Distinguish Organic Compounds
Give a chemical test to distinguish between Ethylene(Ethene) and Acetylene(Ethyne)
1. When acetylene is bubbled through ammoniacal silver nitrate solution, a yellow-white precipitate of silver acetylide would be formed. C2H2 + 2AgNO3 ⟶ Ag2C2 + 2HNO3
2. Similarly, acetylene forms a red precipitate of copper acetylide (Cu2C2) when it is passed through ammoniacal cuprous chloride solution.
C2H2 + 2CuCl + 2NH3 ⟶ Cu2C2 + 2NH4Cl
Ethylene (C2H4) does not react with AgNO3 or Cu2Cl2solution.
C2H4 + 2CuCl + 2NH3 ⟶ No Reaction
C2H4 + 2AgNO3 ⟶ No Reaction
1. When acetylene is bubbled through ammoniacal silver nitrate solution, a yellow-white precipitate of silver acetylide would be formed. C2H2 + 2AgNO3 ⟶ Ag2C2 + 2HNO3
2. Similarly, acetylene forms a red precipitate of copper acetylide (Cu2C2) when it is passed through ammoniacal cuprous chloride solution.
C2H2 + 2CuCl + 2NH3 ⟶ Cu2C2 + 2NH4Cl
Ethylene (C2H4) does not react with AgNO3 or Cu2Cl2solution.
C2H4 + 2CuCl + 2NH3 ⟶ No Reaction
C2H4 + 2AgNO3 ⟶ No Reaction
Give a chemical test to distinguish between Formic Acid and Acetic Acid
Formic acid gives Tollens test whereas acetic acid does not give this test. Formic acid (HCOOH) is unique because it contains both an aldehyde (-CHO) group as well as carboxyl (-COOH) group. Hence it can act as a reducing agent. It reduces Tollens reagent. But acetic acid (CH3COOH) has no aldehydic group hence it does not reduces Tollens reagent.
HCOOH + Ag2O ⟶ H2O + CO2 + 2Ag↓
Acetic acid does not give this test.
CH3COOH + Ag2O ⟶ No Reaction
Formic acid gives Tollens test whereas acetic acid does not give this test. Formic acid (HCOOH) is unique because it contains both an aldehyde (-CHO) group as well as carboxyl (-COOH) group. Hence it can act as a reducing agent. It reduces Tollens reagent. But acetic acid (CH3COOH) has no aldehydic group hence it does not reduces Tollens reagent.
HCOOH + Ag2O ⟶ H2O + CO2 + 2Ag↓
Acetic acid does not give this test.
CH3COOH + Ag2O ⟶ No Reaction
Give a chemical test to distinguish between Acetaldehyde and Acetone
Tollens' reagent oxidizes acetaldehyde to acetic acid. Silver ions are reduced to silver metal. This gives silver mirror. Such silver mirror is not obtained with acetone. Hence, Tollens' reagent is used to distinguish acetaldehyde and acetone.
CH3CHO + 2[Ag(NO3)2]+ + 3OH− ⟶ CH3COO− + 2Ag↓ + 4NH3 + 2H2O
CH3COCH3 ---Tollens' Reagent---> No Ag↓
Tollens' reagent oxidizes acetaldehyde to acetic acid. Silver ions are reduced to silver metal. This gives silver mirror. Such silver mirror is not obtained with acetone. Hence, Tollens' reagent is used to distinguish acetaldehyde and acetone.
CH3CHO + 2[Ag(NO3)2]+ + 3OH− ⟶ CH3COO− + 2Ag↓ + 4NH3 + 2H2O
CH3COCH3 ---Tollens' Reagent---> No Ag↓
Give a chemical test to distinguish between Phenol and Ethanol
Phenol is aromatic alcohol, while ethanol is aliphatic alcohol.
Ethanol gives iodoform test and forms the yellow precipitate, while phenol does not give this test.
Phenol gives violet color with FeCl3 solution, and white ppts with Br2/H2O while ethanol does not give this test.
3C6H5OH + FeCl3 ⟶ (C6H5O)3Fe + 3HCl
Phenol undergoes diazo coupling reaction, while ethanol does not give this test.
Phenol is aromatic alcohol, while ethanol is aliphatic alcohol.
Ethanol gives iodoform test and forms the yellow precipitate, while phenol does not give this test.
Phenol gives violet color with FeCl3 solution, and white ppts with Br2/H2O while ethanol does not give this test.
3C6H5OH + FeCl3 ⟶ (C6H5O)3Fe + 3HCl
Phenol undergoes diazo coupling reaction, while ethanol does not give this test.
Give a chemical test to distinguish among Primary, Secondary and Tertiary Alcohols
Lucas Test: Anhydrous solution of ZnCl2 in conc. HCl is known as Lucas reagent.
Primary Alcohols:
The solution of primary alcohols remains colourless unless it is subjected to heat. The solution forms an oily layer when heated.
RCH2OH + ZnCl2 ⟶ No Reaction
Secondary Alcohols:
The solution of secondary alcohols turns turbid and forms an oily layer in within five minutes.
R2CHOH + ZnCl2 ⟶ R2CHCl↓ + H2O
Tertiary Alcohols:
The solution of tertiary alcohols turns turbid and forms an oily layer immediately.
R3CHOH + ZnCl2 ⟶ R3CCl↓ + H2O
Victor Meyer Test
This test is based on the different behaviour of nitroalkanes obtained from rimary, Secondary and Tertiary alcohols.
This test is carried out by the following steps-
step-1: The given alcohol is converted to iodide by treatment with cold HI.
step-2: The iodide is then treated with silver nitrate so as to get the corresponding nitroalkane.
step-3: The nitroalkane is finally treated with nitrous acid and the solution is made alkaline.
If the blood red color is produced in this way the alcohol is primary, if a blue color is produced, the alcohol is secondary and if there is no color produced, it is tertiary alcohols.
CH3CH2OH + HI ⟶ CH3CH2I ---AgNO2---> CH3CH2NO2 ---HNO2---> CH3C(NO2)=N—OH(Blood Red)
(CH3)2CHOH + HI ⟶ (CH3)2CHI ---AgNO2---> (CH3)2CHNO2 ---HNO2---> (CH3)2C(NO)NO2(Blue)
(CH3)3COH + HI ⟶ (CH3)3CI ---AgNO2---> (CH3)3CNO2 ---HNO2---> No Reaction (Colorless)
Lucas Test: Anhydrous solution of ZnCl2 in conc. HCl is known as Lucas reagent.
Primary Alcohols:
The solution of primary alcohols remains colourless unless it is subjected to heat. The solution forms an oily layer when heated.
RCH2OH + ZnCl2 ⟶ No Reaction
Secondary Alcohols:
The solution of secondary alcohols turns turbid and forms an oily layer in within five minutes.
R2CHOH + ZnCl2 ⟶ R2CHCl↓ + H2O
Tertiary Alcohols:
The solution of tertiary alcohols turns turbid and forms an oily layer immediately.
R3CHOH + ZnCl2 ⟶ R3CCl↓ + H2O
Victor Meyer Test
This test is based on the different behaviour of nitroalkanes obtained from rimary, Secondary and Tertiary alcohols.
This test is carried out by the following steps-
step-1: The given alcohol is converted to iodide by treatment with cold HI.
step-2: The iodide is then treated with silver nitrate so as to get the corresponding nitroalkane.
step-3: The nitroalkane is finally treated with nitrous acid and the solution is made alkaline.
If the blood red color is produced in this way the alcohol is primary, if a blue color is produced, the alcohol is secondary and if there is no color produced, it is tertiary alcohols.
CH3CH2OH + HI ⟶ CH3CH2I ---AgNO2---> CH3CH2NO2 ---HNO2---> CH3C(NO2)=N—OH(Blood Red)
(CH3)2CHOH + HI ⟶ (CH3)2CHI ---AgNO2---> (CH3)2CHNO2 ---HNO2---> (CH3)2C(NO)NO2(Blue)
(CH3)3COH + HI ⟶ (CH3)3CI ---AgNO2---> (CH3)3CNO2 ---HNO2---> No Reaction (Colorless)
Give a chemical test to distinguish among Primary, Secondary and Tertiary Amines
Hinsberg's Test:
Primary amine gives N-methyl benzene sulphonamide (soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
RNH2 + C6H5SO2Cl ⟶ RNH—SO2C6H5
Secondary amine gives N,N-dimethyl benzene sulphonamide (not soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
R2NH + C6H5SO2Cl ⟶ R2N—SO2C6H5
Tertiary amine does not react with hinsberg's reagent (C6H5SO2Cl).
R3N + C6H5SO2Cl ⟶ No Reaction
Hinsberg's Test:
Primary amine gives N-methyl benzene sulphonamide (soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
RNH2 + C6H5SO2Cl ⟶ RNH—SO2C6H5
Secondary amine gives N,N-dimethyl benzene sulphonamide (not soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
R2NH + C6H5SO2Cl ⟶ R2N—SO2C6H5
Tertiary amine does not react with hinsberg's reagent (C6H5SO2Cl).
R3N + C6H5SO2Cl ⟶ No Reaction
Give a chemical test to distinguish between Acetic Acid and Acetaldehyde
On adding aqueous NaHCO3, only acetic acid gives brisk effervescence of CO2. Acetaldehyde does not give this test.
CH3COOH + NaHCO3 ⟶ CH3COONa + CO2 + H2
On adding aqueous FeCl3 to acetic acid, blood red colour is produced.
CH3COOH + FeCl3 ⟶ (CH3COO)3Fe + 3HCl
Acetaldehyde gives Tollens' Reagent and Fehling Solution tests but not acetic acid.
On adding aqueous NaHCO3, only acetic acid gives brisk effervescence of CO2. Acetaldehyde does not give this test.
CH3COOH + NaHCO3 ⟶ CH3COONa + CO2 + H2
On adding aqueous FeCl3 to acetic acid, blood red colour is produced.
CH3COOH + FeCl3 ⟶ (CH3COO)3Fe + 3HCl
Acetaldehyde gives Tollens' Reagent and Fehling Solution tests but not acetic acid.
Give a chemical test to distinguish between Methyl Amine and Dimethyl Amine
Carbyl Amine Test:
On heating with alcoholic KOH and chloroform, methyl amine gives offensive smell of methylisocyanide (Carbyl Amine Test) but dimethyl amine does not give this test.
CH3NH2 + CHCl3 + KOH ⟶ CH3NC + 3KCl + 3H2O
CH3NHCH3 + CHCl3 + KOH ⟶ No Reaction
Hinsberg's Test:
Methyl amine gives N-methyl benzene sulphonamide (soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
CH3NH2 + C6H5SO2Cl ⟶ CH3NH—SO2C6H5
dimethyl amine gives N,N-dimethyl benzene sulphonamide (not soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
(CH3)2NH + C6H5SO2Cl ⟶ (CH3)2N—SO2C6H5
Carbyl Amine Test:
On heating with alcoholic KOH and chloroform, methyl amine gives offensive smell of methylisocyanide (Carbyl Amine Test) but dimethyl amine does not give this test.
CH3NH2 + CHCl3 + KOH ⟶ CH3NC + 3KCl + 3H2O
CH3NHCH3 + CHCl3 + KOH ⟶ No Reaction
Hinsberg's Test:
Methyl amine gives N-methyl benzene sulphonamide (soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
CH3NH2 + C6H5SO2Cl ⟶ CH3NH—SO2C6H5
dimethyl amine gives N,N-dimethyl benzene sulphonamide (not soluble in alkali) when reacts with hinsberg's reagent (C6H5SO2Cl).
(CH3)2NH + C6H5SO2Cl ⟶ (CH3)2N—SO2C6H5
