Chemistry Grade 12 for CBSE

Welcome to the Chemistry Grade 12 book, designed specifically in accordance with the CBSE 2025-26 syllabus.

This book focuses on building conceptual clarity and ensuring you are well-prepared for Competency-Based Questions (CBQs), which are an integral part of the recent CBSE exam patterns.

Book Structure

The book is structured into 10 units as prescribed by the CBSE:

Physical and Inorganic Chemistry:

1. Solutions
2. Electrochemistry
3. Chemical Kinetics
4. d and f Block Elements
5. Coordination Compounds

Organic Chemistry: 6. Haloalkanes and Haloarenes 7. Alcohols, Phenols and Ethers 8. Aldehydes, Ketones and Carboxylic Acids 9. Amines 10. Biomolecules

Each chapter contains theory, necessary mathematical formulas formatted precisely, high-quality diagrams, and a dedicated section at the end for competency-based questions based on the last 5 years of CBSE question papers and official sample papers.

Good luck with your studies!

Unit 1: Solutions

1.1 Types of Solutions

A solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

• Solute: Component present in smaller quantity.
• Solvent: Component present in larger quantity.

Solutions can be classified based on the physical state of the solvent. Our primary focus is on liquid solutions.

1.2 Expression of Concentration of Solutions

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

• Mass Percentage (w/w): \( \text{Mass } % = \frac{\text{Mass of component in solution}}{\text{Total mass of solution}} \times 100 \)
• Volume Percentage (v/v): \( \text{Volume } % = \frac{\text{Volume of component}}{\text{Total volume of solution}} \times 100 \)
• Molarity (M): \( M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} \)
• Molality (m): \( m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \)
• Mole Fraction (x): \( x_A = \frac{n_A}{n_A + n_B} \)

1.3 Solubility of Gases in Liquids

The solubility of gases in liquids is greatly affected by pressure and temperature.

Henry’s Law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution. \[ p = K_H \cdot x \] where \( p \) is the partial pressure of the gas, \( x \) is the mole fraction of the gas in solution, and \( K_H \) is Henry’s law constant.

1.4 Raoult’s Law and Solid Solutions

For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. \[ p_1 = p_1^0 x_1 \] where \( p_1^0 \) is the vapour pressure of pure component 1 at the same temperature.

An Ideal Solution strictly obeys Raoult’s law over the entire range of concentration. Most solutions, however, are non-ideal and show either positive or negative deviation from Raoult’s law.

1.5 Colligative Properties

Properties of dilute solutions which depend only on the number of solute particles (molecules/ions) and not on their nature.

1. Relative Lowering of Vapour Pressure

When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lower than that of the pure solvent. \[ \frac{p_1^0 - p_1}{p_1^0} = x_2 \]

2. Elevation of Boiling Point (\( \Delta T_b \))

The boiling point of a solution containing a non-volatile solute is higher than that of the pure solvent. \[ \Delta T_b = K_b \cdot m \] where \( K_b \) is the molal elevation constant (ebullioscopic constant) and \( m \) is molality.

3. Depression of Freezing Point (\( \Delta T_f \))

The freezing point of a solution is lower than that of the pure solvent. \[ \Delta T_f = K_f \cdot m \] where \( K_f \) is the molal depression constant (cryoscopic constant).

4. Osmosis and Osmotic Pressure (\( \pi \))

Osmosis is the spontaneous flow of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semi-permeable membrane. The osmotic pressure is the excess pressure that must be applied to a solution to prevent osmosis. \[ \pi = C R T \] where \( C \) is the molar concentration (molarity) of the solution, \( R \) is the gas constant, and \( T \) is temperature.

1.6 Abnormal Molecular Masses and van’t Hoff Factor

When a solute undergoes dissociation or association in a solution, the observed molecular mass differs from the normal (expected) molecular mass. The van’t Hoff factor (\( i \)) accounts for the extent of dissociation or association.

\[ i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}} \] \[ i = \frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of particles before association/dissociation}} \]

Modified Colligative Properties Equations:

• Relative lowering of vapour pressure: \( \frac{p_1^0 - p_1}{p_1^0} = i \cdot x_2 \)
• Elevation of Boiling Point: \( \Delta T_b = i \cdot K_b \cdot m \)
• Depression of Freezing Point: \( \Delta T_f = i \cdot K_f \cdot m \)
• Osmotic Pressure: \( \pi = i \cdot C R T \)

Competency-Based Questions (CBQs)

Q1. (CBSE 2023) Based on intermolecular forces and Raoult’s law, a solution of chloroform (\(CHCl_3\)) and acetone (\(CH_3COCH_3\)) shows a specific deviation. Identify the type of deviation and provide the underlying reason.

Answer: The solution of chloroform and acetone shows a negative deviation from Raoult’s law.

Reason: In pure state, both chloroform and acetone have dipole-dipole interactions. When they are mixed, the hydrogen atom of chloroform forms a hydrogen bond with the highly electronegative oxygen atom of acetone: \[ CH_3COCH_3 \cdots H-CCl_3 \] Due to this new intermolecular hydrogen bonding, the interactions between chloroform and acetone molecules are stronger than the pure A-A and B-B interactions. This decreases the escaping tendency of molecules, leading to a lower vapour pressure than predicted by Raoult’s law, and hence a negative deviation.

Q2. (Sample Paper 2023) A 5% (by mass) solution of cane sugar (molar mass 342 g/mol) in water has a freezing point of 271 K. Calculate the freezing point of a 5% (by mass) solution of glucose (molar mass 180 g/mol) in water if the freezing point of pure water is 273.15 K.

Answer: It is given that mass percentage of both solutions is 5%. This means 5 g of solute is present in 100 g of solution, i.e. mass of solvent (water) = 95 g.

For cane sugar solution:

• \( w_2 \) (cane sugar) = 5 g
• \( M_2 \) = 342 g/mol
• \( w_1 \) (water) = 95 g
• \( \Delta T_f \) = Freezing point of pure water - Freezing point of solution = 273.15 - 271 = 2.15 K

Using the formula: \[ \Delta T_f = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1} \] \[ 2.15 = K_f \cdot \frac{5 \times 1000}{342 \times 95} \] \[ K_f = \frac{2.15 \times 342 \times 95}{5000} = 13.99 , \text{K kg mol}^{-1} \]

For glucose solution:

• \( w_2 \) (glucose) = 5 g
• \( M_2 \) = 180 g/mol
• \( w_1 \) (water) = 95 g
• \( K_f \) = 13.99 K kg/mol

\[ \Delta T_f’ = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1} \] \[ \Delta T_f’ = 13.99 \cdot \frac{5 \times 1000}{180 \times 95} = 4.09 , \text{K} \]

The freezing point of the glucose solution = Freezing point of pure water - \(\Delta T_f’\) \[ T_f’ = 273.15 - 4.09 = 269.06 , \text{K} \]

Q3. (CBSE 2020) Why is osmotic pressure considered the preferred colligative property for determining the molecular masses of macromolecules such as proteins and polymers?

Answer: Osmotic pressure is preferred for determining the molecular masses of macromolecules for the following reasons:

1. Magnitude: The osmotic pressure values are reasonably large and measurable even for very dilute solutions of macromolecules. In contrast, the changes in boiling point or freezing point are negligibly small.
2. Temperature: Osmotic pressure is measured at room temperature, which prevents the denaturation or degradation of sensitive biomolecules like proteins that might decompose at higher temperatures (e.g., during boiling point elevation).
3. Molarity vs Molality: Osmotic pressure depends on molarity instead of molality. For large polymers, it is easier to prepare solutions of known volume than known mass of solvent accurately.

Unit 2: Electrochemistry

2.1 Redox Reactions and Galvanic Cells

Electrochemistry is the study of the production of electricity from the energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.

A Galvanic Cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. A typical example is the Daniell Cell:

• Anode (Oxidation): \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)
• Cathode (Reduction): \( Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \)
• Overall Reaction: \( Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s) \)

2.2 Standard Electrode Potential (\( E^\circ \))

Standard electrode potential is the potential difference developed between the metal electrode and the solution of its ions of unit molarity at 298 K and 1 atm pressure. \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}} \]

2.3 Nernst Equation

The Nernst equation relates the cell potential at non-standard conditions to the standard cell potential, temperature, and concentrations of the reacting species.

For the reaction: \( aA + bB \rightarrow cC + dD \) \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \] where:

• \( R \) is the gas constant (8.314 J/(K·mol))
• \( T \) is the temperature in Kelvin
• \( n \) is the number of moles of electrons transferred
• \( F \) is the Faraday constant (~96487 C/mol)
• \( Q \) is the reaction quotient, \( Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

At 298 K, using base 10 logarithm, the Nernst equation simplifies to: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059V}{n} \log Q \]

Equilibrium Constant from Nernst Equation

At equilibrium, \( E_{\text{cell}} = 0 \) and \( Q = K_c \): \[ E^\circ_{\text{cell}} = \frac{0.059V}{n} \log K_c \]

2.4 Gibbs Energy Change

The reversible work done by a galvanic cell is equal to its decrease in Gibbs energy: \[ \Delta_r G = -nFE_{\text{cell}} \] Under standard conditions: \[ \Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -RT \ln K_c \]

2.5 Conductance of Electrolytic Solutions

• Resistance (\(R\)): \( R = \rho \frac{l}{A} \)
• Conductivity (\(\kappa\)): \( \kappa = \frac{1}{\rho} = \frac{1}{R} \left( \frac{l}{A} \right) \), where \( l/A \) is the cell constant (\( G^* \)).
• Molar Conductivity (\(\Lambda_m\)): The conducting power of all the ions produced by dissolving one mole of electrolyte in solution. \[ \Lambda_m = \frac{\kappa \times 1000}{C} \quad (\text{if } \kappa \text{ is in S cm}^{-1} \text{ and } C \text{ is in mol L}^{-1}) \]

2.6 Kohlrausch’s Law of Independent Migration of Ions

According to this law, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. \[ \Lambda^\circ_m = \nu_+ \lambda^\circ_+ + \nu_- \lambda^\circ_- \] where \( \nu_+ \) and \( \nu_- \) are the number of cations and anions respectively, and \( \lambda^\circ_+, \lambda^\circ_- \) are the limiting molar conductivities of the cation and anion.

2.7 Batteries and Corrosion

Primary Batteries: Non-rechargeable (e.g., Dry cell, Button cell). Secondary Batteries: Rechargeable (e.g., Lead storage battery, Nickel-cadmium battery). Fuel Cells: Produce electricity directly from the energy of combustion of fuels like \(H_2\), \(CO\), \(CH_4\). Corrosion: The slow coating of metal surfaces with oxides or other salts. In rusting of iron, a miniature electrochemical cell is formed at the surface of the iron.

Competency-Based Questions (CBQs)

Q1. (CBSE 2022) The standard reduction potentials for \( Zn^{2+}/Zn \), \( Ni^{2+}/Ni \), and \( Fe^{2+}/Fe \) are -0.76 V, -0.23 V, and -0.44 V, respectively. The reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) will be spontaneous when: (A) X = Ni, Y = Fe (B) X = Ni, Y = Zn (C) X = Fe, Y = Zn (D) X = Zn, Y = Ni

Answer: Correct Option: (D) Reasoning and Calculation: For the reaction \( X + Y^{2+} \rightarrow X^{2+} + Y \) to be spontaneous, the standard cell potential \( E^\circ_{\text{cell}} \) must be positive. \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cathode (reduction)}} - E^\circ_{\text{Anode (oxidation)}} \] \[ E^\circ_{\text{cell}} = E^\circ_{Y^{2+}/Y} - E^\circ_{X^{2+}/X} > 0 \] Which means \( E^\circ_{Y^{2+}/Y} > E^\circ_{X^{2+}/X} \). In simpler terms, X must be a stronger reducing agent than Y (X must have a more negative standard reduction potential than Y). Looking at the potentials: \( E^\circ_{Zn} = -0.76 \) V \( E^\circ_{Fe} = -0.44 \) V \( E^\circ_{Ni} = -0.23 \) V

For option (D), X = Zn (-0.76 V) and Y = Ni (-0.23 V). Since -0.23 V > -0.76 V, the reaction is spontaneous.

Q2. (CBSE 2019) Calculate the emf of the following cell at 298 K: \[ Cu(s) | Cu^{2+} (0.130 M) || Ag^+ (1.0 \times 10^{-4} M) | Ag(s) \] Given: \( E^\circ_{Cu^{2+}/Cu} = +0.34 \) V and \( E^\circ_{Ag^+/Ag} = +0.80 \) V.

Answer: Step 1: Write the cell reactions and overall reaction.

• Anode (Oxidation): \( Cu(s) \rightarrow Cu^{2+} + 2e^- \)
• Cathode (Reduction): \( [Ag^+ + e^- \rightarrow Ag(s)] \times 2 \)
• Overall cell reaction: \( Cu(s) + 2Ag^+ \rightarrow Cu^{2+} + 2Ag(s) \)

The number of moles of electrons transferred, \( n = 2 \).

Step 2: Calculate \( E^\circ_{\text{cell}} \). \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cathode}} - E^\circ_{\text{Anode}} \] \[ E^\circ_{\text{cell}} = 0.80 , \text{V} - 0.34 , \text{V} = 0.46 , \text{V} \]

Step 3: Apply the Nernst equation. \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[Cu^{2+}]}{[Ag^+]^2} \] \[ E_{\text{cell}} = 0.46 - \frac{0.059}{2} \log \frac{0.130}{(1.0 \times 10^{-4})^2} \] \[ E_{\text{cell}} = 0.46 - 0.0295 \log \left(\frac{0.130}{1.0 \times 10^{-8}}\right) \] \[ E_{\text{cell}} = 0.46 - 0.0295 \log (1.3 \times 10^7) \] \[ \log (1.3 \times 10^7) = \log(1.3) + 7 = 0.1139 + 7 = 7.1139 \] \[ E_{\text{cell}} = 0.46 - 0.0295 \times 7.1139 \] \[ E_{\text{cell}} = 0.46 - 0.2098 = 0.25 , \text{V} \]

Q3. (Sample Paper 2024) The molar conductivity of 0.025 mol L\(^{-1}\) methanoic acid is 46.1 S cm\(^2\) mol\(^{-1}\). Calculate its degree of dissociation and dissociation constant. Given \(\lambda^\circ (H^+) = 349.6\) S cm\(^2\) mol\(^{-1}\) and \(\lambda^\circ (HCOO^-) = 54.6\) S cm\(^2\) mol\(^{-1}\).

Answer: Step 1: Find the limiting molar conductivity (\( \Lambda^\circ_m \)). According to Kohlrausch’s law: \[ \Lambda^\circ_m (HCOOH) = \lambda^\circ (H^+) + \lambda^\circ (HCOO^-) \] \[ \Lambda^\circ_m = 349.6 + 54.6 = 404.2 , \text{S cm}^2 \text{ mol}^{-1} \]

Step 2: Calculate the degree of dissociation (\( \alpha \)). \[ \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{46.1}{404.2} = 0.114 \]

Step 3: Calculate the dissociation constant (\( K_a \)). For methanoic acid, \( HCOOH \rightleftharpoons HCOO^- + H^+ \) \[ K_a = \frac{C \alpha^2}{1 - \alpha} \] \[ K_a = \frac{0.025 \times (0.114)^2}{1 - 0.114} \] \[ K_a = \frac{0.025 \times 0.013}{0.886} = 3.67 \times 10^{-4} \]

Unit 3: Chemical Kinetics

3.1 Rate of a Chemical Reaction

Chemical kinetics is the branch of chemistry that deals with the study of reaction rates and their mechanisms.

The rate of a reaction is the change in the concentration of any one of the reactants or products per unit time. For a reaction \( R \rightarrow P \):

• Average rate of reaction \( = \frac{-\Delta [R]}{\Delta t} = \frac{+\Delta [P]}{\Delta t} \)
• Instantaneous rate of reaction \( = \frac{-d[R]}{dt} = \frac{+d[P]}{dt} \)

Factors influencing rate of a reaction:

• Concentration of reactants
• Temperature
• Nature of reactants and products
• Presence of a catalyst
• Surface area of reactants (for heterogeneous reactions)
• Exposure to radiation (for photochemical reactions)

3.2 Rate Expression and Rate Constant

The experimental expression of the rate of reaction in terms of the concentration of reactants is known as the rate law or rate expression. For a general reaction: \( aA + bB \rightarrow cC + dD \) \[ \text{Rate} = k [A]^x [B]^y \] where \( x \) and \( y \) are determined experimentally. \( k \) is the rate constant or specific reaction rate.

3.3 Order and Molecularity of a Reaction

Order of a Reaction: The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Let the rate law be Rate = \( k[A]^x[B]^y \). Then, Order = \( x + y \). Order can be 0, 1, 2, 3 and even a fraction.

Molecularity: The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. Molecularity is always a whole number (1, 2, 3…).

3.4 Integrated Rate Equations

Zero Order Reactions

The rate of the reaction is independent of the concentration of reactants. \[ \text{Rate} = \frac{-d[R]}{dt} = k [R]^0 = k \] Integrated form: \[ [R] = -kt + [R]_0 \] where \( [R]_0 \) is the initial concentration and \( [R] \) is the concentration at time \( t \).

First Order Reactions

The rate of the reaction is proportional to the first power of the concentration of the reactant \( R \). \[ \text{Rate} = \frac{-d[R]}{dt} = k [R] \] Integrated form: \[ \ln [R] = -kt + \ln [R]_0 \] Or using base 10 logarithm: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]

3.5 Half-Life of a Reaction (\( t_{1/2} \))

The time in which the concentration of a reactant is reduced to one half of its initial concentration.

• For a zero order reaction: \( t_{1/2} = \frac{[R]_0}{2k} \)
• For a first order reaction: \( t_{1/2} = \frac{0.693}{k} \) (Independent of initial concentration)

3.6 Temperature Dependence of the Rate of a Reaction

Arrhenius Equation describes the effect of temperature on the rate constant (\( k \)) of a reaction: \[ k = A e^{-E_a/RT} \] Taking natural logarithm on both sides: \[ \ln k = -\frac{E_a}{RT} + \ln A \] Taking log base 10: \[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]

where:

• \( A \) is the Arrhenius factor or frequency factor or pre-exponential factor.
• \( R \) is gas constant.
• \( E_a \) is activation energy in joules/mole.

Competency-Based Questions (CBQs)

Q1. (CBSE 2023) The decomposition of \( N_2O_5 \) in \( CCl_4 \) at 318 K has been studied by monitoring the concentration of \( N_2O_5 \) in the solution. Initially, the concentration of \( N_2O_5 \) is 2.33 mol/L and after 184 minutes, it is reduced to 2.08 mol/L. Give the sequence of steps to determine the average rate of this reaction in terms of hours, minutes and seconds.

Answer: Average rate \( = -\frac{\Delta [N_2O_5]}{\Delta t} \) \[ \Delta [N_2O_5] = [\text{Final}] - [\text{Initial}] = 2.08 - 2.33 = -0.25 , \text{mol/L} \] Rate in minutes: \[ \text{Time } \Delta t = 184 \text{ min} \] \[ \text{Rate} = - \frac{-0.25}{184} = 1.36 \times 10^{-3} , \text{mol L}^{-1} \text{min}^{-1} \]

Rate in hours: \[ \Delta t = \frac{184}{60} = 3.067 \text{ hrs} \] \[ \text{Rate} = - \frac{-0.25}{3.067} = 8.15 \times 10^{-2} , \text{mol L}^{-1} \text{h}^{-1} \]

Rate in seconds: \[ \Delta t = 184 \times 60 = 11040 \text{ sec} \] \[ \text{Rate} = - \frac{-0.25}{11040} = 2.26 \times 10^{-5} , \text{mol L}^{-1} \text{s}^{-1} \]

Q2. (Sample Paper 2024) A first order reaction takes 40 min for 30% decomposition. Calculate \( t_{1/2} \).

Answer: For a first order reaction, \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] Here, let initial concentration \( [R]_0 = 100 \). Since 30% decomposes, the remaining concentration \( [R] = 100 - 30 = 70 \). Time \( t = 40 \) min. \[ k = \frac{2.303}{40} \log \frac{100}{70} \] \[ k = \frac{2.303}{40} \log 1.428 \] \[ k = \frac{2.303}{40} \times 0.1548 \] \[ k = 8.91 \times 10^{-3} , \text{min}^{-1} \] Now, calculate the half-life \( t_{1/2} \): \[ t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.91 \times 10^{-3}} = 77.7 , \text{min} \]

Q3. (CBSE 2021) The rate constant of a reaction increases by 5% when its temperature is raised from 27°C to 28°C. Calculate the activation energy of the reaction. (Given: \( R = 8.314 \) J K\(^{-1}\) mol\(^{-1}\))

Answer: Given:

• \( T_1 = 27^\circ \text{C} = 300 , \text{K} \)
• \( T_2 = 28^\circ \text{C} = 301 , \text{K} \)
• Rate constant \( k_2 \) is 5% more than \( k_1 \), i.e., \( k_2 = k_1 + 0.05 k_1 = 1.05 k_1 \).
• This means \( \frac{k_2}{k_1} = 1.05 \).

Using Arrhenius Equation: \[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \] \[ \log (1.05) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{301 - 300}{300 \times 301} \right] \] \[ 0.0212 = \frac{E_a}{19.147} \left[ \frac{1}{90300} \right] \] \[ E_a = 0.0212 \times 19.147 \times 90300 \] \[ E_a = 36653 , \text{J/mol} \] \[ E_a = 36.65 , \text{kJ/mol} \]

Unit 4: d- and f-Block Elements

4.1 The Transition Elements (d-Block)

The d-block of the periodic table contains the elements of the groups 3-12 in which the d orbitals are progressively filled in each of the four long periods (3d, 4d, 5d and 6d series). The transition elements are those elements which have incompletely filled d-subshells in their ground state or in any of their common oxidation states. Zinc, cadmium, and mercury of group 12 have full \( d^{10} \) configuration in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals.

General Properties of the Transition Elements:

1. Metallic Character: Nearly all transition metals exhibit typical metallic properties such as high tensile strength, ductility, malleability, high thermal and electrical conductivity, and metallic luster.
2. Melting and Boiling Points: Transition metals have high melting and boiling points due to strong metallic bonding based on the involvement of both ns and (n-1)d electrons.
3. Atomic and Ionic Radii: The atomic and ionic radii of transition metals decrease from group 3 to group 6 due to an increase in effective nuclear charge, then become almost constant, and slightly increase towards the end of the series due to increased electron-electron repulsion.
4. Ionization Enthalpies: Ionization enthalpies generally increase from left to right along a period due to an increase in nuclear charge which accompanies the filling of inner d orbitals.
5. Oxidation States: Transition metals exhibit variable oxidation states due to the participation of both ns and (n-1)d electrons in bonding, as the energy difference between them is small.
6. Coloured Ions: Many transition metal ions are coloured in solid state or in aqueous solution due to d-d transitions. The excitation of an electron from a lower energy d orbital to a higher energy d orbital requires energy which corresponds to a wavelength in the visible region.
7. Catalytic Properties: Transition metals and their compounds are known for their catalytic activity. This is due to their ability to adopt multiple oxidation states and to form complexes. Examples: V2O5 (Contact Process), finely divided iron (Haber’s Process).
8. Magnetic Properties: Due to the presence of unpaired electrons in the (n-1)d orbitals, most transition metal ions are paramagnetic. (\mu = \sqrt{n(n+2)} , \text{BM}) where n is the number of unpaired electrons.
9. Formation of Complex Compounds: Transition metals form a large number of complex compounds due to the comparatively smaller size of the metal ions, high ionic charges, and availability of vacant d orbitals for bond formation.
10. Formation of Interstitial Compounds: Small atoms like H, C, or N are trapped inside the crystal lattices of metals. They are usually non-stoichiometric (e.g., TiC, Mn4N).
11. Alloy Formation: Alloys are formed by transition metals because their atomic radii are very similar. One metal can easily replace another metal in the crystal lattice.

4.2 Preparation and Properties of \(K_2Cr_2O_7\) and \(KMnO_4\)

Potassium dichromate (\( K_2Cr_2O_7 \)):

• Preparation from chromite ore (\( FeCr_2O_4 \)):
  1. Fusion of chromite ore with sodium carbonate in air: \[ 4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2 \]
  2. Acidification of sodium chromate: \[ 2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O \]
  3. Conversion into potassium dichromate: \[ Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl \]
• Properties: Strong oxidizing agent in acidic medium. \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad (E^\circ = +1.33 \text{V}) \]

Potassium permanganate (\( KMnO_4 \)):

• Preparation from pyrolusite ore (\( MnO_2 \)):
  1. Fusion of \( MnO_2 \) with KOH and an oxidizing agent (\( O_2 \) or \( KNO_3 \)): \[ 2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O \]
  2. Electrolytic oxidation in alkaline solution: \[ MnO_4^{2-} \rightarrow MnO_4^- + e^- \]
• Properties: Very strong oxidizing agent.
  • In acidic medium: \( MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \)

4.3 The Inner Transition Elements (f-Block)

Lanthanides

The 14 elements (Ce to Lu) in which the 4f orbitals are progressively filled.

• Oxidation States: The most common oxidation state is +3. Some elements also show +2 and +4 oxidation states in solutions or in solid compounds to attain the stable \( f^0, f^7 \) or \( f^{14} \) configurations.
• Lanthanide Contraction: The steady decrease in the atomic and ionic radii of lanthanide elements with increasing atomic number.
  • Cause: Imperfect shielding of one 4f electron by another in the same subshell. As the nuclear charge increases, the imperfect shielding is unable to counterbalance the effect of increased nuclear charge.
  • Consequences: Similarity in sizes of elements of second and third transition series (e.g., Zr and Hf have almost identical radii). Difficulty in separating lanthanides.

Actinides

The 14 elements (Th to Lr) in which the 5f orbitals are progressively filled.

• They are all radioactive elements.
• Oxidation States: Like lanthanides, the most common oxidation state is +3. However, they exhibit a larger number of oxidation states (+3, +4, +5, +6, +7) because the energy difference between 5f, 6d and 7s orbitals is very small.

Competency-Based Questions (CBQs)

Q1. (CBSE 2023) Explain why zinc, cadmium, and mercury are not regarded as transition metals.

Answer: A transition element is defined as an element having an incompletely filled d-subshell in either its ground state or in any of its common oxidation states. The outermost electronic configuration of:

• Zinc (Zn, Z=30): \( [Ar] 3d^{10} 4s^2 \)
• Cadmium (Cd, Z=48): \( [Kr] 4d^{10} 5s^2 \)
• Mercury (Hg, Z=80): \( [Xe] 4f^{14} 5d^{10} 6s^2 \)

These elements have completely filled d orbitals (\( d^{10} \)) in their ground states as well as in their most stable oxidation states (which is +2 for all three, attained by losing the two s electrons, leaving the \( d^{10} \) core intact). Since they never possess incompletely filled d orbitals, they are not regarded as transition metals.

Q2. (Sample Paper 2024) Account for the following: The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.

Answer: This phenomenon is due to the Lanthanide Contraction. In the 6th period, the filling of the 4f subshell occurs before the 5d subshell begins to fill. The 14 elements from Cerium (Z=58) to Lutetium (Z=71) are called lanthanides. The shielding effect of the 4f electrons is very poor. As the atomic number increases across the lanthanide series, the effective nuclear charge increases, pulling the valence shell closer and causing a steady decrease in atomic size. This contraction in size (lanthanide contraction) exactly cancels out the expected increase in size that would normally occur upon moving down the group from the 4d to the 5d series. Therefore, the 5d transition metals (like Hf, Ta, W) have atomic sizes almost equal to their 4d counterparts (like Zr, Nb, Mo).

Q3. (CBSE 2020) A green compound (A) is formed when potassium manganate (\( K_2MnO_4 \)) is prepared by fusing \( MnO_2 \) with KOH in the presence of an oxidizing agent. The aqueous solution of (A) upon electrolytic oxidation yields a deep purple compound (B). 1. Identify A and B. 2. Write the balanced chemical equation for the conversion of A to B in acidic medium by disproportionation.

Answer:

1. The green compound (A) is Potassium manganate (\( K_2MnO_4 \)). The deep purple compound (B) is Potassium permanganate (\( KMnO_4 \)).
2. In a neutral or acidic medium, the green manganate ion (\( MnO_4^{2-} \)) disproportionates to purple permanganate (\( MnO_4^- \)) and solid manganese dioxide (\( MnO_2 \)). Let’s write the balanced equation: \[ 3MnO_4^{2-} (aq) + 4H^+ (aq) \rightarrow 2MnO_4^- (aq) + MnO_2 (s) + 2H_2O(l) \] (Note: 3 moles of manganate(+6) disproportionate to give 2 moles of permanganate(+7) and 1 mole of manganese dioxide(+4).)

Unit 5: Coordination Compounds

5.1 Introduction and Basic Terms

Coordination compounds consist of a central metal atom or ion connected to a surrounding array of molecules or anions. These surrounding species are called ligands. Unlike normal salts, coordination compounds retain their identity even in solution.

• Central Atom/Ion: The cation or neutral atom to which ligands are attached. Acts as a Lewis acid (electron pair acceptor).
• Ligand: Ions or molecules bound to the central atom/ion. They act as Lewis bases (electron pair donors).
  • Unidentate: \(Cl^-, H_2O, NH_3\)
  • Didentate: ethylenediamine (en), oxalate (\(C_2O_4^{2-}\))
  • Polydentate: EDTA (hexadentate)
• Coordination Number (CN): The total number of ligand donor atoms to which the metal is directly bonded.

5.2 IUPAC Nomenclature of Coordination Compounds

Rules for naming coordination compounds:

1. The cation is named first in both positively and negatively charged coordination entities.
2. The ligands are named in alphabetical order before the name of the central atom/ion.
3. Names of anionic ligands end in -o (e.g., chlorido, cyanido), neutral ligands have their normal names (exceptions: \(H_2O\) is aqua, \(NH_3\) is ammine, \(CO\) is carbonyl).
4. Prefixes mono, di, tri, etc., indicate the number of individual ligands. When the names of ligands include a numerical prefix, terms like bis, tris, tetrakis are used.
5. Oxidation state of the metal is indicated by a Roman numeral in parentheses.
6. If the complex ion is an anion, the name of the metal ends with the suffix -ate (e.g., ferrate, cobaltate).

5.3 Isomerism in Coordination Compounds

Types of stereoisomerism:

• Geometrical Isomerism: Found mainly in square planar (CN=4) and octahedral (CN=6) complexes. Leads to cis (similar groups adjacent) and trans (similar groups opposite) isomers.
• Optical Isomerism: Observed in octahedral complexes involving didentate ligands (e.g., \([Co(en)_3]^{3+}\)). Isomers are non-superimposable mirror images (d and l forms).

Types of structural isomerism:

• Linkage Isomerism: Arises in complexes containing ambidentate ligands (e.g., \(NO_2^-\), \(SCN^-\)).
• Coordination Isomerism: Interchange of ligands between cationic and anionic entities of different metals.
• Ionisation Isomerism: The counter ion in a complex salt acts as a ligand and the ligand acts as a counter ion.
• Solvate/Hydrate Isomerism: Involves exchange of solvent molecules (e.g., water) between the coordination sphere and the crystal lattice.

5.4 Bonding in Coordination Compounds

Werner’s Theory

Metals possess two types of valencies: primary valency (ionisable, satisfies oxidation state) and secondary valency (non-ionisable, satisfies coordination number and dictates the spatial arrangement of ligands).

Valence Bond Theory (VBT)

The central metal ion uses its (n-1)d, ns, np or ns, np, nd orbitals for hybridization to yield a set of equivalent orbitals of definite geometry (e.g., octahedral, square planar).

• Inner orbital complex: Uses (n-1)d orbitals (e.g., \(d^2sp^3\)). Usually form with strong field ligands; low spin.
• Outer orbital complex: Uses nd orbitals (e.g., \(sp^3d^2\)). Usually form with weak field ligands; high spin.

Crystal Field Theory (CFT)

Consider the ligands as point charges. The five degenerate d-orbitals split into two sets of orbitals of different energies in the presence of the ligand field.

• In an octahedral field: Energy of the \( e_g \) set (\( d_{x^2-y^2} \), \( d_{z^2} \)) is raised while the energy of the \( t_{2g} \) set (\( d_{xy}, d_{yz}, d_{zx} \)) is lowered.
  • Difference in energy is \( \Delta_o \) (Crystal Field Splitting Energy).

• Spectrochemical Series: Arrangement of ligands in increasing order of crystal field splitting strength. \[ I^- < Br^- < SCN^- < Cl^- < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < EDTA^{4-} < NH_3 < en < CN^- < CO \]

Colour: Due to d-d transitions of unpaired electrons in the visible region when light falls on the complex.

5.5 Importance of Coordination Compounds

• Extraction of metals: Ag and Au are extracted using cyanide complexes \([Ag(CN)_2]^-\).
• Estimation of hardness: Ca\(^{2+}\) and Mg\(^{2+}\) can be estimated by complexometric titration with EDTA.
• Biological systems: Chlorophyll (Mg complex), Haemoglobin (Fe complex), Vitamin B\(_{12}\) (Co complex).
• Catalysts: Wilkinson’s catalyst \([RhCl(PPh_3)_3]\) used for hydrogenation of alkenes.
• Medicine: Cisplatin \([Pt(NH_3)_2Cl_2]\) is used in cancer therapy.

Competency-Based Questions (CBQs)

Q1. (CBSE 2024 Pattern) The magnetic moment of \([MnCl_4]^{2-}\) is 5.9 BM whereas for \([Mn(CN)_6]^{3-}\) it is 2.8 BM. Using Valence Bond Theory, explain this difference in magnetic behavior and predict the hybridization in both the complexes.

Answer: For \([MnCl_4]^{2-}\): Oxidation state of Mn is +2. Electronic configuration of Mn\(^{2+}\) is \(3d^5\). Chloride (\(Cl^-\)) is a weak field ligand. It does not force the pairing of electrons against Hund’s rule. So, there are 5 unpaired electrons in the 3d orbitals. Using the formula \(\mu = \sqrt{n(n+2)}\): \[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 , \text{BM} \] The hybridization involves one 4s and three 4p orbitals, forming \(sp^3\) hybridization to accommodate four \(Cl^-\). The complex is tetrahedral and highly paramagnetic.

For \([Mn(CN)_6]^{3-}\): Oxidation state of Mn is +3 (since \(x - 6 = -3 \rightarrow x = +3\)). Electronic configuration of Mn\(^{3+}\) is \(3d^4\). Cyanide (\(CN^-\)) is a strong field ligand. It forces the pairing of electrons in the 3d orbitals to make room for hybridization. Two electrons pair up while the remaining two remain unpaired (so n = 2). \[ \mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.8 , \text{BM} \] Two 3d orbitals are vacant. The hybridization is \(d^2sp^3\) (inner orbital complex). The shape is octahedral and it is paramagnetic but less than \([MnCl_4]^{2-}\).

Q2. (CBSE 2023) Draw the structures of optical isomers of the coordination complex: \( [Co(en)_3]^{3+} \).

Answer: The complex \( [Co(en)_3]^{3+} \) (where ‘en’ stands for ethylenediamine, a didentate ligand) forms an octahedral geometry. As it lacks a plane of symmetry, it exists in two non-superimposable mirror image forms, known as the d- (dextro) and l- (laevo) enantiomers.

Q3. (Sample Paper 2023) Explain on the basis of crystal field theory why \([Co(NH_3)_6]^{3+}\) is an inner orbital and diamagnetic complex, while \([CoF_6]^{3-}\) is an outer orbital and paramagnetic complex.

Answer: Both complexes contain Cobalt in +3 oxidation state, meaning the electronic configuration of \(Co^{3+}\) is \(3d^6\). According to CFT, the 3d orbitals split into lower energy \(t_{2g}\) and higher energy \(e_g\) levels.

• In the presence of \(NH_3\) (a strong field ligand), the crystal field splitting energy (\(\Delta_o\)) is greater than the pairing energy ((P)), i.e. \(\Delta_o > P\). Therefore, electrons prefer to pair up in the lower energy \(t_{2g}\) orbitals rather than jumping to \(e_g\). The configuration becomes \(t_{2g}^6 e_g^0\). Because all electrons are paired, it is a diamagnetic, low-spin inner orbital complex.
• In the presence of \(F^-\) (a weak field ligand), the crystal field splitting energy is less than the pairing energy, \(\Delta_o < P\). Electrons distribute to both \(t_{2g}\) and \(e_g\) orbitals before pairing. The configuration is \(t_{2g}^4 e_g^2\). Because it has four unpaired electrons, it is an outer orbital, highly paramagnetic high-spin complex.

Unit 6: Haloalkanes and Haloarenes

6.1 Nomenclature and Nature of C-X Bond

Haloalkanes (Alkyl halides) and Haloarenes (Aryl halides) are obtained by the replacement of one or more hydrogen atoms of an aliphatic or aromatic hydrocarbon by halogen atoms (F, Cl, Br, I).

Nature of C-X bond: Halogen atoms are more electronegative than carbon. As a result, the carbon-halogen bond is polarized. The carbon atom bears a partial positive charge (\(\delta^+\)) and the halogen atom bears a partial negative charge (\(\delta^-\)). Size of halogens increases from F to I, hence the C-X bond length increases from C-F to C-I and bond enthalpy decreases.

6.2 Haloalkanes: Properties and Reactions

Physical Properties:

• Insoluble in water due to inability to form hydrogen bonds with water molecules.
• Boiling points increase with the size and mass of the halogen atom (RI > RBr > RCl > RF) due to an increase in magnitude of van der Waals forces. For isomeric haloalkanes, the boiling point decreases with branching (due to reduced surface area).

Chemical Reactions: Nucleophilic Substitution

A nucleophile (electron rich species) attacks the partially positive carbon atom of the C-X bond.

1. \(S_N2\) Mechanism (Bimolecular Nucleophilic Substitution):

   • Occurs in a single step (concerted mechanism).
   • Rate depends on the concentration of both alkyl halide and nucleophile: \(\text{Rate} = k[RX][Nu^-]\).
   • Attack of nucleophile occurs from the backside.
   • Leads to inversion of configuration (Walden inversion).
   • Order of reactivity: Primary (\(1^\circ\)) > Secondary (\(2^\circ\)) > Tertiary (\(3^\circ\)) due to steric hindrance in bulkier alkyl groups.

2. \(S_N1\) Mechanism (Unimolecular Nucleophilic Substitution):

   • Occurs in two steps. Step 1: Slow ionization to form a carbocation intermediate. Step 2: Fast attack by the nucleophile.
   • Rate depends only on the concentration of the alkyl halide: \(\text{Rate} = k[RX]\).
   • Leads to racemization if the starting alkyl halide is optically active, because the carbocation intermediate is planar and can be attacked from either side with equal probability.
   • Order of reactivity: Tertiary (\(3^\circ\)) > Secondary (\(2^\circ\)) > Primary (\(1^\circ\)) due to the stability of the carbocation (\(3^\circ > 2^\circ > 1^\circ\)).

Optical Isomerism

• Chiral carbon: A carbon attached to four different groups.
• Enantiomers: Non-superimposable mirror image isomers. They rotate plane-polarized light in opposite directions.
• Dextrorotatory (+ / d): Rotates the plane of polarized light to the right.
• Laevorotatory (- / l): Rotates the plane of polarized light to the left.
• Racemic mixture: An equimolar mixture of d- and l- enantiomers, optically inactive due to external compensation.

6.3 Haloarenes: Nature and Substitution

Aryl halides are extremely less reactive towards Nucleophilic Substitution reactions due to:

1. Resonance effect: The lone pairs on the halogen atom are in conjugation with the (\pi)-electrons of the benzene ring. The C-X bond acquires partial double bond character, making it difficult to cleave.
2. Difference in hybridization of carbon: The sp² hybridized carbon of the C-X bond in haloarenes is more electronegative than the sp³ carbon in haloalkanes, holding the electron pair of the C-X bond more tightly.

Electrophilic Substitution Reactions

Halogens attached to benzene are deactivating but ortho-para directing. While they withdraw electrons through the -I (inductive) effect, their +R (resonance) effect increases electron density slightly more at the ortho and para positions than at the meta position.

• Halogenation (with \(X_2/FeX_3\))
• Nitration (with \(HNO_3 / H_2SO_4\))
• Sulphonation (with concentrated \(H_2SO_4\))
• Friedel-Crafts alkylation and acylation

6.4 Environmental Effects of Polyhalogen Compounds

• Dichloromethane (\(CH_2Cl_2\)): Harms human central nervous system.
• Trichloromethane (Chloroform, \(CHCl_3\)): Slowly oxidized by air in the presence of light to form poisonous phosgene (\(COCl_2\)). Hence, stored in dark bottles.
• Tetrachloromethane (Carbon tetrachloride, \(CCl_4\)): Causes liver cancer in humans. O-zone depletion.
• Iodoform (\(CHI_3\)): Used as an antiseptic due to the liberation of free iodine.
• Freons (CFCs): E.g., \(CCl_2F_2\). Used as refrigerants. Responsible for severe ozone layer depletion in the stratosphere.
• DDT: An insecticide. Banned in many countries due to its high toxicity to fish, chemical stability, and fat solubility leading to biological magnification.

Competency-Based Questions (CBQs)

Q1. (CBSE 2022) Among the isomeric alkanes of molecular formula \(C_5H_{12}\), identify the one that on photochemical chlorination yields: (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides.

Answer: The possible isomers of pentane (\(C_5H_{12}\)) are:

1. n-pentane: \(CH_3-CH_2-CH_2-CH_2-CH_3\)
2. isopentane: \(CH_3-CH(CH_3)-CH_2-CH_3\)
3. neopentane: \(C(CH_3)_4\)

(i) A single monochloride: Isomer: neopentane (2,2-dimethylpropane). All 12 hydrogen atoms are equivalent. Replacement of any of them yields the same product, 1-chloro-2,2-dimethylpropane.

(ii) Three isomeric monochlorides: Isomer: n-pentane. It has three different sets of equivalent hydrogen atoms (at C1, C2, and C3). Substitution yields 1-chloropentane, 2-chloropentane, and 3-chloropentane.

(iii) Four isomeric monochlorides: Isomer: isopentane (2-methylbutane). It has four different types of hydrogen atoms. Substitution yields 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, and 1-chloro-3-methylbutane.

Q2. (Sample Paper 2023) You are provided with two compounds: 1-bromobutane and 2-bromobutane. One of these undergoes an \(S_N1\) reaction faster, while the other undergoes an \(S_N2\) reaction faster. Identify them and explain your reasoning.

Answer: 1-bromobutane (\(CH_3CH_2CH_2CH_2Br\)) is a primary (\(1^\circ\)) alkyl halide. 2-bromobutane (\(CH_3CH_2CH(Br)CH_3\)) is a secondary (\(2^\circ\)) alkyl halide.

• \(S_N2\) Reaction: Proceeds via a transition state where nucleophile attacks from the backside. This attack is highly sensitive to steric hindrance. Since a \(1^\circ\) alkyl halide has less steric hindrance around the carbon atom bearing the halogen, 1-bromobutane undergoes \(S_N2\) reaction faster.
• \(S_N1\) Reaction: Proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the reaction rate. The secondary carbocation formed by 2-bromobutane is more stable than the primary carbocation formed by 1-bromobutane due to greater +I effect and hyperconjugation. Thus, 2-bromobutane undergoes \(S_N1\) reaction faster.

Q3. (CBSE 2021) Explain why haloarenes are extremely less reactive towards nucleophilic substitution reactions as compared to haloalkanes. Support your answer with resonance structures of chlorobenzene.

Answer: Haloarenes are much less reactive towards nucleophilic substitution compared to haloalkanes due to the following reasons:

1. Resonance Effect: In haloarenes (e.g., chlorobenzene), the lone pairs of electrons on the halogen atom are in conjugation with the (\pi)-electrons of the benzene ring. This delocalization imparts a partial double bond character to the C-Cl bond, making it shorter and stronger, and difficult to break by a nucleophile.

2. sp² Hybridization representing the C-X bond: In haloarenes, the carbon atom attached to the halogen is sp² hybridized, whereas in haloalkanes it is sp³ hybridized. sp² carbons have more s-character and are more electronegative, thus holding the electron pair of the C-X bond more tightly, reducing the bond length and making it harder to break.
3. Instability of phenyl cation: If an \(S_N1\) mechanism were to occur, the resulting phenyl cation formed by self-ionization would not be stabilized by resonance.

Unit 7: Alcohols, Phenols and Ethers

7.1 Classification and Nomenclature

• Alcohols: Compounds containing one or more hydroxyl (-OH) groups attached to an aliphatic carbon atom. Classified as primary (\(1^\circ\)), secondary (\(2^\circ\)), or tertiary (\(3^\circ\)) depending on the carbon attached to the -OH group.
• Phenols: Compounds containing an -OH group directly attached to a benzene ring.
• Ethers: Compounds with an oxygen atom bonded to two alkyl or aryl groups (R-O-R’ or Ar-O-Ar’).

7.2 Alcohols

Methods of Preparation

1. From Alkenes:
   • By acid catalyzed hydration: Alkenes react with water in the presence of acid as a catalyst to form alcohols (Markovnikov’s addition).
   • By hydroboration-oxidation: Diborane (\(B_2H_6\)) reacts with alkenes followed by oxidation with \(H_2O_2\) in aqueous sodium hydroxide. Resulting alcohol appears as anti-Markovnikov addition of water.
2. From Carbonyl Compounds:
   • By reduction of aldehydes (to \(1^\circ\) alcohols) and ketones (to \(2^\circ\) alcohols) using \(LiAlH_4\), \(NaBH_4\), or catalytic hydrogenation.
   • By reduction of carboxylic acids and esters.
3. From Grignard Reagents: Reaction of aldehydes and ketones with RMgX followed by hydrolysis.
   • Formaldehyde yields \(1^\circ\) alcohols.
   • Other aldehydes yield \(2^\circ\) alcohols.
   • Ketones yield \(3^\circ\) alcohols.

Identification of Primary, Secondary, and Tertiary Alcohols (Lucas Test)

The reaction of alcohols with Lucas reagent (conc. HCl + anhydrous \(ZnCl_2\)) produces alkyl chlorides, which are insoluble in water and cause turbidity.

• \(3^\circ\) Alcohols: Produce turbidity immediately at room temperature.
• \(2^\circ\) Alcohols: Produce turbidity after about 5 minutes.
• \(1^\circ\) Alcohols: Do not produce turbidity at room temperature (requires heating).

Mechanism of Dehydration of Alcohols

Heating an alcohol with concentrated sulfuric acid at 443 K results in dehydration to form an alkene. Step 1: Protonation of alcohol. \[ CH_3-CH_2-\ddot{O}H + H^+ \rightleftharpoons CH_3-CH_2-\underset{+}{\ddot{O}}H_2 \quad (\text{Fast}) \] Step 2: Formation of carbocation: Cleavage of C-O bond. \[ CH_3-CH_2-\underset{+}{\ddot{O}}H_2 \rightarrow CH_3-\overset{+}{CH_2} + H_2O \quad (\text{Slow / Rate determining}) \] Step 3: Formation of ethene by elimination of a proton. \[ CH_3-\overset{+}{CH_2} \rightarrow CH_2=CH_2 + H^+ \quad (\text{Fast}) \]

7.3 Phenols

Methods of Preparation

1. From haloarenes (Dow’s Process): Chlorobenzene is fused with NaOH at 623K and 300 atm, followed by acidification.
2. From benzene sulphonic acid: Fused with molten NaOH followed by acidification.
3. From diazonium salts: Warming aqueous diazonium salt solution.
4. From cumene: Oxidation of cumene (isopropylbenzene) in air followed by acid hydrolysis gives phenol and acetone.

Acidic Nature of Phenol

Phenol is a stronger acid than aliphatic alcohols. Reason: The phenoxide ion left after the release of a proton is stabilized by resonance (delocalization of the negative charge over the benzene ring). In contrast, alkoxide ions are not resonance stabilized and instead destabilized by the +I effect of the alkyl group.

Effect of substituents on acidity:

• Electron Withdrawing Groups (EWG) like \(-NO_2\) increase the acidity by stabilizing the phenoxide ion through resonance/inductive effect.
• Electron Donating Groups (EDG) like an alkyl group decrease the acidity.

Electrophilic Substitution Reactions

The -OH group is a highly activating and ortho-para directing group.

• Nitration: Dilute \(HNO_3\) yields a mixture of ortho and para nitrophenols. Concentrated \(HNO_3\) with conc. \(H_2SO_4\) yields 2,4,6-trinitrophenol (picric acid).
• Halogenation: With aqueous bromine, it gives a white precipitate of 2,4,6-tribromophenol. With bromine in \(CS_2\) at low temperature, it gives ortho and para bromophenols.
• Kolbe’s Reaction: Phenoxide ion treated with \(CO_2\) followed by acidification yields salicylic acid (2-hydroxybenzoic acid).
• Reimer-Tiemann Reaction: Treating phenol with chloroform in the presence of aq. NaOH yields salicylaldehyde.

7.4 Ethers

• Preparation: Dehydration of alcohols (at 413 K) or Williamson’s synthesis (reaction of an alkyl halide with sodium alkoxide).
• Cleavage: Ethers are cleaved by strong acids like HI to give alcohols and alkyl iodides. Cleavage of the C-O bond in mixed ethers generally happens such that the halogen attaches to the smaller/less sterically hindered alkyl group for primary/secondary groups.

Competency-Based Questions (CBQs)

Q1. (CBSE 2021) How will you distinguish between propan-1-ol and propan-2-ol?

Answer: Propan-1-ol is a primary (\(1^\circ\)) alcohol, whereas propan-2-ol is a secondary (\(2^\circ\)) alcohol. They can be distinguished using the Lucas Test.

• Add a few drops of Lucas reagent (anhydrous \(ZnCl_2\) in concentrated HCl) to the unknown alcohol.
• If turbidity (cloudiness) appears after about 5 minutes, it is propan-2-ol. \[ CH_3-CH(OH)-CH_3 + HCl \xrightarrow{ZnCl_2} CH_3-CH(Cl)-CH_3 \downarrow (\text{turbidity in 5 min}) + H_2O \]
• If no turbidity appears at room temperature (even after 10-15 minutes), it is propan-1-ol. \[ CH_3-CH_2-CH_2-OH + HCl \xrightarrow{ZnCl_2} \text{No reaction at room temp.} \]

Q2. (Sample Paper 2024) Arrange the following compounds in decreasing order of their acid strength: Phenol, 4-nitrophenol, 3-nitrophenol, and 4-methylphenol. Give reason.

Answer: Decreasing order of acid strength: 4-nitrophenol > 3-nitrophenol > Phenol > 4-methylphenol

Reason:

• 4-nitrophenoI: The \(-NO_2\) group is a strong electron-withdrawing group (-R and -I effects). At the para position, it can stabilize the phenoxide ion exceptionally well through extended conjugation (resonance). Thus, it is the strongest acid.
• 3-nitrophenol: The \(-NO_2\) group at the meta position also exerts a localized electron-withdrawing (-I) effect (but no resonance or -R effect onto the phenoxide oxygen). Hence, it increases acidity compared to phenol, but not as much as in 4-nitrophenol.
• Phenol: The reference compound.
• 4-methylphenol (p-Cresol): The methyl group (\(-CH_3\)) is an electron-donating group (+I and hyperconjugation). It intensifies the negative charge on the phenoxide ion, destabilizing it and decreasing the acidic strength below that of phenol.

Q3. (CBSE 2019) Anisole (methoxybenzene) on reaction with HI gives phenol and methyl iodide, and not iodobenzene and methanol. Explain why.

Answer: Anisole has the structure \(C_6H_5-O-CH_3\). When it reacts with HI, it first undergoes protonation to form a methylphenyl oxonium ion: \[ C_6H_5-\ddot{O}-CH_3 + HI \rightarrow C_6H_5-\underset{+}{\ddot{O}}(H)-CH_3 + I^- \]

The bond between the oxygen atom and the phenyl ring (\(C_6H_5-O\)) is much stronger and harder to break because of the partial double bond character due to resonance and the \(sp^2\) hybridized nature of the carbon of the benzene ring. In contrast, the bond between oxygen and the methyl group (\(O-CH_3\)) is a weaker single bond (between \(A \) sp³ carbon and Oxygen). Therefore, the nucleophile (\(I^-\) ion) attacks the less sterically hindered \(CH_3\) group (via \(S_N2\) mechanism), cleaving the \(O-CH_3\) bond to give methyl iodide (\(CH_3I\)) and phenol (\(C_6H_5OH\)).

Unit 8: Aldehydes, Ketones and Carboxylic Acids

8.1 Aldehydes and Ketones

Aldehydes have the carbonyl group (\(-C=O\)) bonded to a carbon and hydrogen, whereas in ketones, it is bonded to two carbon atoms. Both involve an \(sp^2\) hybridized carbon atom bonded to oxygen through a double bond (one \(\sigma\) and one \(\pi\) bond). Because oxygen is much more electronegative than carbon, the carbonyl group is highly polarized with a partial positive charge on carbon (\(\delta^+\)) and a partial negative charge on oxygen (\(\delta^-\)).

Methods of Preparation

1. From Alcohols:
   • Oxidation of primary alcohols gives aldehydes (using PCC/CrO\(_3\)).
   • Oxidation of secondary alcohols gives ketones.
   • Dehydrogenation over heated copper at 573 K.
2. From Hydrocarbons:
   • Ozonolysis of alkenes.
   • Hydration of alkynes in the presence of Hg\(^{2+}\) and dil. H\(_2\)SO\(_4\).
3. From Acid Chlorides (Rosenmund Reduction): Hydrogenation of acyl chlorides over Pd on \(BaSO_4\) yields aldehydes.
4. From Nitriles:
   • Stephen reaction: Reduction of nitriles with \(SnCl_2/HCl\).
   • Reaction with Grignard reagents gives ketones.
5. From Benzene: Friedel-Crafts acylation to prepare aromatic ketones. Etard reaction to prepare benzaldehyde from toluene.

Physical and Chemical Properties

• Boiling Points: Higher than hydrocarbons and ethers of comparable molecular masses due to weak molecular association arising from dipole-dipole interactions. Lower than alcohols (as they do not form intermolecular hydrogen bonds).
• Solubility: Lower members are miscible with water because they can form hydrogen bonds with water molecules.

Mechanism of Nucleophilic Addition

The electrophilic carbonyl carbon is attacked by a nucleophile from a direction perpendicular to the plane of \(sp^2\) hybridized orbitals of carbonyl carbon. The carbon changes from \(sp^2\) to \(sp^3\) creating a tetrahedral alkoxide intermediate.

• Addition of HCN: Gives cyanohydrins.
• Addition of \(NaHSO_3\): Gives bisulphite addition product.
• Addition of Alcohols: Aldehydes form acetals; ketones form ketals.
• Addition of Ammonia Derivatives: Forms imine derivatives with elimination of water (e.g., hydrazine gives hydrazone).

Relative Reactivity: Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic (inductive) reasons. The two alkyl groups in ketones hinder the approach of nucleophiles and reduce the positive charge on the carbonyl carbon (due to their +I effect).

Reactivity of Alpha (\(\alpha\)) Hydrogen

The \(\alpha\)-hydrogen of aldehydes and ketones is acidic due to the electron-withdrawing effect of the carbonyl group and resonance stabilization of the resulting enolate anion.

1. Aldol Condensation: Aldehydes/ketones having at least one \(\alpha\)-hydrogen undergo a reaction in the presence of dilute alkali to form \(\beta\)-hydroxy aldehydes (aldol) or \(\beta\)-hydroxy ketones (ketol). Heating results in dehydration to form \(\alpha,\beta\)-unsaturated carbonyl compounds.
2. Cannizzaro Reaction: Aldehydes which do not have an \(\alpha\)-hydrogen (e.g., formaldehyde, benzaldehyde) undergo self-oxidation and reduction (disproportionation) on treatment with concentrated alkali. \[ 2HCHO + NaOH \rightarrow CH_3OH + HCOONa \]

8.2 Carboxylic Acids

Carbon compounds containing a carboxyl functional group \(-COOH\).

Methods of Preparation

1. From primary alcohols and aldehydes: Using strong oxidizing agents like alkaline \(KMnO_4\) or acidified \(K_2Cr_2O_7\).
2. From alkylbenzenes: Vigorous oxidation of alkylbenzenes with chromic acid or alkaline/acidic \(KMnO_4\) gives benzoic acid.
3. From nitriles and amides: Hydrolysis catalyzed by \(H^+\) or \(OH^-\).
4. From Grignard Reagents: Reaction with solid \(CO_2\) (dry ice) followed by acid hydrolysis.

Chemical Properties and Acidic Nature

Carboxylic acids are weaker than mineral acids but stronger than alcohols and phenols. \[ R-COOH \rightleftharpoons R-COO^- + H^+ \] The carboxylate ion is highly stabilized by two equivalent resonance structures where the negative charge is delocalized over two highly electronegative oxygen atoms.

• Effect of substituents on acidity: Electron-withdrawing groups (EWG) like \(-CF_3, -NO_2, -CN, -Cl\) stabilize the carboxylate ion through the -I / -R effect and increase acidity. Electron-donating groups (EDG) like alkyl groups destabilize the carboxylate ion and decrease acidity.

Important Reactions of Carboxylic Acids

1. Formation of Anhydrides: Heating with \(P_2O_5\) or concentrated \(H_2SO_4\).
2. Esterification: Reaction with alcohols/phenols in the presence of a mineral acid catalyst.
3. Reactions with \(PCl_5, PCl_3, SOCl_2\): The -OH group is replaced by -Cl to form acyl chlorides.
4. Reduction: Reduced to primary alcohols by \(LiAlH_4\) (but not by \(NaBH_4\)).
5. Decarboxylation: Heating sodium salts with soda lime (NaOH + CaO) yields hydrocarbons.
6. 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 small amounts of red phosphorus.

Competency-Based Questions (CBQs)

Q1. (CBSE 2022) Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: Ethanal, Propanal, Propanone, Butanone.

Answer: Increasing order of reactivity: Butanone < Propanone < Propanal < Ethanal

Reason: Nucleophilic addition to a carbonyl compound depends on:

1. Steric Hindrance: Bulky groups hinder the attack of the nucleophile. Ketones have two alkyl groups (bulkier) compared to aldehydes, making ketones less reactive. Butanone has one methyl and one ethyl group; Propanone has two methyl groups. Ethanal has a methyl group, and propanal has an ethyl group.
2. Electronic Factor (+I effect): Alkyl groups are electron-donating. More alkyl groups reduce the electrophilicity (positive charge) of the carbonyl carbon, making it less susceptible to nucleophilic attack. Ketones possess two alkyl groups, reducing the partial positive charge on carbon more than aldehydes do. Combining both effects, larger ketones are the least reactive, and smaller aldehydes are the most reactive.

Q2. (Sample Paper 2024) Predict the products of the Aldol condensation of Ethanal (Acetaldehyde) and name them. Give the overall chemical equation.

Answer: Ethanal (\(CH_3CHO\)) possesses three \(\alpha\)-hydrogen atoms. In the presence of a dilute base (e.g., dilute NaOH), it undergoes self-condensation.

Step 1: Aldol formation. \[ CH_3-CHO + CH_3-CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO \] The product is 3-hydroxybutanal (an aldol).

Step 2: Dehydration. Upon heating, an \(\alpha, \beta\)-elimination of water occurs due to the stability gained by forming a conjugated double bond. \[ CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO + H_2O \] The final product is But-2-enal (Crotonaldehyde).

Q3. (CBSE 2023) An organic compound (A) with molecular formula \(C_8H_8O\) forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollen’s or Fehling’s reagent, nor does it decolourize bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula \(C_7H_6O_2\). Identify the compounds (A) and (B) and write the reactions involved.

Answer:

1. Forms an orange-red ppt with 2,4-DNP: It contains a carbonyl group (aldehyde or ketone).
2. Gives yellow ppt with \(I_2\) and NaOH (Iodoform test): It is a methyl ketone containing the \(CH_3CO-\) group.
3. Does NOT reduce Tollen’s or Fehling’s reagents: It is not an aldehyde. Thus, it must be a ketone.
4. Does NOT decolorize bromine water: It does not have an isolated carbon-carbon double/triple bond (degree of unsaturation belongs to an aromatic ring).
5. Oxidation gives \(C_7H_6O_2\): Compound B is Benzoic Acid (\(C_6H_5COOH\)).

Combining these observations, Compound (A) is Acetophenone (\(C_6H_5COCH_3\)). Molecular formula check for A: \(C_6 + C_2 = 8\) carbons, \(H_5 + H_3 = 8\) hydrogens, 1 oxygen. Perfect match.

Reactions:

1. Iodoform Reaction: \[ C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow (\text{yellow ppt, iodoform}) + 3NaI + 3H_2O \]
2. Oxidation: \[ C_6H_5COCH_3 \xrightarrow{[O] / \text{Chromic acid}, \Delta} C_6H_5COOH , (\text{Compound B, Benzoic acid}) + CO_2 + H_2O \]

Unit 9: Amines

9.1 Classification and Nomenclature

Amines constitute an important class of organic compounds derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl/aryl group(s).

• Primary (\(1^\circ\)) amines: One hydrogen replaced by R/Ar. (e.g., \(CH_3-NH_2\))
• Secondary (\(2^\circ\)) amines: Two hydrogens replaced by R/Ar. (e.g., \(CH_3-NH-CH_3\))
• Tertiary (\(3^\circ\)) amines: All three hydrogens replaced by R/Ar. (e.g., \((CH_3)_3N\))

Like ammonia, the nitrogen atom in amines is \(sp^3\) hybridized, and the geometry is pyramidal due to the presence of an unshared pair of electrons on the nitrogen atom.

9.2 Methods of Preparation

1. Reduction of nitro compounds: \(-NO_2\) to \(-NH_2\) using \(H_2/Pd\) or \(Sn/HCl\) or \(Fe/HCl\).
2. Ammonolysis of alkyl halides: Reaction of alkyl halides with alcoholic ammonia. Leads to a mixture of primary, secondary, tertiary amines and quaternary ammonium salts.
3. Reduction of nitriles: \(R-C \equiv N\) to \(R-CH_2-NH_2\) using \(LiAlH_4\) or \(H_2/Ni\).
4. Reduction of amides: \(R-CO-NH_2\) to \(R-CH_2-NH_2\) using \(LiAlH_4\).
5. Gabriel phthalimide synthesis: Used for the preparation of pure primary aliphatic amines. (Phthalimide + KOH \(\rightarrow\) Potassium phthalimide \(\xrightarrow{RX}\) N-alkylphthalimide \(\xrightarrow{aq. NaOH}\) Primary amine).
6. Hoffmann bromamide degradation reaction: Migration of an alkyl or aryl group from carbonyl carbon of the amide to the nitrogen atom to form a primary amine containing one carbon atom less than the original amide. \[ R-CO-NH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O \]

9.3 Physical and Chemical Properties

• Boiling Points: Primary and secondary amines are engaged in intermolecular hydrogen bonding. Boiling points are \(1^\circ > 2^\circ > 3^\circ\) isomeric amines. Furthermore, they are lower than corresponding alcohols.
• Basic Character: Amines are Lewis bases (electron pair donors). The basicity of amines is influenced by +I (inductive) effect, solvation (hydration) effect, and steric hindrance.
  • Due to +I effect, alkylamines are stronger bases than ammonia.
  • Arylamines (e.g., aniline) are much weaker bases than ammonia. The lone pair of electrons on nitrogen is delocalized over the benzene ring through resonance, making it less available for protonation.

Identification of Primary, Secondary, and Tertiary Amines (Hinsberg Test)

Hinsberg’s Reagent: Benzenesulphonyl chloride (\(C_6H_5SO_2Cl\)).

1. Primary amine: Reacts to form N-alkylbenzenesulphonamide, which has an acidic hydrogen attached to nitrogen. It is soluble in alkali.
2. Secondary amine: Reacts to form N,N-dialkylbenzenesulphonamide, which does not have any acidic hydrogen. It is insoluble in alkali.
3. Tertiary amine: Does not react with Hinsberg’s reagent.

Other Important Chemical Reactions

1. Carbylamine reaction (Isocyanide test): Only primary aliphatic and aromatic amines on heating with chloroform and ethanolic KOH form foul-smelling isocyanides or carbylamines.
2. Reaction with Nitrous acid (\(HNO_2\)):
   • Primary aliphatic amines form highly unstable aliphatic diazonium salts, which decompose to give alcohols and nitrogen gas.
   • Primary aromatic amines (aniline) form relatively stable arenediazonium salts at low temperatures (273-278 K).
3. Electrophilic substitution in Aniline: The \(-NH_2\) group is strongly activating and ortho, para directing.
   • Bromination with aqueous \(Br_2\) yields 2,4,6-tribromoaniline instantly.
   • To get a monosubstituted product, the \(-NH_2\) group must be protected by acetylation (with acetic anhydride) to form acetanilide before electrophilic substitution, and then hydrolyzed back.

9.4 Diazonium Salts

General formula: \(R-N_2^+ X^-\) Preparation (Diazotisation): \[ C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O \]

Chemical Reactions:

• Sandmeyer’s Reaction: Treatment with CuCl/HCl, CuBr/HBr or CuCN/KCN yields chlorobenzene, bromobenzene, or cyanobenzene respectively.
• Gattermann Reaction: Done with Cu powder / HCl or HBr.
• Replacement by iodide ion: Warming with KI gives iodobenzene.
• Replacement by fluoride ion (Balz-Schiemann Reaction): Reaction with \(HBF_4\) followed by heating gives fluorobenzene.
• Coupling Reactions: Diazonium ions act as weak electrophiles and couple with activated aromatic rings such as phenols (giving orange dye) and anilines (giving yellow dye).

Competency-Based Questions (CBQs)

Q1. (CBSE 2022) Arrange the following compounds in decreasing order of their basic strength in aqueous solution: (a) \(C_6H_5NH_2\), \(C_2H_5NH_2\), \((C_2H_5)_2NH\), \(NH_3\) (b) Give the chemical equation for the Hinsberg test of a secondary amine.

Answer: (a) Decreasing order of basic strength in aqueous solution: \((C_2H_5)_2NH\) > \(C_2H_5NH_2\) > \(NH_3\) > \(C_6H_5NH_2\)

Reasoning:

• Aliphatic amines are generally more basic than ammonia due to the electron-donating inductive (+I) effect of alkyl groups, which increases electron density on the nitrogen atom.
• Among ethylamines in aqueous solution, the secondary amine is more basic than the primary amine because the +I effect of two ethyl groups is greater than one, while the steric hindrance is not huge for the ethyl group compared to the +I and hydration effects. (Note: For methylamines, it’s 2° > 1° > 3°; for ethylamines, it’s 2° > 3° > 1° but here only 1° and 2° are given).
• Aniline (\(C_6H_5NH_2\)) is the weakest base because the lone pair of electrons on nitrogen is delocalized into the benzene ring via resonance.

(b) Hinsberg test for a secondary amine (e.g., diethylamine): \[ (C_2H_5)_2NH + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2-N(C_2H_5)_2 + HCl \] (N,N-diethylbenzenesulphonamide precipitates out. Since there is no replaceable hydrogen on the nitrogen atom of the product, it is insoluble in aq. KOH/NaOH).

Q2. (CBSE 2021) How can you convert aniline to chlorobenzene via a diazonium salt? Provide the sequences of reactions.

Answer: This conversion is achieved in two steps using the Sandmeyer Reaction.

Step 1: Diazotisation of Aniline. Aniline is treated with nitrous acid (prepared in situ from sodium nitrite and hydrochloric acid) at a low temperature (0 - 5 °C or 273 - 278 K) to form benzene diazonium chloride. \[ C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O \]

Step 2: Sandmeyer’s Reaction. The freshly prepared benzene diazonium chloride solution is mixed with cuprous chloride (CuCl) dissolved in HCl. The diazonium group is replaced by chlorine. \[ C_6H_5N_2^+Cl^- + CuCl/HCl \xrightarrow{\Delta} C_6H_5-Cl , (\text{Chlorobenzene}) + N_2 \uparrow \]

Q3. (Sample Paper 2024) Explain why Gabriel phthalimide synthesis is preferred for synthesizing primary amines, and why it cannot be used to prepare primary aromatic amines (like aniline).

Answer:

1. Preference for Primary Amines: Gabriel phthalimide synthesis produces purely primary aliphatic amines. It involves the \(S_N2\) attack of the phthalimide nucleophile on an alkyl halide. Because the nitrogen in phthalimide is bonded to two bulky carbonyl groups, it can only attack one molecule of alkyl halide. Over-alkylation (which commonly occurs and leads to a mixture of \(1^\circ\), \(2^\circ\), and \(3^\circ\) amines in ammonolysis) is prevented.
2. Inability to prepare Primary Aromatic Amines: To prepare an aromatic amine like aniline using this method, the nucleophilic potassium phthalimide would have to undergo nucleophilic substitution with an unreactive aryl halide (like chlorobenzene). Since haloarenes do not undergo nucleophilic substitution easily (due to partial double bond character of the C-X bond and other factors), Gabriel synthesis fails for producing aromatic primary amines.

Unit 10: Biomolecules

10.1 Carbohydrates

Carbohydrates are optically active polyhydroxy aldehydes or ketones, or the compounds which produce such units on hydrolysis. They are primarily produced by plants. General formula: \(C_x(H_2O)_y\)

Classification

1. Monosaccharides: Cannot be further hydrolysed. E.g., Glucose (an aldohexose), Fructose (a ketohexose), Ribose.
2. Oligosaccharides: Yield 2 to 10 monosaccharide units on hydrolysis. Further classified as disaccharides (sucrose, maltose, lactose), trisaccharides, etc.
   • Sucrose (non-reducing sugar) yields \(\alpha\)-D-glucose and \(\beta\)-D-fructose.
   • Maltose (reducing sugar) yields two \(\alpha\)-D-glucose units.
   • Lactose (reducing sugar) yields \(\beta\)-D-galactose and \(\beta\)-D-glucose.
3. Polysaccharides: Yield a large number of monosaccharide units on hydrolysis. They are not sweet in taste (non-sugars). E.g., Starch, Cellulose, Glycogen. All are polymers of glucose.
   • Starch: Polymer of \(\alpha\)-D-glucose. Consists of Amylose (water-soluble, linear chain of C1-C4 glycosidic linkage) and Amylopectin (water-insoluble, branched chain with C1-C6 linkages).
   • Cellulose: Polymer of \(\beta\)-D-glucose with linear \(\beta\)-1,4-glycosidic linkages.

Reducing and Non-Reducing Sugars: Carbohydrates that reduce Fehling’s solution and Tollen’s reagent are reducing sugars (all monosaccharides, maltose, lactose). Sucrose is a non-reducing sugar because its reducing groups (hemiacetal -OH) are involved in glycosidic bond formation.

D- and L- Configuration: Relates the structure to glyceraldehyde. If the -OH group on the lowest chiral carbon atom is on the right, it is D-configuration; if on the left, it is L-configuration.

10.2 Proteins

Polymers of \(\alpha\)-amino acids. They are required for growth and maintenance of the body.

Amino Acids and Peptides

Amino acids contain amino (\(-NH_2\)) and carboxyl (\(-COOH\)) functional groups. Except glycine, all naturally occurring \(\alpha\)-amino acids are optically active (L-series).

• Zwitterion: In aqueous solution, the carboxyl group loses a proton (\(H^+\)) and the amino group gains a proton, forming a dipolar ion called a Zwitterion.
• Peptide bond (\(-CO-NH-\) link): Formed between the carboxyl group of one amino acid and the amino group of another with the elimination of water.

Structure of Proteins

1. Primary Structure: The specific sequence of amino acids in the polypeptide chain.
2. Secondary Structure: The shape in which a long polypeptide chain can exist due to regular folding (e.g., \(\alpha\)-helix and \(\beta\)-pleated sheet), stabilized by hydrogen bonds between the \(-NH\) and \(-C=O\) groups of the peptide bond.
3. Tertiary Structure: The overall 3D shape of an entire protein molecule resulting from further folding of the secondary structure. Stabilized by hydrogen bonds, disulphide linkages, van der Waals forces, and electrostatic forces.
4. Quaternary Structure: The spatial arrangement of two or more polypeptide chains (subunits) with respect to each other (e.g., Haemoglobin).

Denaturation of Proteins

When a protein in its native form is subjected to physical change (temperature) or chemical change (pH), hydrogen bonds are disturbed. Globules unfold and helices get uncoiled, and the protein loses its biological activity. Note: During denaturation, secondary and tertiary structures are broken, but the primary structure remains intact. Example: Coagulation of egg white on boiling, curdling of milk.

10.3 Enzymes, Vitamins, and Hormones

• Enzymes: Biological catalysts. Almost all are globular proteins. They are highly specific for particular reactions. Example: Invertase hydrolyses sucrose; Maltase hydrolyses maltose.
• Vitamins: Organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health.
  • Water-soluble: Vitamins B and C (must be supplied regularly as they are excreted in urine).
  • Fat-soluble: Vitamins A, D, E, K (stored in liver and adipose tissues).
• Hormones: Chemical messengers secreted directly into the bloodstream by endocrine glands. They coordinate various physiological processes.

10.4 Nucleic Acids

Polymers of nucleotides (polynucleotides). They are responsible for the transmission of inherent characters from one generation to the next (heredity) and protein synthesis.

• DNA (Deoxyribonucleic Acid): Contains sugar (\(\beta\)-D-2-deoxyribose), a phosphate group, and nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T). Has a double-helical structure.
• RNA (Ribonucleic Acid): Contains sugar (\(\beta\)-D-ribose). Nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), and Uracil (U). Usually short, single-stranded.

Competency-Based Questions (CBQs)

Q1. (CBSE 2024 Pattern) Two samples of carbohydrates (A and B) are analyzed. Sample A reduces Fehling’s solution, whereas Sample B does not. Furthermore, upon hydrolysis in an acidic medium, Sample B yields an equimolar mixture of a dextrorotatory sugar and a laevorotatory sugar, which when combined has a net laevoritatory rotation. Identify sugars A and B with proper justification.

Answer: Sample A: is a reducing sugar. It could be any monosaccharide (like glucose or fructose) or a reducing disaccharide (like maltose or lactose) since they have a free aldehyde/ketone group. Sample B: is a non-reducing sugar. Since Sample B does not reduce Fehling’s solution, it strongly suggests Sucrose. Justification: Sucrose (A dextrorotatory disaccharide, \(+66.5^\circ\)) on hydrolysis gives an equimolar mixture of D-(+)-glucose (\(+52.5^\circ\)) and D-(-)-fructose (\(-92.4^\circ\)). Since the laevorotation of fructose \(-92.4^\circ\) is greater in magnitude than the dextrorotation of glucose \(+52.5^\circ\), the resulting hydrolysed mixture is laevorotatory. This change in specific rotation from dextro to laevo is called inversion of sugar, and the mixture is called invert sugar, exactly matching the description of Sample B.

Q2. (Sample Paper 2023) Hard-boiled eggs and curd are typical examples of protein chemistry in everyday life. Explain the biochemical process involved when an egg is boiled in water.

Answer: When an egg is boiled in water, the heat causes a physical change in the protein present in the egg white (albumin). This process is known as Denaturation of Protein.

1. Due to the high temperature, the hydrogen bonds, disulphide linkages, and other stabilizing forces holding the intricate 3D structure of the protein are disrupted.
2. The globules unfold (tertiary and quaternary structure break down) and the helices uncoil (secondary structure is lost).
3. The primary structure (the sequence of amino acids linked by peptide bonds) remains completely intact.
4. The uncoiled protein chains get entangled and form a massive network via new intermolecular cross-links, causing the water soluble, translucent egg white to coagulate into a firm, water-insoluble opaque white mass. Denaturation renders the protein biologically inactive.

Q3. (CBSE 2020) Distinguish between the following pairs on the basis of their chemical structure/composition: (i) DNA and RNA (ii) Amylose and Amylopectin

Answer: (i) DNA and RNA:

• Sugar: DNA contains \(\beta\)-D-2-deoxyribose sugar; RNA contains \(\beta\)-D-ribose sugar.
• Nitrogenous Bases: DNA contains Adenine (A), Guanine (G), Cytosine (C), and Thymine (T). RNA contains Adenine, Guanine, Cytosine, and Uracil (U) instead of Thymine.
• Structure: DNA usually exists as a very long double-stranded helix. RNA usually exists as a shorter single-stranded molecule (though it can fold onto itself).

(ii) Amylose and Amylopectin: Both are components of starch and are polymers of \(\alpha\)-D-glucose.

• Amylose: Water-soluble component of starch (approx 15-20%). It is a long unbranched chain consisting of 200-1000 \(\alpha\)-D-(+)-glucose units held together by C1-C4 glycosidic linkages.
• Amylopectin: Water-insoluble fraction of starch (approx 80-85%). It is a highly branched chain polymer. The straight chains are formed by C1-C4 glycosidic linkages, whereas the branching occurs by C1-C6 glycosidic linkages.