CBSE Class 12 Chemistry – Complete Notes & Question Bank (All Chapters)

📚 CBSE Class 12 Chemistry
Complete Notes & Question Bank
All Chapters Combined

📖 Master Table of Contents

Chapter 1: Solutions
Chapter 2: Electrochemistry
Chapter 4: d- and f-Block Elements
Chapter 4.2: f-Block Elements (Detailed)
Chapter 5: Coordination Compounds
Chapter 6: Haloalkanes and Haloarenes
Chapter 6.2: Haloarenes, Conversions & Polyhalogen Compounds
Chapter 7: Alcohols, Phenols and Ethers
Chapter 8: Aldehydes, Ketones and Carboxylic Acids
Chapter 9: Amines
Chapter 10: Biomolecules

Chapter 1: Solutions

📚 CBSE Class 12 Chemistry
Chapter 1: Solutions (Complete Notes & Question Bank)

1. Solutions and its Expression of Concentration, Solubility of Gases
2. Raoult’s Law, Ideal and Non-Ideal Solutions
3. Colligative Properties, Determination of Molecular Mass, Abnormal Molecular Mass, Van’t Hoff Factor

1. Solutions and its Expression of Concentration, Solubility of Gases

Basic Concepts

Solution: A homogeneous mixture of two or more pure substances.

Solute: A substance that is dissolved in another substance in lesser amount.
Solvent: A substance in which another substance is dissolved in larger amount. Solvent determines the physical state of the solution.

Types of Solutions

Depending upon the physical states of solute and solvent, there are nine different types of solutions:

S. No.	Types of Solutions	Solute	Solvent	Examples
1.	Solid – Solid	Solid	Solid	Alloys like brass, bronze, amalgam
2.	Solid – Liquid	Solid	Liquid	Solution of sugar, salt, urea in water
3.	Solid – Gas	Solid	Gas	Dust or smoke particles in air
4.	Liquid – Solid	Liquid	Solid	Hydrated salts, mercury in amalgamated zinc
5.	Liquid – Liquid	Liquid	Liquid	Alcohol in water, benzene in toluene
6.	Liquid – Gas	Liquid	Gas	Aerosol, water vapour in air
7.	Gas – Solid	Gas	Solid	Hydrogen adsorbed in palladium
8.	Gas – Liquid	Gas	Liquid	Aerated drinks
9.	Gas – Gas	Gas	Gas	Mixture of gases (e.g., air)

Aqueous solution: A solution containing water as solvent (e.g., sugar solution).
Non-aqueous solution: A solution containing solvent other than water (e.g., iodine in alcohol).
Saturated solution: A solution in which no more solute can be dissolved at the same temperature.
Unsaturated solution: A solution in which more amount of solute can be dissolved at the same temperature.

Visual Summary: Types of Solutions

Solid Solvent

Liquid Solvent

Gaseous Solvent

• Alloys (Solid/Solid) • Hydrated salts (Liquid/Solid) • H2 in Pd (Gas/Solid)

• Sugar in water (Solid/Liquid) • Alcohol in water (Liquid/Liquid) • Aerated drinks (Gas/Liquid)

• Smoke (Solid/Gas) • Fog (Liquid/Gas) • Air (Gas/Gas)

Methods of Expressing Concentration of Solution

(i) Mass percentage (w/w%)

\text{Mass\%} = \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100

(ii) Volume percentage (v/v%)

\text{Volume\%} = \frac{\text{Volume of solute}}{\text{Total volume of solution}} \times 100

(iii) Mass by volume percentage (w/V%)

\text{Mass by volume\%} = \frac{\text{Mass of solute}}{\text{Volume of solution}} \times 100

Commonly used in medicine and pharmacy.

(iv) Parts per million (ppm)

\text{ppm} = \frac{\text{Number of parts of component}}{\text{Total number of parts of all components}} \times 10^6

Can be expressed as: (a) Mass to mass, (b) Volume to volume, (c) Mass to volume.

(v) Mole Fraction ($\chi$)

\chi_A = \frac{n_A}{n_A + n_B}$$ $$\chi_A + \chi_B = 1

(vi) Molarity (M)

M = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{W_B \times 1000}{M_B \times V_{\text{(mL)}}}

Unit: mol L⁻¹. Depends on temperature.

(vii) Molality (m)

m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{W_B \times 1000}{M_B \times W_{A\text{(g)}}}

Unit: mol kg⁻¹. Independent of temperature.

(viii) Normality (N)

N = \frac{\text{Gram equivalents of solute}}{\text{Volume of solution in L}} = \frac{W_B \times 1000}{E_B \times V_{\text{(mL)}}}

Relationships Between Concentration Terms

\text{Molarity (M) and Molality (m): } m = \frac{1000 \times M}{1000 \times d – M \times M_B}$$ $$\text{Mole fraction } (\chi_B) \text{ and Molality (m): } m = \frac{1000 \times \chi_B}{(1 – \chi_B) M_A}

Solubility & Henry’s Law

Solubility: Maximum amount of solute that can be dissolved in 100 g of solvent to form a saturated solution at a given temperature.

Factors affecting solubility:

Nature of Solute and Solvent: “Like dissolves like” (polar dissolves polar, non-polar dissolves non-polar).
Temperature: Increases for endothermic reactions, decreases for exothermic reactions.
Pressure: Does not affect solids/liquids significantly, but greatly affects gases.

Henry’s Law

p = K_H \cdot \chi

The partial pressure of the gas is proportional to its mole fraction in the solution.

Applications: (i) Sealing soda bottles under high pressure. (ii) Scuba diving tanks diluted with He to avoid N₂ toxicity. (iii) Anoxia at high altitudes.

Limitations: Applicable only when pressure is not too high, temperature is not too low, and gas doesn’t undergo chemical change, association, or dissociation.

Example 1 [Board 2023]

If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.48 k bar.

Solution:

\begin{aligned} \chi &= \frac{P(\text{nitrogen})}{K_H} \\ &= \frac{0.987 \text{ bar}}{76480 \text{ bar}} \\ &= 1.29 \times 10^{-5} \end{aligned}$$ $$\text{Since } 1 \text{ mol} = 1000 \text{ mmol}$$ $$\chi = 1.29 \times 10^{-5} \times 1000 = 0.0129 \text{ mmol/L of water}

2. Raoult’s Law, Ideal and Non-Ideal Solutions

Vapour Pressure & Raoult’s Law

Vapour pressure: Pressure exerted by vapours over a liquid at equilibrium at constant temperature. Depends on nature of liquid and temperature.

Raoult’s law for volatile liquids:

p_A = p_A^\circ \cdot \chi_A \quad \text{and} \quad p_B = p_B^\circ \cdot \chi_B$$ $$p_{\text{total}} = p_A + p_B = p_A^\circ \chi_A + p_B^\circ \chi_B

Raoult’s law as a special case of Henry’s law:

Comparing $p_A = p_A^\circ \chi_A$ with Henry’s law $p_A = K_H \chi_A$, we get $p_A^\circ = K_H$.

Raoult’s law for non-volatile solute:

\frac{p_A^\circ – p_A}{p_A^\circ} = \chi_B

Relative lowering of vapour pressure is equal to the mole fraction of solute.

Ideal and Non-Ideal Solutions

Property	Ideal Solution	Non-Ideal Solution
Raoult’s Law	Obeys at all concentrations	Does not obey
Δ_mixH	= 0	≠ 0
Δ_mixV	= 0	≠ 0
Interactions	A-A ≈ B-B ≈ A-B	A-A ≠ A-B
Examples	n-hexane + n-heptane, Benzene + Toluene	Water + Ethanol, Chloroform + Acetone

Visual Summary: Deviations from Raoult’s Law

Negative Deviation A-B interactions stronger than A-A/B-B Mole Fraction ($\chi$) VP Forms Maximum Boiling Azeotrope

Azeotropes

Azeotropes: Liquid mixtures that distil over without change in composition (constant boiling mixtures).

Minimum boiling azeotropes: Formed by solutions showing large positive deviation (e.g., water + benzene, chloroform + methanol).
Maximum boiling azeotropes: Formed by solutions showing large negative deviation (e.g., HNO₃ + H₂O).

Example 2 [Board 2023]

Suppose a solution is prepared by mixing two volatile liquids, A and B. Let $\chi_A$ and $\chi_B$ respectively be their mole fractions, and let $p_A$ and $p_B$ be their partial vapour pressures, respectively, in the solution at a particular temperature. Calculate the composition of the vapour phase in equilibrium with the solution.

Solution:

\begin{aligned} \text{According to Raoult’s law:} \\ p_A &= p_A^\circ \chi_A \\ p_B &= p_B^\circ \chi_B \\ \text{By Dalton’s law of partial pressures:} \\ p_{\text{total}} &= p_A + p_B \\ &= \chi_A p_A^\circ + \chi_B p_B^\circ \\ &= (1 – \chi_B) p_A^\circ + \chi_B p_B^\circ \\ &= p_A^\circ + (p_B^\circ – p_A^\circ)\chi_B \end{aligned}$$ $$\begin{aligned} \text{Composition of vapour phase } (\chi_A’, \chi_B’): \\ p_A &= \chi_A’ p_{\text{total}} \implies \chi_A’ = \frac{p_A}{p_{\text{total}}} \\ p_B &= \chi_B’ p_{\text{total}} \implies \chi_B’ = \frac{p_B}{p_{\text{total}}} \end{aligned}

3. Colligative Properties, Determination of Molecular Mass, Abnormal Molecular Mass, Van’t Hoff Factor

Colligative Properties

Properties of solutions that depend only on the number of particles of solute and not on the nature of the solute.

Visual Summary: Colligative Properties

Relative Lowering of Vapour Pressure

Elevation of Boiling Point

Depression of Freezing Point

Osmotic Pressure

Depend ONLY on number of solute particles

(i) Relative Lowering of Vapour Pressure

\frac{\Delta p}{p_A^\circ} = \frac{p_A^\circ – p_A}{p_A^\circ} = \chi_B = \frac{n}{n + N}

(ii) Elevation of Boiling Point

\Delta T_b = K_b \cdot m = \frac{K_b \times w_2 \times 1000}{M_2 \times w_1}

$K_b$ = Molal elevation constant (Ebullioscopic constant).

(iii) Depression of Freezing Point

\Delta T_f = K_f \cdot m = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}

$K_f$ = Molal depression constant (Cryoscopic constant).

(iv) Osmotic Pressure

\pi = C R T = \frac{n_B}{V} R T

Osmosis: Net flow of solvent to the solution through a semipermeable membrane.

Reverse Osmosis: When applied pressure > osmotic pressure, pure solvent flows out of solution. Used in water purification.

Abnormal Molecular Mass & Van’t Hoff Factor

Abnormal molecular mass: When molecular mass calculated from colligative properties differs from theoretical value due to association or dissociation of solute.

i = \frac{\text{Observed colligative property}}{\text{Normal colligative property}} = \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}}

Modified colligative property equations:

\Delta T_b = i \cdot K_b \cdot m \quad ; \quad \Delta T_f = i \cdot K_f \cdot m \quad ; \quad \pi = i \cdot C R T

Example 3

The freezing point is reduced from 5.51 to 5.03°C when 0.721 g of a compound is added to 75 mL of benzene. (density of benzene = 0.879 g/mL, $K_f$ for benzene = 5.12 K kg mol⁻¹). Calculate the molecular mass of the compound.

Solution:

\begin{aligned} \text{Mass of benzene } (w_1) &= 75 \text{ mL} \times 0.879 \text{ g/mL} \\ &= 65.925 \text{ g} = 0.06593 \text{ kg} \\ \Delta T_f &= 5.51 – 5.03 = 0.48 \text{ K} \\ \text{We know, } \Delta T_f &= \frac{K_f \times w_2 \times 1000}{M_2 \times w_1} \\ 0.48 &= \frac{5.12 \times 0.721 \times 1000}{M_2 \times 65.925} \\ M_2 &= \frac{5.12 \times 0.721 \times 1000}{0.48 \times 65.925} \\ M_2 &= 116.65 \text{ g/mol} \end{aligned}

End of Chapter 1: Solutions

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 2: Electrochemistry

📚 CBSE Class 12 Chemistry
Chapter 2: Electrochemistry (Complete Notes & Question Bank)

1. Electrolytic Conductivity and Kohlrausch’s Law
2. Redox Reactions and Electrochemical Cells, Electrode Potential and Nernst Equation
3. Electrolysis, Law of Electrolysis, Batteries, Fuel Cells and Corrosion

1. Electrolytic Conductivity and Kohlrausch’s Law

Basic Concepts of Conductance

Electrolytic conduction: The flow of electric current through an electrolytic solution.
Electrolyte: A substance that dissociates in the solution to produce ions and hence conducts electricity in dissolved state or molten state.
Degree of ionisation: Ratio of number of ions produced to the total number of molecules in electrolyte.

Mnemonics for Electrolytes:
• SEDC: Strong Electrolytes Dissociate-Completely (e.g., NaCl, HCl, NaOH).
• WED Prime: Weak Electrolytes Dissociate Partially (e.g., H₂CO₃, CH₃COOH, HCN).
• NEND: Non Electrolyte No dissociation.

Resistance (R) and Resistivity (ρ)

Resistance is defined as the property of given substance to obstruct the flow of charge. It is directly proportional to the length ($l$) and inversely proportional to its area of cross-section ($A$).

R \propto \frac{l}{A} \implies R = \rho \frac{l}{A}$$ $$\text{Resistivity } (\rho) = R \frac{A}{l}

SI Unit of Resistivity: Ohm-m ($\Omega \cdot m$) or Ohm-cm ($\Omega \cdot cm$).

Conductance (C) and Conductivity ($\kappa$)

Conductance: The ease with which the current flows through a conductor. It is the reciprocal of resistance.

C = \frac{1}{R} = \frac{1}{\rho} \frac{A}{l}

SI Unit: Siemens (S).

Conductivity ($\kappa$): It is the reciprocal of resistivity.

\kappa = C \times \frac{l}{A}

SI Unit: $S \cdot m^{-1}$ or $S \cdot cm^{-1}$. Depends on: (i) Nature of material, (ii) Temperature, (iii) Size of ions and their solvation.

Cell Constant (G) and Molar Conductivity ($\Lambda_m$)

\text{Cell Constant } (G) = \frac{l}{A} \quad (\text{Unit: } cm^{-1} \text{ or } m^{-1})$$ $$\Lambda_m = \frac{\kappa \times 1000}{C}

SI Unit of $\Lambda_m$: $S \cdot m^2 \cdot mol^{-1}$. Defined as the conducting power of all the ions produced by one mole of an electrolyte in a solution.

Kohlrausch’s Law & Debye Huckel Onsager Equation

\text{Debye Huckel Onsager Eq: } \Lambda_m = \Lambda_m^\circ – A\sqrt{C} \quad (\text{for strong electrolytes})

Kohlrausch’s law of independent migration of ions: Limiting molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of contributions of its individual ions.

\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ

Applications of Kohlrausch’s Law

Calculation of molar conductivities of weak electrolytes at infinite dilution.
Calculation of degree of dissociation ($\alpha$) of weak electrolytes: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$
Determination of dissociation constant ($K_a$): $K_a = \frac{C \alpha^2}{1 – \alpha}$
Determination of solubility of sparingly soluble salts: $S = \frac{\kappa \times 1000}{\Lambda_m^\circ}$

Example 1 [Board 2024]

A conductivity cell with 0.1 mol/L KCl solution has a resistance of 100 $\Omega$. If the resistance changed to 520 $\Omega$ when the KCl solution concentration changed to 0.02 mol/L, what will be the conductivity and molar conductivity of 0.02 mol/L KCl solution? (Conductivity of 0.1 mol/L KCl is 1.29 S/m).

Solution:

\begin{aligned} \kappa_1 &= 1.29 \text{ S/m} = 1.29 \times 10^{-2} \text{ S/cm} \\ \text{Cell constant } (G) &= \kappa_1 \times R_1 = 1.29 \times 10^{-2} \times 100 = 1.29 \text{ cm}^{-1} \\ \kappa_2 &= \frac{G}{R_2} = \frac{1.29}{520} = 0.00248 \text{ S/cm} = 2.48 \times 10^{-3} \text{ S/cm} \\ C &= 0.02 \text{ mol/L} = 2 \times 10^{-5} \text{ mol/cm}^3 \\ \Lambda_m &= \frac{\kappa_2 \times 1000}{C} = \frac{2.48 \times 10^{-3} \times 1000}{2 \times 10^{-5}} = 1.24 \times 10^2 \text{ S cm}^2 \text{mol}^{-1} \end{aligned}

2. Redox Reactions and Electrochemical Cells, Electrode Potential and Nernst Equation

Redox Reactions & Galvanic Cells

Redox reaction: A chemical reaction in which oxidation and reduction take place simultaneously.

Galvanic (Voltaic) cell: A device in which the redox reaction is carried out indirectly and chemical energy is converted to electrical energy.

Mnemonic: eRROR
Interpretation: Redox reaction involves both oxidation and reduction.

Visual Summary: Daniell Cell (Galvanic Cell)

Cathode (+) Cu CuSO₄

Salt Bridge (KCl in Agar-agar)

e⁻ flow

Zn → Zn²⁺ + 2e⁻ (Oxidation) Cu²⁺ + 2e⁻ → Cu (Reduction) Net: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Salt Bridge and its Functions

It is an inverted U-shaped glass tube containing a suitable salt in the form of a thick paste made in agar-agar.
Completes the inner cell circuit.
Prevents transference of electrolyte from one half-cell to the other.
Maintains the electrical neutrality of the electrolytes in the two half-cells.

Electrode Potential & Nernst Equation

Electrode Potential: Potential developed by the electrode w.r.t the standard reference electrode (SHE, $E^\circ = 0$ V).

Standard Electrode Potential ($E^\circ$): Electrode potential at 25°C, 1 bar pressure, and 1 M solution.

EMF of the cell: $E_{cell} = E_{cathode} – E_{anode}$

\text{Nernst Equation: } E_{(M^{n+}/M)} = E^\circ_{(M^{n+}/M)} – \frac{RT}{nF} \ln \frac{[M(s)]}{[M^{n+}(aq)]}$$ $$\text{At 298 K: } E_{cell} = E^\circ_{cell} – \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]}$$ $$\text{At equilibrium: } E^\circ_{cell} = \frac{0.059}{n} \log K_c

Mnemonics:
• OPIIEc: Oxidising Power Increases With Increase In $E^\circ$ Value.
• GRAA / GOAC: Good Reducing Agents form Anodes / Good Oxidising Agents form Cathodes.

Gibbs Energy

\Delta_r G^\circ = -nFE^\circ_{cell} \quad ; \quad \Delta_r G = -nFE_{cell}$$ $$\Delta G^\circ = -2.303 RT \log K_c

Note: For a cell reaction to be spontaneous, $\Delta G$ must be negative.

Example 2 [Board 2023]

Calculate the $\Delta G^\circ$ for the reaction: $Zn(s) + Cu^{2+} \rightarrow Zn^{2+}(aq) + Cu(s)$. The standard electrode potential of cell is +1.1V.

Solution:

\begin{aligned} n &= 2 \\ \Delta G^\circ &= -nFE^\circ_{cell} \\ &= -2 \times 96500 \times 1.1 \\ &= -212300 \text{ J/mol} = -212.3 \text{ kJ/mol} \end{aligned}

3. Electrolysis, Law of Electrolysis, Batteries, Fuel Cells and Corrosion

Faraday’s Laws of Electrolysis

\text{First Law: } m = Z \times I \times t$$ $$\text{Second Law: } \frac{m_1}{m_2} = \frac{E_1}{E_2}

Products of electrolysis depend on: (i) Physical state of material, (ii) Types of electrodes being used.

Batteries

Type	Description	Examples
Primary Batteries	Non-chargeable, reaction cannot be reversed.	Dry cell (Leclanche), Mercury cell
Secondary Batteries	Chargeable, involve reversible reactions.	Lead storage battery, Ni-Cd cell

Lead Storage Battery

Anode: Spongy lead ($Pb$)
Cathode: Lead packed with lead dioxide ($PbO_2$)
Electrolyte: Aqueous solution of $H_2SO_4$ (38%)

\text{Discharge Anode: } Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$$ $$\text{Discharge Cathode: } PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$$ $$\text{Overall Discharge: } Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$$ $$\text{Recharge: } 2PbSO_4(s) + 2H_2O(l) \xrightarrow{\text{charge}} Pb(s) + PbO_2(s) + 2H_2SO_4(aq)

Visual Summary: Lead Storage Battery

Anode (Pb) (-) Terminal

Cathode (PbO₂) (+) Terminal

Separators

Fuel Cells

Fuel cells: Electrical cells designed to convert the energy from the combustion of fuels (e.g., $H_2$, $CO$, $CH_4$) directly into electrical energy.

Mnemonic: FCCEE
Interpretation: Fuel Cell Converts Chemical Energy of Fuel Into Electrical Energy.

\text{Anode: } [H_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^-] \times 2$$ $$\text{Cathode: } O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)$$ $$\text{Net Reaction: } 2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

Corrosion and its Prevention

Corrosion: The process of slow conversion of metals into their undesirable compounds (usually oxides) by reaction with moisture and other gases present in the atmosphere.

Rusting of Iron

\text{Anode: } Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-$$ $$\text{Cathode: } O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$$ $$\text{Overall: } 2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$$ $$4Fe^{2+} + O_2 + 4H_2O \rightarrow 2Fe_2O_3 + 8H^+$$ $$Fe_2O_3 + xH_2O \rightarrow Fe_2O_3 \cdot xH_2O \text{ (Rust)}

Visual Summary: Prevention of Corrosion

Sacrificial Protection Galvanisation, Cathodic Protection Uses Zn or Al which corrode sacrificially

Alloying Mixing metals to improve properties e.g., Stainless Steel (Fe + Cr + Ni)

Example 3 [Board 2023]

(a) What are the different types of batteries? Give examples.
(b) Give a brief account of lead storage battery.
(c) State all the reactions involved at cathode and anode and overall reactions in lead storage battery.

Solution:

(a) Batteries are of two types: Primary and secondary. Leclanche cells and dry cells are examples of primary batteries. Lead storage batteries and Nickel-cadmium cells are examples of secondary batteries.

(b) Lead storage battery is a type of secondary cell. Spongy lead acts as anode and lead packed with lead dioxide acts as cathode. An aqueous solution of sulphuric acid (38%) is an electrolyte.

(c)
Discharge:
At anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At cathode: $PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Complete cell reaction: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$

Recharge:
At cathode: $PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq)$
At anode: $PbSO_4(s) + 2H_2O \rightarrow PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^-$
Complete cell reaction: $2PbSO_4(s) + 2H_2O(l) \xrightarrow{\text{charge}} Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$

End of Chapter 2: Electrochemistry

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 4: d- and f-Block Elements

📚 CBSE Class 12 Chemistry
Chapter 4: d- and f-Block Elements (Complete Notes & Question Bank)

1. d-Block Elements, Their Properties and Compounds
2. Potassium Dichromate ($K_2Cr_2O_7$) and Potassium Permanganate ($KMnO_4$)
3. f-Block Elements: Lanthanoids and Actinoids

1. d-Block Elements, Their Properties and Compounds

Introduction to d-Block and Transition Elements

d-block elements: The elements in which the last electron enters the d-subshell (penultimate shell). They lie in the middle of the periodic table belonging to groups 3 to 12.
Transition elements: Defined as elements which have incompletely filled d-orbitals in their ground states or in their most common oxidation state. They possess properties transitional between s-block and p-block elements.

General Electronic Configuration: $(n-1)d^{1-10} ns^{1-2}$

Different Transition Series

Series	Elements	Orbitals Filled
First (3d)	Sc (21) to Zn (30)	3d
Second (4d)	Y (39) to Cd (48)	4d
Third (5d)	La (57), Hf (72) to Hg (80)	5d
Fourth (6d)	Ac (89), Rf (104) onwards	6d

Visual Summary: Transition Series

4d Series Y to Cd (39 – 48)

5d Series La, Hf to Hg (57, 72 – 80)

6d Series Ac, Rf onwards (89, 104+)

General Configuration: $(n-1)d^{1-10} ns^{1-2}$

Mnemonics for Transition Elements:
• 3d Series: Scary Tiny Vicious Creatures Mingle (with) Fellow Cow Nilgai Cougar Zebra. (Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn)
• 4d Series: Yesterday Zora Nabbed a Monkey Tricking her Rheumatic Padosan Agnes Cadillac. (Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd)
• 5d Series: Late Harry Took Walk, Reached Office In Pajamas After an Hour. (La, Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg)

General Characteristics of Transition Elements

Physical Properties

Metallic Nature: All are metals, malleable, ductile (except Hg which is liquid), with high thermal and electrical conductivity, metallic lustre, and sonorous.
Atomic & Ionic Radii: Atomic radii are smaller than s-block but larger than p-block in a period. Radii decrease till the middle, remain constant, and then increase. Down the group, size increases, but 4d and 5d elements have almost similar sizes due to Lanthanoid Contraction. Ionic radii decrease with an increase in oxidation state.
Density & Ionisation Enthalpy: Density increases from left to right. Ionisation enthalpy generally increases across a series but shows irregular trends due to electronic configuration changes. IE decreases from 3d to 4d but increases from 4d to 5d (due to lanthanoid contraction).
Melting/Boiling Points & Enthalpy of Atomisation: High melting/boiling points (except Zn, Cd, Hg) due to strong metallic bonds and partially filled d-orbitals. High enthalpy of atomisation, maximum in the middle of the series.

Chemical Properties

Variable Oxidation States: Due to comparable energies of $ns$ and $(n-1)d$ electrons, both take part in bonding.
Catalytic Properties: Act as catalysts due to multiple oxidation states and large surface area (e.g., Fe, Co, Ni, Pt, $V_2O_5$, $MnO_2$).
Magnetic Properties: Paramagnetic substances have unpaired electrons. Magnetic moment $\mu = \sqrt{n(n+2)}$ B.M. (where $n$ = number of unpaired electrons).
Formation of Coloured Compounds: Due to $d-d$ transition of unpaired electrons absorbing visible light.
Complex Formation: Due to high charge density and availability of vacant d-orbitals.
Alloy & Interstitial Compound Formation: Similar atomic sizes allow alloy formation. Small atoms (H, B, C, N) occupy interstitial voids forming hard, high-melting, chemically inert compounds.

Oxides of Transition Metals

They form oxides like $MO, M_2O_3, MO_2, M_2O_5, MO_6$. Lower oxidation states are basic, while higher are amphoteric or acidic.

\text{MnO (Basic)} \xrightarrow{\text{Oxidation State +2}} \text{Mn}_2\text{O}_3 \text{ (Amphoteric)} \xrightarrow{\text{+3}} \text{MnO}_2 \text{ (Amphoteric)} \xrightarrow{\text{+4}} \text{Mn}_2\text{O}_7 \text{ (Acidic)}

2. Potassium Dichromate ($K_2Cr_2O_7$) and Potassium Permanganate ($KMnO_4$)

Potassium Dichromate ($K_2Cr_2O_7$)

Preparation

\text{(i) } 4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \xrightarrow{\Delta} 2Fe_2O_3 + 8Na_2CrO_4 + 8CO_2 \uparrow$$ $$\text{(ii) } 2Na_2CrO_4 + 2H^+ \xrightarrow{H^+} Na_2Cr_2O_7 + 2Na^+ + H_2O$$ $$\text{(iii) } Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl

Properties & Oxidising Action

Effect of pH: Chromate (yellow) in alkali $\rightleftharpoons$ Dichromate (orange-red) in acid.

Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O \text{ (Yellow)}$$ $$2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O \text{ (Orange-red)}

In acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Oxidises $Fe^{2+}$ to $Fe^{3+}$: $Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
Oxidises $Sn^{2+}$ to $Sn^{4+}$: $Cr_2O_7^{2-} + 3Sn^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O$
Oxidises $H_2S$ to $S$: $Cr_2O_7^{2-} + 3H_2S + 8H^+ \rightarrow 2Cr^{3+} + 3S + 7H_2O$
Oxidises $I^-$ to $I_2$: $Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$

Structures: Chromate vs Dichromate Ions

+ 2H⁺ – H₂O

Cr O O O O

Cr O O O

Dichromate ($Cr_2O_7^{2-}$) – Orange

Potassium Permanganate ($KMnO_4$)

Preparation

\text{(i) } 2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O \text{ (Green Manganate)}$$ $$\text{(ii) } 3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$$ $$\text{(iii) Lab: } 2Mn^{2+} + 5S_2O_8^{2-} + 8H_2O \rightarrow 2MnO_4^- + 10SO_4^{2-} + 16H^+

Oxidising Action

In Acidic Medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

$2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O$
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
$2MnO_4^- + 5NO_2^- + 6H^+ \rightarrow 2Mn^{2+} + 5NO_3^- + 3H_2O$

In Neutral/Alkaline Medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

$2MnO_4^- + I^- + H_2O \rightarrow IO_3^- + 2MnO_2 + 2OH^-$
$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$

Structures: Manganate vs Permanganate Ions

Oxidation

Mn O O O O Permanganate ($MnO_4^-$) – Purple

Example 1 [Board 2018/2023]

(i) For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems, $E^\circ$ values for some metals are as follows:
$Cr^{3+}/Cr^{2+} = -0.4\text{ V}$, $Mn^{3+}/Mn^{2+} = +1.5\text{ V}$, $Fe^{3+}/Fe^{2+} = +0.8\text{ V}$
Use this data to comment upon:
(a) the stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ and $Mn^{3+}$.
(b) the ease with which iron can be oxidised as compared to the similar process for either Cr or Mn metals.
(ii) What can be inferred from the magnetic moment of the complex $K_4[Mn(CN)_6]$? (Magnetic moment: 2.2 BM)

Solution:

(i)(a) $Cr^{3+}/Cr^{2+}$ has a negative reduction potential. Hence, $Cr^{3+}$ cannot be reduced to $Cr^{2+}$. $Cr^{3+}$ is most stable. $Mn^{3+}/Mn^{2+}$ have large positive $E^\circ$ values. Hence, $Mn^{3+}$ can be easily reduced to $Mn^{2+}$ which has $d^5$ state and is stable. Thus, $Mn^{3+}$ is least stable. $Fe^{3+}/Fe^{2+}$ couple has a positive $E^\circ$ value but is small. Thus, the stability of $Fe^{3+}$ is more than $Mn^{3+}$ but less stable than $Cr^{3+}$.

(i)(b) Comparing reduction potential values, $Mn^{2+}/Mn$ has the most negative value, i.e., its oxidation potential value is most positive. Thus, it is most easily oxidised. Therefore, the decreasing order for their ease of oxidation is $Mn > Cr > Fe$.

(ii) In $K_4[Mn(CN)_6]$, Mn is in +2 oxidation state ($3d^5$). Magnetic moment 2.2 indicates that it has one unpaired electron and hence forms inner orbital or low-spin complex. In the presence of $CN^-$ which is a strong ligand, hybridisation involved is $d^2sp^3$ (octahedral complex).

3. f-Block Elements: Lanthanoids and Actinoids

Introduction to f-Block Elements

f-block elements: Elements in which filling of electrons takes place in $(n-2)$ f-subshell (anti-penultimate shell). Also known as inner transition elements.

General Electronic Configuration: $(n-2)f^{1-14}(n-1)d^{0-1} ns^2$

Lanthanoids

The series involves the filling of 4f orbitals following Lanthanum $La (Z=57)$. There are 14 elements from $Ce (Z=58)$ to $Lu (Z=71)$.

Mnemonic for Lanthanoids:
Late CEO Promoted Nadia Palmer Smart to Europe so she said Goodbye to Toby a Day before Hoarding Eroding Timber in Yard of Lu.
Interpretations: La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu

Properties of Lanthanoids

Physical: Highly dense, soft, malleable, ductile metals with high melting points.
Magnetic: $La^{3+}$ and $Lu^{3+}$ are diamagnetic ($4f^0$ and $4f^{14}$). All other trivalent ions are paramagnetic.
Oxidation State: Mainly +3. Some show +2 and +4.
Colour: Trivalent ions are coloured due to $f-f$ transitions.

Lanthanoid Contraction

Due to the poor shielding effect of 4f orbitals, there is a steady decrease in the atomic and ionic radii of lanthanoids with an increase in atomic number. This is known as Lanthanoid Contraction.

Consequences:

Leads to similar physical and chemical properties among lanthanoids, making their separation difficult.
Elements of the second and third transition series (e.g., Zr and Hf) have almost similar atomic radii and hence similar properties.

Visual Summary: Lanthanoid Contraction

Atomic Number (Z = 58 to 71) Atomic Radius

Steady Decrease in Radii (Poor 4f Shielding)

Uses of Lanthanoids

Misch Metal: Alloy of Cerium (~55%), other lanthanoids (~40-43%), and traces of Fe, S, C, Si, Ca, Al. Used in lighter flints (pyrophoric).
Lanthanoid oxides used for polishing glass.
Cerium salts used in dyeing cotton and as catalysts.

Actinoids

The series involves the filling of 5f orbitals from Actinium $Ac (Z=89)$ to Lawrencium $Lr (Z=103)$. Elements beyond Uranium are transuranic (man-made, radioactive).

Mnemonic for Actinoids:
Active Thor Paid Ur Nephew for Pumpkins. Amy Came Back from California. Einstein and Fermi Made No. of Laws.
Interpretations: Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr

Properties of Actinoids

Physical: Silvery white, highly dense, electropositive metals with high melting points.
Actinoid Contraction: Gradual decrease in atomic and ionic radii due to poor shielding of 5f electrons.
Oxidation States: Common +3, but exhibit higher states (+4, +5, +6, +7) more readily than lanthanoids.
Radioactivity: All are radioactive.
Chemical Reactivity: More reactive than lanthanoids. Less reactive towards acids.

Uses of Actinoids

Thorium: Treatment of cancer, incandescent gas mantles.
Uranium: Nuclear fuel, glass industry, medicines.
Plutonium: Atomic reactors and bombs.

Differences Between Lanthanoids and Actinoids

S. No.	Lanthanoids	Actinoids
1.	4f orbital is progressively filled.	5f orbital is progressively filled.
2.	+3 is most common, along with +2 and +4.	+3 is common, but exhibits higher oxidation states (+4, +5, +6, +7).
3.	Except Pm, all are non-radioactive.	All are radioactive.
4.	Less tendency of complex formation.	Strong tendency of complex formation.
5.	Chemically less reactive.	More reactive than lanthanoids.

Example 2 [Board 2017/2023]

(i) How would you account for the following:
(a) Actinoid contraction is greater than lanthanoid contraction.
(b) Transition metals form coloured compounds.
(ii) Complete the following equation:
$2MnO_4^- + 6H^+ + 5NO_2^- \rightarrow$

Solution:

(i)(a) 5f orbitals have a poorer shielding effect than 4f orbitals. Actinoid contraction is greater than lanthanoid contraction due to less effective shielding by intervening 5f electrons.

(i)(b) Transition elements generally form coloured compounds on account of $d-d$ transition. When visible light falls on the compounds, they absorb certain radiations and reflect others. The colour observed corresponds to the complementary colour of the absorbed light.

(ii) $2MnO_4^- + 6H^+ + 5NO_2^- \rightarrow 5NO_3^- + 2Mn^{2+} + 3H_2O$

End of Chapter 4: d- and f-Block Elements

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 4.2: f-Block Elements (Detailed)

📚 CBSE Class 12 Chemistry
Chapter 4.2: f-Block Elements (Complete Notes & Question Bank)

1. General Introduction of f-Block Elements
2. Lanthanoids: Properties, Contraction, and Uses
3. Actinoids: Properties, Contraction, and Uses
4. Differences Between Lanthanoids and Actinoids
5. Solved Board Questions & Examples

1. General Introduction of f-Block Elements

f-block elements: The elements in which filling of electrons takes place in $(n-2)$ f-sub-shell, which belongs to anti-penultimate (third to the outermost) energy shell.

This block consists of two series of elements known as Lanthanoids and Actinoids. These elements are also known as inner transition elements.

General Electronic Configuration:

(n-2)f^{1-14}(n-1)d^{0-1}ns^2

For lanthanoids, $n$ is 6, while its value is 7 for actinoids. There are many exceptions in the electronic configuration.

Visual Summary: f-Block Elements Overview

Actinoids (5f) Th (Z=90) to Lr (Z=103) Filling of 5f orbitals n = 7

Inner Transition Elements (f-Block)

2. Lanthanoids: Properties, Contraction, and Uses

Lanthanoids: The series involves the filling of 4f orbitals following lanthanum $La (Z=57)$ is called the lanthanoid series. There are 14 elements in this series, starting with $Ce (Z=58)$ to $Lu (Z=71)$.

Electronic Configuration

[Xe] 4f^{1-14} 5d^{0-1} 6s^2

Physical Properties

(i) Highly dense metals, soft, malleable and ductile.
(ii) High melting point.
(iii) Forms alloys easily with other metals.
(iv) Magnetic properties: Among lanthanoids, $La^{3+}$ and $Lu^{3+}$ which have $4f^0$ or $4f^{14}$ electronic configurations are diamagnetic and all other trivalent lanthanoid ions are paramagnetic due to the presence of unpaired electrons.
(v) Atomic and ionic sizes: With increasing atomic number, the atomic and ionic radii decrease from one element to the other, but the decrease is very small. Due to poor shielding effect of f orbitals, a steady decrease in the size of lanthanoids with increase in atomic number is known as lanthanoid contraction.

Consequences of Lanthanoid Contraction

(a) It leads to similar physical and chemical properties among lanthanoids.
(b) Zr and Hf have same properties, due to similar atomic radii.
(c) Chemical separation of lanthanoids becomes difficult.

Visual Summary: Lanthanoid Contraction

Atomic Number (Z = 58 to 71) Atomic Radius

Steady Decrease in Radii (Poor 4f Shielding)

Chemical Properties & Other Features

Oxidation state: They mainly show +3 oxidation state. Some elements show +2 and +4 oxidation states.
Colour: Some of the trivalent ions are coloured. This happens due to the absorption in the visible region of the spectrum, resulting in $f-f$ transitions.
Chemical properties: All lanthanoids are highly electropositive metals and have almost similar chemical reactivity.

Uses of Lanthanoids

(i) Misch metal is the alloy of cerium (about 55%) and various other Lanthanoid elements (40-43%). It also contains iron up to 5% and traces of sulphur, carbon, silicon, calcium and aluminium. It is a pyrophoric material, hence, it is used in lighter flints.
(ii) Lanthanoid oxides are used for polishing glass.
(iii) Cerium salts are used in dyeing cotton and also as catalysts.
(iv) Lanthanoid compounds are used as catalysts for hydrogenated dehydrogenation and petroleum cracking.
(v) Pyrophoric alloys are used for making tracer bullets and shells.

Mnemonic for Lanthanoids Series:
Late CEO Promoted Nadia Palmer Smart to Europe so she said Goodbye to Toby a Day before Hoarding Eroding Timber in Yard of Lu.
Interpretations: La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu

3. Actinoids: Properties, Contraction, and Uses

Actinoids: The series involving the filling of 5f orbitals from actinium, $Ac (Z=89)$ up to lawrencium, $Lr (Z=103)$, comprises actinoids. All the elements beyond uranium are known as transuranic or man-made elements. These elements do not occur in nature because their half-life periods are so short.

Electronic Configuration

[Rn] 5f^{1-14} 6d^{0-1} 7s^2

Physical Properties

(i) Highly dense metals and form alloys with other metals.
(ii) Silvery white metals.
(iii) Highly electropositive.
(iv) High melting point.
(v) Ionic and atomic radii: The atomic and ionic size decreases with an increase in atomic number due to actinoid contraction. The electrons are added to 5f shell, resulting in an increase in the nuclear charge, causing the shell to shrink inwards. This is known as actinoid contraction.
(vi) Colour: Many actinoid ions are coloured.
(vii) Magnetic properties: Many actinoid ions are paramagnetic.
(viii) Oxidation State: The common oxidation state exhibited is +3. They also exhibit oxidation states of +4, +5, +6 and +7.
(ix) Many elements are radioactive.

Chemical Reactivity & Uses

Chemical reactivity: Less reactive towards acids.
Uses:
- (i) Thorium is used in the treatment of cancer and in incandescent gas mantles.
- (ii) Uranium is used in the glass industry, in medicines and as nuclear fuel.
- (iii) Plutonium is used in atomic reactors and in atomic bombs.

Mnemonic for Actinoids Series:
Active Thor Paid Ur Nephew for Pumpkins.
Amy Came Back from California.
Einstein and Fermi Made No. of Laws.
Interpretations: Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr

4. Differences Between Lanthanoids and Actinoids

S. No.	Lanthanoids	Actinoids
(i)	4f orbital is progressively filled.	5f orbital is progressively filled.
(ii)	+3 oxidation state is most common along with +2 and +4.	+3 oxidation state is most common but exhibits higher oxidation states of +4, +5, +6 and +7.
(iii)	Except for promethium, all are non-radioactive.	All are radioactive.
(iv)	Less tendency of complex formation.	Strong tendency of complex formation.
(v)	Chemically less reactive than actinoids.	More reactive than lanthanoids.

5. Solved Board Questions & Examples

Example 2 [Board 2017]

ANS

(i)(a) 5f orbitals have poor shielding effect than 4f orbitals. Actinoid contraction is greater than lanthanoid contraction due to less effective shielding by intervening 5f electrons.

(i)(b) Transition elements generally form coloured compounds on account of $d-d$ transition. When the visible light falls on the compounds, they absorb certain radiations and reflect others. The colour observed corresponds to absorbed light.

(ii) The balanced chemical equation is:

2MnO_4^- + 6H^+ + 5NO_2^- \rightarrow 5NO_3^- + 2Mn^{2+} + 3H_2O

End of Chapter 4.2: f-Block Elements

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 5: Coordination Compounds

📚 CBSE Class 12 Chemistry
Chapter 5: Coordination Compounds (Complete Notes & Question Bank)

1. Introduction, Werner’s Theory & Important Definitions
2. Nomenclature of Coordination Compounds (IUPAC)
3. Isomerism in Coordination Compounds
4. Valence Bond Theory (VBT)
5. Crystal Field Theory (CFT)
6. Stability of Coordination Compounds
7. Applications of Coordination Compounds
8. Solved Board Questions & Examples

1. Introduction, Werner’s Theory & Important Definitions

Introduction

Coordination compounds (or complex compounds) are those compounds which retain their identity in the solid state as well as in the dissolved state. They contain a central metal atom or ion surrounded by a fixed number of ions or molecules called ligands.

Examples: $K_4[Fe(CN)_6]$, $[Cu(NH_3)_4]SO_4$, $[Ni(CO)_4]$, $[Co(NH_3)_6]Cl_3$

These compounds do not ionise completely in solution. The complex ion retains its identity and does not give the tests of its constituent ions.

Werner’s Theory of Coordination Compounds (1893)

Alfred Werner proposed the following postulates:

Primary valency: These are ionisable and are satisfied by negative ions. They correspond to the oxidation state of the central metal ion. They are non-directional.
Secondary valency: These are non-ionisable and are satisfied by negative ions or neutral molecules. They correspond to the coordination number of the central metal ion. They are directional in nature and determine the geometry of the complex.
Every metal has a fixed number of secondary valencies. The ions or molecules satisfying the secondary valencies are directly attached to the metal and are shown within square brackets $[\;]$.
The primary valencies are shown outside the square brackets and are ionisable.

Visual Summary: Werner’s Theory Illustrated with $[Co(NH_3)_6]Cl_3$

Co³⁺

NH₃

Cl⁻ Cl⁻ Cl⁻

Primary Valency (Ionisable, Oxidation State = 3)

Secondary Valency (Non-ionisable, Coordination No. = 6)

Important Definitions

Coordination entity: A central metal atom or ion bonded to a fixed number of ions or molecules. e.g., $[Co(NH_3)_6]^{3+}$, $[Fe(CN)_6]^{4-}$.
Central atom/ion: The metal atom or ion to which ligands are bonded. e.g., Co in $[Co(NH_3)_6]^{3+}$.
Ligands: Ions or molecules bonded to the central metal atom/ion. They are Lewis bases (electron pair donors).
- Unidentate: Donate one electron pair (e.g., $Cl^-$, $NH_3$, $H_2O$, $CN^-$)
- Didentate: Donate two electron pairs (e.g., ethylenediamine (en), oxalate $C_2O_4^{2-}$)
- Polydentate: Donate multiple electron pairs (e.g., EDTA is hexadentate)
- Ambidentate: Can donate through two different atoms (e.g., $NO_2^-$ via N or O; $SCN^-$ via S or N; $CN^-$ via C or N)
- Chelating ligand: Didentate or polydentate ligand forming a ring structure with the metal ion. e.g., en, EDTA.
Coordination number: The total number of ligand donor atoms directly bonded to the central metal ion. e.g., in $[PtCl_6]^{2-}$, CN = 6; in $[Ni(NH_3)_4]^{2+}$, CN = 4.
Coordination sphere: The central metal ion and the ligands enclosed in square brackets $[\;]$.
Coordination polyhedron: The spatial arrangement of ligand atoms around the central metal. Common: octahedral, tetrahedral, square planar.
Oxidation number: The charge the central atom would carry if all ligands were removed along with the electron pairs shared. e.g., in $[Fe(CN)_6]^{4-}$: $x + 6(-1) = -4 \Rightarrow x = +2$.
Homoleptic complex: Metal bonded to only one kind of ligand. e.g., $[Co(NH_3)_6]^{3+}$.
Heteroleptic complex: Metal bonded to more than one kind of ligand. e.g., $[Co(NH_3)_4Cl_2]^+$.

Double Salts vs Coordination Compounds

Property	Double Salt	Coordination Compound
Dissociation	Completely dissociates into simple ions in solution	Does not completely dissociate; complex ion retains identity
Tests for ions	Gives tests for all constituent ions	Does not give tests for ions within coordination sphere
Example	$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ (Potash alum)	$K_4[Fe(CN)_6]$ (does not give test for $Fe^{2+}$)

2. Nomenclature of Coordination Compounds (IUPAC)

Rules for Writing Formulae

The central atom is listed first.
Ligands are listed in alphabetical order (based on ligand name, not prefix).
The entire coordination entity is enclosed in square brackets.
Charge of the complex ion is shown as a superscript outside the bracket.

Rules for Naming

Cation named first, then anion (as in simple salts).
Ligands named first (in alphabetical order), then the central metal.
Anionic ligands end in -o: $Cl^-$ → chlorido, $CN^-$ → cyanido, $OH^-$ → hydroxido, $C_2O_4^{2-}$ → oxalato.
Neutral ligands: retain their name. Exceptions: $H_2O$ → aqua, $NH_3$ → ammine, $CO$ → carbonyl, $NO$ → nitrosyl.
Prefixes: di-, tri-, tetra-, penta-, hexa- for simple ligands. bis-, tris-, tetrakis- for ligands already containing a numerical prefix or polydentate ligands.
Oxidation state of the metal is indicated by Roman numeral in parentheses.
If the complex is an anion, the metal name ends in -ate: Fe → ferrate, Cu → cuprate, Ag → argentate, Co → cobaltate, Ni → nickelate, Pt → platinate, Au → aurate.

Common Ligand Names Table

Ligand Formula	Ligand Name	Type
$H_2O$	Aqua	Neutral, unidentate
$NH_3$	Ammine	Neutral, unidentate
$CO$	Carbonyl	Neutral, unidentate
$en$	Ethylenediamine	Neutral, didentate
$Cl^-$	Chlorido	Anionic, unidentate
$CN^-$	Cyanido	Anionic, ambidentate
$OH^-$	Hydroxido	Anionic, unidentate
$C_2O_4^{2-}$	Oxalato	Anionic, didentate
$EDTA^{4-}$	Ethylenediaminetetraacetato	Anionic, hexadentate
$NO_2^-$	Nitrito-N (via N) / Nitrito-O (via O)	Anionic, ambidentate
$SCN^-$	Thiocyanato-S / Thiocyanato-N	Anionic, ambidentate

Example 1: IUPAC Naming

Write the IUPAC names of the following:

(a) $[Co(NH_3)_6]Cl_3$ (b) $K_3[Fe(CN)_6]$ (c) $[Pt(NH_3)_2Cl_2]$ (d) $[Ni(CO)_4]$

Solution:

(a) Hexaamminecobalt(III) chloride

(b) Potassium hexacyanidoferrate(III)

(d) Tetracarbonylnickel(0)

3. Isomerism in Coordination Compounds

Visual Summary: Types of Isomerism in Coordination Compounds

Structural Isomerism

• Ionisation • Linkage • Coordination • Solvate / Hydrate

Stereoisomerism

• Geometrical (cis-trans) • Optical (enantiomers)

Geometrical: $[Pt(NH_3)_2Cl_2]$ cis (same side) and trans (opposite side) Shown by CN=4 (sq. planar) & CN=6

Optical: $[Co(en)_3]^{3+}$ Non-superimposable mirror images (d- and l- forms, chiral complexes)

Structural Isomerism

(a) Ionisation Isomerism

Arises when the counter ion in a complex salt is itself a potential ligand and can exchange places with a ligand.

Example: $[Co(NH_3)_5Br]SO_4$ (violet, gives test for $SO_4^{2-}$) and $[Co(NH_3)_5SO_4]Br$ (red, gives test for $Br^-$).

(b) Linkage Isomerism

Arises from the presence of an ambidentate ligand.

Example: $[Co(NH_3)_5(NO_2)]Cl_2$ (yellow, nitro via N) and $[Co(NH_3)_5(ONO)]Cl_2$ (red, nitrito via O).

(c) Coordination Isomerism

Arises from interchange of ligands between cationic and anionic entities of different metal ions.

Example: $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$.

(d) Solvate (Hydrate) Isomerism

Differs in the number of water molecules directly coordinated to the metal ion vs. present as free solvent.

Example: $CrCl_3 \cdot 6H_2O$ has three isomers:

$[Cr(H_2O)_6]Cl_3$ (violet, 3 Cl⁻ ions precipitated)
$[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ (blue-green, 2 Cl⁻ ions)
$[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$ (dark green, 1 Cl⁻ ion)

Stereoisomerism

(a) Geometrical Isomerism

Arises due to different spatial arrangements of ligands around the central metal. Shown by:

Square planar complexes (CN = 4): Type $MA_2B_2$ shows cis and trans. e.g., $[Pt(NH_3)_2Cl_2]$ (cisplatin and transplatin).
Octahedral complexes (CN = 6): Types $MA_4B_2$, $MA_3B_3$ (fac and mer isomers).

Note: Tetrahedral complexes do NOT show geometrical isomerism because all positions are equivalent.

(b) Optical Isomerism

Arises when a complex has non-superimposable mirror images (enantiomers). The complex must lack a plane of symmetry (chiral).

Examples: $[Co(en)_3]^{3+}$, $[PtCl_2(en)_2]^{2+}$ (cis form only), $[Fe(C_2O_4)_3]^{3-}$.

d-form (dextrorotatory): Rotates plane-polarised light to the right.
l-form (laevorotatory): Rotates plane-polarised light to the left.
Racemic mixture: Equal amounts of d and l forms; optically inactive.

4. Valence Bond Theory (VBT)

Proposed by Linus Pauling. Main postulates:

The central metal ion provides empty orbitals for bonding with ligands.
The number of empty orbitals provided equals the coordination number.
These orbitals undergo hybridisation to form hybrid orbitals of equivalent energy.
Ligands donate electron pairs into these hybrid orbitals forming coordinate bonds.
If inner d-orbitals ($(n-1)d$) are used → inner orbital complex (low spin). If outer d-orbitals ($nd$) are used → outer orbital complex (high spin).

Hybridisation and Geometry

Coordination No.	Hybridisation	Geometry	Example
2	$sp$	Linear	$[Ag(NH_3)_2]^+$
4	$sp^3$	Tetrahedral	$[NiCl_4]^{2-}$, $[Zn(NH_3)_4]^{2+}$
4	$dsp^2$	Square planar	$[Ni(CN)_4]^{2-}$, $[PtCl_4]^{2-}$
6	$d^2sp^3$	Octahedral (inner)	$[Fe(CN)_6]^{4-}$, $[Co(NH_3)_6]^{3+}$
6	$sp^3d^2$	Octahedral (outer)	$[FeF_6]^{3-}$, $[CoF_6]^{3-}$

Magnetic Properties from VBT

\mu = \sqrt{n(n+2)} \text{ B.M.}

where $n$ = number of unpaired electrons. If $n = 0$ → diamagnetic; if $n > 0$ → paramagnetic.

Limitations of VBT

Does not explain the colour of complexes.
Does not give quantitative interpretation of magnetic data.
Does not distinguish between weak and strong ligands.
Does not explain the relative energies of different geometries.

Example 2: VBT Application

Using VBT, explain the geometry and magnetic behaviour of $[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$.

Solution:

$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand, so it causes pairing of electrons. Two $3d$ electrons pair up, freeing one $3d$ orbital. Hybridisation: $dsp^2$ → Square planar. All electrons paired → Diamagnetic ($\mu = 0$ BM).

$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand, so no pairing occurs. Hybridisation: $sp^3$ → Tetrahedral. Two unpaired electrons → Paramagnetic ($\mu = \sqrt{2(2+2)} = 2.83$ BM).

5. Crystal Field Theory (CFT)

CFT is an electrostatic model that considers the metal-ligand bond as purely ionic. Ligands are treated as point charges (anionic) or point dipoles (neutral).

The five d-orbitals of the metal ion, which are degenerate in a free ion, split into sets of different energies in the presence of ligands. This is called crystal field splitting.

Crystal Field Splitting in Octahedral Complexes

The five d-orbitals split into two sets:

$t_{2g}$ set (lower energy): $d_{xy}$, $d_{yz}$, $d_{zx}$ — directed between the axes.
$e_g$ set (higher energy): $d_{x^2-y^2}$, $d_{z^2}$ — directed along the axes (towards ligands).

The energy difference between the two sets is called crystal field splitting energy ($\Delta_o$).

Crystal Field Splitting in Octahedral Complex

$e_g$ $d_{x²-y²}, d_{z²}$

Barycentre

$t_{2g}$ $d_{xy}, d_{yz}, d_{zx}$

$\Delta_o$

+0.6$\Delta_o$ −0.4$\Delta_o$

Crystal Field Splitting in Tetrahedral Complexes

The splitting pattern is reversed compared to octahedral:

$e$ set (lower energy): $d_{x^2-y^2}$, $d_{z^2}$
$t_2$ set (higher energy): $d_{xy}$, $d_{yz}$, $d_{zx}$

\Delta_t = \frac{4}{9} \Delta_o

Since $\Delta_t$ is small, tetrahedral complexes are almost always high spin.

Spectrochemical Series

I^- < Br^- < S^{2-} < SCN^- < Cl^- < N_3^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < CH_3CN < py < NH_3 < en < bipy < phen < NO_2^- < PPh_3 < CN^- < CO

Increasing field strength →

Weak field ligands (left): Small $\Delta_o$, high spin complexes, outer orbital.
Strong field ligands (right): Large $\Delta_o$, low spin complexes, inner orbital.

High Spin vs Low Spin Complexes (Octahedral, $d^4$ to $d^7$)

Condition	High Spin (Weak Field)	Low Spin (Strong Field)
Energy comparison	$\Delta_o < P$ (pairing energy)	$\Delta_o > P$
Electron filling	Electrons occupy $e_g$ before pairing in $t_{2g}$	Electrons pair in $t_{2g}$ before entering $e_g$
Unpaired electrons	Maximum	Minimum
Magnetic moment	Higher	Lower
Example ($d^6$)	$[FeF_6]^{3-}$ (5 unpaired e⁻)	$[Fe(CN)_6]^{3-}$ (1 unpaired e⁻)

Colour of Coordination Compounds (CFT Explanation)

When visible light falls on a complex, electrons in the lower $t_{2g}$ level absorb a photon of energy equal to $\Delta_o$ and get excited to the $e_g$ level. This is called a $d-d$ transition. The colour observed is the complementary colour of the absorbed wavelength.

\Delta_o = h\nu = \frac{hc}{\lambda}

Complexes with $d^0$ or $d^{10}$ configuration are colourless (no $d-d$ transition possible). e.g., $Sc^{3+}$ ($d^0$), $Zn^{2+}$ ($d^{10}$), $Ti^{4+}$ ($d^0$).

6. Stability of Coordination Compounds

The stability of a complex in solution is expressed by its stability constant (formation constant) $K$ or $\beta$.

For the reaction: $M + nL \rightleftharpoons [ML_n]$

\beta_n = \frac{[ML_n]}{[M][L]^n}

Higher the value of $\beta$, greater the stability of the complex.

Factors Affecting Stability

Charge on metal ion: Higher charge → greater stability.
Size of metal ion: Smaller size → greater stability (for same charge).
Nature of ligand: Strong field ligands form more stable complexes. Chelating ligands form more stable complexes than unidentate ligands (chelate effect).
CFSE: Higher crystal field stabilisation energy → greater stability.

Chelate Effect: Complexes containing chelate rings (formed by polydentate ligands) are more stable than similar complexes with unidentate ligands. This is due to an increase in entropy ($\Delta S > 0$) when a chelating ligand replaces multiple unidentate ligands.

7. Applications of Coordination Compounds

S. No.	Application	Example
(i)	Electroplating	$K[Ag(CN)_2]$ for silver plating; $K[Au(CN)_2]$ for gold plating
(ii)	Estimation of hardness of water	EDTA is used for estimation of $Ca^{2+}$ and $Mg^{2+}$ in hard water
(iii)	Extraction of metals	Silver and gold extracted by treating Zn with their cyanide complexes
(iv)	Medicine / Cancer treatment	cis-platin $[Pt(NH_3)_2Cl_2]$ used as anti-tumour agent
(v)	Removal of toxic metals	EDTA removes Pb by forming Pb-EDTA complex, eliminated in urine
(vi)	Qualitative / Quantitative analysis	$Ni^{2+}$ tested and estimated by DMG (dimethylglyoxime); $Fe^{3+}$ by thiocyanate
(vii)	Oxidation reactions	Some ligands oxidise $Co^{2+}$ to $Co^{3+}$ ion
(viii)	Photography	$[Ag(S_2O_3)_2]^{3-}$ formed during fixing of photographs
(ix)	Biological systems	Chlorophyll (Mg complex), Haemoglobin (Fe complex), Vitamin B₁₂ (Co complex)

8. Solved Board Questions & Examples

Example 3 [Board 2023]

For the complex $[Fe(en)_2Cl_2]Cl$, identify the following:
(i) Oxidation number of iron
(ii) Hybrid orbitals and shape of the complex
(iii) Magnetic behaviour
(iv) Number of geometrical isomers
(v) Is there any optical isomer?
(vi) Name of the complex
(Atomic number of Fe = 26)

Solution:

(i) Let oxidation state of Fe = $x$. $en$ is neutral, $Cl^-$ has $-1$ charge.

x + 0 + 2(-1) + (-1) = 0 \implies x = +3

(ii) $Fe^{3+}$ is $3d^5$. $en$ is a strong field ligand → pairing occurs. One $3d$ orbital is freed. Hybridisation: $d^2sp^3$ → Octahedral shape.

(iii) After pairing in $d^5$: one unpaired electron remains → Paramagnetic. $\mu = \sqrt{1(1+2)} = 1.73$ BM.

(iv) Two geometrical isomers: cis and trans.

(v) The cis isomer is optically active (exists as d and l forms). The trans isomer is optically inactive (has a plane of symmetry).

(vi) IUPAC name: Dichloridobis(ethylenediamine)iron(III) chloride.

Example 4 [Board 2024]

(a) What is the spin-only magnetic moment of $[MnBr_4]^{2-}$? Predict its geometry. (Atomic number of Mn = 25)
(b) Write the IUPAC name of $[Co(NH_3)_5Cl]Cl_2$.

Solution:

(a) $Mn^{2+}$ is $3d^5$. $Br^-$ is a weak field ligand → no pairing. All 5 electrons remain unpaired.

\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} = 5.92 \text{ BM}

Coordination number = 4, no pairing → $sp^3$ hybridisation → Tetrahedral geometry.

(b) Pentaamminechloridocobalt(III) chloride.

Example 5 [Board 2023]

Explain the following:
(a) $[NiCl_4]^{2-}$ is paramagnetic while $[Ni(CO)_4]$ is diamagnetic though both are tetrahedral.
(b) $[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic whereas $[Fe(CN)_6]^{3-}$ is weakly paramagnetic.

Solution:

(a) In $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand → no pairing → 2 unpaired electrons → paramagnetic. In $[Ni(CO)_4]$: $Ni$ is in zero oxidation state ($3d^8 4s^2$). $CO$ is a strong field ligand → causes pairing of all electrons and $4s$ electrons shift to $3d$ → $3d^{10}$ → 0 unpaired electrons → diamagnetic.

(b) In $[Fe(H_2O)_6]^{3+}$: $Fe^{3+}$ is $3d^5$. $H_2O$ is a weak field ligand → no pairing → 5 unpaired electrons → strongly paramagnetic ($\mu = 5.92$ BM). In $[Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand → pairing occurs → 1 unpaired electron → weakly paramagnetic ($\mu = 1.73$ BM).

Example 6 [Board 2023]

What type of isomerism is shown by the following complexes? Write their isomers.
(a) $[Co(NH_3)_5(NO_2)]^{2+}$
(b) $[Co(en)_3]^{3+}$

Solution:

(a) Linkage isomerism (due to ambidentate ligand $NO_2^-$):

$[Co(NH_3)_5(NO_2)]^{2+}$ → pentaamminenitrito-N-cobalt(III) (yellow)
$[Co(NH_3)_5(ONO)]^{2+}$ → pentaamminenitrito-O-cobalt(III) (red)

(b) Optical isomerism. $[Co(en)_3]^{3+}$ exists as two non-superimposable mirror images: d-form (dextrorotatory) and l-form (laevorotatory).

Example 7 [Board 2024]

Using Crystal Field Theory, explain why $[Ti(H_2O)_6]^{3+}$ is coloured. What happens when it is heated?

Solution:

$Ti^{3+}$ has $3d^1$ configuration. In an octahedral field, the single d-electron occupies a $t_{2g}$ orbital. When visible light falls on the complex, this electron absorbs energy equal to $\Delta_o$ and gets excited to the $e_g$ level ($d-d$ transition). The absorbed wavelength corresponds to green-yellow light, so the transmitted (observed) colour is purple/violet.

On heating, the complex loses water molecules and the crystal field splitting changes, leading to a change in colour. If all water is removed, the anhydrous salt may become colourless as the octahedral field is destroyed.

Example 8 [Board 2023]

Compare the stability of $[Co(NH_3)_6]^{3+}$ and $[Co(en)_3]^{3+}$. Give reason.

Solution:

$[Co(en)_3]^{3+}$ is more stable than $[Co(NH_3)_6]^{3+}$.

Reason: $en$ (ethylenediamine) is a didentate chelating ligand, while $NH_3$ is a unidentate ligand. Chelate rings formed by $en$ provide extra stability due to the chelate effect. The chelate effect is an entropy-driven phenomenon: when three $en$ molecules replace six $NH_3$ molecules, the total number of particles in solution increases, leading to an increase in entropy ($\Delta S > 0$), which makes $\Delta G$ more negative and the complex more stable.

Example 9 [Board 2024]

Write the formula for the following coordination compounds:
(a) Tetraamminediaquacobalt(III) chloride
(b) Potassium tetracyanidonickelate(II)
(c) Tris(ethylenediamine)cobalt(III) chloride
(d) Amminebromidochloridonitrito-N-platinate(II)

Solution:

(a) $[Co(NH_3)_4(H_2O)_2]Cl_3$

(b) $K_2[Ni(CN)_4]$

(d) $[Pt(NH_3)BrCl(NO_2)]^-$ → as a salt, e.g., $K[Pt(NH_3)BrCl(NO_2)]$

Example 10 [Board 2023]

What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of $d^4$ in terms of $t_{2g}$ and $e_g$ in an octahedral field when:
(i) $\Delta_o > P$
(ii) $\Delta_o < P$

Solution:

Crystal field splitting energy ($\Delta_o$): The energy difference between the $t_{2g}$ and $e_g$ sets of d-orbitals in an octahedral crystal field.

(i) When $\Delta_o > P$ (strong field / low spin): Electrons prefer to pair in $t_{2g}$ rather than jump to $e_g$.

t_{2g}^4 \; e_g^0 \quad \text{(2 unpaired electrons)}

(ii) When $\Delta_o < P$ (weak field / high spin): Electrons occupy $e_g$ before pairing in $t_{2g}$.

t_{2g}^3 \; e_g^1 \quad \text{(4 unpaired electrons)}

End of Chapter 5: Coordination Compounds

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 6: Haloalkanes and Haloarenes

📚 CBSE Class 12 Chemistry
Chapter 6: Haloalkanes and Haloarenes (Complete Notes & Question Bank)

1. Haloalkanes and Haloarenes: Classifications
2. Nomenclature of Halides
3. Methods of Preparation of Haloalkanes
4. Nature of C-X Bond and Physical Properties
5. Chemical Properties of Haloalkanes
6. Mechanism of Nucleophilic Substitution (SN1 & SN2)
7. Optical Rotation and Stereoisomerism
8. Haloarenes: Preparation and Properties
9. Nature of C-X Bond in Haloarenes
10. Important Conversions and Name Reactions
11. Polyhalogen Compounds
12. Solved Board Questions & Examples

1. Haloalkanes and Haloarenes: Classifications

Haloalkanes: Aliphatic hydrocarbons where a hydrogen atom is replaced by a halogen. The halogen atom is attached to an $sp^3$ hybridised carbon atom.

Haloarenes: Aromatic hydrocarbons where a hydrogen atom in the benzene ring is replaced by a halogen. The halogen atom is attached to an $sp^2$ hybridised carbon atom.

Classification Based on Number of Halogen Atoms

Monohalides: One halogen atom (e.g., $CH_3Cl$).
Dihalides: Two halogen atoms.
- Geminal dihalides: Both halogens on the same carbon (e.g., 1,1-Dichloroethane).
- Vicinal dihalides: Halogens on adjacent carbons (e.g., 1,2-Dichloroethane).
Polyhalides: Three or more halogen atoms (e.g., $CHCl_3$, $CCl_4$).

Classification Based on Hybridisation

$sp^3$ Hybridised Carbon:
- Allylic halides: Halogen attached to $sp^3$ carbon next to $C=C$ double bond.
- Benzylic halides: Halogen attached to $sp^3$ carbon next to an aromatic ring.
- Primary (1°), Secondary (2°), Tertiary (3°) haloalkanes.
$sp^2$ Hybridised Carbon:
- Vinylic halides: Halogen attached directly to $C=C$ double bond.
- Aryl halides: Halogen attached directly to aromatic ring.

Visual Summary: Classification of Halides

sp3 Hybridised

sp2 Hybridised

Allylic Halides

Benzylic Halides

1°, 2°, 3° Alkyl

Vinylic Halides

Aryl Halides

Reactivity Order: Allylic/Benzylic > 1° > 2° > 3° (SN2) Vinylic/Aryl Halides are least reactive towards Nucleophilic Substitution

2. Nomenclature of Halides

Structure	IUPAC Name	Common Name
$CH_3Cl$	Chloromethane	Methyl chloride
$CH_3CH_2Cl$	Chloroethane	Ethyl chloride
$CH_3CH_2CH_2Cl$	1-Chloropropane	n-Propyl chloride
$CH_3CH(Cl)CH_3$	2-Chloropropane	Isopropyl chloride
$CH_2Cl_2$	Dichloromethane	Methylene dichloride
$CHCl_3$	Trichloromethane	Chloroform
$CCl_4$	Tetrachloromethane	Carbon tetrachloride
$CH_2=CHCl$	Chloroethene	Vinyl chloride
$CH_2=CH-CH_2Br$	3-Bromopropene	Allyl bromide
$C_6H_5Cl$	Chlorobenzene	Chlorobenzene
$C_6H_5CH_2Cl$	1-Chloro-1-phenylmethane	Benzyl chloride

3. Methods of Preparation of Haloalkanes

(a) From Alcohols

R-OH + HX \xrightarrow{ZnCl_2} R-X + H_2O \quad \text{(Lucas Reagent for 3°/2°)}$$ $$3R-OH + PX_3 \rightarrow 3R-X + H_3PO_3$$ $$R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl$$ $$R-OH + SOCl_2 \xrightarrow{\Delta} R-Cl + SO_2\uparrow + HCl\uparrow \quad \text{(Best method)}

(b) From Hydrocarbons (Free Radical Halogenation)

CH_3CH_2CH_2CH_3 + Cl_2 \xrightarrow{UV \text{ light}} CH_3CH_2CH_2CH_2Cl + CH_3CH_2CHClCH_3

(c) From Alkenes

\text{Addition of HX (Markovnikov): } R-CH=CH_2 + HBr \rightarrow R-CH(Br)-CH_3$$ $$\text{Addition of HBr (Anti-Markovnikov/Peroxide): } R-CH=CH_2 + HBr \xrightarrow{Peroxide} R-CH_2-CH_2Br$$ $$\text{Addition of X}_2: \quad R-CH=CH_2 + Br_2/CCl_4 \rightarrow R-CH(Br)-CH_2(Br)

(d) Halide Exchange

\text{Finkelstein: } R-X + NaI \xrightarrow{\text{Dry acetone}} R-I + NaX \quad (X = Cl, Br)$$ $$\text{Swarts: } R-Cl + AgF \rightarrow R-F + AgCl

Visual Summary: Preparation of Haloalkanes

From Alcohols HX/ZnCl₂, PX₃, PCl₅, SOCl₂

From Alkenes HX (Markovnikov), X₂

From Alkanes X₂ / UV Light

4. Nature of C-X Bond and Physical Properties

Nature of C-X Bond

The C-X bond is polar due to the difference in electronegativity. Carbon holds a partial positive charge ($\delta^+$) and the halogen holds a partial negative charge ($\delta^-$).

Bond Length & Enthalpy: Size of halogen increases down the group, so bond length increases ($C-F < C-Cl < C-Br < C-I$) and bond enthalpy decreases.

Physical Properties

Boiling/Melting Points: Higher than parent hydrocarbons due to dipole-dipole interactions. Order: $RF < RCl < RBr < RI$. For isomers, B.P. decreases with branching.
Density: Increases with increase in carbon atoms, halogen atoms, and atomic mass of halogen.
Solubility: Insoluble in water but soluble in organic solvents. [Board 2023]

5. Chemical Properties of Haloalkanes

(a) Nucleophilic Substitution Reactions

R-X + OH^- \rightarrow R-OH + X^-$$ $$R-X + NH_3 \rightarrow R-NH_2 + HX$$ $$R-X + KCN \rightarrow R-CN + KX \quad \text{(Major product: Alkyl cyanide)}$$ $$R-X + AgCN \rightarrow R-NC + AgX \quad \text{(Major product: Alkyl isocyanide)}

(b) Elimination Reactions (Dehydrohalogenation)

CH_3-CH(Br)-CH_3 + KOH_{(alc)} \xrightarrow{\Delta} CH_3-CH=CH_2 + KBr + H_2O

(c) Reaction with Metals

\text{Grignard Reagent: } R-X + Mg \xrightarrow{\text{Dry ether}} R-Mg-X$$ $$\text{Wurtz Reaction: } 2R-X + 2Na \xrightarrow{\text{Dry ether}} R-R + 2NaX

(d) Reduction

R-X + 2[H] \xrightarrow{Zn/HCl} R-H + HX

6. Mechanism of Nucleophilic Substitution (SN1 & SN2)

Feature	SN1 Mechanism	SN2 Mechanism
Steps	Two steps (Carbocation intermediate)	One step (Concerted, Transition state)
Rate Law	Rate = k[RX] (Unimolecular)	Rate = k[RX][Nu⁻] (Bimolecular)
Stereochemistry	Racemisation (Inversion + Retention)	Complete Inversion (Walden Inversion)
Order of Reactivity	3° > 2° > 1° (Carbocation stability)	1° > 2° > 3° (Steric hindrance)
Solvent	Polar protic (Water, Alcohols)	Polar aprotic (Acetone, DMSO)

Visual Summary: SN1 vs SN2 Mechanism

Step 2: Fast R⁺ + Nu⁻ → R-Nu

Results in Racemisation Favours 3° Haloalkanes

SN2 Mechanism Single Step (Concerted) Nu⁻ + R-X → [Nu···R···X]‡ (Transition State)

Back-side Attack → Nu-R + X⁻

Results in Walden Inversion Favours 1° Haloalkanes

7. Optical Rotation and Stereoisomerism

Key Concepts

Plane-Polarised Light (PPL): Light oscillating in a single plane, obtained using a Nicol prism.
Optical Activity: Ability of a substance to rotate the plane of PPL.
Dextrorotatory (d or +): Rotates PPL to the right (clockwise).
Laevorotatory (l or -): Rotates PPL to the left (anticlockwise).

Specific Rotation

[\alpha]_{\lambda}^{t} = \frac{\text{observed rotation } (\alpha_{obs})}{\text{length of tube (dm)} \times \text{concentration of solution (g mol}^{-1}\text{)}}

Chirality and Enantiomers

Chiral Carbon: Carbon attached to four different groups.
Enantiomers: Non-superimposable mirror images (e.g., d- and l-2-butanol).
Racemic Mixture: 50:50 mixture of enantiomers; optically inactive due to external compensation.
Meso Compounds: Have chiral centers but possess an internal plane of symmetry; optically inactive due to internal compensation.

Visual Summary: Chirality and Enantiomers

d-Isomer (Dextrorotatory)

l-Isomer (Laevorotatory)

Non-Superimposable

8. Haloarenes: Preparation and Properties

Preparation of Haloarenes

\text{Electrophilic Substitution: } C_6H_6 + Cl_2 \xrightarrow{FeCl_3} C_6H_5Cl + HCl$$ $$\text{From Diazonium Salts (Sandmeyer): } ArN_2^+Cl^- + CuCl/HCl \rightarrow ArCl + N_2$$ $$\text{From Diazonium Salts (Gattermann): } ArN_2^+Cl^- + Cu/HCl \rightarrow ArCl + N_2

Physical Properties

Isomeric haloarenes have similar boiling points. However, the para-isomer has a higher melting point than ortho and meta isomers due to its symmetrical structure, which fits better into the crystal lattice.

Chemical Properties

Nucleophilic Substitution: Very difficult due to partial double bond character of C-X bond. Requires extreme conditions (e.g., Dow’s Process: Chlorobenzene + NaOH at 623K, 300 atm → Phenol).
Electrophilic Substitution: Halogens are ortho/para directing but deactivating. Reactions include Halogenation, Nitration, Sulphonation, and Friedel-Crafts reactions.

9. Nature of C-X Bond in Haloarenes

Why is the C-X bond in haloarenes less reactive than in haloalkanes?

Resonance Effect: Lone pairs on halogen delocalise with the benzene ring, giving C-X partial double bond character.
Difference in Hybridisation: Carbon in haloarenes is $sp^2$ hybridised (more electronegative) compared to $sp^3$ in haloalkanes.
Instability of Phenyl Cation: Phenyl cation formed by loss of halogen is highly unstable.
Electronic Repulsion: Electron-rich nucleophile is repelled by the electron-rich benzene ring.

Visual Summary: Resonance in Chlorobenzene

Cl Structure I

↔

Cl δ⁻ Structure II

↔

Cl δ⁻ Structure III

→ Partial Double Bond

10. Important Conversions and Name Reactions

Important Name Reactions

Sandmeyer Reaction: $ArN_2^+Cl^- + CuCl/HCl \rightarrow ArCl + N_2$
Finkelstein Reaction: $R-X + NaI \xrightarrow{\text{Dry acetone}} R-I + NaX$
Wurtz Reaction: $2R-X + 2Na \xrightarrow{\text{Dry ether}} R-R + 2NaX$
Wurtz-Fittig Reaction: $R-X + Ar-X + 2Na \xrightarrow{\text{Dry ether}} R-Ar + 2NaX$
Fittig Reaction: $2Ar-X + 2Na \xrightarrow{\text{Dry ether}} Ar-Ar + 2NaX$
Friedel-Crafts Alkylation: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl$
Dow’s Process: $C_6H_5Cl + NaOH \xrightarrow{623K, 300atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$
Hunsdiecker Reaction: $RCOOAg + Br_2 \xrightarrow{CCl_4} RBr + CO_2 + AgBr$
Gattermann Reaction: $ArN_2^+Cl^- + Cu/HCl \rightarrow ArCl + N_2$

Important Conversions

\text{(i) Propene to Propan-1-ol:}$$ $$CH_3CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2Br \xrightarrow{\text{aq. KOH, }\Delta} CH_3CH_2CH_2OH

\text{(ii) Ethanol to But-2-yne:}$$ $$CH_3CH_2OH \xrightarrow{P/I_2} CH_3CH_2I \xrightarrow{\text{alc. KOH}} CH_2=CH_2 \xrightarrow{Br_2/CCl_4} CH_2Br-CH_2Br$$ $$\xrightarrow{\text{alc. KOH}} HC\equiv CH \xrightarrow{NaNH_2} Na-C\equiv C-Na \xrightarrow{2CH_3I} CH_3-C\equiv C-CH_3

\text{(iii) 1-Bromopropane to 2-Bromopropane:}$$ $$CH_3CH_2CH_2Br \xrightarrow{\text{alc. KOH}} CH_3CH=CH_2 \xrightarrow{HBr} CH_3CH(Br)CH_3

\text{(iv) Toluene to Benzyl Alcohol:}$$ $$C_6H_5CH_3 \xrightarrow{Cl_2/h\nu} C_6H_5CH_2Cl \xrightarrow{\text{aq. KOH}} C_6H_5CH_2OH

\text{(v) Benzene to 4-Bromonitrobenzene:}$$ $$C_6H_6 \xrightarrow{Br_2/FeBr_3} C_6H_5Br \xrightarrow{HNO_3/H_2SO_4} p-Br-C_6H_4-NO_2

\text{(vi) Benzyl Alcohol to 2-Phenylethanoic Acid:}$$ $$C_6H_5CH_2OH \xrightarrow{SOCl_2} C_6H_5CH_2Cl \xrightarrow{KCN} C_6H_5CH_2CN \xrightarrow{H_3O^+} C_6H_5CH_2COOH

\text{(vii) Ethanol to Propanenitrile:}$$ $$CH_3CH_2OH \xrightarrow{P/I_2} CH_3CH_2I \xrightarrow{KCN/\Delta} CH_3CH_2CN

\text{(viii) Aniline to Chlorobenzene:}$$ $$C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 273-278K} C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl

\text{(ix) 2-Chlorobutane to 3,4-Dimethylhexane:}$$ $$2CH_3CH(Cl)CH_2CH_3 \xrightarrow{2Na/\text{dry ether}} CH_3CH(CH_3)CH(CH_3)CH_2CH_3

\text{(x) 2-Methyl-1-propene to 2-Chloro-2-methylpropane:}$$ $$CH_3-C(CH_3)=CH_2 \xrightarrow{HCl} CH_3-C(Cl)(CH_3)-CH_3

\text{(xi) Ethyl Chloride to Propanoic Acid:}$$ $$CH_3CH_2Cl \xrightarrow{KCN} CH_3CH_2CN \xrightarrow{H_3O^+} CH_3CH_2COOH

\text{(xii) But-1-ene to n-Butyl Iodide:}$$ $$CH_3CH_2CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2CH_2Br \xrightarrow{NaI/\text{acetone}} CH_3CH_2CH_2CH_2I

\text{(xiii) 2-Chloropropane to 1-Propanol:}$$ $$CH_3CH(Cl)CH_3 \xrightarrow{\text{alc. KOH}} CH_3CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2Br \xrightarrow{\text{aq. KOH}} CH_3CH_2CH_2OH

\text{(xiv) Isopropyl Alcohol to Iodoform:}$$ $$(CH_3)_2CHOH \xrightarrow{I_2/NaOH, \Delta} CHI_3\downarrow + CH_3COONa

\text{(xv) Chlorobenzene to p-Nitrophenol:}$$ $$C_6H_5Cl \xrightarrow{HNO_3/H_2SO_4} p-Cl-C_6H_4-NO_2 \xrightarrow{NaOH, 623K} p-O_2N-C_6H_4-ONa \xrightarrow{H^+} p-O_2N-C_6H_4-OH

\text{(xvi) 2-Bromopropane to 1-Bromopropane:}$$ $$CH_3CH(Br)CH_3 \xrightarrow{\text{alc. KOH}} CH_3CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2Br

\text{(xvii) Chloroethane to Butane:}$$ $$2CH_3CH_2Cl \xrightarrow{2Na/\text{dry ether}} CH_3CH_2CH_2CH_3 + 2NaCl

\text{(xviii) Benzene to Diphenyl:}$$ $$2C_6H_5Br \xrightarrow{2Na/\text{dry ether}} C_6H_5-C_6H_5 + 2NaBr \quad \text{(Fittig Reaction)}

\text{(xix) tert-Butyl Bromide to iso-Butyl Bromide:}$$ $$(CH_3)_3CBr \xrightarrow{\text{alc. KOH}} (CH_3)_2C=CH_2 \xrightarrow{HBr/\text{peroxide}} (CH_3)_2CH-CH_2Br

\text{(xx) Aniline to Phenyl Isocyanide:}$$ $$C_6H_5NH_2 \xrightarrow{CHCl_3/KOH(alc)} C_6H_5NC \quad \text{(Carbylamine Reaction)}

11. Polyhalogen Compounds

Dichloromethane ($CH_2Cl_2$): Prepared by chlorination of methane. Used as a solvent in pharmaceuticals and food industry.
Chloroform ($CHCl_3$): Prepared by chlorination of methane. Used as a solvent. Formerly used as an anaesthetic, but forms poisonous phosgene gas ($COCl_2$) on exposure to light and air.
Iodoform ($CHI_3$): Prepared by heating ethanol or acetone with $I_2$ and $NaOH$ (Iodoform test). Used as an antiseptic.
Carbon Tetrachloride ($CCl_4$): Prepared from $CH_4$ or $CS_2$. Used as a solvent and in fire extinguishers (Pyrene).
DDT (p,p’-Dichlorodiphenyltrichloroethane): Prepared by condensing chlorobenzene with chloral in presence of $H_2SO_4$. Potent insecticide, but non-biodegradable and accumulates in fat tissues.

12. Solved Board Questions & Examples

Example 1 [NCERT]

Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of $(CH_3)_2CHCH_2CH_3$.

ANS

In the given molecule, there are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following isomers:

\text{(i) } (CH_3)_2CHCH_2CH_2Cl$$ $$\text{(ii) } (CH_3)_2CHCH(Cl)CH_3$$ $$\text{(iii) } (CH_3)_2C(Cl)CH_2CH_3$$ $$\text{(iv) } CH_3CH(CH_2Cl)CH_2CH_3

Example 2

Write the products of the following reactions:

(i) $CH_3-CH=CH_2 + HCl \rightarrow$

(ii) $CH_3-CH_2-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}}$

ANS

\text{(i) } CH_3-CH(Cl)-CH_3 \quad \text{(Markovnikov addition)}$$ $$\text{(ii) } CH_3-CH_2-CH_2-CH_2Br \quad \text{(Anti-Markovnikov addition)}

Example 3 [NCERT]

Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.

ANS

KCN is predominantly ionic and provides cyanide ions ($CN^-$) in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through the carbon atom and not through the nitrogen atom since the C-C bond is more stable than the C-N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate an electron pair, forming isocyanide as the main product.

Example 4 [NCERT]

In the following pairs of halogen compounds, which would undergo $S_N2$ reaction faster?

(i) 1-Chlorobutane or 2-Chlorobutane

(ii) 1-Chloro-2-methylpropane or 2-Chloro-2-methylpropane

ANS

$S_N2$ reactions are favoured by primary halides due to less steric hindrance.

(i) 1-Chlorobutane (1° halide) reacts faster than 2-Chlorobutane (2° halide).

(ii) 1-Chloro-2-methylpropane (1° halide) reacts faster than 2-Chloro-2-methylpropane (3° halide).

Example 5 [NCERT]

Predict the order of reactivity of the following compounds in $S_N1$ and $S_N2$ reactions:

The four isomeric bromobutanes: $CH_3CH_2CH_2CH_2Br$, $(CH_3)_2CHCH_2Br$, $CH_3CH_2CH(Br)CH_3$, $(CH_3)_3CBr$

ANS

\text{For } S_N1 \text{ (depends on carbocation stability):}$$ $$(CH_3)_3CBr > CH_3CH_2CH(Br)CH_3 > (CH_3)_2CHCH_2Br > CH_3CH_2CH_2CH_2Br$$ $$\text{(3° > 2° > 1°)}$$ $$\text{For } S_N2 \text{ (depends on steric hindrance):}$$ $$CH_3CH_2CH_2CH_2Br > (CH_3)_2CHCH_2Br > CH_3CH_2CH(Br)CH_3 > (CH_3)_3CBr$$ $$\text{(1° > 1° branched > 2° > 3°)}

Example 6

Account for the following:

(i) Aromatic carboxylic acids do not undergo Friedel-Crafts reactions. [Note: This is from Chapter 8, but often asked in context of deactivating groups. Let’s provide a haloarene specific example instead based on the text]

Correction based on text: Why is the C-X bond in haloarenes less reactive than in haloalkanes?

ANS

The C-X bond in haloarenes is less reactive due to:

Resonance effect: Delocalisation of lone pairs on halogen gives partial double bond character to C-X.
Difference in hybridisation: Carbon in haloarenes is $sp^2$ hybridised (more electronegative) than $sp^3$ carbon in haloalkanes.
Instability of phenyl cation: Phenyl cation formed by loss of halogen is highly unstable.
Electronic repulsion: Electron-rich nucleophile is repelled by the electron-rich benzene ring.

End of Chapter 6: Haloalkanes and Haloarenes

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 6.2: Haloarenes, Conversions & Polyhalogen Compounds

📚 CBSE Class 12 Chemistry
Chapter 6.2: Haloarenes, Conversions & Polyhalogen Compounds

1. Solved Examples: IUPAC, Isomerism & Reactions
2. Preparation of Haloarenes
3. Physical & Chemical Properties of Haloarenes
4. Nature of C-X Bond in Haloarenes
5. Important Conversions (Board Special)
6. Important Name Reactions
7. Polyhalogen Compounds
8. Solved Examples: SN1/SN2 & Nucleophilic Attack

1. Solved Examples: IUPAC, Isomerism & Reactions

Example 1 [NCERT]

Write IUPAC names of the following:

(i) 4-Bromopent-2-ene (ii) 3-Bromo-2-methylbut-1-ene (iii) 1-Bromo-3-methylpent-3-ene

(iv) 1-Bromo-2-methylbut-2-ene (v) 1-Bromobut-2-ene (vi) 1-Bromo-2-methylpropen-2-ene

ANS

The IUPAC names are as given in the question prompt based on standard IUPAC nomenclature rules for alkenyl halides.

Example 2 [NCERT]

Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of $(CH_3)_2CHCH_2CH_3$.

ANS

In the given molecule (2-methylbutane), there are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following isomers:

\text{(i) } (CH_3)_2CHCH_2CH_2Cl$$ $$\text{(ii) } (CH_3)_2CHCH(Cl)CH_3$$ $$\text{(iii) } (CH_3)_2C(Cl)CH_2CH_3$$ $$\text{(iv) } CH_3CH(CH_2Cl)CH_2CH_3

Example 3

Write the products of the following reactions:

(i) $\text{Cyclohexene} + HBr \rightarrow$

(ii) $CH_3-CH_2-CH=CH_2 + HCl \rightarrow$

(iii) $CH_2=CH-CH_2 + HBr \xrightarrow{\text{Peroxide}}$

ANS

\text{(i) Bromocyclohexane (Markovnikov addition)}$$ $$\text{(ii) } CH_3-CH_2-CH(Cl)-CH_3 \text{ (2-Chlorobutane)}$$ $$\text{(iii) } CH_3-CH_2-CH_2-CH_2Br \text{ (1-Bromobutane, Anti-Markovnikov)}

2. Preparation of Haloarenes

(i) From Benzene (Electrophilic Substitution)

C_6H_6 + X_2 \xrightarrow{Fe \text{ or } FeX_3, \text{ dark}} C_6H_5X + HX \quad (X = Cl, Br)$$ $$C_6H_6 + I_2 \xrightarrow{HNO_3 \text{ or } HIO_3} C_6H_5I + HI

(ii) From Diazonium Salts

Sandmeyer Reaction: $ArN_2^+Cl^- + Cu_2Cl_2/HCl \rightarrow ArCl + N_2 \uparrow$
Gattermann Reaction: $ArN_2^+Cl^- + Cu/HCl \rightarrow ArCl + N_2 \uparrow$
With KI: $ArN_2^+Cl^- + KI \xrightarrow{\text{warm}} ArI + KCl + N_2 \uparrow$
Balz-Schiemann Reaction: $ArN_2^+Cl^- + HBF_4 \rightarrow ArN_2^+BF_4^- \xrightarrow{\Delta} ArF + BF_3 + N_2 \uparrow$

(iii) Substitution Reactions from Chlorobenzene

Dow’s Process (-OH): $C_6H_5Cl + NaOH \xrightarrow{623K, 300atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$
Substitution by -CN: $C_6H_5Cl + CuCN \xrightarrow{DMF, 673K} C_6H_5CN + CuCl$
Substitution by -NH₂: $C_6H_5Cl + NH_3 \xrightarrow{Cu_2O, 475K, 60atm} C_6H_5NH_2 + HCl$

(iv) Reaction with Metals

Fittig Reaction: $2C_6H_5Br + 2Na \xrightarrow{\text{Dry Ether}} C_6H_5-C_6H_5 + 2NaBr$
Wurtz-Fittig Reaction: $C_6H_5Cl + CH_3Cl + 2Na \xrightarrow{\text{Dry Ether}} C_6H_5CH_3 + 2NaCl$
Grignard Reagent: $C_6H_5Cl + Mg \xrightarrow{\text{Dry Ether}} C_6H_5MgCl$
Ullmann Reaction: $2C_6H_5Cl + 2Cu \xrightarrow{\Delta} C_6H_5-C_6H_5 + 2CuCl$

Visual Summary: Preparation of Haloarenes from Diazonium Salts

Ar-Cl (Sandmeyer)

Ar-Br (Sandmeyer)

Ar-CN (CuCN/KCN)

Ar-F (Balz-Schiemann)

N₂ gas is released in all these reactions Driving force: Stability of N₂ molecule

3. Physical & Chemical Properties of Haloarenes

Physical Properties

Chemical Properties

(a) Nucleophilic Substitution Reactions

Nucleophilic substitution is very difficult in haloarenes due to the partial double bond character of the C-X bond, difference in hybridization ($sp^2$ vs $sp^3$), instability of the phenyl cation, and electronic repulsion between the nucleophile and the electron-rich benzene ring.

(b) Electrophilic Substitution Reactions

Halogens are ortho/para directing but deactivating due to the -I effect dominating the +R effect in overall reactivity, while +R effect dictates the orientation.

Halogenation: $C_6H_5Cl + Cl_2 \xrightarrow{FeCl_3} o\text{-Dichlorobenzene} + p\text{-Dichlorobenzene}$
Nitration: $C_6H_5Cl + HNO_3 \xrightarrow{H_2SO_4} o\text{-Chloronitrobenzene} + p\text{-Chloronitrobenzene}$
Sulphonation: $C_6H_5Cl + H_2SO_4 \xrightarrow{\Delta} o\text{-Chlorosulphonic acid} + p\text{-Chlorosulphonic acid}$
Friedel-Crafts Alkylation: $C_6H_5Cl + CH_3Cl \xrightarrow{AlCl_3} o\text{-Chlorotoluene} + p\text{-Chlorotoluene}$
Friedel-Crafts Acylation: $C_6H_5Cl + CH_3COCl \xrightarrow{AlCl_3} o\text{-Chloroacetophenone} + p\text{-Chloroacetophenone}$

Visual Summary: Electrophilic Substitution in Haloarenes

o o

m m

Electrophiles attack at Ortho and Para positions Due to +R effect of Halogen (increases electron density at o/p)

4. Nature of C-X Bond in Haloarenes

Why is the C-X bond in haloarenes less reactive than in haloalkanes?

Resonance Effect: The lone pair of electrons on the halogen atom delocalizes with the $\pi$-electrons of the benzene ring (+R effect), giving the C-X bond partial double bond character.
Difference in Hybridisation: The carbon atom in haloarenes is $sp^2$ hybridized (more electronegative) compared to the $sp^3$ hybridized carbon in haloalkanes, making the C-X bond shorter and stronger.
Instability of Phenyl Cation: The phenyl cation formed by the loss of the halogen atom is highly unstable, so the SN1 mechanism is not favored.
Electronic Repulsion: The benzene ring is electron-rich, which repels the incoming electron-rich nucleophile.

5. Important Conversions (Board Special)

\text{(i) Propene to Propan-1-ol:}$$ $$CH_3CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2Br \xrightarrow{\text{aq. KOH, }\Delta} CH_3CH_2CH_2OH

\text{(ii) Ethanol to But-2-yne:}$$ $$CH_3CH_2OH \xrightarrow{P/I_2, \Delta} CH_3CH_2I \xrightarrow{\text{alc. KOH}} CH_2=CH_2 \xrightarrow{Br_2/CCl_4} CH_2Br-CH_2Br$$ $$\xrightarrow{\text{alc. KOH}} HC\equiv CH \xrightarrow{NaNH_2} Na-C\equiv C-Na \xrightarrow{2CH_3I} CH_3-C\equiv C-CH_3

\text{(iii) 1-Bromopropane to 2-Bromopropane:}$$ $$CH_3CH_2CH_2Br \xrightarrow{\text{alc. KOH}} CH_3CH=CH_2 \xrightarrow{HBr} CH_3CH(Br)CH_3

\text{(iv) Toluene to Benzyl Alcohol:}$$ $$C_6H_5CH_3 \xrightarrow{Cl_2/h\nu} C_6H_5CH_2Cl \xrightarrow{\text{aq. KOH}} C_6H_5CH_2OH

\text{(v) Benzene to 4-Bromonitrobenzene:}$$ $$C_6H_6 \xrightarrow{Br_2/FeBr_3} C_6H_5Br \xrightarrow{HNO_3/H_2SO_4} p-Br-C_6H_4-NO_2

\text{(vi) Benzyl Alcohol to 2-Phenylethanoic Acid:}$$ $$C_6H_5CH_2OH \xrightarrow{SOCl_2} C_6H_5CH_2Cl \xrightarrow{KCN} C_6H_5CH_2CN \xrightarrow{H_3O^+} C_6H_5CH_2COOH

\text{(vii) Ethanol to Propanenitrile:}$$ $$CH_3CH_2OH \xrightarrow{P/I_2} CH_3CH_2I \xrightarrow{KCN/\Delta} CH_3CH_2CN

\text{(viii) Aniline to Chlorobenzene:}$$ $$C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 273-278K} C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl

\text{(ix) 2-Chlorobutane to 3,4-Dimethylhexane:}$$ $$2CH_3CH(Cl)CH_2CH_3 \xrightarrow{2Na/\text{dry ether}} CH_3CH(CH_3)CH(CH_3)CH_2CH_3

\text{(x) 2-Methyl-1-propene to 2-Chloro-2-methylpropane:}$$ $$CH_3-C(CH_3)=CH_2 \xrightarrow{HCl} CH_3-C(Cl)(CH_3)-CH_3

\text{(xi) Ethyl Chloride to Propanoic Acid:}$$ $$CH_3CH_2Cl \xrightarrow{KCN} CH_3CH_2CN \xrightarrow{H_3O^+} CH_3CH_2COOH

\text{(xii) But-1-ene to n-Butyl Iodide:}$$ $$CH_3CH_2CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2CH_2Br \xrightarrow{NaI/\text{acetone}} CH_3CH_2CH_2CH_2I

\text{(xiii) 2-Chloropropane to 1-Propanol:}$$ $$CH_3CH(Cl)CH_3 \xrightarrow{\text{alc. KOH}} CH_3CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2Br \xrightarrow{\text{aq. KOH}} CH_3CH_2CH_2OH

\text{(xiv) Isopropyl Alcohol to Iodoform:}$$ $$(CH_3)_2CHOH \xrightarrow{I_2/NaOH, \Delta} CHI_3\downarrow + CH_3COONa

\text{(xv) Chlorobenzene to p-Nitrophenol:}$$ $$C_6H_5Cl \xrightarrow{HNO_3/H_2SO_4} p-Cl-C_6H_4-NO_2 \xrightarrow{NaOH, 623K} p-O_2N-C_6H_4-ONa \xrightarrow{H^+} p-O_2N-C_6H_4-OH

\text{(xvi) 2-Bromopropane to 1-Bromopropane:}$$ $$CH_3CH(Br)CH_3 \xrightarrow{\text{alc. KOH}} CH_3CH=CH_2 \xrightarrow{HBr/\text{peroxide}} CH_3CH_2CH_2Br

\text{(xvii) Chloroethane to Butane:}$$ $$2CH_3CH_2Cl \xrightarrow{2Na/\text{dry ether}} CH_3CH_2CH_2CH_3 + 2NaCl

\text{(xviii) Benzene to Diphenyl:}$$ $$2C_6H_5Br \xrightarrow{2Na/\text{dry ether}} C_6H_5-C_6H_5 + 2NaBr \quad \text{(Fittig Reaction)}

\text{(xix) tert-Butyl Bromide to iso-Butyl Bromide:}$$ $$(CH_3)_3CBr \xrightarrow{\text{alc. KOH}} (CH_3)_2C=CH_2 \xrightarrow{HBr/\text{peroxide}} (CH_3)_2CH-CH_2Br

\text{(xx) Aniline to Phenyl Isocyanide:}$$ $$C_6H_5NH_2 \xrightarrow{CHCl_3/KOH(alc)} C_6H_5NC \quad \text{(Carbylamine Reaction)}

6. Important Name Reactions

(i) Sandmeyer Reaction: $ArN_2^+Cl^- + CuCl/HCl \rightarrow ArCl + N_2$
(ii) Finkelstein Reaction: $R-X + NaI \xrightarrow{\text{Dry acetone}} R-I + NaX$
(iii) Wurtz Reaction: $2R-X + 2Na \xrightarrow{\text{Dry ether}} R-R + 2NaX$
(iv) Wurtz-Fittig Reaction: $R-X + Ar-X + 2Na \xrightarrow{\text{Dry ether}} R-Ar + 2NaX$
(v) Fittig Reaction: $2Ar-X + 2Na \xrightarrow{\text{Dry ether}} Ar-Ar + 2NaX$
(vi) Friedel-Crafts Alkylation: $C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl$
(vii) Dow’s Process: $C_6H_5Cl + NaOH \xrightarrow{623K, 300atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$
(viii) Hunsdiecker Reaction: $RCOOAg + Br_2 \xrightarrow{CCl_4} RBr + CO_2 + AgBr$
(ix) Gattermann Reaction: $ArN_2^+Cl^- + Cu/HCl \rightarrow ArCl + N_2$

7. Polyhalogen Compounds

Dichloromethane ($CH_2Cl_2$): Prepared by chlorination of methane. Used as a solvent in pharmaceuticals and food industry.
Chloroform ($CHCl_3$): Prepared by chlorination of methane. Used as a solvent. Formerly used as an anaesthetic, but forms poisonous phosgene gas ($COCl_2$) on exposure to light and air.
Iodoform ($CHI_3$): Prepared by heating ethanol or acetone with $I_2$ and $NaOH$ (Iodoform test). Used as an antiseptic.
Carbon Tetrachloride ($CCl_4$): Prepared from $CH_4$ or $CS_2$. Used as a solvent and in fire extinguishers (Pyrene).
DDT (p,p’-Dichlorodiphenyltrichloroethane): Prepared by condensing chlorobenzene with chloral in presence of $H_2SO_4$. Potent insecticide, but non-biodegradable and accumulates in fat tissues.

Visual Summary: Common Polyhalogen Compounds

CHCl₃ Chloroform Forms Phosgene on oxidation (COCl₂)

CHI₃ Iodoform Yellow ppt Antiseptic

CCl₄ Carbon Tet. Fire Extinguisher (Pyrene)

DDT Insecticide Non-bio degradable

8. Solved Examples: SN1/SN2 & Nucleophilic Attack

Example 4 [NCERT]

In the following pairs of halogen compounds, which would undergo $S_N2$ reaction faster?

(i) 1-Chlorobutane or 2-Chlorobutane

(ii) 1-Chloro-2-methylpropane or 2-Chloro-2-methylpropane

ANS

$S_N2$ reactions are favoured by primary halides due to less steric hindrance.

(i) 1-Chlorobutane (1° halide) reacts faster than 2-Chlorobutane (2° halide).

(ii) 1-Chloro-2-methylpropane (1° halide) reacts faster than 2-Chloro-2-methylpropane (3° halide).

Example 5 [NCERT]

Predict the order of reactivity of the following compounds in $S_N1$ and $S_N2$ reactions:

The four isomeric bromobutanes: $CH_3CH_2CH_2CH_2Br$, $(CH_3)_2CHCH_2Br$, $CH_3CH_2CH(Br)CH_3$, $(CH_3)_3CBr$

ANS

\text{For } S_N1 \text{ (depends on carbocation stability):}$$ $$(CH_3)_3CBr > CH_3CH_2CH(Br)CH_3 > (CH_3)_2CHCH_2Br > CH_3CH_2CH_2CH_2Br$$ $$\text{(3° > 2° > 1°)}$$ $$\text{For } S_N2 \text{ (depends on steric hindrance):}$$ $$CH_3CH_2CH_2CH_2Br > (CH_3)_2CHCH_2Br > CH_3CH_2CH(Br)CH_3 > (CH_3)_3CBr$$ $$\text{(1° > 1° branched > 2° > 3°)}

Example 6 [NCERT]

Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.

ANS

KCN: KCN is predominantly ionic and provides cyanide ions ($CN^-$) in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through the carbon atom and not through the nitrogen atom since the C-C bond is more stable than the C-N bond. Hence, alkyl cyanide is the main product.

AgCN: AgCN is mainly covalent in nature and nitrogen is free to donate an electron pair, forming isocyanide as the main product.

End of Chapter 6.2: Haloarenes, Conversions & Polyhalogen Compounds

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 7: Alcohols, Phenols and Ethers

📚 CBSE Class 12 Chemistry
Chapter 7: Alcohols, Phenols and Ethers (Complete Notes & Question Bank)

1. Classification and Nomenclature of Alcohols and Phenols
2. Methods of Preparation of Alcohols
3. Preparation of Phenols
4. Physical Properties of Alcohols and Phenols
5. Chemical Properties of Alcohols and Phenols
6. Important Name Reactions (Kolbe’s, Reimer-Tiemann, Fries)
7. Ethers: Nomenclature and Preparation
8. Physical and Chemical Properties of Ethers
9. Solved Board Questions & Examples

1. Classification and Nomenclature of Alcohols and Phenols

Classification of Alcohols

Alcohols: Compounds where one or more hydrogen atoms of an alkane are replaced by an $-OH$ group. General formula: $C_nH_{2n+1}OH$.

(a) Based on the number of hydroxyl (-OH) groups:

Monohydric: One $-OH$ group (e.g., Ethanol).
Dihydric (Diols): Two $-OH$ groups (e.g., Ethane-1,2-diol).
Trihydric (Triols): Three $-OH$ groups (e.g., Propane-1,2,3-triol).
Polyhydric: More than three $-OH$ groups.

(b) Based on the carbon atom attached to the -OH group ($sp^3$ hybridised):

Primary (1°): $-OH$ attached to a primary carbon (e.g., $CH_3CH_2OH$).
Secondary (2°): $-OH$ attached to a secondary carbon (e.g., Propan-2-ol).
Tertiary (3°): $-OH$ attached to a tertiary carbon (e.g., 2-Methylpropan-2-ol).

(c) Allylic, Vinylic, and Benzylic Alcohols:

Allylic: $-OH$ attached to an $sp^3$ carbon next to a $C=C$ double bond (e.g., $CH_2=CH-CH_2-OH$).
Vinylic: $-OH$ attached directly to an $sp^2$ hybridised $C=C$ carbon. Highly unstable, tautomerises to aldehydes/ketones.
Benzylic: $-OH$ attached to an $sp^3$ carbon next to an aromatic ring (e.g., Benzyl alcohol, $C_6H_5CH_2OH$).

Visual Summary: Classification of Alcohols

By Number of -OH

By Carbon Type

• Monohydric (1 -OH) • Dihydric (2 -OH) • Trihydric (3 -OH) • Polyhydric (>3 -OH)

• Primary (1°), Secondary (2°), Tertiary (3°) • Allylic (next to C=C) • Benzylic (next to Benzene) • Vinylic (on C=C, unstable)

Nomenclature

Formula	Common Name	IUPAC Name
$CH_3OH$	Methyl alcohol	Methanol
$CH_3CH_2OH$	Ethyl alcohol	Ethanol
$CH_3CH_2CH_2OH$	n-Propyl alcohol	Propan-1-ol
$CH_3CH(OH)CH_3$	Isopropyl alcohol	Propan-2-ol
$(CH_3)_3COH$	tert-Butyl alcohol	2-Methylpropan-2-ol
$C_6H_5OH$	Phenol / Carbolic acid	Phenol
$C_6H_4(OH)_2$ (1,2)	Catechol	Benzene-1,2-diol
$C_6H_4(OH)_2$ (1,3)	Resorcinol	Benzene-1,3-diol
$C_6H_4(OH)_2$ (1,4)	Hydroquinone	Benzene-1,4-diol

2. Methods of Preparation of Alcohols

(1) From Haloalkanes

R-X + KOH_{(aq)} \xrightarrow{\Delta} R-OH + KX

(2) From Aldehydes and Ketones

(i) Reduction: Using $LiAlH_4$, $NaBH_4$, or $H_2$ with $Ni/Pt$ catalyst.

\text{Aldehyde} \xrightarrow{[H]} \text{Primary Alcohol}$$ $$\text{Ketone} \xrightarrow{[H]} \text{Secondary Alcohol}

(ii) Grignard Reagent:

$HCHO + RMgX \rightarrow R-CH_2-OH$ (1° Alcohol)
$R-CHO + R’MgX \rightarrow R-CH(OH)-R’$ (2° Alcohol)
$R-CO-R’ + R”MgX \rightarrow R-C(OH)(R’)-R”$ (3° Alcohol)

(3) From Alkenes

(i) Acid Catalysed Hydration (Markovnikov):

CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3

(ii) Hydroboration-Oxidation (Anti-Markovnikov):

3R-CH=CH_2 + (BH_3)_2 \xrightarrow{THF} (R-CH_2-CH_2)_3B \xrightarrow{H_2O_2, OH^-} 3R-CH_2-CH_2-OH

(iii) Oxymercuration-Demercuration: Markovnikov addition without rearrangement.

(4) Reduction of Carboxylic Acids and Esters

R-COOH \xrightarrow{(i) LiAlH_4 / Ether \ (ii) H_3O^+} R-CH_2OH

3. Preparation of Phenols

(i) From Aryl Halides (Dow’s Process)

C_6H_5Cl + NaOH \xrightarrow{623K, 300 \text{ atm}} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH

(ii) From Benzene Sulphonic Acid

C_6H_6 \xrightarrow{H_2SO_4} C_6H_5SO_3H \xrightarrow{(i) NaOH \ (ii) H^+} C_6H_5OH

(iii) From Diazonium Salts

C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\text{warm}} C_6H_5OH + N_2\uparrow + HCl

(iv) From Cumene (Industrial Method)

Cumene (isopropylbenzene) is oxidised to cumene hydroperoxide, which on acid hydrolysis gives phenol and acetone.

Visual Summary: Industrial Preparation of Phenol from Cumene

O₂, Air

Cumene Hydroperoxide C₆H₅-C(CH₃)₂-O-O-H

H⁺, Δ

Phenol

Acetone

4. Physical Properties of Alcohols and Phenols

Boiling Points

Alcohols and phenols have higher boiling points than hydrocarbons, ethers, and haloalkanes of comparable molecular masses due to intermolecular hydrogen bonding.

BP increases with an increase in the number of carbon atoms (Van der Waals forces).
BP decreases with an increase in branching (decrease in surface area).

Visual Summary: Hydrogen Bonding in Alcohols vs Ethers

H-Bond

O CH₂-CH₃

Strong Intermolecular H-Bonding Higher Boiling Point

Methoxymethane (BP = -24°C) CH₃ O CH₃

No H-Bond

No Intermolecular H-Bonding Lower Boiling Point

Solubility

Lower alcohols and phenols are soluble in water due to their ability to form hydrogen bonds with water molecules. Solubility decreases as the size of the hydrophobic alkyl/aryl group increases.

5. Chemical Properties of Alcohols and Phenols

Reactions Involving Cleavage of O-H Bond

(a) Reaction with Metals

$$2R-OH + 2Na \rightarrow 2R-ONa + H_2\uparrow$$ Alcohols react with active metals (Na, K, Al) to form alkoxides, proving their acidic nature.

(b) Esterification

R-COOH + R’-OH \xrightarrow{H^+} R-COOR’ + H_2O

Reactions Involving Cleavage of C-O Bond

Order of reactivity: 3° Alcohol > 2° Alcohol > 1° Alcohol

(a) Reaction with Hydrogen Halides (Lucas Test)

R-OH + HX \xrightarrow{ZnCl_2} R-X + H_2O

3° Alcohol: Immediate turbidity.
2° Alcohol: Turbidity after 5 minutes.
1° Alcohol: No turbidity at room temperature.

(b) Reaction with Phosphorus Halides and Thionyl Chloride

R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl$$ $$R-OH + SOCl_2 \rightarrow R-Cl + SO_2\uparrow + HCl\uparrow \quad \text{(Best method, pure product)}

Dehydration of Alcohols

Heating alcohols with conc. $H_2SO_4$ at 443 K yields alkenes.

CH_3CH_2OH \xrightarrow{\text{conc. } H_2SO_4, 443K} CH_2=CH_2 + H_2O

Mechanism: (1) Protonation of alcohol, (2) Formation of carbocation (slow, rate-determining step), (3) Elimination of proton to form alkene.

Oxidation of Alcohols

1° Alcohol: $\xrightarrow{[O]}$ Aldehyde $\xrightarrow{[O]}$ Carboxylic Acid
2° Alcohol: $\xrightarrow{[O]}$ Ketone
3° Alcohol: No reaction under mild conditions. Under vigorous conditions, C-C bonds break to form carboxylic acids with fewer carbon atoms.

Acidity of Phenols

Phenols are more acidic than alcohols and water. The phenoxide ion is stabilised by resonance (delocalisation of negative charge over the benzene ring), whereas the alkoxide ion has a localised negative charge.

\text{Acidity Order:}$$ $$\text{Carboxylic Acids} > \text{Phenols} > \text{Water} > \text{Alcohols}

Effect of Substituents:

Electron-Withdrawing Groups (e.g., $-NO_2$): Increase acidity by stabilising the phenoxide ion.
Electron-Donating Groups (e.g., $-CH_3$): Decrease acidity by destabilising the phenoxide ion.

Tests for Alcohols and Phenols

Iodoform Test: Ethanol and secondary alcohols with a methyl group adjacent to the $-OH$ carbon give a yellow precipitate of $CHI_3$ with $I_2/NaOH$.
Ferric Chloride Test: Phenols give a violet/green/red complex with neutral $FeCl_3$.

6. Important Name Reactions

(i) Kolbe’s Reaction

Phenol reacts with sodium hydroxide and carbon dioxide to form salicylic acid.

C_6H_5OH + NaOH \rightarrow C_6H_5ONa \xrightarrow{CO_2, 400K, 4-7 \text{ atm}} \text{Sodium Salicylate} \xrightarrow{H^+} \text{Salicylic Acid}

Note: Salicylic acid reacts with acetic anhydride to form Aspirin (2-Acetoxybenzoic acid).

(ii) Reimer-Tiemann Reaction

Phenol reacts with chloroform in the presence of NaOH to form salicylaldehyde.

C_6H_5OH + CHCl_3 + NaOH \xrightarrow{\Delta} \text{Salicylaldehyde (o-Hydroxybenzaldehyde)}

(iii) Fries Rearrangement

Phenyl acetate rearranges to ortho- and para-hydroxyacetophenones in the presence of Lewis acids like $AlCl_3$.

7. Ethers: Nomenclature and Preparation

Ethers: Compounds with the general formula $C_nH_{2n+2}O$, represented as $R-O-R’$.

Nomenclature

In the IUPAC system, the smaller alkyl group is treated as an alkoxy substituent, and the larger alkyl group is the parent hydrocarbon.

Compound	Common Name	IUPAC Name
$CH_3-O-CH_3$	Dimethyl ether	Methoxymethane
$CH_3-O-C_2H_5$	Ethyl methyl ether	1-Methoxyethane
$C_6H_5-O-CH_3$	Anisole	Methoxybenzene

Methods of Preparation

(i) Williamson Synthesis

Reaction of an alkyl halide with a sodium alkoxide.

R-X + R’-O^-Na^+ \rightarrow R-O-R’ + NaX

Note: To prepare unsymmetrical ethers, the alkyl halide should be primary to avoid elimination (E2) side reactions.

(ii) Dehydration of Alcohols

2C_2H_5OH \xrightarrow{\text{conc. } H_2SO_4, 413K} C_2H_5-O-C_2H_5 + H_2O

Note: Only suitable for symmetrical ethers from primary alcohols.

8. Physical and Chemical Properties of Ethers

Physical Properties

Colourless, pleasant-smelling, volatile liquids.
Lower boiling points than isomeric alcohols (no intermolecular H-bonding).
Soluble in water due to H-bonding with water molecules.

Chemical Properties

(i) Cleavage with Halogen Acids (HX)

R-O-R’ + HX \xrightarrow{373K} R-OH + R’-X

For Alkyl-Aryl Ethers (e.g., Anisole):

C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH + CH_3I

Reason: The $C-O$ bond attached to the benzene ring has partial double bond character due to resonance and is difficult to break. Hence, phenol and alkyl halide are formed.

Note: With excess HI and high temperature, phenol further reacts to form iodobenzene.

(ii) Electrophilic Substitution in Aromatic Ethers

The alkoxy group ($-OR$) is ortho/para directing and activating due to the +R effect of oxygen.

Bromination: Anisole + $Br_2$ in acetic acid $\rightarrow$ p-Bromoanisole (Major) + o-Bromoanisole (Minor).
Friedel-Crafts Alkylation: Anisole + $CH_3Cl / AlCl_3 \rightarrow$ p-Methoxytoluene (Major).
Nitration: Anisole + $HNO_3 / H_2SO_4 \rightarrow$ p-Nitroanisole (Major).

9. Solved Board Questions & Examples

Example 1 [Board 2023, 24]

The mechanism of dehydration of ethanol involves the following steps. Explain.

ANS

Step I: Formation of protonated alcohol.

CH_3-CH_2-OH + H^+ \xrightarrow{\text{Fast}} CH_3-CH_2-\overset{+}{O}H_2

Step II: Formation of carbocation. (It is the slowest step and hence, the rate-determining step of the reaction.)

CH_3-CH_2-\overset{+}{O}H_2 \xrightarrow{\text{Slow}} CH_3-CH_2^+ + H_2O

Step III: Formation of ethene by elimination of a proton.

CH_3-CH_2^+ \xrightarrow{\text{Fast}} CH_2=CH_2 + H^+

The acid used in step 1 is released in step 3. To drive the equilibrium to the right, ethene is removed as it is formed.

Example 2 [OEB]

An alcohol has the formula $C_5H_{11}OH$. Draw the structural formulae of any one of its isomers that is:

(i) a primary alcohol and has an IUPAC name based on propane

(ii) a secondary alcohol and has an IUPAC name based on butane

(iii) a tertiary alcohol

ANS

\text{(i) Primary alcohol based on propane: } 2,2\text{-Dimethylpropan-1-ol}$$ $$CH_3-C(CH_3)_2-CH_2-OH$$ $$\text{(ii) Secondary alcohol based on butane: } \text{Butan-2-ol}$$ $$CH_3-CH_2-CH(OH)-CH_3$$ $$\text{(iii) Tertiary alcohol: } 2\text{-Methylbutan-2-ol}$$ $$CH_3-C(CH_3)(OH)-CH_2-CH_3

Example 3 [Concept Applied]

To prepare n-propyl ethyl ether, Kavita heats a mixture of n-propyl alcohol and ethyl alcohol in the presence of concentrated sulphuric acid. Is this a good method to prepare the product? Give reasons for your answer.

ANS

This is not a good method for the preparation of n-propyl ethyl ether (an unsymmetrical ether).

Reason: Heating a mixture of two different alcohols with conc. $H_2SO_4$ will produce a mixture of three different ethers (n-propyl propyl ether, ethyl ethyl ether, and n-propyl ethyl ether), which would be very difficult to separate. Williamson synthesis is the preferred method for preparing unsymmetrical ethers.

End of Chapter 7: Alcohols, Phenols and Ethers

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 8: Aldehydes, Ketones and Carboxylic Acids

📚 CBSE Class 12 Chemistry
Chapter 8: Aldehydes, Ketones and Carboxylic Acids (Complete Notes & Question Bank)

1. Aldehydes and Ketones: Introduction and Nomenclature
2. Methods of Preparation of Aldehydes and Ketones
3. Physical Properties of Aldehydes and Ketones
4. Chemical Properties of Aldehydes and Ketones
5. Carboxylic Acids: Structure, Nomenclature and Properties
6. Methods of Preparation of Carboxylic Acids
7. Chemical Properties of Carboxylic Acids
8. Solved Board Questions & Examples

1. Aldehydes and Ketones: Introduction and Nomenclature

Carbonyl Group: Organic compounds containing a carbon-oxygen double bond ($>C=O$) are called carbonyl compounds. They are broadly classified into:

Aldehydes: The carbonyl group is attached to at least one hydrogen atom. General formula: $R-CHO$ (where $R$ can be $H$, alkyl, or aryl).
Ketones: The carbonyl group is attached to two carbon atoms (alkyl or aryl groups). General formula: $R-CO-R’$.

Visual Summary: Structure of Carbonyl Group

R / H R’ / H

Bond angle ≈ 120° (sp² hybridised) Polar: C is δ⁺, O is δ⁻

δ⁻ δ⁺

Nomenclature

Structural Formula	Common Name	IUPAC Name
$HCHO$	Formaldehyde	Methanal
$CH_3CHO$	Acetaldehyde	Ethanal
$CH_3COCH_3$	Acetone	Propanone
$CH_3CH_2CHO$	Propionaldehyde	Propanal
$C_6H_5CHO$	Benzaldehyde	Benzenecarbaldehyde
$C_6H_5COCH_3$	Acetophenone	1-Phenylethanone

2. Methods of Preparation of Aldehydes and Ketones

Preparation of Aldehydes

(i) From Primary Alcohols

\text{Oxidation: } R-CH_2OH \xrightarrow{K_2Cr_2O_7/H_2SO_4 \text{ or } KMnO_4} R-CHO$$ $$\text{Dehydrogenation: } R-CH_2OH \xrightarrow{Cu, 573K} R-CHO + H_2

(ii) From Hydrocarbons

Ozonolysis of alkenes: $R-CH=CH-R’ \xrightarrow{(i) O_3, (ii) Zn/H_2O} R-CHO + R’-CHO$
Hydration of alkynes: $HC \equiv CH \xrightarrow{H_2O, H_2SO_4/HgSO_4} [CH_2=CH-OH] \rightleftharpoons CH_3CHO$ (Acetaldehyde)

(iii) From Acyl Chlorides (Rosenmund Reduction)

R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl

Note: $BaSO_4$ acts as a catalyst poison to prevent further reduction to alcohol.

(iv) From Nitriles and Esters

Stephen’s Reaction: $R-CN \xrightarrow{(i) SnCl_2/HCl, (ii) H_2O} R-CHO$
DIBAL-H Reduction: $R-CN \xrightarrow{(i) DIBAL-H, (ii) H_2O} R-CHO$

Preparation of Ketones

(i) From Secondary Alcohols

\text{Oxidation: } R-CH(OH)-R’ \xrightarrow{K_2Cr_2O_7/H_2SO_4} R-CO-R’$$ $$\text{Dehydrogenation: } R-CH(OH)-R’ \xrightarrow{Cu, 573K} R-CO-R’ + H_2

(ii) From Hydrocarbons

Hydration of alkynes (Kucherov’s Reaction): $CH_3-C \equiv CH \xrightarrow{H_2O, H_2SO_4/HgSO_4} CH_3-CO-CH_3$ (Acetone)
Ozonolysis of alkenes: $R_2C=CR’_2 \xrightarrow{(i) O_3, (ii) Zn/H_2O} R_2C=O + R’_2C=O$

(iii) From Acyl Chlorides

2R-COCl + R’_2Cd \rightarrow 2R-CO-R’ + CdCl_2

(iv) Friedel-Crafts Acylation (Aromatic Ketones)

C_6H_6 + R-COCl \xrightarrow{Anhyd. AlCl_3} C_6H_5-CO-R + HCl

3. Physical Properties of Aldehydes and Ketones

Physical State: Methanal is a gas, ethanal is a volatile liquid. Higher members are liquids or solids.
Boiling Points: Higher than hydrocarbons and ethers of comparable molecular mass due to dipole-dipole interactions. However, lower than alcohols due to the absence of intermolecular hydrogen bonding.
Solubility: Lower members (up to 4 carbons) are miscible with water due to hydrogen bonding with water molecules. Solubility decreases with increasing length of the hydrophobic alkyl chain.

4. Chemical Properties of Aldehydes and Ketones

(A) Nucleophilic Addition Reactions

The carbonyl carbon is electrophilic ($\delta^+$) due to the electron-withdrawing nature of oxygen. Nucleophiles attack this carbon, changing hybridization from $sp^2$ (trigonal planar) to $sp^3$ (tetrahedral).

\text{(i) Addition of HCN: } >C=O + HCN \rightleftharpoons >C(OH)(CN) \quad \text{(Cyanohydrin)}$$ $$\text{(ii) Addition of } NaHSO_3: >C=O + NaHSO_3 \rightleftharpoons >C(OH)(SO_3Na) \quad \text{(Bisulphite adduct)}$$ $$\text{(iii) Addition of Grignard Reagent: } >C=O + R-MgX \rightarrow >C(OMgX)(R) \xrightarrow{H_2O/H^+} >C(OH)(R)$$ $$\text{(iv) Addition of Alcohols: } R-CHO + R’OH \xrightarrow{dry HCl} R-CH(OR’)(OH) \text{ (Hemiacetal)} \xrightarrow{R’OH, dry HCl} R-CH(OR’)_2 \text{ (Acetal)}

Visual Summary: Nucleophilic Addition Mechanism

C O

R / H

C Nu O⁻ R / H

Tetrahedral Intermediate

(B) Reduction Reactions

\text{To Alcohols: } >C=O \xrightarrow{NaBH_4 \text{ or } LiAlH_4 \text{ or } H_2/Ni} >CH-OH$$ $$\text{Clemmensen Reduction: } >C=O \xrightarrow{Zn(Hg), \text{ conc. } HCl} >CH_2$$ $$\text{Wolff-Kishner Reduction: } >C=O \xrightarrow{NH_2NH_2, KOH/\text{ethylene glycol}, \Delta} >CH_2 + N_2

Mnemonic for Reduction to Alkanes: Can Zebra Woo Nightingale
Interpretations:
Clemmensen uses Zn-Hg / HCl
Wolff-Kishner uses $NH_2NH_2$ / $OH^-$

(C) Oxidation Reactions

Tollens’ Test: Aldehydes reduce Tollens’ reagent (ammoniacal $AgNO_3$) to form a silver mirror. Ketones do not respond.
$R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow R-COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O$
Fehling’s Test: Aliphatic aldehydes reduce Fehling’s solution to give a red precipitate of $Cu_2O$. Aromatic aldehydes do not respond.
Haloform Reaction: Methyl ketones ($CH_3-CO-R$) react with sodium hypohalite ($NaOX$) to form a haloform ($CHX_3$) and a carboxylate salt.

(D) Reactions due to $\alpha$-Hydrogen

The $\alpha$-hydrogen atoms in aldehydes and ketones are acidic due to the strong electron-withdrawing effect of the carbonyl group.

(i) Aldol Condensation

Aldehydes/ketones with at least one $\alpha$-hydrogen react in the presence of dilute alkali to form $\beta$-hydroxy aldehydes (aldol) or $\beta$-hydroxy ketones (ketol), which on heating lose water to form $\alpha,\beta$-unsaturated carbonyl compounds.

2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O

(ii) Cannizzaro Reaction

Aldehydes without $\alpha$-hydrogen atoms undergo self-oxidation and reduction (disproportionation) on heating with concentrated alkali to yield a mixture of alcohol and carboxylic acid salt.

2HCHO + \text{conc. } KOH \xrightarrow{\Delta} HCOOK + CH_3OH$$ $$2C_6H_5CHO + \text{40\% } NaOH \xrightarrow{\Delta} C_6H_5COONa + C_6H_5CH_2OH

Mnemonic for Cannizzaro: ALCO AHA CAR
Interpretations: Aldehyde Containing No Alpha Hydrogen undergoes Cannizzaro reaction.

5. Carboxylic Acids: Structure, Nomenclature and Properties

Carboxyl Group: Consists of a carbonyl group ($>C=O$) and a hydroxyl group ($-OH$) attached to the same carbon atom. The carboxylic carbon is less electrophilic than a typical carbonyl carbon due to resonance stabilization.

Visual Summary: Resonance in Carboxyl Group

↔

C O δ⁻ O H Structure II (Resonance)

Physical Properties

Boiling Points: Higher than aldehydes, ketones, and even alcohols of comparable molecular mass due to extensive intermolecular hydrogen bonding (forming dimers).
Solubility: Lower members (up to 4 carbons) are miscible with water. Solubility decreases with increasing hydrophobic chain length.

6. Methods of Preparation of Carboxylic Acids

(i) Oxidation of Primary Alcohols and Aldehydes

R-CH_2OH \xrightarrow{KMnO_4/KOH, \text{ then } H_3O^+} R-COOH$$ $$R-CHO \xrightarrow{K_2Cr_2O_7/H_2SO_4} R-COOH

(ii) From Nitriles and Amides

R-CN \xrightarrow{H_3O^+, \Delta} R-COOH + NH_4^+$$ $$R-CONH_2 \xrightarrow{H_3O^+, \Delta} R-COOH + NH_4^+

(iii) From Grignard Reagents

R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH + Mg(OH)X

(iv) Hydrolysis of Esters

R-COOR’ + NaOH \xrightarrow{\Delta} R-COONa + R’OH \xrightarrow{H_3O^+} R-COOH

(v) Oxidation of Alkyl Benzenes

Vigorous oxidation of alkyl benzenes with chromic acid or acidic/alkaline $KMnO_4$ oxidizes the entire side chain to a carboxyl group, regardless of its length.

C_6H_5-CH_3 \xrightarrow{KMnO_4/KOH, \Delta, \text{ then } H_3O^+} C_6H_5-COOH

7. Chemical Properties of Carboxylic Acids

(A) Reactions Involving Cleavage of O-H Bond (Acidic Nature)

2R-COOH + 2Na \rightarrow 2R-COO^-Na^+ + H_2\uparrow$$ $$R-COOH + NaOH \rightarrow R-COO^-Na^+ + H_2O$$ $$R-COOH + NaHCO_3 \rightarrow R-COO^-Na^+ + H_2O + CO_2\uparrow \quad \text{(Effervescence)}

Effect of Substituents: Electron-withdrawing groups (e.g., $-NO_2$, $-Cl$) increase acidity by stabilizing the carboxylate anion via the $-I$ effect. Electron-donating groups (e.g., $-CH_3$) decrease acidity via the $+I$ effect.

(B) Reactions Involving Cleavage of C-OH Bond

(i) Formation of Anhydrides

2R-COOH \xrightarrow{P_2O_5, \Delta} (R-CO)_2O + H_2O

(ii) Esterification

R-COOH + R’-OH \xrightleftharpoons{H^+} R-COOR’ + H_2O

(iii) Reaction with $PCl_5$, $PCl_3$, and $SOCl_2$

R-COOH + PCl_5 \rightarrow R-COCl + POCl_3 + HCl$$ $$R-COOH + SOCl_2 \rightarrow R-COCl + SO_2\uparrow + HCl\uparrow \quad \text{(Best method)}

(C) Hell-Volhard-Zelinsky (HVZ) Reaction

Carboxylic acids having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus.

R-CH_2-COOH \xrightarrow{(i) X_2 / \text{Red P}, (ii) H_2O} R-CH(X)-COOH \quad (X = Cl, Br)

(D) Electrophilic Substitution (Aromatic Carboxylic Acids)

The $-COOH$ group is deactivating and meta-directing. Aromatic carboxylic acids do not undergo Friedel-Crafts reactions because the carboxyl group binds to the Lewis acid catalyst ($AlCl_3$), deactivating it.

\text{Nitration: } C_6H_5COOH \xrightarrow{conc. HNO_3 / conc. H_2SO_4} \text{m-Nitrobenzoic acid}$$ $$\text{Halogenation: } C_6H_5COOH \xrightarrow{X_2 / FeX_3} \text{m-Halobenzoic acid}

8. Solved Board Questions & Examples

Example 1 [Board 2019]

Write balanced chemical equations for the following reactions:
(i) Two molecules of propanone are treated with dilute $Ba(OH)_2$.
(ii) Acetophenone is treated with $Zn(Hg)$/conc. $HCl$.
(iii) Benzoyl chloride is hydrogenated in the presence of $Pd/BaSO_4$.

ANS

\text{(i) Aldol Condensation: } 2CH_3COCH_3 \xrightarrow{\text{dil. } Ba(OH)_2} (CH_3)_2C(OH)CH_2COCH_3$$ $$\text{(ii) Clemmensen Reduction: } C_6H_5COCH_3 \xrightarrow{Zn(Hg), \text{ conc. } HCl} C_6H_5CH_2CH_3 + H_2O$$ $$\text{(iii) Rosenmund Reduction: } C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl

Example 2 [SQP 2023-24]

Write the name of the reaction, structure, and IUPAC name of the product formed when phenol reacts with $CHCl_3$ in the presence of $NaOH$ followed by hydrolysis.

ANS

Name of Reaction: Reimer-Tiemann Reaction

\text{Product Structure: } \text{Salicylaldehyde (2-Hydroxybenzaldehyde)}$$ $$\text{IUPAC Name: 2-Hydroxybenzaldehyde}

Example 3 [Board 2018]

Account for the following:
(i) Aromatic carboxylic acids do not undergo Friedel-Crafts reactions.
(ii) The $pK_a$ value of 4-nitrobenzoic acid is lower than that of benzoic acid.

ANS

(i) The carboxyl group ($-COOH$) is strongly deactivating. Furthermore, it acts as a Lewis base and forms a complex with the Lewis acid catalyst ($AlCl_3$), thereby deactivating the catalyst and preventing the Friedel to occur.

(ii) The nitro group ($-NO_2$) is a strong electron-withdrawing group ($-I$ and $-R$ effects). It stabilizes the carboxylate anion by dispersing the negative charge, making it easier to release $H^+$. Hence, 4-nitrobenzoic acid is a stronger acid than benzoic acid, resulting in a lower $pK_a$ value.

Example 4 [Board 2020]

Write the structures of the main products when benzene diazonium chloride reacts with the following reagents:
(i) $CuCN/KCN$
(ii) $H_2O$

ANS

\text{(i) Sandmeyer Reaction: } C_6H_5N_2^+Cl^- + CuCN/KCN \rightarrow C_6H_5CN \text{ (Benzonitrile)} + N_2\uparrow + CuCl$$ $$\text{(ii) Hydrolysis: } C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH \text{ (Phenol)} + N_2\uparrow + HCl

End of Chapter 8: Aldehydes, Ketones and Carboxylic Acids

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 9: Amines

📚 CBSE Class 12 Chemistry
Chapter 9: Amines (Complete Notes & Question Bank)

1. Introduction and Classification of Amines
2. Nomenclature of Amines
3. Methods of Preparation of Amines
4. Physical and Chemical Properties (Basic Strength)
5. Chemical Reactions of Amines
6. Diazonium Salts and their Reactions
7. Identification of 1°, 2°, and 3° Amines
8. Solved Board Questions & Examples

1. Introduction and Classification of Amines

Amines: Amines are considered as amino derivatives of hydrocarbons or alkyl derivatives of ammonia. In these compounds, one, two, or three hydrogen atoms of ammonia ($NH_3$) are replaced by alkyl or aryl groups.

Aliphatic amino compounds: Called amino alkanes.
Aromatic amines: Called amino arenes (e.g., Aniline, $C_6H_5NH_2$).

Classification

Primary (1°): One H atom replaced by alkyl/aryl group ($R-NH_2$).
Secondary (2°): Two H atoms replaced ($R_2NH$).
Tertiary (3°): Three H atoms replaced ($R_3N$).

Structure of Amines

Nitrogen orbitals in amines are $sp^3$ hybridised and the geometry is pyramidal. Due to the presence of an unshared pair of electrons, the bond angle $C-N-E$ (where E is C or H) is less than the ideal tetrahedral angle of $109.5^\circ$ (e.g., in trimethylamine, it is $108^\circ$).

Visual Summary: Classification of Amines

Secondary (2°) R₂NH e.g., Dimethylamine

Tertiary (3°) R₃N e.g., Trimethylamine

2. Nomenclature of Amines

Amine	Common Name	IUPAC Name
$CH_3CH_2NH_2$	Ethylamine	Ethanamine
$CH_3CH_2CH_2NH_2$	n-Propylamine	Propan-1-amine
$CH_3CH(NH_2)CH_3$	Isopropylamine	Propan-2-amine
$CH_3-NH-CH_2CH_3$	Ethylmethylamine	N-Methylethanamine
$(CH_3)_3N$	Trimethylamine	N,N-Dimethylmethanamine
$C_6H_5NH_2$	Aniline	Benzenamine / Aniline
$C_6H_5N(CH_3)_2$	N,N-Dimethylaniline	N,N-Dimethylbenzenamine

3. Methods of Preparation of Amines

(i) By Reduction of Nitro Compounds

R-NO_2 \xrightarrow{LiAlH_4 \text{ or } H_2/Pd \text{ or } Sn+HCl} R-NH_2

(ii) By Ammonolysis of Alkyl Halides (Hoffmann Ammonolysis)

Nucleophilic substitution reaction. Yields a mixture of 1°, 2°, 3° amines and quaternary ammonium salts.

R-X + NH_3 \rightarrow R-NH_3^+X^- \xrightarrow{NaOH} R-NH_2 + NaX + H_2O

(iii) By Reduction of Nitriles

Used to obtain amines containing one carbon atom more than the starting compound.

R-C \equiv N \xrightarrow{LiAlH_4 \text{ or } H_2/Ni} R-CH_2-NH_2

Note: Aromatic primary amines cannot be prepared by this method.

(iv) Gabriel Phthalimide Synthesis

Used exclusively for the synthesis of primary (1°) amines.

\text{Phthalimide} \xrightarrow{KOH(alc)} \text{Potassium phthalimide} \xrightarrow{R-X} \text{N-alkyl phthalimide} \xrightarrow{NaOH(aq)} R-NH_2

(v) By Reduction of Amides

R-CO-NH_2 \xrightarrow{(i) LiAlH_4, \text{ ether} \ (ii) H_2O} R-CH_2-NH_2

(vi) Hoffmann Bromamide Degradation

Treating an amide with bromine in aqueous/ethanolic NaOH results in a primary amine containing one carbon atom less than the parent amide.

R-CO-NH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O

4. Physical and Chemical Properties (Basic Strength)

Physical Properties

Boiling Point: Increases with molecular weight. Order: $1^\circ > 2^\circ > 3^\circ$ (due to intermolecular H-bonding). Branching decreases B.P.
Solubility: Lower members are soluble in water due to H-bonding. Solubility decreases as the hydrophobic alkyl/aryl group increases.

Basic Strength of Amines

Amines are basic due to the lone pair of electrons on the nitrogen atom. They act as Lewis bases.

(a) In Gaseous Phase

Basicity depends only on the +I (inductive) effect of alkyl groups.

\text{Order: } 3^\circ > 2^\circ > 1^\circ > NH_3

(b) In Aqueous Phase

Basicity depends on +I effect, solvation effect (H-bonding with water), and steric hindrance.

\text{For Ethyl-substituted: } (C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$$ $$\text{For Methyl-substituted: } (CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3

(c) Aromatic Amines

Aniline and other aromatic amines are less basic than ammonia and aliphatic amines because the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring due to resonance, making it less available for protonation.

Electron-releasing groups (e.g., $-CH_3, -OCH_3$) increase basic strength.
Electron-withdrawing groups (e.g., $-NO_2, -X$) decrease basic strength.

Visual Summary: Basic Strength of Amines

In Aqueous Phase 2° > 3° > 1° > NH₃ (Ethyl) +I, Solvation & Steric effects

Aromatic Amines Aliphatic > NH₃ > Aromatic

5. Chemical Reactions of Amines

(i) Alkylation

R-NH_2 + R-X \rightarrow R_2NH \xrightarrow{R-X} R_3N \xrightarrow{R-X} R_4N^+X^-

(ii) Acylation

Reaction with acid chlorides or anhydrides to form N-substituted amides. (Only 1° and 2° amines undergo this).

R-NH_2 + CH_3COCl \rightarrow R-NH-CO-CH_3 + HCl

(iii) Carbylamine Reaction

Specific test for primary amines (both aliphatic and aromatic).

R-NH_2 + CHCl_3 + 3KOH_{(alc)} \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O

(iv) Bromination of Aniline

Aniline reacts with aqueous bromine to give a white precipitate of 2,4,6-tribromoaniline.

C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2Br_3NH_2\downarrow + 3HBr

(v) Nitration

Direct nitration of aniline yields m-nitroaniline because in acidic medium, aniline is protonated to the m-directing anilinium ion.

(vi) Sulphonation

Heating aniline with conc. $H_2SO_4$ at 453-473 K yields sulphanilic acid (zwitter ion).

6. Diazonium Salts and their Reactions

Diazonium Salts: General formula $RN_2^+X^-$. Primary aromatic amines form arene diazonium salts which are stable at low temperatures (273-278 K).

Diazotisation: The conversion of primary aromatic amines into diazonium salts.

C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

Reactions Involving Displacement of Nitrogen

Sandmeyer Reaction: $ArN_2^+Cl^- + Cu_2Cl_2/HCl \rightarrow ArCl + N_2$
Gattermann Reaction: $ArN_2^+Cl^- + Cu/HCl \rightarrow ArCl + N_2$
Replacement by CN: $ArN_2^+Cl^- + CuCN/KCN \rightarrow ArCN + N_2$
Replacement by I: $ArN_2^+Cl^- + KI \rightarrow ArI + KCl + N_2$
Balz-Schiemann Reaction (F): $ArN_2^+Cl^- + HBF_4 \rightarrow ArN_2^+BF_4^- \xrightarrow{\Delta} ArF + BF_3 + N_2$
Replacement by H: $ArN_2^+Cl^- + H_3PO_2 + H_2O \rightarrow ArH + N_2 + H_3PO_3 + HCl$
Replacement by OH: $ArN_2^+Cl^- + H_2O \xrightarrow{\Delta} ArOH + N_2 + HCl$

Reactions Involving Retention of Diazo Group

Coupling Reaction

Formation of azo dyes.

With Phenol (pH 9-10): $C_6H_5N_2^+Cl^- + C_6H_5OH \rightarrow p-HO-C_6H_4-N=N-C_6H_5$ (Orange dye)
With Aniline (pH 4-5): $C_6H_5N_2^+Cl^- + C_6H_5NH_2 \rightarrow p-H_2N-C_6H_4-N=N-C_6H_5$ (Yellow dye)

Visual Summary: Reactions of Benzene Diazonium Chloride

Cu₂Cl₂/HCl → C₆H₅Cl

CuCN/KCN → C₆H₅CN

H₂O/Δ → C₆H₅OH

H₃PO₂ → C₆H₆

C₆H₅OH (pH 9) → Azo Dye

HBF₄/Δ → C₆H₅F

7. Identification of 1°, 2°, and 3° Amines

Test	Primary (1°) Amine	Secondary (2°) Amine	Tertiary (3°) Amine
Carbylamine Test ($CHCl_3 + KOH_{alc}$)	Foul smelling isocyanide (R-NC) formed.	No reaction.	No reaction.
Hinsberg Test ($C_6H_5SO_2Cl$)	Forms sulphonamide soluble in alkali.	Forms sulphonamide insoluble in alkali.	No reaction.
Nitrous Acid Test ($HNO_2$)	Gives alcohol with effervescence of $N_2$ gas.	Gives oily nitrosoamine.	Forms salt, decomposes on heating.

8. Solved Board Questions & Examples

Example 1

Write chemical equations for the following reactions:
(i) Reaction of ethanolic $NH_3$ with $C_2H_5Cl$.
(ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of $CH_3Cl$.

ANS

\text{(i) } C_2H_5Cl + NH_3 \xrightarrow{\text{ethanol}} C_2H_5NH_3^+Cl^- \xrightarrow{NaOH} C_2H_5NH_2$$ $$C_2H_5NH_2 + C_2H_5Cl \rightarrow (C_2H_5)_2NH + HCl$$ $$(C_2H_5)_2NH + C_2H_5Cl \rightarrow (C_2H_5)_3N + HCl$$ $$(C_2H_5)_3N + C_2H_5Cl \rightarrow (C_2H_5)_4N^+Cl^-

\text{(ii) } C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2 \text{ (Benzylamine)}$$ $$C_6H_5CH_2NH_2 + 2CH_3Cl \rightarrow C_6H_5CH_2N(CH_3)_2 + 2HCl$$ $$\text{(N,N-Dimethylphenylmethanamine)}

Example 2

Write chemical equations for the following conversions:
(i) $CH_3-CH_2-Cl$ into $CH_3-CH_2-CH_2-NH_2$
(ii) $C_6H_5-CH_2-Cl$ into $C_6H_5-CH_2-CH_2-NH_2$

ANS

\text{(i) } CH_3CH_2Cl \xrightarrow{\text{Ethanolic NaCN}} CH_3CH_2CN \xrightarrow{LiAlH_4 / H_2/Ni} CH_3CH_2CH_2NH_2

\text{(ii) } C_6H_5CH_2Cl \xrightarrow{\text{Ethanolic NaCN}} C_6H_5CH_2CN \xrightarrow{LiAlH_4 / H_2/Ni} C_6H_5CH_2CH_2NH_2

Example 3

Write structures and IUPAC names of:
(i) The amide which gives propanamine by Hoffmann bromamide reaction.
(ii) The amine produced by the Hoffmann degradation of benzamide.

ANS

(i) Propanamine contains 3 carbon atoms. Hence, the parent amide must contain 4 carbon atoms.

\text{Structure: } CH_3-CH_2-CH_2-CO-NH_2$$ $$\text{IUPAC Name: Butanamide}

(ii) Benzamide is an aromatic amide containing 7 carbon atoms. The amine formed will contain 6 carbon atoms.

\text{Structure: } C_6H_5NH_2$$ $$\text{IUPAC Name: Benzenamine (or Aniline)}

End of Chapter 9: Amines

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chapter 10: Biomolecules

📚 CBSE Class 12 Chemistry
Chapter 10: Biomolecules (Complete Notes & Question Bank)

1. Carbohydrates: Classification and Importance
2. Monosaccharides: Glucose and Fructose
3. Polysaccharides: Starch, Cellulose, and Glycogen
4. Proteins: Amino Acids and Structure
5. Nucleic Acids: DNA and RNA
6. Vitamins and Hormones

1. Carbohydrates: Classification and Importance

Biomolecules: The naturally occurring organic compounds present as essential constituents of living organisms in different cells. E.g., polysaccharides, proteins, etc.

Carbohydrates: Optically active polyhydroxy aldehydes or ketones or compounds that produce such units on hydrolysis. E.g., cellulose, glycogen, starch, etc.

Visual Summary: Classification of Carbohydrates

Monosaccharides Non-hydrolysable (Glucose)

Oligosaccharides 2-10 units (Sucrose, Maltose)

Polysaccharides Many units (Starch, Cellulose)

Reducing and Non-Reducing Sugars

Reducing sugars: Contain free aldehydic or ketonic group and reduce Fehling’s solution and Tollen’s reagents. E.g., all monosaccharides, maltose.
Non-reducing sugars: Do not have free aldehydic or ketonic group. E.g., sucrose.

2. Monosaccharides: Glucose and Fructose

Glucose

Glucose occurs in nature in free as well as in combined form. It is present in sweet fruits and honey.

Preparation of Glucose

\text{(i) From Sucrose: } C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$$ $$\text{(ii) From Starch: } (C_6H_{10}O_5)_n + nH_2O \xrightarrow{H^+, 393K, 2-3 \text{ bar}} nC_6H_{12}O_6

Structure of Glucose

It is a six-carbon straight-chain aldose with one aldehydic group ($-CHO$), one primary hydroxyl group ($-CH_2OH$), and four secondary hydroxyl groups ($-CHOH$).

Reactions of Glucose

(i) With HI: Forms n-Hexane (suggests all 6 carbons are in a straight chain).
(ii) Presence of Carbonyl Group ($>C=O$):
- With $NH_2OH \rightarrow$ Glucose oxime
- With $HCN \rightarrow$ Glucose cyanohydrin
- With Bromine water $\rightarrow$ Gluconic acid (confirms aldehyde group)
(iii) Presence of five $-OH$ groups: Reacts with acetic anhydride to form Glucose pentaacetate.
(iv) Presence of primary alcoholic $-OH$ group: Oxidation with $HNO_3$ yields Saccharic acid.

Visual Summary: Chemical Reactions of Glucose

HI, Δ → n-Hexane (Straight chain of 6 C)

Br₂ water → Gluconic acid (Confirms -CHO group)

(CH₃CO)₂O → Pentaacetate (Confirms five -OH groups)

HNO₃ → Saccharic acid (Confirms primary -OH)

Fructose

It is a ketohexose obtained by hydrolysis of disaccharide sucrose. It is a laevorotatory compound, written as D-(-) fructose.

Open Chain Structure: Contains a ketonic group at C-2.
Cyclic Structure: Exists as a five-membered ring (furanose) containing one oxygen and five carbons.

3. Polysaccharides: Starch, Cellulose, and Glycogen

(i) Starch

Main storage polysaccharide of plants. Consists of two components:

Amylose (15-20%): Water-soluble, long unbranched chain of $\alpha$-D-glucose units held by $C_1-C_4$ glycosidic linkage.
Amylopectin (80-85%): Insoluble in water, branched chain polymer of $\alpha$-D-glucose. Chain formed by $C_1-C_4$ linkage, branching by $C_1-C_6$ linkage.

(ii) Cellulose

Predominant constituent of plant cell walls. Straight-chain polysaccharide composed only of $\beta$-D-glucose units joined by glycosidic linkage between $C_1$ of one glucose unit and $C_4$ of the next.

(iii) Glycogen

Carbohydrates stored in the animal body (liver, muscles, brain). Structurally similar to amylopectin but more highly branched.

Distinction between Glucose, Sucrose, and Starch

S. No.	Test	Glucose (Monosaccharide)	Sucrose (Disaccharide)	Starch (Polysaccharide)
1.	Fehling’s solution	Red precipitate (Reducing)	No precipitate	No precipitate
2.	Tollens’ reagent	Silver mirror formed	No silver mirror	No silver mirror
3.	Phenylhydrazine	Yellow osazone crystals	No osazone	No osazone
4.	Iodine solution	No colour	No colour	Blue-violet colour

4. Proteins: Amino Acids and Structure

Proteins: Complex polyamides formed from amino acids. Essential for proper growth and maintenance of the body. They contain many peptide ($-CO-NH-$) bonds.

Amino Acids

Compounds containing a carboxylic acid ($-COOH$) group and an amino group ($-NH_2$). $\alpha$-amino acids (where both groups are attached to the same carbon) are obtained by hydrolysis of proteins.

Neutral, Acidic, Basic: Depending on the relative number of amino and carboxyl groups.
Essential: Cannot be synthesised in the body (must be obtained by diet). E.g., Valine, Leucine.
Non-essential: Can be synthesised in the body. E.g., Glycine, Alanine.

Classification of Proteins

Based on molecular shape:
- Fibrous proteins: Thread-like molecules (e.g., Keratin, Collagen).
- Globular proteins: Folded into compact spheroidal shapes (e.g., Insulin, Albumin).
Based on structure:
- Primary structure: Specific sequence of amino acids.
- Secondary structure: $\alpha$-helix (intramolecular H-bonds, e.g., Keratin) or $\beta$-pleated sheet (intermolecular H-bonds, e.g., Silk).
- Tertiary structure: Overall folding of polypeptide chains (H-bonds, disulphide linkages, etc.).
- Quaternary structure: Spatial arrangement of multiple polypeptide chains.

Visual Summary: Levels of Protein Structure

Secondary α-helix or β-pleated sheet

Tertiary 3D folding (Globular)

Quaternary Multiple subunits

Denaturation of Proteins

When a protein in its native form is subjected to changes like temperature or pH, hydrogen bonds are disturbed. Globules unfold and helix gets uncoiled, leading to a loss of biological activity. E.g., coagulation of egg white on boiling, curdling of milk.

5. Nucleic Acids: DNA and RNA

Nucleic Acids: Polymers of nucleotides that help in the synthesis of proteins and transfer of genetic traits. They are of two types: DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid).

Constituents of Nucleic Acids

Pentose Sugar: Deoxyribose in DNA, Ribose in RNA.
Phosphoric Acid.
Nitrogenous Bases:
- Purines: Adenine (A) and Guanine (G) (Present in both DNA and RNA).
- Pyrimidines: Cytosine (C) and Thymine (T) in DNA; Cytosine (C) and Uracil (U) in RNA.

Nucleosides and Nucleotides

Nucleoside: Base + Sugar (Linked at 1′-position).
Nucleotide: Nucleoside + Phosphoric acid (Linked at 5′-position).

Visual Summary: DNA vs RNA

RNA Single Stranded, Ribose Bases: A, U, G, C

Mnemonics for Bases:
• DNA: ATGC (Adenine, Thymine, Guanine, Cytosine)
• RNA: Area Under The Growth Curve (Adenine, Uracil, Guanine, Cytosine)

6. Vitamins and Hormones

Vitamins

Organic molecules essential in small quantities for proper metabolism. Classified based on solubility:

Fat-Soluble Vitamins (A, D, E, K): Soluble in fats/oils, stored in liver and adipose tissues.
Water-Soluble Vitamins (B group, C): Must be supplied regularly in the diet as they are readily excreted in urine (except $B_{12}$).

Hormones

Molecules that act as intercellular messengers. Produced by endocrine glands and poured directly into the bloodstream.

Chemical Nature: Steroids (oestrogens, androgens), Polypeptides (insulin), Amino acid derivatives (epinephrine).
Functions: Maintain balance of biological activities. E.g., Insulin regulates blood glucose levels.

Important Endocrine Glands and Hormones

S. No.	Endocrine Gland	Hormone Secreted	Functions
1.	Pituitary gland	Growth hormone	Regulates normal growth of a person.
2.	Thyroid gland	Thyroxine	Controls body metabolism. Lack causes Goitre.
3.	Pancreas	Insulin	Controls carbohydrate metabolism. Lack causes Diabetes.
4.	Adrenal gland	Adrenaline	Maintains correct balance of salt and water in the blood.

End of Chapter 10: Biomolecules

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

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CBSE Class 12 Chemistry – Complete Notes & Question Bank

📚 CBSE Class 12 ChemistryComplete Notes & Question BankAll Chapters Combined

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📚 CBSE Class 12 ChemistryChapter 1: Solutions (Complete Notes & Question Bank)

Table of Contents

1. Solutions and its Expression of Concentration, Solubility of Gases

Basic Concepts

Types of Solutions

Visual Summary: Types of Solutions

Methods of Expressing Concentration of Solution

(i) Mass percentage (w/w%)

(ii) Volume percentage (v/v%)

(iii) Mass by volume percentage (w/V%)

(iv) Parts per million (ppm)

(v) Mole Fraction ($\chi$)

(vi) Molarity (M)

(vii) Molality (m)

(viii) Normality (N)

Relationships Between Concentration Terms

Solubility & Henry’s Law

Henry’s Law

Example 1 [Board 2023]

Solution:

2. Raoult’s Law, Ideal and Non-Ideal Solutions

Vapour Pressure & Raoult’s Law

Raoult’s law for volatile liquids:

Raoult’s law as a special case of Henry’s law:

Raoult’s law for non-volatile solute:

Ideal and Non-Ideal Solutions

Visual Summary: Deviations from Raoult’s Law

Azeotropes

Example 2 [Board 2023]

Solution:

3. Colligative Properties, Determination of Molecular Mass, Abnormal Molecular Mass, Van’t Hoff Factor

Colligative Properties

Visual Summary: Colligative Properties

(i) Relative Lowering of Vapour Pressure

(ii) Elevation of Boiling Point

(iii) Depression of Freezing Point

(iv) Osmotic Pressure

Abnormal Molecular Mass & Van’t Hoff Factor

Example 3

Solution:

📚 CBSE Class 12 ChemistryChapter 2: Electrochemistry (Complete Notes & Question Bank)

Table of Contents

1. Electrolytic Conductivity and Kohlrausch’s Law

Basic Concepts of Conductance

Resistance (R) and Resistivity (ρ)

Conductance (C) and Conductivity ($\kappa$)

Cell Constant (G) and Molar Conductivity ($\Lambda_m$)

Kohlrausch’s Law & Debye Huckel Onsager Equation

Applications of Kohlrausch’s Law

Example 1 [Board 2024]

Solution:

2. Redox Reactions and Electrochemical Cells, Electrode Potential and Nernst Equation

Redox Reactions & Galvanic Cells

Visual Summary: Daniell Cell (Galvanic Cell)

Salt Bridge and its Functions

Electrode Potential & Nernst Equation

Gibbs Energy

Example 2 [Board 2023]

Solution:

3. Electrolysis, Law of Electrolysis, Batteries, Fuel Cells and Corrosion

Faraday’s Laws of Electrolysis

Batteries

Lead Storage Battery

Visual Summary: Lead Storage Battery

Fuel Cells

Corrosion and its Prevention

Rusting of Iron

Visual Summary: Prevention of Corrosion

Example 3 [Board 2023]

Solution:

📚 CBSE Class 12 ChemistryChapter 4: d- and f-Block Elements (Complete Notes & Question Bank)

Table of Contents

1. d-Block Elements, Their Properties and Compounds

Introduction to d-Block and Transition Elements

Different Transition Series

Visual Summary: Transition Series

General Characteristics of Transition Elements

📚 CBSE Class 12 Chemistry
Complete Notes & Question Bank
All Chapters Combined

📚 CBSE Class 12 Chemistry
Chapter 1: Solutions (Complete Notes & Question Bank)

📚 CBSE Class 12 Chemistry
Chapter 2: Electrochemistry (Complete Notes & Question Bank)

📚 CBSE Class 12 Chemistry
Chapter 4: d- and f-Block Elements (Complete Notes & Question Bank)

📚 CBSE Class 12 Chemistry
Chapter 4.2: f-Block Elements (Complete Notes & Question Bank)

📚 CBSE Class 12 Chemistry
Chapter 5: Coordination Compounds (Complete Notes & Question Bank)