Step 1: Nature of potassium sulphate.

Potassium sulphate (K\(_2\)SO\(_4\)) consists of potassium cations (K\(^+\)) and sulphate anions (SO\(_4^{2-}\)). These ions are held together by strong electrostatic forces of attraction.


Step 2: Classify the solid.

- Ionic solids are characterized by a lattice structure of cations and anions.

- Metallic solids consist of metal atoms with delocalized electrons.

- Covalent solids are formed by atoms connected through covalent bonds in a network.

- Molecular solids are made up of molecules that are held together by weak forces like van der Waals or hydrogen bonds.


Step 3: Conclusion.

Since K\(_2\)SO\(_4\) is composed of ions held together by ionic bonds, it is classified as an ionic solid.


Final Answer: \[ \boxed{(A) Ionic solid} \] Quick Tip: Ionic solids are formed by strong electrostatic forces between cations and anions; they have high melting points and conduct electricity in molten state.

Step 1: Formula for number of moles.

The number of moles is given by: \[ Number of moles = \frac{Given mass}{Molar mass} \]

Step 2: Molar mass of water.

The molar mass of H\(_2\)O is calculated as: \[ Molar mass of H\(_2\)O = 2 \times 1 + 16 = 18 \, g/mol \]

Step 3: Substitution.

Now, substitute the values into the formula: \[ Number of moles = \frac{180}{18} = 10 \]

Step 4: Conclusion.

Thus, 180 g of water contains 10 moles.


Final Answer: \[ \boxed{10 \, moles} \] Quick Tip: Always divide the given mass of a substance by its molar mass to calculate moles.

Step 1: Understanding molecularity.

Molecularity refers to the number of reacting species (molecules, atoms, or ions) involved in the elementary step of a chemical reaction.


Step 2: Reaction given.

The given reaction is: \[ A + 2B \longrightarrow Product \]
This indicates that one molecule of A and two molecules of B participate in the reaction.


Step 3: Calculate molecularity.

The total number of species involved in the reaction is: \[ 1 + 2 = 3 \]

Step 4: Conclusion.

Thus, the molecularity of this reaction is 3.


Final Answer: \[ \boxed{3} \] Quick Tip: Molecularity is always a whole number and is determined by adding the number of reactant species in an elementary reaction step.

Step 1: Recall Cannizzaro’s reaction.

Cannizzaro’s reaction occurs with aldehydes that lack an \(\alpha\)-hydrogen atom. In this reaction, one molecule of the aldehyde is reduced to an alcohol, while another molecule is oxidized to a carboxylic acid, in the presence of a strong base like concentrated NaOH or KOH.


Step 2: Applying to formaldehyde.

Formaldehyde (HCHO) does not have an \(\alpha\)-hydrogen atom. Thus, it undergoes Cannizzaro’s reaction to produce:
- Methanol (CH\(_3\)OH), and
- Formic acid (HCOOH).


Step 3: Analyzing the options.

- (A) Methane: Not produced.

- (B) Methyl alcohol: Correct, formaldehyde forms methyl alcohol (methanol).

- (C) Methyl cyanide: This is not a product of this reaction.

- (D) Ethyl amine: Not formed in this reaction.


Step 4: Conclusion.

Hence, formaldehyde forms methyl alcohol in Cannizzaro’s reaction.


Final Answer: \[ \boxed{(B) Methyl alcohol} \] Quick Tip: Aldehydes without \(\alpha\)-hydrogens undergo Cannizzaro’s reaction to give one molecule of alcohol and one molecule of acid.

Step 1: Recall the carbylamine reaction.

In the carbylamine reaction, a primary amine (R–NH\(_2\)) reacts with chloroform (CHCl\(_3\)) and alcoholic KOH to form isocyanides (R–NC), which are known for their foul smell.


Step 2: Match with the given reaction.

The given equation: \[ R–NH_{2} + CHCl_{3} + 3KOH \longrightarrow RNC + 3KCl + 3H_{2}O \]
matches exactly with the carbylamine test.


Step 3: Analyzing options.

- (A) Coupling reaction: This involves diazonium salts and is not relevant to the current reaction.

- (B) Carbylamine reaction: Correct. This reaction forms isocyanides (R–NC).

- (C) Hoffmann bromamide reaction: This converts amides to amines, which is not applicable here.

- (D) Schmidt reaction: This involves hydrazoic acid and carboxylic acids, a different type of reaction.


Step 4: Conclusion.

Thus, the given reaction is the carbylamine reaction.


Final Answer: \[ \boxed{(B) Carbylamine reaction} \] Quick Tip: Carbylamine test is used to detect primary amines; the product is an isocyanide (R–NC) with a foul odour.

Step 1: Structure of glucose.

Glucose is an aldohexose with the molecular formula C\(_6\)H\(_{12}\)O\(_6\). Its open-chain structure is: \[ HOCH_2 – (CHOH)_4 – CHO \]

Step 2: Identify alcoholic groups.

- The four middle carbons each have a hydroxyl (–OH) group, making them secondary alcohols.

- The terminal –CH\(_2\)OH group at C-6 is a primary alcoholic group.


Step 3: Conclusion.

Thus, glucose has only one primary alcoholic group (at carbon-6).


Final Answer: \[ \boxed{(A) One} \] Quick Tip: In glucose, the –CH\(_2\)OH group at the end of the chain is the only primary alcohol; the rest are secondary alcohols.

Step 1: Definition.

Packing efficiency is defined as the percentage of total space in a crystal structure that is actually occupied by the constituent particles (atoms, ions, or molecules).


Step 2: General Expression.
\[ Packing Efficiency = \frac{Volume occupied by particles in the unit cell}{Total volume of unit cell} \times 100 \]

Step 3: Example.

In face-centered cubic (FCC) structure:
- 74% of the volume is occupied by spheres.

- 26% of the volume remains vacant (void space).



Conclusion:

Packing efficiency determines how closely the particles are packed in different crystal lattice structures. Quick Tip: Among simple cubic, body-centered cubic (BCC), and face-centered cubic (FCC), the FCC lattice has the highest packing efficiency (74%).

(i) Reaction of chlorine with sulphur dioxide:

When chlorine reacts with sulphur dioxide in the presence of water, sulphuric acid and hydrochloric acid are formed. \[ Cl_2 + SO_2 + 2H_2O \longrightarrow H_2SO_4 + 2HCl \]

(ii) Reaction of chlorine with turpentine oil:

Chlorine reacts violently with turpentine oil (a mixture of terpenes) producing fumes. This reaction is used in the laboratory as a test for chlorine gas. \[ Cl_2 + Turpentine oil \longrightarrow Chlorinated products (with dense fumes) \]


Conclusion:

- With sulphur dioxide, chlorine forms sulphuric acid and hydrochloric acid.

- With turpentine oil, chlorine gives a vigorous reaction producing white fumes. Quick Tip: Turpentine oil is often used as a qualitative test for chlorine gas since it produces dense white fumes on reaction.

Step 1: Nature of reaction.

When a copper rod is dipped into AgNO\(_3\) solution, a redox reaction takes place. Copper is more reactive than silver and displaces it from silver nitrate solution.


Step 2: Chemical reaction.
\[ Cu (s) + 2AgNO_3 (aq) \rightarrow Cu(NO_3)_2 (aq) + 2Ag (s) \]

Step 3: Explanation of colour change.

- Cu(NO\(_3\))\(_2\) is formed in the solution, which contains Cu\(^{2+}\) ions.

- Cu\(^{2+}\) ions impart a blue colour to the solution.

- Silver metal (Ag) is deposited on the surface of the copper rod.



Conclusion:

The solution turns blue because Cu\(^{2+}\) ions are formed in solution, and silver gets deposited on the copper rod. Quick Tip: Always remember: In displacement reactions, the more reactive metal displaces the less reactive one from its salt solution.

Step 1: Definition.

A Schottky defect is a type of point defect in ionic crystals where equal numbers of cations and anions are missing from their lattice sites.


Step 2: Characteristics.


It reduces the density of the crystal.
It maintains electrical neutrality because both cations and anions are missing in equal numbers.
It is common in highly ionic compounds with high coordination numbers such as NaCl, KCl, CsCl.


Step 3: Diagram representation.
\[ \begin{array}{|c|c|c|c|} \hline + & - & + & -
\hline - & \square & - & +
\hline + & - & \square & -
\hline - & + & - & +
\hline \end{array} \]
Here, \(\square\) indicates missing ions representing the defect.



Conclusion:

A Schottky defect is a vacancy defect where cations and anions are absent from lattice sites, leading to decreased density of the solid. Quick Tip: Schottky defect reduces density, whereas Frenkel defect does not.

Step 1: Definition.

Packing efficiency is defined as the percentage of total space in a crystal structure that is actually occupied by the constituent particles (atoms, ions, or molecules).


Step 2: General Expression.
\[ Packing Efficiency = \frac{Volume occupied by particles in the unit cell}{Total volume of unit cell} \times 100 \]

Step 3: Example.

In face-centered cubic (FCC) structure:
- 74% of the volume is occupied by spheres.

- 26% of the volume remains vacant (void space).



Conclusion:

Packing efficiency determines how closely the particles are packed in different crystal lattice structures. Quick Tip: Among simple cubic, body-centered cubic (BCC), and face-centered cubic (FCC), the FCC lattice has the highest packing efficiency (74%).

(i) Reaction of chlorine with sulphur dioxide:

When chlorine reacts with sulphur dioxide in the presence of water, sulphuric acid and hydrochloric acid are formed. \[ Cl_2 + SO_2 + 2H_2O \longrightarrow H_2SO_4 + 2HCl \]

(ii) Reaction of chlorine with turpentine oil:

Chlorine reacts violently with turpentine oil (a mixture of terpenes) producing fumes. This reaction is used in the laboratory as a test for chlorine gas. \[ Cl_2 + Turpentine oil \longrightarrow Chlorinated products (with dense fumes) \]


Conclusion:

- With sulphur dioxide, chlorine forms sulphuric acid and hydrochloric acid.

- With turpentine oil, chlorine gives a vigorous reaction producing white fumes. Quick Tip: Turpentine oil is often used as a qualitative test for chlorine gas since it produces dense white fumes on reaction.

Step 1: Identify ligands and counter ions.

The given compound is \([Cu(NH_3)_4]SO_4\).
- The coordination sphere is \([Cu(NH_3)_4]^{2+}\).
- Sulphate ion \((SO_4^{2-})\) is outside the coordination sphere.


Step 2: Charge balance.

Let oxidation state of Cu = \(x\).

Ammonia \((NH_3)\) is a neutral ligand, so its charge = 0. \[ x + (4 \times 0) = +2 \] \[ x = +2 \]


Conclusion:

The oxidation number of copper in the compound \([Cu(NH_3)_4]SO_4\) is: \[ \boxed{+2} \] Quick Tip: Neutral ligands like NH\(_3\), H\(_2\)O, CO do not affect oxidation number; only counter ions (like SO\(_4^{2-}\), Cl\(^-\)) matter.

Step 1: Nature of glucose.

Glucose is an aldohexose and contains an aldehyde group (–CHO). Bromine water is a mild oxidising agent.


Step 2: Reaction.

Bromine water oxidises the –CHO group of glucose to –COOH, forming gluconic acid.
\[ C_6H_{12}O_6 \,(Glucose) + Br_2 + H_2O \;\longrightarrow\; C_6H_{12}O_7 \,(Gluconic \, acid) + 2HBr \]


Step 3: Observation.

The reddish-brown colour of bromine water disappears, confirming the presence of the aldehyde group in glucose.


Conclusion:

Glucose reacts with bromine water to form gluconic acid, showing that glucose contains an aldehyde group. Quick Tip: Bromine water oxidises only aldehydes (–CHO) to acids but does not oxidise ketones.

Step 1: General form of the Nernst equation.

The Nernst equation relates the electrode potential of a half-cell to the standard electrode potential, temperature, and the activities (or concentrations) of the chemical species involved.

\[ E = E^\circ - \frac{2.303RT}{nF} \log \frac{[Products]}{[Reactants]} \]

At \(298 \, K\), the equation simplifies to: \[ E = E^\circ - \frac{0.0591}{n} \log \frac{[Products]}{[Reactants]} \]

where,
\(E\) = electrode potential
\(E^\circ\) = standard electrode potential
\(R\) = gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\))
\(T\) = temperature (in Kelvin)
\(n\) = number of electrons involved in reaction
\(F\) = Faraday constant (96500 C mol\(^{-1}\))


Step 2: One application.

The Nernst equation is used to calculate the electrode potential of a cell under non-standard conditions. For example, the potential of a Daniell cell: \[ Zn | Zn^{2+} (aq) || Cu^{2+} (aq) | Cu \]
can be calculated under different concentrations of Zn\(^{2+}\) and Cu\(^{2+}\).


Conclusion:

The Nernst equation provides a direct relation between electrode potential and ionic concentrations, helping in predicting the feasibility and direction of redox reactions. Quick Tip: At \(298 \, K\), always remember the simplified form: \(E = E^\circ - \frac{0.0591}{n} \log Q\).

Step 1: Understanding colloids.

Colloids are classified into two categories based on the interaction of dispersed phase with dispersion medium: lyophilic (solvent-loving) and lyophobic (solvent-hating).


Step 2: Differences.


\begin{tabular{|c|c|c|
\hline
Property & Lyophilic Colloids & Lyophobic Colloids

\hline
Affinity for medium & High affinity for dispersion medium & Little or no affinity for medium

\hline
Stability & Very stable and not easily coagulated & Unstable and easily coagulated

\hline
Reversibility & Reversible sols & Irreversible sols

\hline
Examples & Starch sol, Gum, Gelatin & Gold sol, Sulphur sol, As\(_2\)S\(_3\) sol

\hline
Preparation & Form spontaneously & Require special methods

\hline
\end{tabular


Conclusion:

Lyophilic colloids are stable and reversible due to strong attraction for solvent, while lyophobic colloids are unstable, irreversible and need stabilizers for existence. Quick Tip: Lyophilic = “solvent loving” (stable, reversible); Lyophobic = “solvent hating” (unstable, irreversible).

Step 1: Definition.

Acetylation is the process of introducing an acetyl functional group (\(-COCH_3\)) into a molecule. It usually involves the replacement of a hydrogen atom with an acetyl group.


Step 2: Types of Acetylation.

1. Chemical Acetylation: Reaction of compounds with acetic anhydride or acetyl chloride. Example: Conversion of salicylic acid to acetylsalicylic acid (aspirin).

2. Biological Acetylation: In living organisms, enzymes like acetyltransferases transfer acetyl groups, such as acetylation of histone proteins that regulate gene expression.



Conclusion:

Acetylation is important both in organic synthesis (e.g., formation of aspirin) and in biological systems (e.g., regulation of gene activity). Quick Tip: Histone acetylation in cells plays a vital role in turning genes “on” or “off” by altering chromatin structure.

Step 1: Definition.

Denaturation of proteins is the process in which proteins lose their native three-dimensional structure without breaking peptide bonds. This leads to the loss of their biological activity.


Step 2: Causes of Denaturation.

Denaturation can be caused by heat, strong acids or bases, heavy metal ions, organic solvents, or radiation.


Step 3: Effects of Denaturation.

- Hydrogen bonds and other non-covalent interactions are disrupted.

- The protein unfolds, losing its specific shape and biological function.

- Example: When egg white (albumin) is boiled, it turns from transparent to opaque due to protein denaturation.



Conclusion:

Denaturation of proteins is an irreversible process in most cases, resulting in the loss of their natural activity and function. Quick Tip: Always remember: Denaturation affects the shape and function of proteins but does not break peptide bonds of the primary structure.

Step 1: Formula for osmotic pressure.
\[ \pi = C R T \]
where, \(C\) = molar concentration = \(\dfrac{M}{10} = 0.1 \, mol \, L^{-1}\)
\(R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}\(
\(T = 27^\circ C = 27 + 273 = 300 \, K\)


Step 2: Substitution.
\[ \pi = 0.1 \times 0.0821 \times 300 \] \[ \pi = 2.463 \, atm \]


Conclusion:

The osmotic pressure of the solution is: \[ \boxed{2.46 \, atm} \] Quick Tip: Always convert temperature to Kelvin when using gas constant \(R\).

Step 1: Formula for first-order rate constant.
\[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]
where, \([R]_0 = 1.24 \times 10^{-2} \, mol L^{-1}\) (initial concentration)
\([R] = 0.20 \times 10^{-2} \, mol L^{-1}\) (concentration after 60 min)
\(t = 60 \, min = 60 \times 60 = 3600 \, s\)


Step 2: Substitution.
\[ k = \frac{2.303}{3600} \log \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} \] \[ k = \frac{2.303}{3600} \log (6.2) \] \[ \log (6.2) \approx 0.792 \] \[ k = \frac{2.303 \times 0.792}{3600} \] \[ k \approx 5.06 \times 10^{-4} \, s^{-1} \]


Conclusion:

The rate constant of the reaction at 318 K is: \[ \boxed{5.06 \times 10^{-4} \, s^{-1}} \] Quick Tip: For first-order reactions, the half-life is independent of concentration: \(t_{1/2} = \dfrac{0.693}{k}\).

Step 1: Definition.

Transition elements are d-block elements that have partially filled d-orbitals in their ground state or in one of their oxidation states.


Step 2: Characteristics.

Some important characteristics of transition elements are:


Variable oxidation states: Transition metals commonly exhibit more than one oxidation state (e.g., Fe\(^{2+}\) and Fe\(^{3+}\)).
Formation of coloured compounds: Due to d–d electronic transitions, many transition metal salts and complexes are coloured.
Catalytic properties: They act as good catalysts because of variable oxidation states and ability to provide active sites (e.g., V\(_2\)O\(_5\) in contact process).
Formation of complex compounds: Transition metals form coordination complexes with ligands due to availability of vacant d-orbitals (e.g., [Fe(CN)\(_6\)]\(^{3-}\)).



Conclusion:

Thus, transition elements show variable oxidation states, coloured ions, catalytic activity, and complex formation as their main characteristics. Quick Tip: The unique properties of transition metals are due to the presence of partially filled d-orbitals.

Step 1: Rules of IUPAC nomenclature.

The IUPAC system provides a standard way of naming coordination compounds. The rules include:


Name of cation is written before anion.

For example: [Co(NH\(_3\))\(_6\)]Cl\(_3\) is named as hexaamminecobalt(III) chloride.

Ligands are named first, followed by the central atom.

Ligands are named in alphabetical order irrespective of charge.

Prefixes are used for number of ligands.

Mono-, di-, tri-, etc. are used. For complex ligands, bis-, tris-, tetrakis- are used.

Oxidation state of the central metal ion is shown in Roman numerals.

Example: [Fe(CN)\(_6\)]\(^{4-}\) → hexacyanoferrate(II).

Anionic complexes end with -ate.

Example: [PtCl\(_6\)]\(^{2-}\) → hexachloroplatinate(IV).



Step 2: Example.

For [Cr(H\(_2\)O)\(_4\)Cl\(_2\)]Cl: Name is tetraaquadichlorochromium(III) chloride.


Conclusion:

The IUPAC nomenclature system ensures clarity and uniformity in naming coordination compounds by following systematic rules. Quick Tip: Always name ligands first (alphabetical order), then the central atom with oxidation state, and finally the counter ion.

Step 1: Preparation of Ozone.

Ozone is prepared by passing silent electric discharge through pure, dry oxygen. \[ 3O_2 \xrightarrow{electric\ discharge} 2O_3 \]

Step 2: Oxidizing Properties of Ozone.

1. Oxidation of Lead Sulphide: \[ PbS + O_3 \longrightarrow PbSO_4 \]

2. Oxidation of Potassium Iodide: \[ 2KI + O_3 + H_2O \longrightarrow 2KOH + I_2 + O_2 \]

3. Oxidation of Ferrous Salts: \[ 2Fe^{2+} + O_3 + 2H^+ \longrightarrow 2Fe^{3+} + O_2 + H_2O \]


Conclusion:

Ozone is a powerful oxidizing agent and can oxidize sulphides, iodides, and ferrous salts effectively. Quick Tip: Ozone is stronger than oxygen as an oxidizing agent due to the presence of weak oxygen-oxygen bonds.

Step 1: Definition of Aqua Regia.

Aqua regia is a freshly prepared mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3:1. It is a powerful oxidizing and dissolving medium, capable of dissolving noble metals like gold and platinum.

Step 2: Reaction with Gold.

Gold dissolves in aqua regia forming chloroauric acid. \[ Au + HNO_3 + 4HCl \longrightarrow H[AuCl_4] + NO + 2H_2O \]

Step 3: Reaction with Platinum.

Platinum dissolves in aqua regia forming chloroplatinic acid. \[ Pt + 4HNO_3 + 6HCl \longrightarrow H_2[PtCl_6] + 4NO_2 + 4H_2O \]


Conclusion:

Aqua regia is a unique mixture that dissolves even noble metals like gold and platinum, which do not react with single acids. Quick Tip: The name “Aqua Regia” means “Royal Water” because of its ability to dissolve noble metals like gold and platinum.

Part 1: Two methods of preparation of monohydric alcohol.


Method 1: By hydration of alkenes.

Ethene reacts with water in presence of conc. H\(_2\)SO\(_4\) or phosphoric acid catalyst to give ethanol: \[ CH_2=CH_2 + H_2O \;\xrightarrow{H_2SO_4}\; CH_3CH_2OH \]

Method 2: By reduction of aldehydes.

Acetaldehyde on reduction with hydrogen gives ethanol: \[ CH_3CHO + H_2 \;\xrightarrow{Ni/Pt}\; CH_3CH_2OH \]


Part 2: Reactions of ethyl alcohol with conc. H\(_2\)SO\(_4\) at different temperatures.


(i) At 443 K (140\(^\circ\)C): Formation of diethyl ether.
\[ C_2H_5OH + C_2H_5OH \;\xrightarrow[443K]{conc. H_2SO_4}\; C_2H_5OC_2H_5 + H_2O \]

(ii) At 443–463 K: Dehydration to form ethene.
\[ C_2H_5OH \;\xrightarrow[443-463K]{conc. H_2SO_4}\; CH_2=CH_2 + H_2O \]

(iii) As oxidising agent: Formation of acetaldehyde.
\[ CH_3CH_2OH \;\xrightarrow{conc. H_2SO_4}\; CH_3CHO + H_2 \]


Conclusion:

Monohydric alcohols can be prepared by hydration of alkenes or reduction of aldehydes. Ethanol reacts with conc. H\(_2\)SO\(_4\) differently depending on temperature, giving ether, ethene, or acetaldehyde. Quick Tip: Remember the temperature effect: \(140^\circ C \to\) ether, \(170^\circ C \to\) ethene, higher \(T \to\) oxidation products.

(i) Reimer–Tiemann Reaction:

The Reimer–Tiemann reaction is used to introduce a formyl group (\(-CHO\)) at the ortho position of phenol.

When phenol is treated with chloroform (CHCl\(_3\)) and aqueous sodium hydroxide, it gives salicylaldehyde (o-hydroxybenzaldehyde).
\[ C_6H_5OH + CHCl_3 + 3NaOH \; \longrightarrow \; o\!-\!HOC_6H_4CHO + 3NaCl + 2H_2O \]
This reaction is important for preparing aromatic aldehydes.



(ii) Kolbe Reaction:

The Kolbe reaction is the carboxylation of phenol to produce salicylic acid.

Phenol is first converted to sodium phenoxide, which reacts with carbon dioxide under high pressure and temperature, followed by acidification.
\[ C_6H_5ONa + CO_2 \; \xrightarrow{373K, \; 4-7 atm} \; o\!-\!HOC_6H_4COONa \] \[ o\!-\!HOC_6H_4COONa + HCl \; \longrightarrow \; o\!-\!HOC_6H_4COOH \]
This method is widely used for the synthesis of salicylic acid, a precursor of aspirin.


Conclusion:

Reimer–Tiemann introduces an aldehyde group at the ortho position of phenol, while Kolbe introduces a carboxyl group, both being important named reactions in aromatic chemistry. Quick Tip: Remember: Reimer–Tiemann → \(-CHO\) group (aldehyde), Kolbe → \(-COOH\) group (acid).

(i) Chlorobenzene to Picric Acid:
\[ C_6H_5Cl \; \xrightarrow{Conc. HNO_3 / Conc. H_2SO_4} \; O_2N\!-\!C_6H_4Cl \; \xrightarrow{Further \; Nitration} \; (NO_2)_3C_6H_2OH \; (Picric Acid) \]



(ii) Chlorobenzene to Dichlorobenzene:
\[ C_6H_5Cl \; \xrightarrow{Cl_2 / FeCl_3} \; C_6H_4Cl_2 \; (o- and p-dichlorobenzene) \]



(iii) Chlorobenzene to Chlorobenzene Sulphonic Acid:
\[ C_6H_5Cl \; \xrightarrow{Conc. H_2SO_4, \; Heat} \; ClC_6H_4SO_3H \]


Conclusion:

Chlorobenzene undergoes nitration to give picric acid, halogenation to give dichlorobenzene, and sulphonation to give chlorobenzene sulphonic acid. Quick Tip: Electrophilic substitution reactions of chlorobenzene give nitro, chloro, and sulpho derivatives depending on the reagents used.

Step 1: Introduction.

Haloalkanes (alkyl halides) undergo nucleophilic substitution reactions in which the halogen atom (a good leaving group) is replaced by a nucleophile (electron-rich species such as OH\(^-\), CN\(^-\), NH\(_3\), etc.). These reactions are important for the preparation of a wide variety of organic compounds.


Step 2: Types of nucleophilic substitution.



SN1 mechanism (Unimolecular):
- Occurs in two steps.
- First, the C–X bond breaks heterolytically to form a carbocation.
- Then, the nucleophile attacks the carbocation.
- Favoured in tertiary haloalkanes due to carbocation stability.

Example:
\[ (CH_3)_3C{-}Br \;\xrightarrow{H_2O}\; (CH_3)_3C{-}OH + HBr \]

SN2 mechanism (Bimolecular):
- Occurs in one step (simultaneous bond breaking and bond forming).
- Nucleophile attacks from the backside, leading to inversion of configuration (Walden inversion).
- Favoured in primary haloalkanes.

Example:
\[ CH_3{-}CH_2{-}Br + OH^- \;\longrightarrow\; CH_3{-}CH_2{-}OH + Br^- \]



Step 3: Importance.

Nucleophilic substitution in haloalkanes is a key synthetic method for preparing alcohols, amines, nitriles, ethers, etc.


Conclusion:

Haloalkanes undergo nucleophilic substitution reactions mainly by SN1 or SN2 pathways depending on the structure of the alkyl group. These reactions form the basis for a wide range of organic syntheses. Quick Tip: Remember: Primary haloalkanes → SN2, Tertiary haloalkanes → SN1, Secondary can follow either depending on conditions.

(i) From Toluene:

Toluene is oxidized with alkaline or acidic KMnO\(_4\) to form benzoic acid.
\[ C_6H_5CH_3 \; \xrightarrow{[O], KMnO_4} \; C_6H_5COOH \]



(ii) From Benzamide:

Benzamide undergoes hydrolysis in the presence of dilute acid or alkali to give benzoic acid.
\[ C_6H_5CONH_2 + H_2O \; \xrightarrow{H^+ / OH^-} \; C_6H_5COOH + NH_3 \]



(iii) From Ethyl Benzoate:

Ethyl benzoate undergoes acidic or basic hydrolysis to give benzoic acid.
\[ C_6H_5COOC_2H_5 + H_2O \; \xrightarrow{H^+ / OH^-} \; C_6H_5COOH + C_2H_5OH \]


Conclusion:

Benzoic acid can be prepared by oxidation of toluene, hydrolysis of benzamide, and hydrolysis of ethyl benzoate. Quick Tip: Oxidation of side chains (toluene) and hydrolysis of amides/esters are standard methods to prepare benzoic acid.

(i) Aldol Condensation:

Aldol condensation is a reaction in which aldehydes or ketones having \(\alpha\)-hydrogen atoms undergo self-condensation in the presence of a dilute base (NaOH, KOH) to form \(\beta\)-hydroxy aldehydes (aldols) or \(\beta\)-hydroxy ketones. On heating, these dehydrate to form \(\alpha,\beta\)-unsaturated aldehydes or ketones.
\[ 2CH_3CHO \; \xrightarrow{NaOH} \; CH_3CH(OH)CH_2CHO \; \xrightarrow{\Delta} \; CH_3CH=CHCHO \]



(ii) Decarboxylation:

Decarboxylation is the elimination of a carboxyl group (\(-COOH\)) from carboxylic acids as carbon dioxide. This generally takes place in the presence of soda lime (NaOH + CaO).
\[ C_6H_5COOH + NaOH \; \xrightarrow{CaO, \; \Delta} \; C_6H_6 + Na_2CO_3 \]
This reaction is useful for decreasing the carbon chain length by one carbon atom.



(iii) Cannizzaro Reaction:

Aldehydes which do not have \(\alpha\)-hydrogen undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali to form a primary alcohol and a carboxylate salt.
\[ 2HCHO \; \xrightarrow{Conc. NaOH} \; CH_3OH + HCOONa \]
Example: Formaldehyde gives methanol and sodium formate.


Conclusion:

Aldol condensation forms C–C bonds, decarboxylation removes \(-COOH\) groups, and Cannizzaro reaction gives alcohols and salts from aldehydes without \(\alpha\)-hydrogen. Quick Tip: Aldol = \(\alpha\)-H present, Cannizzaro = \(\alpha\)-H absent, Decarboxylation = loss of \(CO_2\).