Organic Chemistry (Techniques Of Purification And Analysis)
Methods Of Purification Of Organic Compounds
Purification of organic compounds is essential to remove impurities and obtain them in a pure form for further study or application. Various techniques are employed based on the physical and chemical properties of the compound and its impurities.
Sublimation:
Principle: This method is used for compounds that sublime, i.e., change directly from solid to gas on heating and then back to solid on cooling, without passing through the liquid phase. The impure compound is heated, and the volatile component deposits on a cold surface.
Used for: Naphthalene, camphor, benzoic acid, anthracene, iodine.
Process: The impure solid is heated in a dish covered with a cold surface (e.g., an inverted funnel with its stem dipped in cold water or ice). The pure substance sublimes and deposits as crystals on the cold surface.
Crystallisation:
Principle: This is a widely used method for purifying solid organic compounds. It is based on the difference in solubility of the compound and its impurities in a given solvent at different temperatures. The impure solid is dissolved in a minimum amount of hot solvent. As the solution cools, the pure compound crystallizes out, while impurities that are either more soluble or less soluble remain in the solution or precipitate out.
Steps:
- Solvent selection: Choose a solvent in which the compound is sparingly soluble at room temperature but highly soluble at high temperatures, while impurities are either very soluble or insoluble.
- Dissolution: Dissolve the impure solid in a minimum amount of hot solvent.
- Filtration (if necessary): If impurities are insoluble, filter the hot solution.
- Cooling: Cool the filtrate slowly to allow the formation of pure crystals.
- Filtration: Separate the crystals by filtration.
- Washing: Wash the crystals with a small amount of cold solvent to remove any adhering impurities.
- Drying: Dry the crystals.
Used for: Purification of most solid organic compounds.
Distillation:
Principle: Used for purifying liquids. It is based on the difference in boiling points of the components in a mixture.
Types:
- Simple Distillation: Used when the boiling points of the components differ significantly (by at least 25 K). The liquid is heated, vaporized, and then condensed.
- Fractional Distillation: Used when the boiling points of the components are close. The mixture is passed through a fractionating column, where repeated vaporization and condensation cycles lead to a better separation.
- Distillation under Reduced Pressure (Vacuum Distillation): Used for liquids that decompose at their atmospheric boiling points. Lowering the pressure lowers the boiling point, allowing distillation at a lower temperature.
- Steam Distillation: Used for purifying temperature-sensitive compounds that are immiscible with water but volatile with steam. The mixture is heated with steam, and the compound distills over at a temperature lower than its boiling point.
Used for: Purification of liquids like ethanol, acetone, aniline, etc.
Differential Extraction:
Principle: This technique is used when the compound to be purified has different solubilities in two immiscible solvents (e.g., an organic solvent and water). The compound preferentially dissolves in one solvent, leaving the impurities in the other.
Process: The mixture is shaken with two immiscible solvents in a separating funnel. The component that is more soluble in one solvent is extracted into that solvent layer. The layers are separated, and the desired component can be further purified from its solvent.
Used for: Separating organic products from aqueous reaction mixtures or isolating natural products.
Chromatography:
Principle: Chromatography is a powerful technique for separating components of a mixture based on their differential partitioning between a stationary phase and a mobile phase.
Types:
- Column Chromatography: The stationary phase (e.g., silica gel or alumina) is packed in a column, and the mobile phase (solvent) carries the mixture through it. Components separate based on their differential adsorption and solubility.
- Thin Layer Chromatography (TLC): Similar to column chromatography but uses a thin layer of adsorbent on a glass plate.
- Paper Chromatography: Uses filter paper as the stationary phase and a suitable solvent as the mobile phase.
- Gas Chromatography (GC): Used for volatile compounds. The mobile phase is a gas, and the stationary phase is a liquid or solid support.
Used for: Separation and purification of complex mixtures, qualitative and quantitative analysis.
Qualitative Analysis Of Organic Compounds
Qualitative analysis of organic compounds aims to identify the elements present in the compound and to determine the functional groups. It involves a series of tests to detect the presence or absence of specific elements.
Detection Of Carbon And Hydrogen:
Procedure: A small amount of the organic compound is heated in a hard glass test tube with copper(II) oxide (CuO). CuO acts as an oxidizing agent.
Reactions:
- If carbon is present, it gets oxidized to carbon dioxide ($CO_2$).
- If hydrogen is present, it gets oxidized to water ($H_2O$).
Detection of Carbon (as $CO_2$): The gases evolved are passed through limewater (calcium hydroxide solution, $Ca(OH)_2$). If $CO_2$ is present, it turns limewater milky due to the formation of insoluble calcium carbonate ($CaCO_3$).
$CO_2(g) + Ca(OH)_2(aq) \rightarrow CaCO_3(s) \downarrow + H_2O(l)$
Detection of Hydrogen (as $H_2O$): Water vapour evolved is detected by passing the gases over anhydrous copper(II) sulphate ($CuSO_4$), which turns blue in the presence of water.
$CuSO_4(s) (\text{white}) + 4H_2O(g) \rightarrow CuSO_4 \cdot 4H_2O(s) (\text{blue})$
Detection Of Other Elements (N, S, Halogens, P):
Lassaigne's Test (Sodium Fusion Test): This is a general test for detecting Nitrogen (N), Sulphur (S), Halogens (X), and Phosphorus (P). The organic compound is fused with metallic sodium in a fusion tube.
Principle: During fusion, all the elements like N, S, X, P present in the organic compound are converted into ionic compounds of sodium, such as sodium cyanide ($NaCN$), sodium sulphide ($Na_2S$), sodium halide ($NaX$), and sodium phosphide ($Na_3P$), which are soluble in water.
Procedure:
- A small amount of the organic compound is heated with a small piece of metallic sodium in a fusion tube until the compound is strongly heated.
- The hot fusion mixture is quickly dropped into distilled water in an evaporating dish.
- The mixture is boiled and then filtered. The filtrate is called Sodium Fusion Extract (SFE) or Lassaigne's filtrate. This filtrate is used for testing the presence of N, S, X, and P.
1. Test for Nitrogen (N):
Test: To a part of the SFE, a freshly prepared solution of ferrous sulphate ($FeSO_4$) is added, and the mixture is boiled. Then, a few drops of ferric chloride ($FeCl_3$) solution are added, and the mixture is cooled.
Positive Result: Formation of a Prussian blue (ferric ferrocyanide) precipitate indicates the presence of nitrogen.
Reactions:
- Formation of sodium cyanide: $Na + C + N \rightarrow NaCN$
- Formation of ferrous ferrocyanide: $6NaCN + FeSO_4 \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4$
- $Na_4[Fe(CN)_6] + Fe^{3+} \rightarrow \text{Prussian Blue}$ (complex formation)
If only ferrous sulphate is used, a greenish precipitate of ferrous ferrocyanide is formed, which turns Prussian blue on adding ferric chloride.
2. Test for Sulphur (S):
Test: To a part of the SFE, sodium nitroprusside ($Na_2[Fe(CN)_5NO]$) solution is added.
Positive Result: Formation of a violet colour indicates the presence of sulphur.
Reaction:
- Formation of sodium sulphide: $2Na + S \rightarrow Na_2S$
- $Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]$ (Sodium nitroprusside) (Violet colour)
Alternative Test (using Lead Acetate): To a part of the SFE, lead acetate solution ($Pb(CH_3COO)_2$) is added, and the solution is heated.
Positive Result: Formation of a black precipitate of lead sulphide ($PbS$) indicates the presence of sulphur.
$Na_2S + Pb(CH_3COO)_2 \rightarrow PbS(s) (\text{black}) + 2CH_3COONa$
3. Test for Nitrogen and Sulphur Together:
If both N and S are present in the organic compound, they form sodium thiocyanate ($NaSCN$) during Lassaigne's test.
Test: To the SFE, ferrous sulphate ($FeSO_4$) solution is added and boiled, followed by a few drops of $FeCl_3$.
Positive Result: Formation of blood-red colour indicates the presence of both N and S.
Reaction:
- Formation of sodium thiocyanate: $Na + C + N + S \rightarrow NaSCN$
- $NaSCN + Fe^{3+} \rightarrow [\text{Fe}(SCN)(H_2O)_5]^{2+} + \text{complexes leading to blood-red colour}$
4. Test for Halogens (X = Cl, Br, I):
Test: To a part of the SFE, freshly prepared silver nitrate ($AgNO_3$) solution is added.
Positive Result:
- Formation of a white precipitate (soluble in $NH_4OH$) indicates Chlorine. ($AgCl$)
- Formation of a pale yellow precipitate (sparingly soluble in $NH_4OH$) indicates Bromine. ($AgBr$)
- Formation of a yellow precipitate (insoluble in $NH_4OH$) indicates Iodine. ($AgI$)
Reactions:
- Formation of sodium halide: $Na + X \rightarrow NaX$
- $NaX + AgNO_3 \rightarrow AgX(s) + NaNO_3$
To distinguish between Br and I if precipitate is pale yellow or yellow: Add $CS_2$ to the precipitate. $AgBr$ is insoluble in $CS_2$, while $AgI$ dissolves in $CS_2$.
Note: If Nitrogen is also present with Halogens, it forms $NaCN$ which reacts with $AgNO_3$ to form $AgCN$, a white precipitate insoluble in dilute $NH_4OH$. Therefore, before testing for halogens, the SFE must be boiled with excess ferrous sulphate to destroy cyanide ions.
5. Test for Phosphorus (P):
Test: Heat the organic compound with dry sodium carbonate ($Na_2CO_3$) to fuse it. The fused mass is dissolved in water, filtered, and acidified with nitric acid ($HNO_3$). To this, ammonium molybdate ($ (NH_4)_2MoO_4$) solution is added, followed by heating.
Positive Result: Formation of a yellow precipitate (ammonium phosphomolybdate, $(NH_4)_3[P(MoO_3)_6]$) indicates the presence of phosphorus.
Reaction:
- $P + Na_2CO_3 \xrightarrow{\Delta} Na_3PO_4$ (Sodium phosphate)
- $Na_3PO_4 + 12(NH_4)_2MoO_4 + HNO_3 \rightarrow (NH_4)_3[P(MoO_3)_6] (\text{yellow ppt}) + ...$
Quantitative Analysis
Quantitative analysis of organic compounds involves determining the percentage composition by mass of each element present in the compound.
Carbon And Hydrogen:
Method: Combustion Analysis (Dumas' Method for N, but conceptually similar for C & H).
Principle: A known weight of the organic compound is completely burnt in excess of oxygen. The carbon present in the compound is converted to carbon dioxide ($CO_2$), and the hydrogen is converted to water ($H_2O$). These products are absorbed by specific reagents, and their amounts are determined gravimetrically.
Apparatus: A combustion train consisting of:
- A furnace for heating the compound.
- A pre-heating tube containing copper turnings to ensure complete combustion.
- A vessel containing fused $CaCl_2$ (anhydrous) to absorb water vapour.
- A vessel containing caustic potash ($KOH$) solution to absorb carbon dioxide.
Procedure:
- A known mass of the organic compound ($w$ grams) is heated strongly in a combustion tube in a current of dry oxygen.
- The evolved $CO_2$ and $H_2O$ are passed through a weighed absorption tube containing anhydrous $CaCl_2$ (which absorbs $H_2O$) and then through a weighed absorption tube containing $KOH$ solution (which absorbs $CO_2$).
- Let the increase in mass of the $CaCl_2$ tube be $w_1$ grams (mass of $H_2O$).
- Let the increase in mass of the $KOH$ tube be $w_2$ grams (mass of $CO_2$).
Calculations:
- Mass of Hydrogen:
- Mass of Carbon:
- Percentage of Oxygen: Can be calculated by subtracting the percentages of all other elements from 100%.
Molar mass of $H_2O = 18$ g/mol. Mass of H in $H_2O = 2$ g/mol.
Mass of $H = \frac{2}{18} \times w_1 = \frac{1}{9} \times w_1$ grams.
Percentage of Hydrogen $= (\frac{\text{Mass of H}}{\text{Mass of compound}}) \times 100 = (\frac{w_1/9}{w}) \times 100 \%$
Molar mass of $CO_2 = 44$ g/mol. Mass of C in $CO_2 = 12$ g/mol.
Mass of $C = \frac{12}{44} \times w_2 = \frac{3}{11} \times w_2$ grams.
Percentage of Carbon $= (\frac{\text{Mass of C}}{\text{Mass of compound}}) \times 100 = (\frac{3w_2/11}{w}) \times 100 \%$
Percentage of Oxygen $= 100\% - (\%C + \%H + \%N + \%S + \%P + ...)$
Nitrogen:
Dumas Method:
Principle: A known weight of the organic compound is heated with excess copper(II) oxide (CuO) in a combustion tube. All the nitrogen in the compound is converted into nitrogen gas ($N_2$). The evolved gases ($N_2$, $CO_2$, $H_2O$, etc.) are passed through a heated copper spiral to reduce any nitrogen oxides formed to $N_2$. The gases are then collected over mercury, and the volume of nitrogen gas evolved is measured. $CO_2$ and $H_2O$ are absorbed by $KOH$ and anhydrous $CaCl_2$, respectively.
Procedure:
- A known mass of the organic compound ($w$ grams) is mixed with excess CuO and heated strongly.
- The evolved gases are passed through a heated copper spiral and then collected over mercury in an eudiometer tube.
- The volume of $N_2$ gas is measured at a known temperature and pressure. Let it be $V$ mL.
Calculations:
- Convert the measured volume of $N_2$ at room temperature and pressure to standard temperature and pressure (STP) using the gas laws, or calculate the mass of $N_2$.
- Mass of $N_2$ evolved $= \frac{\text{Volume of } N_2 \text{ at STP}}{22400 \text{ mL/mol}} \times \text{Molar mass of } N_2$ (which is 28 g/mol).
- Percentage of Nitrogen $= (\frac{\text{Mass of N}}{\text{Mass of compound}}) \times 100$
Halogens:
Carius Method:
Principle: A known weight of the organic compound containing halogens is heated with fuming nitric acid ($HNO_3$) in a sealed glass tube (Carius tube) in the presence of silver nitrate ($AgNO_3$). The halogen is converted into silver halide ($AgX$), which is filtered, dried, and weighed.
Procedure:
- A known mass of the organic compound ($w$ grams) is taken in a Carius tube along with excess of fuming $HNO_3$ and a pinch of $AgNO_3$.
- The tube is sealed and heated in a furnace for about 4-5 hours.
- After cooling, the tube is broken, and the contents are transferred to a dish. The precipitated silver halide ($AgX$) is filtered, washed, dried, and weighed. Let its mass be $w_1$ grams.
Calculations:
- For Chlorine: Molar mass of $AgCl = 143.5$ g/mol. Mass of Cl in $AgCl = 35.5$ g/mol.
- For Bromine: Molar mass of $AgBr = 188$ g/mol. Mass of Br in $AgBr = 80$ g/mol.
- For Iodine: Molar mass of $AgI = 235$ g/mol. Mass of I in $AgI = 127$ g/mol.
Mass of $Cl = \frac{35.5}{143.5} \times w_1$ grams.
Percentage of Chlorine $= (\frac{35.5}{143.5} \times \frac{w_1}{w}) \times 100 \%$
Mass of $Br = \frac{80}{188} \times w_1$ grams.
Percentage of Bromine $= (\frac{80}{188} \times \frac{w_1}{w}) \times 100 \%$
Mass of $I = \frac{127}{235} \times w_1$ grams.
Percentage of Iodine $= (\frac{127}{235} \times \frac{w_1}{w}) \times 100 \%$
Sulphur:
Carius Method:
Principle: A known weight of the organic compound is heated with fuming nitric acid ($HNO_3$) in a sealed Carius tube. Sulphur is oxidized to sulphuric acid ($H_2SO_4$). The sulphuric acid is then precipitated as barium sulphate ($BaSO_4$) by adding barium chloride ($BaCl_2$) solution. The $BaSO_4$ precipitate is filtered, dried, and weighed.
Procedure:
- A known mass of the organic compound ($w$ grams) is heated with excess fuming $HNO_3$ in a Carius tube.
- After cooling, the contents are diluted with water and acidified with $HCl$.
- Then, $BaCl_2$ solution is added, and the mixture is heated. A precipitate of $BaSO_4$ is formed.
- The precipitate is filtered, washed, dried, and weighed. Let its mass be $w_1$ grams.
Calculations:
- Molar mass of $BaSO_4 = 233.4$ g/mol. Mass of S in $BaSO_4 = 32$ g/mol.
- Mass of $S = \frac{32}{233.4} \times w_1$ grams.
- Percentage of Sulphur $= (\frac{32}{233.4} \times \frac{w_1}{w}) \times 100 \%$
Phosphorus:
Carius Method:
Principle: A known weight of the organic compound is heated with nitric acid in a sealed Carius tube. Phosphorus is oxidized to phosphoric acid ($H_3PO_4$). Phosphoric acid is then precipitated as ammonium phosphomolybdate ($ (NH_4)_3[P(MoO_3)_6]$) by adding ammonium molybdate solution in the presence of nitric acid. The precipitate is filtered, dried, and weighed.
Procedure:
- A known mass of the organic compound ($w$ grams) is heated with concentrated $HNO_3$ in a Carius tube.
- After cooling, the solution is treated with ammonium molybdate solution.
- The yellow precipitate of ammonium phosphomolybdate is filtered, washed, dried, and weighed. Let its mass be $w_1$ grams.
Calculations:
- Molar mass of Ammonium phosphomolybdate $(NH_4)_3[P(MoO_3)_6]$ $= 1876.4$ g/mol.
- Mass of P in $(NH_4)_3[P(MoO_3)_6]$ $= 31$ g/mol.
- Mass of $P = \frac{31}{1876.4} \times w_1$ grams.
- Percentage of Phosphorus $= (\frac{31}{1876.4} \times \frac{w_1}{w}) \times 100 \%$
Oxygen:
Method: Indirect Calculation
Principle: The percentage of oxygen cannot be determined directly by combustion analysis because it is not directly measured. It is calculated by subtracting the percentages of all other elements (C, H, N, S, P, halogens) from 100%.
Percentage of Oxygen $= 100\% - (\%C + \%H + \%N + \%S + \%P + \%X)$
Note: If the organic compound contains halogens or sulphur, they might interfere with the direct determination of oxygen. Special combustion methods exist, but indirect calculation is the most common approach in introductory contexts.