SO₄

3Na₄[Fe(CN)₆] + 2Fe₂(SO₄)₃  → Fe₄[Fe(CN)₆]₃ + 6Na₂SO_{4
                                                                         Purssian Blue}
Test for Sulphur

This element as sulphide ions may be tested by either of the following two methods.

1. The filtered fusion solution dil. Acetic acid followed by lead acetate solution is added. A black precipitate of lead sulphide indicates the presence of sulphur.

2. To the solution 2-3 drops of freshly prepared dilute solution of disodium pentacyanonitroxyl ferrate Na₂[Fe(CN)₅NO] is added. A purple colouration indicates sulphur, the colouration slowly fades on standing.

Test for Halogens (Nitrogen and Sulphur absent)

A portion of the fusion solution is acidified with dilute nitric acid and then excess of silver nitrate solution is added. A precipitate indicates the presence of a halogen. The mother liquor is decanted and the precipitate is treated with dilute aqueous ammonia solution. If the precipitate is white and readily soluble in ammonia solution, chlorine is present; if it is pale yellow and difficulty soluble, bromine is present; if it yellow and insoluble, then iodine and bromine may be confirmed by the following test.

Fusion solution is acidified with dil. sulphuric acid and then chlorine water is added drop-wise with vigorous shaking to the mixture. If iodine is present the organic phase becomes people in colour. As the addition of chlorine water is continued, the purple colour disappears (owing to oxidation of iodine to iodate) and if bromine is present, is present, is replaced by a brown or reddish colour. If bromine is absent, the organic layer will be colourless.

Test for Halogens (Nitrogen and/or Sulphur present)

Cyanide and sulphide ions both interfere with this test for halide by forming silver cyanide and silver sulphide precipitates. If nitrogen or sulphur has been detected, therefore, the interferim ions must be removed. To remove cyanide and sulphide ions, fusion solution is acidified with dilute nitric acid and evaporated to half of the original volume in order to expel hydrogen cyanide and/or hydrogen sulphide which may be present.

Alternatively to the fusion solution 1-2 drops of 5% nickel (M) nitrate solution is added, nickel (II) cyanide and/or nickel sulphide is filtered OH. The filtrate is acidified with 2M nitric acid and then test for halogens is performed an mentioned above.
 

Detection of Functional Groups

Alcoholic –OH group: The alcoholic group can be detected by the following tests:

1. Sodium metal test

2. Acetyl chloride

3. Cerric ammonium

1. Sodium Metal Test: This test is based in the appearance of effervescence due to liberation of hydrogen gas when the alcohol is reacted with active metals like sodium.

2R – OH + 2Na →  2RONa + H₂­

The alcohol used should be dry as water also reacts with sodium metal to evolve hydrogen gas.

2. Acetyl Chloride Test: Acetyl chloride reacts vigorously with primary and secondary alcohols with the evolution of hydrogen chloride. The hydrogen chloride formed gives white fumes of ammonium chloride with ammonium hydroxide.

Cerric Ammonium Test: To the sample few drops of cerric ammonium nitrate is added and shaken well. Appearance of pink or red colour indicates the presence of an alcoholic group.

 2ROH + (NH₄)₂Ce(NO₃)₆ →  (ROH)₂Ce(NO₃)₄ + 2NH₄NO_{3
                                                                Red}

Phenolic –OH Group

The phenolic group can be detected by the following tests:

1. Litmus Test

2. Ferric chloride test

3. Liebermann’s test

Litmus Test: A drop of the given liquid or a crystal (if solid) is placed on a moist blue litmus paper. If the colour changes to red, phenolic group may be present.

Neutral Ferric Chloride Test: If the compound is sparingly soluble in water a hot saturated aqueous solution is prepare and filtered. Then to the cold filtrate drops of neutral 1% ferric chloride solution is added. If a transient or permanent colouration (usually purple, blue or green) other than yellow or orange is observed, the substance probably a phenol (or an enol). If no colour is observed the test is repeated as above but water substituted by absolute ethanol as solvent.

Neutral FeCl₃ (i.e. free from HCl) is prepared by adding dilute NaOH solution to 1% FeCl₃ solution until a slight precipitate of iron (III) hydroxide is formed. The precipitate is filtered OH and clear filtrate is used for the filtrate.

FeCl₃+ 6C₆H₅OH → [Fe(OC₆H₅)₆]^3- + 3HCl

Liebermann’s Test: When a drop of phenol is heated a few crystals of sodium nitrite and 1 ml concentrated sulphuric acid a green or blue colouration develops. When this solution is powered into less of water the colour of the aqueous solution is red, when excess of NaOH solution is added to the aqueous solution the red colour changes to blue or green.

Carbonyls (Aldehydes and Ketones)

2,4-dinitrophenyl hydrazine test: Small amount (2 drops or 0.05 – 0.1g) of the substance is added to 3 ml of 2,4-dinitrophenyl hydrazine reagent and shacked well. If no precipitate forms immediately allow it is allowed to stand for 5-10 minutes. A crystalline precipitate indicates the presence of a carbonyl compound. Occasionally the precipitate is oily at first but this becomes crystalline upon standing.

Differentially Tests for Aldehydes

Schiff’s Test: Given compound is dissolved in alcohol and then 1-2ml of Schiff’s reagent is added. Appearance of pink, red or magenta colour confirms the presence of aldehyde group. While ketones are without effect. Some aromatic aldehyde (e.g Vannillin) give a negative result. Schiff’s reagent can be prepared by dissoting 0.2g of pure p-rosaniline hydrochloride in 20 ml of a cold, freshly prepared, saturated aqueous solution of sulphur dioxide. The solution is allowed to stand for a few hours until it becomes colourless or pale yellow. Then the solution is diluted to 200 ml and keep in a slightly stoppered bottle.

Tollen’s Test (Silver Mirror Test): Tollen’s reagent is prepared as follows 3g of silver nitrate is dissolved in 30 ml of water (solution A) and 3g of sodium hydroxide in 30 ml of water (solution B). When the reagent is required, equal volumes of solution A and B are mixed and then dil. Ammonia solution is added drop by drop until the silver oxide is just dissolved.

To the tollen’s reagent 3-4 drops of the given liquid (or 0.5 – 1g of solid) is added contained in a clear test tube and warm the test tube is warmed or a water both for about five minutes. A shining mirror confirms the presence of the aldehyde. 

2Ag(NH₃)₂⁺ + RCHO + 3OH^– →  RCOO^– + 2Ag¯ + 4NH₃ + 2H₂O

 Fehling’s Test: Preparation of Fehlings solution No.1: 34.64g of copper (II) sulphate crystals is dissolved in water containing a few drops of dilute sulphuric acid and the solution is diluted to 500ml.

Fehlings solution No.II: 60g of pure NaOH and 173g of pure Rochelle salt (sodium potassium tartarate) is dissolved in water and the volume is made up to 5CN ml.

Two solutions separately keeped in tightly stoppered bottles and mixed exactly equal volumes immediately before use.

Carboxyl Group 

Carboxylic acid can be identified by the following tests

1. Sodium bicarborate test

2. Ester test

Sodium Bicarbonate test: Sodium bicarbonate (NaHCO₃) to the 1 ml of the sample a pinch of effervescence indicates the presence of carboxylic group.

RCO₂H + NaHCO₃ --> RCOONa + CO₂­ + H₂O

Both phenol and carboxylic acid turns moist blue litmus paper to red put as phenol gives negative result with NaHCO₃ the above mentioned test is used for differentiating carboxylic acids from phenols.

Ester Test: A small amount of the acid with two parts of absolute ethanol and one pare of concentrated sulphuric acid is warmed for 2 minutes. Solution is cooled and poured continuously into aqueous Na₂CO₃ solution contained in an evaporating dish. A sweet, fruity smell of an ester confirms the presence of ester.

Amino Group

The most important basic nitrogen compounds are the primary, secondary and tertiary amines and they dissolve in mineral acids and change red litmus to blue.

Chemical classification of the amine function: The classification of primary, secondary or tertiary amines should be carried out by means of the reaction with nitrous acid.

Nitrous Acid Test: 2g of the substance of 5 ml of 2 M HCl acid is dissolved and cooled and then 2 ml of ice-cold 10% aqueous NaNO₂ solution slowly by means of a dropper and stirred. If a clear solution is obtained with a continuous evolution of nitrogen gas the substance is a primary amine.

RNH₂ + HONO →  ROH + H₂O + N₂

If there is apparently no evolution of nitrogen gas from clear solution. One half of the solution is added to a cold solution of b-napthol in NaOH. The formation of a coloured (eg. Orange – red) azo dye indicates the presence of a primary aromatic amine

the formation of N-nitroamines which normally separate as orange-yellow oils or low melting solids indicates the presence of a secondary amine. Formation of nitroamine is also confirmed by the libormann nitroso test.

If a colourless solution is obtained which gives an immediate and sustained positive test with starch indicate paper when only a little sodium nitrite solution has been added the compound is a tertiary aliphatic amine.

If the compound is a tertiary aromatic amine the treatment with HNO₂ will yield a dark orange – red solution or an orange crystalline resulting from the formation of the hydrochloride of the C-nitrosoamine.

Conformations Tests for Primary Amines

Carbylamine Test: This test is given by both aliphatic and aromatic primary mines. Solid KOH (OH-Zg) is taken in a clear dry test tube and then 2 ml of ethanol is added. The test tube is warmed until the pallets dissolve. To this a few drops of chloroform small amount of the given compound is added and warmed gently. An unleagetn odour confirms the presence of primary amine.

RNH₂ + CHCl₃ + 3KOH →  RNC + 3KCl + 3H₂O

The test is extremely delicate and will often detect traces of primary amines in secondary and tertiary amines.

Hinsberg Test: Sample is kept in a clear test tube to this dil. NaOH solution and benzene sulphonyl chloride is added. The mixture is warmed on a water both to complete the reaction and then conc. HCl is added.

A clear solution in NaOH solution which on acidification gives an insoluble material indicates primary amine.

Nitro Group

Mullikeno Test: A little of the sample of dissolved in 2 ml alcohol and is reduced with Zn/NH₄Cl or CaCl₂ solution and Zn dust by boiling for five minutes. It is then filtered, cooled and treated with ammoniacal silver nitrate solution and heated in a water bath; white to grey and then black precipitate formation confirms the presence of the nitro group.

RNO₂ + 4[H] RNHOH + H₂O

The presence of nitro group also be deleted by reducing the compound with Sn/HCl and then conducting the carbyl amine test.
 

Separation Based on Differences in the Chemical Properties of the Compounds

Toluene and Aniline: A mixture of toluene may be separated by extraction with dilute HCl; the aniline passes into the aqueous layer in the form of the salt aniline hydrochloride and may be recovered by neutralization.

Phenol and Toluene: They can be separated by treatment with dil. NaOH

Dibutyl Ether and Chlorobenzene: Concentrated H₂SO₄ dissolves only the dibutyl ether and it may be recovered from solution by dilution with water.

 The above examples are simple applications of the fact that the various components full into different solubility groups.

Classification of Organic Compounds According to Solubility behaviour

Phenols and any compound containing -COOH group (e.g. o-cresol from benzoic acid): Phenols may be separated from acids by a dilute solution of NaHCO₃; the weakly acidic phenols (and also enols) are not converted into salts by this reagent and removed by ether extraction; the acids pass into solution as the sodium salts and may be recovered after acidification.

Separation of mixture of amines: Mixture of primary, secondary and tertiary amines can usually be separated by Hinsberg method. The mixture of amines is treated with NaOH solution and toluene-p-sulphonyl chloride is warmed in a water bath and finally acidified with dil. HCl the sulphonamides of primary and secondary amines are precipitated the solid is filtered and washed with a little cold water the tertiary amine will be present in the filtrate. The solid is then treated with alkli again the sulphonamide from the primary amine is filtered OH. The filtrate is acidified with dil. HCl to precipitate the derivative of the primary amine.

Aldehydes and Ketones from toehr liquid hydrocarbons and other neutral liquid compound (e.g. ethanal and ethanol): The mixture is shaken with a solution of NaHCO₃, the carbonyl compound forms a solid bisulphite compound which is filtered OH and decomposed with dil. acid in order to recover the aldehyde or ketone.

Separate based on Differences in the vaolalities of the component in aqueous solution:

1. Diethylamine and butan-1-ol: They can be separated by adding sufficient dilute H₂SO₄ to neutralise the base steam distillation will remove the alcohol. The amine can be recovered by adding NaOH to the residue and repeating the distillation.

2. Diethyl Ketone and Acetic Acid: The mixture is treated with sufficient dilute NaOH solution to transform the acid into sodium acetate and distilling the aqueous mixture. The ketone will pass over in the steam and the non-volatile stable salt will remain in the flask. Acidification with dilute H₂SO₄ liberates acetic acid which can be isolated by steam distillation or by extraction.
    

Quantitative Analysis of Organic Compounds (Estimation of Elements)

After qualitative analysis of elements, the next step in the determination of molecular formula of an organic compound is the estimation of various elements by mass, i.e. finding the percentage composition of the elements by mass. The various methods commonly employed for the estimation of principal elements are discussed in the table.

1.  Physical methods for volatile compounds

Victor Meyer's method : Molecular mass of volatile liquids and solids can be easily determined from the application of Avogadro hypothesis according to which the mass of 22.4 litres or 22400ml of the vapour of any volatile substance at NTP is equal to the molecular mass of the substance.

In Victor Meyer's method, a known mass of the volatile substance is vaporised in a Victor Meyer's tube. The vapours formed displace an equal volume of air into a graduated tube. The volume of air collected in graduated tube is measured under experimental conditions. This volume is converted to NTP conditions.    

Calculations: Mass of the organic substance 

Let the volume of the air displaced be ; 

Temperature 

Pressure (after deducting aqueous tension) 

 Let the volume at NTP be 

Applying gas equation,  

  ml of vapours weight at NTP = Wg

    22400 ml of vapour weight at NTP =  

Alternate method : Vapour density of substance

 or  V. D.     ( Mass of 1 ml of  at NTP

 or )

 or  V. D. ;

Mol. Mass, 

Hofmann's method: The method is applied to those substances which are not stable at their boiling points, but which may be volatilised without decomposition under reduced pressure. A known mass of the substance is vaporised above a mercury column in a barometric tube and the volume of the vapour formed is recorded. It is then reduced to NTP conditions. The molecular mass of the organic substance can be calculated by the application of following relationship,

Mol. Mass 

(ii)   Physical methods for Non-volatile substances :

The molecular mass of a non-volatile organic compound can be determined by noting either the elevation in boiling point of the solvent (Ebullioscopic method) or the depression in freezing point of the solvent (Cryoscopic method) produced by dissolving a definite mass of the substance in a known mass of the solvent. The molecular mass of the compound can be calculated from the following mathematical relationships:

(a) Elevation in boiling point: Mol. Mass 

Where,  Molal elevation constant of the solvent,  Mass of the compound,  Mass of the solvent

Elevation in boiling point of the solvent (determined experimentally)   

(b)     Depression in freezing point: Mol. Mass 

Where, Molal depression constant of the solvent,  Mass of the compound, Mass of the solvent

Depression in freezing point of the solvent (determined experimentally)

  3. Chemical methods:

Silver salt method for acids: It is based on the fact that silver salt of an organic acid on heating gives residue of metallic silver.

From the mass of the silver salt taken and the mass of the silver residue obtained, the equivalent mass of the silver salt can be calculated.

Knowing the equivalent mass of silver salt, the equivalent mass of the acid can be obtained. The molecular mass of an acid can be determined with the help of the following relationship,

Mol. mass of the acid = Equivalent mass of the acid  basicity

Calculations: 

Mass of silver salt taken  

Mass of metallic silver   

;  

Eq. mass of silver salt 

Let the equivalent mass of the acid be E. In the preparation of silver salt, a hydrogen atom of the carboxylic group is replaced by a silver atom.

Thus, Equivalent mass of silver salt  

 Thus, or   

If n be the basicity of the acid, then  Mol. Mass of the acid 

Platinichloride method for bases: Organic bases combine with chloroplatinic acid, to form insoluble platinichlorides, which, on ignition, leave a residue of metallic platinum. Knowing the mass of platinum salt and the mass of metallic platinum, the molecular mass of the platinum salt can be determined. Let B represents one molecule of the base. If the base is mono-acidic, the formula of the salt will be .

 Let E be the equivalent mass of the base.

Molecular mass of the salt

 So ;

  ;  

 Mol. mass of the base  Eq. mass  acidity 

 where n is the acidity of the base.
 

Calculation of Empirical and Molecular Formula

Empirical formula: Empirical formula of a substance gives the simplest whole number ratio between the atoms of the various elements present in one molecule of the substance. For example, empirical formula of glucose is , i.e. for each carbon atom, there are two H-atoms and one oxygen atom. Its molecular formula is however, .

Calculation of empirical formula: The steps involved in the calculation are as follows,

1. Divide the percentage of each element by its atomic mass. This gives the relative number of atoms.

2. Divide the figures obtained in step (i) by the lowest one. This gives the simplest ratio of the various elements present.

3. If the simplest ratio obtained in step (ii) is not a whole number ratio, then multiply all the figures with a suitable integer i.e., 2, 3, etc. to make it simplest whole number ratio.

4. Write down the symbols of the various elements side by side with the above numbers at the lower right corner of each. This gives the empirical or the simplest formula.

Molecular formula: Molecular formula of a substance gives the actual number of atoms present in one molecule of the substance.

 Molecular formula = Empirical formula

 Where, n is a simple integer 1, 2, 3, ...... etc. given by the equation,

where the molecular mass of the compound is determined experimentally by any one of the methods discussed former, empirical formula mass is calculated by adding the atomic masses of all the atoms present in the empirical formula.

Molecular formula of gaseous hydrocarbons (Eudiometry):

Eudiometry is a direct method for determination of molecular formula of gaseous hydrocarbons without determining the percentage composition of various elements in it and without knowing the molecular weight of the hydrocarbon. The actual method used involves the following steps,

1. A known volume of the gaseous hydrocarbon is mixed with an excess (known or unknown volume) of oxygen in the eudiometer tube kept in a trough of mercury.   

2. The mixture is exploded by passing an electric spark between the platinum electrodes. As a result, carbon and hydrogen of the hydrocarbon are oxidised to and  vapours respectively.

3. The tube is allowed to cool to room temperature when water vapours condense to give liquid water which has a negligible volume as compared to the volume of water vapours, Thus, the gaseous mixture left behind in the eudiometer tube after explosion and cooling consists of only  and unused .

4. Caustic potash or caustic soda solution is then introduced into the eudiometer tube which absorbs  completely and only unused  is left behind.          

Thus, the decrease in volume on introducing NaOH or KOH solution gives the volume of  formed. Sometimes, the volume of  left unused is found by introducing pyrogallol and noting the decrease in volume.

Calculation: From the volume of  formed and the total volume of  used, it is possible to calculate the molecular formula of gaseous hydrocarbon with the help of the following equation.

(Negligible volume on condensation)

From the above equation, it is evident that for one volume of hydrocarbon,

1.  volume of  is used

2.  x volume of  is produced

3. y/2 volume of  vapours is produced which condense to give iquid  with negligible volume.

4. Contraction on explosion and cooling 

By equating the experimental values with the theoretical values from the above combustion equation, the values of x and y and  hence the molecular formula of the gaseous hydrocarbon can be easily determined.

Refer the below mentioned links to get an immediate solution to all queries on Organic chemistry:

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