Chapter 12.21 - Qualitative Analysis

In the previous section, we saw chromatography. In this section, we will see qualitative analysis of organic compounds.

• In chemistry, qualitative analysis involves the determination of the various elements present in a compound.
• In organic chemistry, most compounds will contain carbon and hydrogen.
• Besides C and H, some organic compounds will contain O, N, S, halogens and P.
• Our aim is to confirm the presence of these elements in the given compound.

Detection of Carbon and Hydrogen

This can be explained in 3 steps:
1. The presence of C and H can be detected by heating the compound with copper(II) oxide
• Note that, copper(II) oxide is different from copper(I) oxide.
   ♦ Copper(II) oxide is CuO. The oxidation state of Cu in this compound is +2  
   ♦ Copper(I) oxide is Cu₂O. The oxidation state of Cu in this compound is +1
2. The C in the given compound reacts with CuO to give CO₂.
The equation is: C + 2CuO ⟶ 2Cu + CO₂
• The presence of CO₂ can be detected as follows:
(i) The vapors formed from the reaction mixture is passed through lime water.
(ii) The lime water becomes turbid due the the precipitation of CaCO₃.
(iii) The turbidity confirms the presence of CO₂.
The equation is: CO₂ + Ca(OH)₂ ⟶ CaCO₃ + H₂O
• A video can be seen here.
3. The presence of water can be detected as follows:
(i) The vapors formed from the reaction is passed through anhydrous copper sulphate.
(ii) The H₂O molecules get entrapped in the crystal structure of copper sulphate. The color of the copper sulphate becomes blue.
(iii) The formation of blue color confirms the presence of H₂O
The equation is: 5H₂O + CuSO₄ ⟶ CuSO₄.5H₂O
• A video can be seen here.
• We have seen 'water of crystallization' in a previous section. See interstitial water below fig.9.3 in section 9.3.
4. We can write the inference:
• If the presence of CO₂ is confirmed in step (2), we can say:
The given compound contains C    
• If the presence of H₂O is confirmed in step (3), we can say:
The given compound contains H

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Detection of other elements

Nitrogen can be detected by Lassaigne’s test. It can be explained in 5 steps:
1. The given organic compound is heated together with sodium. They melt (fusion) and combine together to give new compounds.
• If nitrogen is present, the equation is: Na + C + N ⟶ NaCN
• We know that, in the organic compound, the N will be covalently bonded with other atoms. But NaCN, is in an ionic compound.
2. The fused mass is boiled with distilled water. The resulting solution is filtered to obtain the sodium fusion extract (SFE)
• Since SFE is an aqueous solution, the ionic compound NaCN will be present as Na⁺ and CN^- ions.
3. To the SFE, freshly prepared FeSO₄ solution is added. This solution will contain Fe²⁺ ions.
• The Fe²⁺ ions will react with CN^- ions to give hexacyanoferrate(II) [Fe(CN)₆]^4- ions.
• The equation is: 6CN^- + Fe²⁺ ⟶ [Fe(CN)₆]^4-
4. To this solution, we add a few drops of FeCl₃. This will dissociate into Fe³⁺ and Cl^- ions.
• The Fe³⁺ ions react with hexacyanoferrate(II) ions to give ferriferrocyanide (Fe₄[Fe(CN)₆]₃.xH₂O]).
• The equation is: 3[Fe(CN)₆]^4- + 4Fe³⁺ ⟶ Fe₄[Fe(CN)₆]₃.xH₂O]
• Ferriferrocyanide is prussian blue in color.
5. We can write the inference:
• The formation of prussian blue precipitate confirms the presence of CN^- ions in the SFE.
• The presence of CN^- ions in the SFE confirms the presence of N in the given organic compound.

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Sulfur can also be detected by Lassaigne’s test. It can be explained in 7 steps:

1. The given organic compound is heated together with sodium. They melt (fusion) and combine together to give new compounds.
• If sulfur is present, the equation is: 2Na + S ⟶ Na₂S
• We know that, in the organic compound, the S will be covalently bonded with other atoms. But Na₂S, is in an ionic compound.
2. The fused mass is boiled with distilled water. The resulting solution is filtered to obtain the sodium fusion extract (SFE).
• Since SFE is an aqueous solution, the ionic compound Na₂S will be present as Na⁺ and S^2- ions.
3. The SFE is acidified with acetic acid and lead acetate is added. The lead acetate will contain Pb²⁺ ions.
• The Pb²⁺ ions will react with S^2- ions to give lead sulphide (PbS).
• The equation is: S^2- + Pb²⁺ ⟶ Pbs
• Pbs is black in color
4. Another method is to treat the SFE with sodium nitroprusside Na₂[Fe(CN)₅NO]2H₂O.
• In aqueous solution, this will give [Fe(CN)₅NO]^2- ions.
• These ions will react with the S^2- ions to give [Fe(CN)₅NOS]^4- ions.
• The equation is: S^2- + [Fe(CN)₅NO]^2- ⟶ [Fe(CN)₅NOS]^4-
• [Fe(CN)₅NOS]^4- is violet in color.
5. We can write the inference:
• The formation of black precipitate when lead acetate is added to the SFE, confirms the presence of S^2- in the SFE.
• The appearance of violet color when sodium nitroprusside is added to SFE, also confirms the presence of S^2- in the SFE.
• The presence of S^2- ions in the SFE confirms the presence of S in the given organic compound.
6. Suppose that, both N and S are present in the given organic compound. Then the test can be written in 4 steps:
(i) Consider the first reaction that we wrote in the case of N:
Na + C + N ⟶ NaCN
• If S is also present, this equation will become:
Na + C + N + S ⟶ NaSCN
(ii) So if both N and S are present, the SFE will contain SCN^- in the place of CN^-.
(iii) In the case of N, we added Fe³⁺ in the form of FeCl₃
• So if both N and S are present, the Fe³⁺ ions will be reacting with SCN^- ions.
• The product is [Fe(SCN)]²⁺. This is blood red in color.
• The equation is: Fe³⁺ + SCN^- ⟶ [Fe(SCN)]²⁺
(iv) We can write the inference in 2 steps:
    ♦ While testing for N, we add FeCl₃.
    ♦ If blood red color is obtained instead of prussian blue color, we can confirm that, both N and S are present.
7. Let us see the procedure when both N and S are present. It can be written in steps:
(i) In the above step (6), we saw that, if both N and S are present, NaSCN will be produced.
(ii) If the sodium fusion is carried out with excess of sodium, the NaSCN produced, will decompose to NaCN and Na₂S
• The equation is: NaSCN + 2Na ⟶ NaCN + Na₂S
(iii) So the SFE will contain CN^- ions and S^2- ions.
• Thus we can avoid SCN^- ions.
    ♦ The CN^- ions can be detected by the method that we saw for N.
    ♦ The S^2- ions can be detected by the methods that we saw for S.

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Halogens can also be detected by Lassaigne’s test. It can be explained in 6 steps:

1. The given organic compound is heated together with sodium. They melt (fusion) and combine together to give new compounds.
• If any halogens (Cl, Br or I) is present, the equations are:
    ♦ 2Na + Cl₂ ⟶ 2NaCl
    ♦ 2Na + Br₂ ⟶ 2NaBr
    ♦ 2Na + I₂ ⟶ 2NaI
• We know that, in the organic compound, Cl, Br or I will be covalently bonded with other atoms. But NaCl, NaBr and NaI are ionic compounds.
2. The fused mass is boiled with distilled water. The resulting solution is filtered to obtain the sodium fusion extract (SFE).
• Since SFE is an aqueous solution, the ionic compounds NaCl, NaBr and NaI will be present as Na⁺ and Cl^-, Br^-, I^- ions.
3. The SFE is acidified with nitric acid and then treated with silver nitrate (AgNO₃). Thus we will get Ag⁺ ions in the solution.
4. The Ag⁺ ions will react with the Cl^-, Br^- or I^- ions.
• If Cl^- is present in the SFE, we will get silver chloride (AgCl).
    ♦ AgCl is obtained as a white precipitate. It is soluble in ammonium hydroxide.
• If Br^- is present in the SFE, we will get silver bromide (AgBr).
    ♦ AgBr is obtained as a yellowish precipitate. It is sparingly soluble in ammonium hydroxide.
• If I^- is present in the SFE, we will get silver iodide (AgI).
    ♦ AgI is obtained as a yellow precipitate. It is insoluble in ammonium hydroxide.
5. So we can write the inference:
(i) White precipitate soluble in ammonium hydroxide, confirms the presence of Cl^- in the SFE.
• The presence of Cl^- ions in the SFE confirms the presence of Cl in the given organic compound.
(ii) Yellowish precipitate sparingly soluble in ammonium hydroxide, confirms the presence of Br^- in the SFE.
• The presence of Br^- ions in the SFE confirms the presence of Br in the given organic compound.
(iii) Yellow precipitate insoluble in ammonium hydroxide, confirms the presence of I^- in the SFE.
• The presence of I^- ions in the SFE confirms the presence of I in the given organic compound.
6. Suppose that, in addition to any halogens, N and S are also present in the given organic compound.
• Then, as we saw before, NaCN or Na₂S will also be present in the SFE.
• The CN^- or S^2- ions thus produced, will interfere with the silver nitrate test.
• To avoid this, we boil the SFE with concentrated nitric acid before beginning the silver nitrate test.
• When the SFE is boiled in this way, the CN^- and S^- ions will be decomposed.

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Test for phosphorus can be explained in 4 steps:
1. The given organic compound is heated with sodium peroxide (Na₂O₂).
• The phosphorus in the compound will be oxidized to phosphate.
• Thus we get sodium phosphate (Na₃PO₄).
2. The solution is boiled with nitric acid.
• We get phosphoric acid (H₃PO₄).
• The equation is: Na₃PO₄ + 3HNO₃ ⟶ H₃PO₄ + 3NaNO₃
3. To this solution, we add ammonium molybdate (NH₄)₂MoO₄.
• The phosphoric acid reacts with ammonium molybdate to give ammonium phosphomolybdate (NH₄)₃PO₄.12MoO₃.
• The equation is:
H₃PO₄ + 12(NH₄)₂MoO₄ + 21HNO₃ ⟶ (NH₄)₃PO₄.12MoO₃ + 21NH₄NO₃ + 12H₂O
• Ammonium phosphomolybdate is yellow in color.
4. So we can write the inference:
A yellow coloration or precipitate indicates the presence of phosphorus.

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In the next section we will see quantitative analysis of organic compounds.

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