Step 1: Definition of molarity.

Molarity (M) is defined as the ratio of the number of moles of solute to the volume of solution in liters. \[ Molarity (M) = \dfrac{Number of moles of solute}{Volume of solution in liters}. \]

Step 2: Find moles of water.

- The molar mass of water is 18 g mol\(^{-1}\).

- The density of water is 1000 g L\(^{-1}\).

Thus, the number of moles in 1 liter of water is: \[ \frac{1000}{18} = 55.56 \, mol. \]

Step 3: Conclusion.

Therefore, the molarity of pure water is 55.56 mol L\(^{-1}\).


Final Answer: \[ \boxed{55.56 \ mol L^{-1}} \] Quick Tip: For molarity of pure liquids, use density and molar mass: \(M = \dfrac{density \times 1000}{molar mass}\).

Step 1: Understanding specific conductance.

Specific conductance (or conductivity, \(\kappa\)) refers to the conductance of 1 cm\(^3\) of solution placed between two electrodes, each having an area of 1 cm\(^2\) and separated by 1 cm.


Step 2: Relation of conductance and resistance.

Conductance is the reciprocal of resistance: \[ Conductance = \dfrac{1}{Resistance} = ohm^{-1}. \]
Since specific conductance takes into account the distance (cm) and area (cm\(^2\)), the unit of specific conductance becomes: \[ Unit of specific conductance = ohm^{-1} \ cm^{-1}. \]

Step 3: Conclusion.

Thus, the unit of specific conductance is cm\(^{-1}\) ohm\(^{-1}\).


Final Answer: \[ \boxed{cm^{-1} \ ohm^{-1}} \] Quick Tip: Specific conductance (κ) is the reciprocal of specific resistance and has the unit ohm\(^{-1}\) cm\(^{-1}\).

Step 1: General reactivity of noble gases.

Noble gases are typically inert because of their stable electronic configuration. However, heavier noble gases, such as xenon, can form compounds due to their larger atomic size and the availability of vacant d-orbitals.


Step 2: Known compounds.

- Neon (Ne), Argon (Ar), and Helium (He) are largely inert and seldom form stable compounds.

- Xenon (Xe), on the other hand, can form several stable compounds, including XeF\(_2\), XeF\(_4\), XeF\(_6\), XeO\(_3\), and XeO\(_4\).


Step 3: Conclusion.

Therefore, Xenon forms the maximum number of compounds among the noble gases.


Final Answer: \[ \boxed{(B) Xe} \] Quick Tip: Xenon is the most reactive noble gas and forms stable fluorides and oxides under specific conditions.

Step 1: Why ions show colours.

The colour of ions, particularly in transition metals, arises from the presence of unpaired d-electrons. These electrons can undergo d–d electronic transitions when light is absorbed, which results in the appearance of colour.


Step 2: Analyzing options.

- (A) Unpaired electrons: Correct. Unpaired electrons are responsible for colour through electronic transitions.

- (B) Paired electrons: Do not contribute to colour as no electronic transitions can occur.

- (C) Non-bonded electrons: While they influence other properties, they do not result in coloured ions.

- (D) None of the above: Incorrect, as unpaired electrons are the cause of colour.


Step 3: Conclusion.

Thus, the colour of ions is due to the presence of unpaired electrons.


Final Answer: \[ \boxed{(A) unpaired electrons} \] Quick Tip: Transition metal ions are coloured because of d–d transitions of unpaired electrons in partially filled d-orbitals.

Step 1: Recall the Hunsdiecker reaction.

The Hunsdiecker reaction involves the reaction of silver salts of carboxylic acids (RCOOAg) with bromine in the presence of CCl\(_4\). This reaction results in the formation of alkyl bromides (RBr) with the loss of one carbon atom.


Step 2: Analyzing the options.

- (A) Hunsdiecker reaction: Correct. This reaction matches the one described in the problem.

- (B) Schmidt reaction: This reaction converts carboxylic acids to amines using hydrazoic acid, which is not relevant here.

- (C) Hell-Volhard-Zelinsky reaction: This reaction involves halogenation at the \(\alpha\)-position of carboxylic acids, which is different from the given case.

- (D) Tishchenko reaction: This reaction forms esters from aldehydes, which is unrelated to the current reaction.


Step 3: Conclusion.

Therefore, the reaction described is the Hunsdiecker reaction.


Final Answer: \[ \boxed{(A) Hunsdiecker reaction} \] Quick Tip: Hunsdiecker reaction shortens the carbon chain by one carbon when converting carboxylic acids to alkyl halides.

Step 1: Understanding amino acids.

Amino acids are organic compounds that contain both an amino group (-NH\(_2\)) and a carboxyl group (-COOH). They serve as the fundamental building blocks of proteins.


Step 2: Protein structure.

Proteins are formed by long chains of amino acids connected by peptide bonds. The sequence of amino acids in a protein determines its unique structure and function.


Step 3: Analyzing the options.

- (A) Carbohydrates: These are made from monosaccharides, not amino acids.

- (B) Proteins: Correct. Proteins are polymers made up of amino acids.

- (C) Lipids: These are made from fatty acids and glycerol, not amino acids.

- (D) Vitamins: These are organic micronutrients and not polymers of amino acids.


Step 4: Conclusion.

Thus, amino acids are the structural units that form proteins.


Final Answer: \[ \boxed{(B) Proteins} \] Quick Tip: Proteins are polymers of amino acids linked by peptide bonds; they are essential biomolecules for structure and function of living cells.

Step 1: Formula for density.

The density of a crystal can be calculated using the formula: \[ d = \frac{Z \times M}{N_A \times a^3} \]
where, \(d\) = density of the crystal = \(10.5 \, gm cm^{-3}\)
\(Z\) = number of atoms per unit cell = 4 (for f.c.c.)
\(M\) = molar mass of silver (to be calculated)
\(N_A\) = Avogadro’s number = \(6.022 \times 10^{23} \, mol^{-1}\)
\(a\) = edge length = \(4.077 \times 10^{-8}\) cm


Step 2: Calculate unit cell volume.

The volume of the unit cell is calculated as: \[ a^3 = (4.077 \times 10^{-8})^3 \, cm^3 = 6.79 \times 10^{-23} \, cm^3 \]


Step 3: Apply values in formula.

Substituting the known values into the formula for density: \[ 10.5 = \frac{4 \times M}{6.022 \times 10^{23} \times 6.79 \times 10^{-23}} \]


Step 4: Simplify.

Simplifying the equation: \[ 10.5 = \frac{4M}{40.9} \]
Solving for \(M\): \[ M = \frac{10.5 \times 40.9}{4} \approx 107.3 \]


Conclusion:

The atomic mass of silver is: \[ \boxed{107.3 \, g mol^{-1}} \] Quick Tip: For f.c.c. lattices, always use \(Z = 4\). For b.c.c., \(Z = 2\) and for simple cubic, \(Z = 1\).

Step 1: Apply Raoult’s Law.

Raoult's Law relates the vapor pressure of the solution to the mole fraction of the solvent: \[ \frac{P}{P^\circ} = X_{solvent} \]
Given: \( \frac{P}{P^\circ} = 0.8 \)
Thus, \( X_{solvent} = 0.8 \)


Step 2: Relation between mole fractions.

The mole fraction of the solvent is given by: \[ X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}} = 0.8 \]


Step 3: Calculate moles of solvent.

The molar mass of octane is \( 114 \, g/mol \), and the mass of octane is \( 114 \, g \).
Thus, the number of moles of solvent is: \[ n_{solvent} = \frac{114}{114} = 1 \, mol \]


Step 4: Substitute values.

Substituting the values into the equation for the mole fraction: \[ \frac{1}{1 + n_{solute}} = 0.8 \]
Solving for \(n_{solute}\): \[ 1 = 0.8 (1 + n_{solute}) \] \[ 1 = 0.8 + 0.8n_{solute} \] \[ 0.2 = 0.8n_{solute} \] \[ n_{solute} = 0.25 \, mol \]


Step 5: Calculate mass of solute.

The molar mass of the solute is \( 40 \, g/mol \), so the mass of the solute is: \[ Mass = n \times M = 0.25 \times 40 = 10 \, g \]


Conclusion:

Thus, the required mass of solute is: \[ \boxed{10 \, g} \] Quick Tip: In Raoult’s law problems, start with mole fraction formula and use the relation \(\frac{P}{P^\circ} = X_{solvent}\).

Step 1: Identify electrodes.

- Oxidation occurs at the anode (Zn electrode).

- Reduction occurs at the cathode (Cu electrode).


Step 2: Formula for standard e.m.f.
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Step 3: Substitute values.
\[ E^\circ_{cell} = (+0.34 \, V) - (-0.76 \, V) \] \[ E^\circ_{cell} = 0.34 + 0.76 = 1.10 \, V \]


Conclusion:

The standard e.m.f. of the given cell is: \[ \boxed{1.10 \, V} \] Quick Tip: Always apply the formula \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\) for standard cell potential calculations.

Step 1: Understanding colloids.

Colloids are heterogeneous systems where one substance is dispersed in another. Based on affinity between dispersed phase and dispersion medium, colloids are of two types – lyophilic and lyophobic.


Step 2: Key differences.


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

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

\hline
Stability & Very stable & Less stable, easily coagulated

\hline
Reversibility & Reversible sols & Irreversible sols

\hline
Examples & Starch, proteins, gum & Gold sol, sulphur sol, As\(_2\)S\(_3\) sol

\hline
Preparation & Form spontaneously & Need special methods

\hline
\end{tabular


Conclusion:

Lyophilic colloids are stable and reversible due to high affinity with solvent, while lyophobic colloids are unstable and need stabilizing agents. Quick Tip: Remember: Lyophilic = “solvent loving” (stable and reversible); Lyophobic = “solvent hating” (unstable and irreversible).

Step 1: Statement of Kohlrausch’s Law.

Kohlrausch’s law of independent migration of ions states that at infinite dilution, each ion makes a definite contribution to the molar conductivity of the electrolyte, independent of the other ions present.
\[ \Lambda_m^\infty = \lambda^0_+ + \lambda^0_- \]

where,
- \(\Lambda_m^\infty\) = molar conductivity at infinite dilution

- \(\lambda^0_+\) = contribution of cation

- \(\lambda^0_-\) = contribution of anion


Step 2: Application.

It is used to calculate the molar conductivity of weak electrolytes at infinite dilution by using strong electrolytes.

Example:

For acetic acid (\(CH_3COOH\)), which is a weak electrolyte: \[ \Lambda_m^\infty (CH_3COOH) = \Lambda_m^\infty (CH_3COONa) + \Lambda_m^\infty (HCl) - \Lambda_m^\infty (NaCl) \]


Conclusion:

Kohlrausch’s law helps in determining dissociation constants of weak electrolytes and studying ionic conductivities. Quick Tip: Remember: The law is most useful for weak electrolytes, since their molar conductivity at infinite dilution cannot be measured directly.

Step 1: Nature of Sulphuric Acid.

Concentrated sulphuric acid is a strong dehydrating agent. It has a strong affinity for water and removes water molecules from many compounds.

Step 2: Two Dehydrating Properties.

1. It dehydrates carbohydrates such as sugar: \[ C_{12}H_{22}O_{11} \xrightarrow{H_2SO_4} 12C + 11H_2O \]
(Sugar is dehydrated to carbon.)

2. It removes water from hydrated salts: \[ CuSO_4 \cdot 5H_2O \xrightarrow{H_2SO_4} CuSO_4 + 5H_2O \]
(Copper sulphate pentahydrate is converted to anhydrous copper sulphate.)


Conclusion:

Concentrated \(H_2SO_4\) acts as a strong dehydrating agent, removing water from organic compounds and hydrated salts. Quick Tip: Always remember: Concentrated sulphuric acid is not only a strong acid but also a powerful dehydrating and oxidizing agent.

Step 1: Case of FeSO\(_4\) + (NH\(_4\))\(_2\)SO\(_4\).

A 1:1 molar mixture of ferrous sulfate and ammonium sulfate forms Mohr’s salt \((FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O)\). In this double salt, the \(Fe^{2+}\) ions are free and ionize completely in aqueous solution. Hence, Mohr’s salt gives the usual tests of \(Fe^{2+}\) ions.


Step 2: Case of CuSO\(_4\) + NH\(_3\) (aq).

When CuSO\(_4\) is treated with excess ammonia (1:4 ratio), a complex compound \([Cu(NH_3)_4]^{2+}\) is formed. In this complex, copper ions are coordinated with ammonia molecules and are not free in solution. Hence, it does not give the normal tests of \(Cu^{2+}\) ions.


Conclusion:
\[ Mohr’s salt → Double salt (ions free) \quad but \quad [Cu(NH_3)_4]SO_4 \to Complex (ions not free) \] Quick Tip: Double salts ionize completely and show individual ion tests, but complex salts do not dissociate completely and thus fail to give usual ionic tests.

Step 1: General case of ketones.

Normally, ketones cannot reduce Fehling’s solution or Tollen’s reagent because they cannot be easily oxidized under mild conditions. Only aldehydes undergo such mild oxidation.


Step 2: Special case of fructose.

Fructose is a ketohexose, but in alkaline medium it undergoes tautomerization (Lobry de Bruyn–van Ekenstein transformation) to form glucose and mannose, both of which are aldehydes.
\[ Fructose \;\xrightarrow{OH^-}\; Glucose + Mannose \]


Step 3: Reaction with reagents.

Since glucose and mannose contain aldehyde groups, they reduce Fehling’s solution and Tollen’s reagent. Thus, fructose indirectly gives positive tests.


Conclusion:

Ordinary ketones do not reduce Fehling’s or Tollen’s reagent, but fructose does due to tautomerization into aldoses in alkaline medium. Quick Tip: Remember: Fructose is an exception because it tautomerizes to glucose/mannose in base, making it act like an aldehyde.

Step 1: Understanding the classification.

Solids are broadly classified as:
- Ionic solids: Composed of cations and anions held by electrostatic forces.

- Metallic solids: Composed of metal atoms with mobile electrons (metallic bonding).

- Molecular solids: Composed of discrete molecules held by van der Waals forces or hydrogen bonds.


Step 2: Classify each given substance.


\begin{tabular{|c|c|
\hline
Substance & Type of Solid

\hline
P\(_4\) & Molecular solid

\hline
Ammonium phosphate & Ionic solid

\hline
Brass & Metallic solid (alloy of Cu and Zn)

\hline
Tetraphosphorus decoxide (P\(_4\)O\(_{10}\)) & Molecular solid

\hline
Rb (Rubidium) & Metallic solid

\hline
I\(_2\) & Molecular solid (van der Waals forces)

\hline
LiBr & Ionic solid

\hline
\end{tabular


Conclusion:

- Ionic solids: Ammonium phosphate, LiBr

- Metallic solids: Brass, Rb

- Molecular solids: P\(_4\), P\(_4\)O\(_{10}\), I\(_2\)
Quick Tip: Molecular solids are usually soft and volatile, ionic solids are hard and brittle with high melting points, and metallic solids conduct electricity due to free electrons.

(i) The colloidal solution of Fe(OH)\(_3\) obtained from FeCl\(_3\) is positively charged:

When FeCl\(_3\) is hydrolyzed, Fe(OH)\(_3\) sol is formed. The Fe\(^{3+}\) ions from FeCl\(_3\) get adsorbed on the surface of Fe(OH)\(_3\) particles.

This leads to the development of a positive charge on the colloidal particles. Hence, the sol of Fe(OH)\(_3\) is positively charged.



(ii) Alum is used for the purification of water:

Alum (K\(_2\)SO\(_4 \cdot\) Al\(_2\)(SO\(_4\))\(_3 \cdot\) 24H\(_2\)O) produces Al\(^{3+}\) ions in water. These ions neutralize the negative charges on colloidal impurities present in water.

As a result, the colloidal particles coagulate, settle down, and can be easily removed. Hence, alum acts as a coagulant in water purification.



(iii) Rivers form a delta while joining a sea/ocean:

Rivers carry colloidal particles like clay and silt. When river water meets sea water (which contains electrolytes like NaCl, MgCl\(_2\)), the electrolytes coagulate the colloidal particles.

These coagulated particles settle at the river mouth, gradually forming delta regions.


Conclusion:

The above examples demonstrate the role of adsorption and coagulation in colloidal chemistry and natural phenomena. Quick Tip: Remember: The charge on colloids depends on the ions adsorbed, and coagulation by electrolytes is the key principle in both water purification and delta formation.

(i) Variation of atomic radii across a series:

Across a transition series, atomic radii decrease slightly from left to right. This happens because as we move across the period, the nuclear charge increases due to the addition of protons.

However, the added electrons enter the penultimate \((n-1)d)\) orbital, providing poor shielding of the outer electrons. Thus, the increased nuclear charge pulls electrons closer, reducing atomic size.

After the mid-series, the effect of electron-electron repulsion in the \(d\)-orbitals balances the nuclear pull, so the decrease becomes very small.



(ii) Paramagnetic nature of transition metal ions:

Transition metal ions generally contain unpaired electrons in their \((n-1)d)\) orbitals.

Paramagnetism arises due to the presence of these unpaired electrons, which have magnetic moments associated with their spins. Hence, most transition metal ions exhibit paramagnetism.



(iii) Variable oxidation states in transition elements:

Transition elements show variable oxidation states because the energy difference between the \((n-1)d)\) and \(ns\) orbitals is very small.

As a result, both \(ns\) and \((n-1)d)\) electrons can take part in bonding.

Thus, transition metals can exhibit multiple oxidation states (e.g., Fe shows +2 and +3, Mn shows +2 to +7).


Conclusion:

The unique electronic configuration of transition metals (\((n-1)d^{1-10}ns^{0-2}\)) explains their small variation in radii, paramagnetism, and variable oxidation states. Quick Tip: Remember: Transition metals are defined by their partially filled \(d\)-orbitals, which account for their magnetic properties, size trends, and multiple oxidation states.

Step 1: Presence of five –OH groups.

When glucose is acetylated with acetic anhydride, it forms penta-acetate, showing the presence of five –OH groups. \[ C_6H_{12}O_6 + 5 (CH_3CO)_2O \;\longrightarrow\; C_6H_7O(OCOCH_3)_5 + 5CH_3COOH \]
Thus, glucose contains five hydroxyl groups.


Step 2: Presence of –CHO group.

Glucose reacts with hydroxylamine (NH\(_2\)OH) to form oxime, and with hydrogen cyanide (HCN) to form cyanohydrin. These reactions prove the presence of the –CHO group in glucose.
\[ C_6H_{12}O_6 + NH_2OH \;\longrightarrow\; C_6H_{11}O_6N (Oxime) \] \[ C_6H_{12}O_6 + HCN \;\longrightarrow\; C_6H_{11}O_6CN (Cyanohydrin) \]


Step 3: Silver mirror test.

When glucose is warmed with Tollen’s reagent (ammoniacal AgNO\(_3\)), the –CHO group is oxidised to –COOH, and metallic silver is deposited on the walls of the test tube as a shining silver mirror.
\[ C_6H_{12}O_6 + 2[Ag(NH_3)_2]^+ + 3OH^- \;\longrightarrow\; C_6H_{12}O_7 + 2Ag \downarrow + 4NH_3 + H_2O \]


Conclusion:

Glucose contains five alcoholic –OH groups and one aldehydic –CHO group. The aldehyde group is confirmed by Tollen’s reagent test, giving a silver mirror. Quick Tip: Remember: Glucose → Penta-acetate (5 –OH groups) and reacts with Tollen’s reagent (–CHO group). This is the classical proof of its structure.

Step 1: Definitions.


Molar elevation constant (K\(_b'\)): It is the elevation of boiling point produced when one mole of solute is dissolved in 1 litre of solvent.
Molal elevation constant (K\(_b\)): It is the elevation of boiling point produced when one mole of solute is dissolved in 1000 g (1 kg) of solvent.



Step 2: Formula with molal elevation constant.
\[ \Delta T_b = K_b \times m \]
where, \(m = \frac{1000 \times w}{M \times W}\)

Thus, \[ M = \frac{1000 \times K_b \times w}{\Delta T_b \times W} \]
Here, \(w\) = mass of solute, \(W\) = mass of solvent in g, \(M\) = molar mass of solute.


Step 3: Formula with molar elevation constant.
\[ \Delta T_b = K_b' \times C \]
where, \(C = \frac{w}{M \times V}\) (molar concentration)

Thus, \[ M = \frac{K_b' \times w}{\Delta T_b \times V} \]
Here, \(V\) = volume of solvent in litres.


Conclusion:

Molar elevation constant is defined with respect to litre of solvent, while molal elevation constant is defined with respect to kg of solvent. Both can be used to calculate molar mass of solute. Quick Tip: Always remember: \(K_b\) → kg of solvent; \(K_b'\) → litre of solvent.

Step 1: Ostwald’s process.

Ostwald’s process is used for large-scale manufacture of nitric acid from ammonia.


Step 2: Reactions involved.
\[ 4NH_3 + 5O_2 \;\xrightarrow{Pt/Rh,\, 500^\circ C,\, 9\, atm}\; 4NO + 6H_2O \] \[ 2NO + O_2 \;\longrightarrow\; 2NO_2 \] \[ 4NO_2 + 2H_2O + O_2 \;\longrightarrow\; 4HNO_3 \]


Step 3: Detection of nitrate ion (NO\(_3^-\)).

Brown ring test is used. The solution containing nitrate is treated with freshly prepared FeSO\(_4\) solution and concentrated H\(_2\)SO\(_4\) is added along the sides of the test tube. A brown ring appears at the junction, indicating the presence of NO\(_3^-\).

Reaction: \[ NO_3^- + 3Fe^{2+} + 4H^+ \;\longrightarrow\; NO + 3Fe^{3+} + 2H_2O \] \[ Fe^{2+} + NO \;\longrightarrow\; [Fe(NO)]^{2+} \quad (brown ring complex) \]


Conclusion:

Ostwald’s process produces nitric acid from ammonia using catalytic oxidation. The presence of nitrate ions can be confirmed by the brown ring test. Quick Tip: Remember the sequence: \(NH_3 \to NO \to NO_2 \to HNO_3\) in Ostwald’s process.

(i) [Co(NH\(_3\))\(_6\)]Cl\(_3\):

Ligand: NH\(_3\) = ammine (neutral)

Central atom: Cobalt

Oxidation state of Co: \(x + 0 = +3 \implies x = +3\)

Name: Hexaamminecobalt(III) chloride



(ii) K\(_3\)[Cr(C\(_2\)O\(_4\))\(_3\)]:

Ligand: C\(_2\)O\(_4^{2-}\) = oxalato (bidentate)

Central atom: Chromium

Oxidation state of Cr: \(x + 3(-2) = -3 \implies x = +3\)

Name: Potassium trioxalatochromate(III)



(iii) [Pt(NH\(_3\))\(_2\)Cl\(_2\)]:

Ligands: NH\(_3\) = ammine (neutral), Cl\(^-\) = chloro

Central atom: Platinum

Oxidation state of Pt: \(x + 0 + (-1)\times 2 = 0 \implies x = +2\)

Name: Diamminedichloroplatinum(II)



(iv) [Ni(CO)\(_4\)]:

Ligand: CO = carbonyl (neutral)

Central atom: Nickel

Oxidation state of Ni: \(x + 0 = 0 \implies x = 0\)

Name: Tetracarbonylnickel(0)


Conclusion:

The names of coordination compounds are given by mentioning ligands first (in alphabetical order), followed by the central metal with its oxidation state in Roman numerals. Quick Tip: Always name ligands before the central atom, use alphabetical order for multiple ligands, and indicate the oxidation state of the central metal in Roman numerals.

Step 1: Identify compound A.

Since the product (C) has molecular formula C\(_6\)H\(_7\)N, it must be an aromatic amine (aniline). The intermediate step suggests that compound B is amide, which upon Hoffmann bromamide reaction gives amine. Hence, A must be an aromatic acid derivative.

Thus, compound A is benzamide (C\(_6\)H\(_5\)CONH\(_2\)).


Step 2: Reaction with aqueous ammonia.

An aromatic acid chloride (benzoyl chloride, C\(_6\)H\(_5\)COCl) reacts with aqueous ammonia to form benzamide (B).
\[ C_6H_5COCl + NH_3 \rightarrow C_6H_5CONH_2 + HCl \]

Step 3: Hoffmann bromamide reaction.

When benzamide (B) is treated with Br\(_2\) and KOH, Hoffmann bromamide reaction occurs. The –CONH\(_2\) group is converted into –NH\(_2\). Thus, the product is aniline (C\(_6\)H\(_5\)NH\(_2\)).
\[ C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O \]


Step 4: Structures of compounds.



Compound A: Benzoyl chloride (C\(_6\)H\(_5\)COCl)
Compound B: Benzamide (C\(_6\)H\(_5\)CONH\(_2\))
Compound C: Aniline (C\(_6\)H\(_5\)NH\(_2\))



Conclusion:

- A = Benzoyl chloride (C\(_6\)H\(_5\)COCl)

- B = Benzamide (C\(_6\)H\(_5\)CONH\(_2\))

- C = Aniline (C\(_6\)H\(_5\)NH\(_2\)) Quick Tip: Hoffmann bromamide reaction always converts an amide (R–CONH\(_2\)) into a primary amine (R–NH\(_2\)) with one carbon less.

Step 1: Reaction of A with CHCl\(_3\) and KOH.

This is the Reimer–Tiemann reaction. Phenol reacts with chloroform and alkali to form salicylaldehyde (ortho-hydroxy benzaldehyde, B) and para-hydroxy benzaldehyde (C).
\[ A = C_6H_5OH \, (Phenol) \] \[ C_6H_5OH + CHCl_3 + 3KOH \;\longrightarrow\; o-OH{-}C_6H_4{-}CHO \,(B) + p-OH{-}C_6H_4{-}CHO \,(C) + 3KCl + 2H_2O \]


Step 2: Distillation of A with Zn dust.

Phenol on distillation with Zn dust gives benzene.
\[ C_6H_5OH \;\xrightarrow{Zn}\; C_6H_6 + ZnO \] \[ D = C_6H_6 \, (Benzene) \]


Step 3: Oxidation of D.

Benzene on oxidation yields benzoic acid.
\[ C_6H_6 \;\xrightarrow{[O]}\; C_6H_5COOH \] \[ E = C_7H_6O_2 \, (Benzoic \, acid) \]


Final Identification:
\[ A = Phenol, \quad B = o-Hydroxybenzaldehyde, \quad C = p-Hydroxybenzaldehyde, \quad D = Benzene, \quad E = Benzoic \, acid \]


Conclusion:

The aromatic compound A is phenol. Through Reimer–Tiemann reaction it forms aldehydes, through reduction with Zn it forms benzene, and further oxidation of benzene produces benzoic acid. Quick Tip: Phenol is a key starting compound in aromatic chemistry. Remember: Reimer–Tiemann → aldehyde, Zn dust → benzene, oxidation → benzoic acid.

(i) Williamson’s Ether Synthesis:

It is a laboratory method for the preparation of ethers. In this method, an alkoxide ion reacts with a primary alkyl halide to produce an ether.
\[ R\!-\!ONa + R'X \; \longrightarrow \; R\!-\!O\!-\!R' + NaX \]
Example: \[ C_2H_5ONa + CH_3I \; \longrightarrow \; C_2H_5OCH_3 + NaI \]
This method is widely used for the synthesis of symmetrical and unsymmetrical ethers.



(ii) Fermentation:

Fermentation is a biochemical process in which sugars such as glucose and sucrose are converted into ethanol and carbon dioxide by the action of enzymes (zymase) produced by yeast.
\[ C_6H_{12}O_6 \; \xrightarrow{zymase} \; 2C_2H_5OH + 2CO_2 \]
Fermentation is the basis of alcoholic beverages and bioethanol production.



(iii) To obtain methyl alcohol from pyroligneous acid:

Pyroligneous acid is obtained by the destructive distillation of wood. It contains acetic acid, acetone, and methyl alcohol as impurities.

On neutralization and distillation, methyl alcohol is separated as the main product. Thus, wood spirit (methyl alcohol) is commercially obtained from pyroligneous acid.


Conclusion:

These processes highlight important laboratory and industrial methods for preparing ethers, alcohols, and biofuels. Quick Tip: Remember: Williamson’s method = ethers, fermentation = ethanol, pyroligneous acid = source of methanol.

Step 1: General rate expression.

For the reaction: \[ 2A + B \longrightarrow 2C + 3D \]
The rate of reaction is given by: \[ Rate = -\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2}\frac{d[C]}{dt} = \frac{1}{3}\frac{d[D]}{dt} \]

Step 2: Given data.
\[ \frac{d[C]}{dt} = +1.0 \, mol L^{-1}s^{-1} \]

Step 3: Calculate rate of reaction.
\[ Rate = \frac{1}{2}\frac{d[C]}{dt} = \frac{1}{2}(1.0) = 0.5 \, mol L^{-1}s^{-1} \]

Step 4: Calculate rate of change of other species.

- For A: \[ -\frac{1}{2}\frac{d[A]}{dt} = 0.5 \quad \Rightarrow \quad \frac{d[A]}{dt} = -1.0 \, mol L^{-1}s^{-1} \]

- For B: \[ -\frac{d[B]}{dt} = 0.5 \quad \Rightarrow \quad \frac{d[B]}{dt} = -0.5 \, mol L^{-1}s^{-1} \]

- For D: \[ \frac{1}{3}\frac{d[D]}{dt} = 0.5 \quad \Rightarrow \quad \frac{d[D]}{dt} = +1.5 \, mol L^{-1}s^{-1} \]


Conclusion:

- Rate of reaction = \(0.5 \, mol L^{-1}s^{-1}\)

- Rate of change: \(d[A]/dt = -1.0\), \(d[B]/dt = -0.5\), \(d[C]/dt = +1.0\), \(d[D]/dt = +1.5\) mol L\(^{-1}\)s\(^{-1}\). Quick Tip: Use the stoichiometric coefficients in the rate equation to relate the rate of reaction with the rate of change of species.

Step 1: Rate expression.

For the reaction: \[ 2A \longrightarrow Products \] \[ Rate = -\frac{1}{2}\frac{\Delta [A]}{\Delta t} \]

Step 2: Change in concentration of A.

Initial \([A] = 0.5\) mol L\(^{-1}\)

Final \([A] = 0.4\) mol L\(^{-1}\)
\[ \Delta [A] = 0.4 - 0.5 = -0.1 \, mol L^{-1} \]

Step 3: Time interval.
\[ \Delta t = 10 \, min = 600 \, s \]

Step 4: Calculate rate.
\[ Rate = -\frac{1}{2}\frac{\Delta [A]}{\Delta t} = -\frac{1}{2}\frac{-0.1}{600} \] \[ Rate = \frac{0.1}{1200} = 8.33 \times 10^{-5} \, mol L^{-1}s^{-1} \]


Conclusion:

The rate of reaction during the given time interval is: \[ \boxed{8.33 \times 10^{-5} \, mol L^{-1}s^{-1}} \] Quick Tip: Always convert time into seconds when calculating rates in kinetics problems.

Step 1: Write the balanced chemical reaction.
\[ N_2O_5 \longrightarrow 2NO_2 + \tfrac{1}{2} O_2 \]

Step 2: Define the rate of reaction.

The rate of reaction can be expressed as: \[ -\frac{1}{1}\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = \frac{1}{\tfrac{1}{2}} \frac{d[O_2]}{dt} \] \[ -\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = 2\frac{d[O_2]}{dt} \]

Step 3: Substitute the given rate expressions.
\[ K'[N_2O_5] = \frac{1}{2}K''[N_2O_5] = 2K'''[N_2O_5] \]

Step 4: Simplify the relations.
\[ K'' = 2K', \quad K''' = \tfrac{1}{2}K' \]


Conclusion:

The relation between the constants is: \[ \boxed{K'' = 2K' \quad and \quad K''' = \tfrac{1}{2}K'} \] Quick Tip: Always divide the rate of change of concentration by the stoichiometric coefficient to establish relations between rate constants.

Step 1: General rate law for first-order reaction.
\[ Rate = k[A]^1 \]

Step 2: Effect of doubling concentration.

If concentration of \([A]\) is doubled: \[ Rate_{new} = k(2[A]) = 2k[A] \]

Step 3: Compare with original rate.
\[ \frac{Rate_{new}}{Rate_{old}} = \frac{2k[A]}{k[A]} = 2 \]


Conclusion:

For a first-order reaction, doubling the concentration of the reactant doubles the rate of reaction. Quick Tip: In first-order kinetics, rate is directly proportional to concentration. Doubling concentration doubles the rate, tripling it makes the rate three times, and so on.

(i) n-Butyl chloride + alcoholic KOH:

Dehydrohalogenation occurs, producing 1-butene.
\[ CH_3CH_2CH_2CH_2Cl + KOH(alc.) \; \xrightarrow{\Delta} \; CH_3CH_2CH=CH_2 + KCl + H_2O \]



(ii) Bromobenzene + Mg in dry ether:

Grignard reagent is formed.
\[ C_6H_5Br + Mg \; \xrightarrow{dry ether} \; C_6H_5MgBr \]



(iii) Ethyl chloride + aqueous KOH:

Hydrolysis takes place, giving ethyl alcohol.
\[ C_2H_5Cl + KOH(aq) \; \longrightarrow \; C_2H_5OH + KCl \]



(iv) Methyl bromide + Na in dry ether:

Wurtz reaction occurs, forming ethane.
\[ 2CH_3Br + 2Na \; \xrightarrow{dry ether} \; C_2H_6 + 2NaBr \]



(v) Ethyl bromide + KCN (alc.):

Nucleophilic substitution occurs, forming propionitrile.
\[ C_2H_5Br + KCN(alc.) \; \longrightarrow \; C_2H_5CN + KBr \]


Conclusion:

These reactions show elimination (E2), substitution (SN1/SN2), Grignard reagent formation, and carbon–carbon bond formation, highlighting important pathways in organic synthesis. Quick Tip: Remember: Alcoholic KOH → elimination (alkenes), Aqueous KOH → substitution (alcohols), Dry ether + Mg → Grignard reagent, Wurtz reaction → higher alkane, and KCN → alkyl nitrile.

(i) Chlorobenzene to Benzene:

By reduction with Ni/Alkali.
\[ C_6H_5Cl + 2[H] \; \xrightarrow{Ni/NaOH} \; C_6H_6 + HCl \]



(ii) Chlorobenzene to Phenol:

By hydrolysis under high temperature and pressure.
\[ C_6H_5Cl + NaOH \; \xrightarrow{623K, \; 300 atm} \; C_6H_5OH + NaCl \]



(iii) Chlorobenzene to Toluene:

Through Friedel–Crafts alkylation.
\[ C_6H_5Cl + CH_3Cl \; \xrightarrow{AlCl_3} \; C_6H_5CH_3 + HCl \]



(iv) Chlorobenzene to Aniline:

Via nitration followed by reduction.
\[ C_6H_5Cl \; \xrightarrow{HNO_3/H_2SO_4} \; o\!-\!\&\!p\!-\!ClC_6H_4NO_2 \] \[ ClC_6H_4NO_2 \; \xrightarrow{Sn/HCl} \; ClC_6H_4NH_2 \]
Then by Sandmeyer reaction: \[ C_6H_5Cl \; \xrightarrow{NH_3/Cu_2O} \; C_6H_5NH_2 \]



(v) Chlorobenzene to Diphenyl:

Through Ullmann reaction.
\[ 2C_6H_5Cl + 2Na \; \xrightarrow{Cu, \; \Delta} \; C_6H_5\!-\!C_6H_5 + 2NaCl \]


Conclusion:

Chlorobenzene undergoes reduction, hydrolysis, substitution, or coupling reactions under suitable conditions to give benzene, phenol, toluene, aniline, and diphenyl. Quick Tip: Remember: Hydrolysis gives phenol, reduction gives benzene, Friedel–Crafts gives toluene, ammonolysis gives aniline, and Ullmann reaction gives diphenyl.

Step 1: Percentage composition.

Given: \[ %C = 69.77, \quad %H = 11.63, \quad %O = 100 - (69.77 + 11.63) = 18.60 \]


Step 2: Calculate empirical formula.
\[ For C: \frac{69.77}{12} \approx 5.81 \] \[ For H: \frac{11.63}{1} \approx 11.63 \] \[ For O: \frac{18.60}{16} \approx 1.16 \]

Dividing all by 1.16 (smallest): \[ C : H : O = 5.81/1.16 : 11.63/1.16 : 1.16/1.16 \approx 5 : 10 : 1 \]
So, empirical formula = \(C_5H_{10}O\).


Step 3: Molecular formula.

Empirical formula mass = \(5 \times 12 + 10 \times 1 + 16 = 86\).

Since molecular mass = 86, the molecular formula is the same: \[ C_5H_{10}O \]


Step 4: Functional group tests.


Compound does not reduce Tollen’s reagent → Not an aldehyde.
Forms addition compound with NaHSO\(_3\) → Presence of carbonyl group.
Positive iodoform test → Presence of methyl ketone group (\(-COCH_3\)).


Thus, compound must be a ketone with structure \(R-COCH_3\).


Step 5: Oxidation products.

On strong oxidation, compound gives ethanoic acid and propanoic acid. This indicates the structure is pentan-2-one.
\[ CH_3-CO-CH_2-CH_2-CH_3 \quad (Pentan-2-one) \]


Step 6: Reaction equations.

(i) NaHSO\(_3\) addition: \[ CH_3COCH_2CH_2CH_3 + NaHSO_3 \;\longrightarrow\; Addition compound \]

(ii) Iodoform test: \[ CH_3COCH_2CH_2CH_3 + 3I_2 + 4NaOH \;\longrightarrow\; CHI_3 \downarrow + CH_3CH_2COONa + 3NaI + 3H_2O \]

(iii) Oxidation: \[ CH_3COCH_2CH_2CH_3 \;\xrightarrow{[O]}\; CH_3COOH + CH_3CH_2COOH \]


Conclusion:

The possible structure of the compound is: \[ \boxed{CH_3-CO-CH_2-CH_2-CH_3 \quad (Pentan-2-one)} \] Quick Tip: Positive iodoform test always indicates the presence of a \(-COCH_3\) group or a secondary alcohol oxidisable to it.

(i) Claisen Reaction:

The Claisen condensation is the reaction of two esters or one ester and another carbonyl compound in the presence of a strong base (like sodium ethoxide) to form a \(\beta\)-keto ester or a \(\beta\)-diketone.
\[ 2CH_3COOC_2H_5 \; \xrightarrow{NaOC_2H_5} \; CH_3COCH_2COOC_2H_5 + C_2H_5OH \]
This reaction is important in C–C bond formation in organic synthesis.



(ii) Cannizzaro’s Reaction:

Aldehydes without \(\alpha\)-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.



(iii) Rosenmund’s Reaction:

This reaction involves the hydrogenation of acid chlorides in the presence of palladium (Pd) catalyst on barium sulfate (BaSO\(_4\)) to produce aldehydes.
\[ RCOCl + H_2 \; \xrightarrow{Pd/BaSO_4} \; RCHO + HCl \]
This method is used for the controlled synthesis of aldehydes from acyl chlorides.


Conclusion:

Claisen reaction forms \(\beta\)-keto esters, Cannizzaro’s reaction converts aldehydes without \(\alpha\)-H into alcohols and acids, and Rosenmund’s reaction selectively reduces acyl chlorides to aldehydes. Quick Tip: Claisen → condensation (C–C bond), Cannizzaro → disproportionation (no \(\alpha\)-H), Rosenmund → acyl chloride to aldehyde.