Bihar Board Class 12 Chemistry Solved Model Paper 2026
Section-A / Objective Type Questions
Question Nos 1 to 70 have four options, out of which only one is correct, You have to mark your selected option on the OMR Sheet. Answer any 35 questions. 35 x 1 = 35
Section-B
Short Answer Type Questions
Question Nos 1 to 20 are Short Answer Type. Answer any 10 questions. Each question carries 2 marks: 10 x 2 = 20
Q.1 Write Raoult's law for a solution of volatile liquids.
Answer:
For a solution of two volatile liquids A and B, the partial vapour pressure of each component is equal to the vapour pressure of the pure component multiplied by its mole fraction.
PA = P°A × XA and PB = P°B × XB
Total vapour pressure P = PA + PB = P°A × XA + P°B × XB
For a solution of two volatile liquids A and B, the partial vapour pressure of each component is equal to the vapour pressure of the pure component multiplied by its mole fraction.
PA = P°A × XA and PB = P°B × XB
Total vapour pressure P = PA + PB = P°A × XA + P°B × XB
Q.2 What is denaturation of protein?
Answer:
When a protein is subjected to change in temperature, pH, etc., the hydrogen bonds and other weak forces are disturbed and the protein loses its specific three-dimensional structure (native structure) and biological activity. This process is called denaturation of protein. The primary structure remains intact.
When a protein is subjected to change in temperature, pH, etc., the hydrogen bonds and other weak forces are disturbed and the protein loses its specific three-dimensional structure (native structure) and biological activity. This process is called denaturation of protein. The primary structure remains intact.
Q.3 Why does white ZnO becomes yellow upon heating?
Answer:
White zinc oxide (ZnO) becomes yellow upon heating because it loses oxygen, which creates a crystal defect. The excess zinc ions (Zn2+) and electrons move into interstitial sites within the crystal lattice, creating F-centers. These F-centers cause the material to absorb blue light, making it appear yellow because the remaining colors of the light spectrum are reflected.
White zinc oxide (ZnO) becomes yellow upon heating because it loses oxygen, which creates a crystal defect. The excess zinc ions (Zn2+) and electrons move into interstitial sites within the crystal lattice, creating F-centers. These F-centers cause the material to absorb blue light, making it appear yellow because the remaining colors of the light spectrum are reflected.
Q.4 A first order reaction has rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?
Answer:
t = 2.303/k × log [R]₀/[R]
t = 2.303/(1.15 × 10⁻³) × log (5/3)
t = 2004.78 × log 1.6667 ≈ 2004.78 × 0.2219 ≈ 444.6 seconds
∴ Time required ≈ 445 seconds
t = 2.303/k × log [R]₀/[R]
t = 2.303/(1.15 × 10⁻³) × log (5/3)
t = 2004.78 × log 1.6667 ≈ 2004.78 × 0.2219 ≈ 444.6 seconds
∴ Time required ≈ 445 seconds
Q.5 What is the difference between multimolecular and macromolecular colloids?
Answer:
| Multimolecular colloids | Macromolecular colloids |
|---|---|
| Formed by aggregation of large number of atoms or small molecules | Consist of large single molecules (macromolecules) |
| Particle size < 1 nm initially | Particle size itself is in colloidal range |
| Example: Gold sol, S-sol | Example: Starch, protein, cellulose |
Q.6 Why o-nitrophenol has lower boiling point than p-nitrophenol?
Answer:
o-Nitrophenol shows intramolecular H-bonding whereas p-nitrophenol shows intermolecular H-bonding. Due to intermolecular H-bonding, p-nitrophenol exists as associated molecule and has higher boiling point. o-Nitrophenol exists as discrete molecule and has lower boiling point.
o-Nitrophenol shows intramolecular H-bonding whereas p-nitrophenol shows intermolecular H-bonding. Due to intermolecular H-bonding, p-nitrophenol exists as associated molecule and has higher boiling point. o-Nitrophenol exists as discrete molecule and has lower boiling point.
Q.7 Why do transition metals form coloured compounds?
Answer:
Transition metal ions have partially filled d-orbitals. When light falls, electrons get excited from lower to higher d-orbitals (d–d transition) and absorb a particular wavelength of visible light. The complementary colour is transmitted and the compound appears coloured.
Transition metal ions have partially filled d-orbitals. When light falls, electrons get excited from lower to higher d-orbitals (d–d transition) and absorb a particular wavelength of visible light. The complementary colour is transmitted and the compound appears coloured.
Q.8 Why is NH₃ more basic than PH₃?
Answer:
Lone pair of electrons on nitrogen in NH₃ is more available for donation because N–H bond is less polar than P–H bond (due to lower electronegativity of P). Also, smaller size of N makes lone pair more concentrated. Hence NH₃ is stronger base than PH₃.
Lone pair of electrons on nitrogen in NH₃ is more available for donation because N–H bond is less polar than P–H bond (due to lower electronegativity of P). Also, smaller size of N makes lone pair more concentrated. Hence NH₃ is stronger base than PH₃.
Q.9 What are interhalogen compounds? Give one example.
Answer:
Compounds formed by combination of two different halogens are called interhalogen compounds.
Example: ICl, BrF₃, IF₅, ClF₃ etc.
Compounds formed by combination of two different halogens are called interhalogen compounds.
Example: ICl, BrF₃, IF₅, ClF₃ etc.
Q.10 What is acetylation reaction?
Answer:
Introduction of acetyl group (CH₃CO–) into an organic molecule (alcohol, phenol, amine etc.) using acetyl chloride or acetic anhydride is called acetylation.
Example: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
Introduction of acetyl group (CH₃CO–) into an organic molecule (alcohol, phenol, amine etc.) using acetyl chloride or acetic anhydride is called acetylation.
Example: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH
Q.11 Why is C–Cl bond in chlorobenzene difficult to break?
Answer:
Due to resonance, C–Cl bond in chlorobenzene acquires partial double bond character and becomes shorter and stronger than normal C–Cl bond. Hence it is difficult to break.
Due to resonance, C–Cl bond in chlorobenzene acquires partial double bond character and becomes shorter and stronger than normal C–Cl bond. Hence it is difficult to break.
Q.12 What are monosaccharides? Give one example.
Answer:
Carbohydrates which cannot be hydrolysed further into simpler units are called monosaccharides.
Example: Glucose, fructose, ribose etc.
Carbohydrates which cannot be hydrolysed further into simpler units are called monosaccharides.
Example: Glucose, fructose, ribose etc.
Q.13 What is pseudo first order reaction?
Answer:
A reaction which is actually second order but behaves as first order because one of the reactants is taken in very large excess is called pseudo first order reaction.
Example: Hydrolysis of ethyl acetate (CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH) is pseudo first order because water is in excess.
A reaction which is actually second order but behaves as first order because one of the reactants is taken in very large excess is called pseudo first order reaction.
Example: Hydrolysis of ethyl acetate (CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH) is pseudo first order because water is in excess.
Q.14 Write the formula of Potassium tetrachloridopalladate(II).
Answer:
K₂[PdCl₄]
K₂[PdCl₄]
Q.15 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol forms intermolecular hydrogen bonding whereas methoxymethane (dimethyl ether) does not form hydrogen bonding. Due to strong intermolecular H-bonding, ethanol has higher boiling point than methoxymethane.
Ethanol forms intermolecular hydrogen bonding whereas methoxymethane (dimethyl ether) does not form hydrogen bonding. Due to strong intermolecular H-bonding, ethanol has higher boiling point than methoxymethane.
Q.16 Write complete equation for Gabriel phthalimide synthesis.
Answer:
Gabriel phthalimide synthesis is a method for preparing primary amines from phthalimide, an alkyl halide, and a base.
Gabriel phthalimide synthesis is a method for preparing primary amines from phthalimide, an alkyl halide, and a base.
Q.17 Write two differences between fibrous protein and globular protein.
Answer:
| Fibrous protein | Globular protein |
|---|---|
| Polypeptide chains are long and parallel | Polypeptide chains are folded into spherical shape |
| Insoluble in water | Soluble in water |
| Example: Keratin, collagen | Example: Insulin, haemoglobin |
Q.18 Why is sucrose called invert sugar?
Answer:
On hydrolysis, sucrose (dextrorotatory) gives equimolar mixture of glucose (dextrorotatory) and fructose (laevorotatory). The laevorotatory power of fructose (−) fructose is greater than dextrorotatory power of (+) glucose, so the resulting mixture becomes laevorotatory. This change from dextro to laevo is called inversion and the mixture is called invert sugar.
On hydrolysis, sucrose (dextrorotatory) gives equimolar mixture of glucose (dextrorotatory) and fructose (laevorotatory). The laevorotatory power of fructose (−) fructose is greater than dextrorotatory power of (+) glucose, so the resulting mixture becomes laevorotatory. This change from dextro to laevo is called inversion and the mixture is called invert sugar.
Q.19 Write Faraday’s 1st law of electrolysis.
Answer:
The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
m ∝ Q or m = Z × I × t
The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
m ∝ Q or m = Z × I × t
Q.20 What is calcination? Give one example.
Answer:
Heating of carbonate ore in absence or limited supply of air to convert it into oxide is called calcination.
Example: ZnCO₃ → ZnO + CO₂
Heating of carbonate ore in absence or limited supply of air to convert it into oxide is called calcination.
Example: ZnCO₃ → ZnO + CO₂
Section-B
Long Answer Type Questions
Question Nos 21 to 26 are Long Answer Type. Answer any 3 questions. Each question carries 5 marks: 3 x 5 = 15
Q.21 Write Henry's Law. Write three applications of Henry's Law. [2 + 3 = 5 marks]
Answer:
Henry's Law Statement:
At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Mathematically: S = KH × P or P = KH × x where, S = solubility of gas, P = partial pressure of gas, x = mole fraction of gas in solution, KH = Henry's law constant
Three Important Applications of Henry's Law:
1. In production of carbonated beverages (cold drinks): CO₂ gas is dissolved in the drink under high pressure. When the bottle is opened, pressure decreases → solubility decreases → CO₂ bubbles come out.
2. In deep-sea diving (Scuba diving): At high pressure under the sea, more N₂ and O₂ dissolve in blood. If the diver comes up quickly, pressure decreases suddenly → gases form bubbles in blood (causing "the bends" or decompression sickness). So, ascent is done slowly.
3. In oxygen supply for mountaineers and patients: At high altitudes, partial pressure of O₂ is low → less oxygen dissolves in blood. Therefore, oxygen cylinders are used to increase partial pressure so that more oxygen dissolves in blood.
Henry's Law Statement:
At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Mathematically: S = KH × P or P = KH × x where, S = solubility of gas, P = partial pressure of gas, x = mole fraction of gas in solution, KH = Henry's law constant
Three Important Applications of Henry's Law:
1. In production of carbonated beverages (cold drinks): CO₂ gas is dissolved in the drink under high pressure. When the bottle is opened, pressure decreases → solubility decreases → CO₂ bubbles come out.
2. In deep-sea diving (Scuba diving): At high pressure under the sea, more N₂ and O₂ dissolve in blood. If the diver comes up quickly, pressure decreases suddenly → gases form bubbles in blood (causing "the bends" or decompression sickness). So, ascent is done slowly.
3. In oxygen supply for mountaineers and patients: At high altitudes, partial pressure of O₂ is low → less oxygen dissolves in blood. Therefore, oxygen cylinders are used to increase partial pressure so that more oxygen dissolves in blood.
Q.22 What is Lanthanoid Contraction? Write any three consequences of Lanthanoid contraction. [2 + 3 = 5 marks]
Answer:
Definition of Lanthanoid Contraction:
The steady decrease in the atomic and ionic radii of the 14 lanthanoid elements (from Ce to Lu) with increase in atomic number due to poor shielding effect of 4f-electrons is called lanthanoid contraction.
Three Important Consequences of Lanthanoid Contraction:
1. Similarity in size of 4d and 5d series elements (Zr–Hf similarity): Due to lanthanoid contraction, the atomic radius of Hf (5d series) becomes almost equal to that of Zr (4d series). Hence Zr and Hf have almost identical size, chemical properties and are very difficult to separate.
2. Basic strength of lanthanoid hydroxides decreases from La(OH)₃ to Lu(OH)₃: As the size of Ln³⁺ ion decreases from La³⁺ to Lu³⁺, the covalent character of Ln–OH bond increases and basic nature decreases progressively.
3. Difficulty in separation of lanthanoids: Because of very small difference in size and properties of lanthanoid ions (Ln³⁺), their chemical properties are very similar. Hence, separation of lanthanoids from each other is extremely difficult.
Definition of Lanthanoid Contraction:
The steady decrease in the atomic and ionic radii of the 14 lanthanoid elements (from Ce to Lu) with increase in atomic number due to poor shielding effect of 4f-electrons is called lanthanoid contraction.
Three Important Consequences of Lanthanoid Contraction:
1. Similarity in size of 4d and 5d series elements (Zr–Hf similarity): Due to lanthanoid contraction, the atomic radius of Hf (5d series) becomes almost equal to that of Zr (4d series). Hence Zr and Hf have almost identical size, chemical properties and are very difficult to separate.
2. Basic strength of lanthanoid hydroxides decreases from La(OH)₃ to Lu(OH)₃: As the size of Ln³⁺ ion decreases from La³⁺ to Lu³⁺, the covalent character of Ln–OH bond increases and basic nature decreases progressively.
3. Difficulty in separation of lanthanoids: Because of very small difference in size and properties of lanthanoid ions (Ln³⁺), their chemical properties are very similar. Hence, separation of lanthanoids from each other is extremely difficult.
Q.23 What is crystal field splitting energy? Using the valence bond approach, deduce the shape and magnetic character of [Co(NH₃)₆]³⁺. [2 + 1.5 + 1.5 = 5 marks]
Answer:
1. Crystal Field Splitting Energy (Δ):
When a transition metal ion is surrounded by ligands is placed in a crystal field, the five degenerate d-orbitals split into two sets of different energies.
The energy difference between the lower energy set (t₂g) and the higher energy set (eg) in an octahedral field is called crystal field splitting energy or crystal field stabilization energy parameter, denoted by Δ₀ (or simply Δ).
In tetrahedral field it is denoted by Δₜ (Δₜ = 4/9 Δ₀).
2. Valence Bond Theory for [Co(NH₃)₆]³⁺:
Complex: [Co(NH₃)₆]³⁺
Metal ion: Co³⁺
Electronic configuration of Co (Z = 27): [Ar] 3d⁶ 4s²
Co³⁺: 3d⁶
Ligand: NH₃ → strong field ligand
Oxidation state: +3
Coordination number: 6 → octahedral geometry
Hybridisation and electron pairing (using Valence Bond Approach):
Conclusion:
• Shape: Octahedral (due to d²sp³ hybridisation)
• Magnetic character: All 6 electrons are paired in 3d orbitals → no unpaired electron
∴ Diamagnetic.
1. Crystal Field Splitting Energy (Δ):
When a transition metal ion is surrounded by ligands is placed in a crystal field, the five degenerate d-orbitals split into two sets of different energies.
The energy difference between the lower energy set (t₂g) and the higher energy set (eg) in an octahedral field is called crystal field splitting energy or crystal field stabilization energy parameter, denoted by Δ₀ (or simply Δ).
In tetrahedral field it is denoted by Δₜ (Δₜ = 4/9 Δ₀).
2. Valence Bond Theory for [Co(NH₃)₆]³⁺:
Complex: [Co(NH₃)₆]³⁺
Metal ion: Co³⁺
Electronic configuration of Co (Z = 27): [Ar] 3d⁶ 4s²
Co³⁺: 3d⁶
Ligand: NH₃ → strong field ligand
Oxidation state: +3
Coordination number: 6 → octahedral geometry
Hybridisation and electron pairing (using Valence Bond Approach):
| Orbitals → | 3d | 4s | 4p | 4d | ||||||
| Co³⁺ (free ion) | ↑↓ | ↑ | ↑ | ↑ | ↑ | |||||
| In complex (with NH₃) | ↑↓ | ↑↓ | ↑↓ | ×× | ×× | ×× | ×× | |||
| Hybridisation | d²sp³ (inner orbital) | |||||||||
• Shape: Octahedral (due to d²sp³ hybridisation)
• Magnetic character: All 6 electrons are paired in 3d orbitals → no unpaired electron
∴ Diamagnetic.
Q.24
a) What is the name of two poisonous gases which can be prepared from chlorine gas?
b) Why is ICl more reactive than I₂?
c) Balance the following equations:
(i) XeF₆ + H₂O → XeO₂F₂ + HF
(ii) Ag + PCl₅ → AgCl + PCl₃ [1.5 + 1.5 + 1 + 1]
b) Why is ICl more reactive than I₂?
c) Balance the following equations:
(i) XeF₆ + H₂O → XeO₂F₂ + HF
(ii) Ag + PCl₅ → AgCl + PCl₃ [1.5 + 1.5 + 1 + 1]
Answer:
a) Two poisonous gases prepared from chlorine gas
Phosgene and chloropicrin can be prepared from chlorine gas as shown below-
1. Phosgene (COCl₂) – prepared by reacting chlorine with carbon monoxide over activated charcoal in sunlight.
CO + Cl₂ → COCl₂
2. Chlorine monoxide (Cl₂O) or Tear gas (CCl₃NO₂ – chloro picrin) – also obtained using Cl₂.
Chloropicrin (CCl₃NO₂) is prepared by reacting nitromethane (CH₃NO₂) with chlorine gas (Cl₂). The chlorine atoms substitute the hydrogen atoms on the carbon atom of nitromethane.
CH₃NO₂ + 3Cl₂ → CCl₃NO₂ + 3HCl
b) Why ICl is more reactive than I₂:
ICl is more reactive than I₂ because the I–Cl bond is polar due to large electronegativity difference between I (2.5) and Cl (3.0). The chlorine end acquires partial negative charge (δ⁻) and iodine end δ⁺, making it more susceptible to attack by nucleophiles as well as electrophiles. In I₂, the I–I bond is non-polar and symmetrical. Hence, ICl undergoes heterolytic fission easily, whereas I₂ undergoes homolytic fission. Thus, ICl is more reactive than I₂.
c) Balanced equations:
(i) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
(ii) 2Ag + PCl₅ → 2AgCl + PCl₃
a) Two poisonous gases prepared from chlorine gas
Phosgene and chloropicrin can be prepared from chlorine gas as shown below-
1. Phosgene (COCl₂) – prepared by reacting chlorine with carbon monoxide over activated charcoal in sunlight.
CO + Cl₂ → COCl₂
2. Chlorine monoxide (Cl₂O) or Tear gas (CCl₃NO₂ – chloro picrin) – also obtained using Cl₂.
Chloropicrin (CCl₃NO₂) is prepared by reacting nitromethane (CH₃NO₂) with chlorine gas (Cl₂). The chlorine atoms substitute the hydrogen atoms on the carbon atom of nitromethane.
CH₃NO₂ + 3Cl₂ → CCl₃NO₂ + 3HCl
b) Why ICl is more reactive than I₂:
ICl is more reactive than I₂ because the I–Cl bond is polar due to large electronegativity difference between I (2.5) and Cl (3.0). The chlorine end acquires partial negative charge (δ⁻) and iodine end δ⁺, making it more susceptible to attack by nucleophiles as well as electrophiles. In I₂, the I–I bond is non-polar and symmetrical. Hence, ICl undergoes heterolytic fission easily, whereas I₂ undergoes homolytic fission. Thus, ICl is more reactive than I₂.
c) Balanced equations:
(i) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
(ii) 2Ag + PCl₅ → 2AgCl + PCl₃
Q.25
a) Arrange the following compounds in the increasing order of boiling points:
C₂H₅OC₂H₅, C₄H₉COOH, C₄H₉OH
b) What is the name of reaction given below:
RCOCl + H₂ --(Pd/BaSO₄)→ RCHO + HCl
c) Give reason for the following:
(i) Benzoic acid is stronger acid than acetic acid.
(ii) Why Grignard reagents should be prepared under anhydrous conditions? [1+1+1.5+1.5]
C₂H₅OC₂H₅, C₄H₉COOH, C₄H₉OH
b) What is the name of reaction given below:
RCOCl + H₂ --(Pd/BaSO₄)→ RCHO + HCl
c) Give reason for the following:
(i) Benzoic acid is stronger acid than acetic acid.
(ii) Why Grignard reagents should be prepared under anhydrous conditions? [1+1+1.5+1.5]
Answer:
a) Increasing order of boiling points:
C₂H₅OC₂H₅ < C₄H₉OH < C₄H₉COOH
Reason:
Diethyl ether (C₂H₅OC₂H₅) has no H-bonding → lowest b.p. (~34.6°C)
Butan-1-ol (C₄H₉OH) has intermolecular H-bonding → higher b.p. (~117°C)
Butanoic acid (C₄H₉COOH) forms strong dimer due to double H-bonding → highest b.p. (~164°C)
b) Name of the reaction:
Rosenmund Reduction
c) (i) Benzoic acid is stronger than acetic acid:
Benzoic acid is stronger because the carboxylate anion (C₆H₅COO⁻) is stabilized by resonance with the benzene ring. The negative charge is delocalized over the ring, making the anion more stable. In CH₃COO⁻, there is no such resonance stabilization, only +I effect of methyl group which destabilizes the anion.
Hence, benzoic acid (pKa ≈ 4.2) is stronger than acetic acid (pKa ≈ 4.76).
(ii) Grignard reagents are prepared under anhydrous conditions:
Grignard reagents (R–MgX) are very strong nucleophiles and bases. They react instantly react with water or moisture to form alkane:
R–MgX + H₂O → R–H + Mg(OH)X
Even trace moisture destroys the reagent. Therefore, all apparatus and solvents must be completely dry (anhydrous conditions) during preparation and use of Grignard reagent.
a) Increasing order of boiling points:
C₂H₅OC₂H₅ < C₄H₉OH < C₄H₉COOH
Reason:
Diethyl ether (C₂H₅OC₂H₅) has no H-bonding → lowest b.p. (~34.6°C)
Butan-1-ol (C₄H₉OH) has intermolecular H-bonding → higher b.p. (~117°C)
Butanoic acid (C₄H₉COOH) forms strong dimer due to double H-bonding → highest b.p. (~164°C)
b) Name of the reaction:
Rosenmund Reduction
c) (i) Benzoic acid is stronger than acetic acid:
Benzoic acid is stronger because the carboxylate anion (C₆H₅COO⁻) is stabilized by resonance with the benzene ring. The negative charge is delocalized over the ring, making the anion more stable. In CH₃COO⁻, there is no such resonance stabilization, only +I effect of methyl group which destabilizes the anion.
Hence, benzoic acid (pKa ≈ 4.2) is stronger than acetic acid (pKa ≈ 4.76).
(ii) Grignard reagents are prepared under anhydrous conditions:
Grignard reagents (R–MgX) are very strong nucleophiles and bases. They react instantly react with water or moisture to form alkane:
R–MgX + H₂O → R–H + Mg(OH)X
Even trace moisture destroys the reagent. Therefore, all apparatus and solvents must be completely dry (anhydrous conditions) during preparation and use of Grignard reagent.
Q.26
a) What is Carbylamine reaction?
b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzoic acid and Phenol
(ii) Benzaldehyde and Benzoic acid [2 + 3 = 5 marks]
b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzoic acid and Phenol
(ii) Benzaldehyde and Benzoic acid [2 + 3 = 5 marks]
Answer:
a) Carbylamine reaction:
Primary amines (both aliphatic and aromatic) when heated with chloroform (CHCl₃) and alcoholic KOH give foul/offensive-smelling isocyanide (carbylamine).
R–NH₂ + CHCl₃ + 3KOH (alc.) → R–NC (isocyanide) + 3KCl + 3H₂O
C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O
Very offensive smell (like rotting fish) is produced. This reaction is given only by primary amines and is used as a test for primary amines.
b) Distinction tests:
(i) Benzoic acid and Phenol:
(ii) Benzaldehyde and Benzoic acid:
Most commonly used and easiest test:
NaHCO₃ test is sufficient for both pairs.
a) Carbylamine reaction:
Primary amines (both aliphatic and aromatic) when heated with chloroform (CHCl₃) and alcoholic KOH give foul/offensive-smelling isocyanide (carbylamine).
R–NH₂ + CHCl₃ + 3KOH (alc.) → R–NC (isocyanide) + 3KCl + 3H₂O
C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O
Very offensive smell (like rotting fish) is produced. This reaction is given only by primary amines and is used as a test for primary amines.
b) Distinction tests:
(i) Benzoic acid and Phenol:
| Test | Benzoic acid (C₆H₅COOH) | Phenol (C₆H₅OH) |
|---|---|---|
| NaHCO₃ test (or brisk effervescence test) | Gives brisk effervescence of CO₂ C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂↑ + H₂O | No reaction, no effervescence |
| FeCl₃ test (Neutral ferric chloride) | No colour (or very pale) | Violet/purple colour |
| Test | Benzaldehyde (C₆H₅CHO) | Benzoic acid (C₆H₅COOH) |
|---|---|---|
| NaHCO₃ test or Litmus test | No reaction with NaHCO₃ Does not turn blue litmus red | Brisk effervescence of CO₂ Turns blue litmus red (acidic) |
| Tollens’ reagent test | Gives silver mirror C₆H₅CHO + 2[Ag(NH₃)₂]⁺ → C₆H₅COOH + 2Ag↓ | No silver mirror |
| Iodoform test | Gives yellow ppt of CHI₃ | No yellow ppt |
NaHCO₃ test is sufficient for both pairs.
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