Board Question Paper: March 2022
(i) The co-ordination number of atoms in body centred cubic structure (bcc) is _______.
(c) 8
Explanation: In a body-centred cubic (BCC) structure, the central atom touches 8 corner atoms.
(ii) In calculating osmotic pressure, the concentration of solute is expressed in _______.
(a) molarity
Explanation: The formula for osmotic pressure is \(\pi = MRT\), where M is Molarity.
(iii) The enthalpy change for the chemical reaction H₂O(s) ⟶ H₂O(l) is called enthalpy of _______.
(b) fusion
Explanation: The phase change from solid to liquid is called fusion.
(iv) Which of the following transition element shows maximum oxidation state?
(c) Mn
Explanation: Manganese (Mn) exhibits oxidation states from +2 to +7, which is the highest among the 3d series elements listed.
(v) The correct formula for the complex compound, sodium hexacyanoferrate (III) is _______.
(c) Na₃ [Fe(CN)₆]
Explanation: Iron is in +3 state, CN is -1. The complex ion is \([Fe(CN)_6]^{3-}\). To balance charges with Na⁺, we need 3 sodium ions.
(vi) Isopropylbenzene on air oxidation followed by decomposition by dilute acid gives _______.
(a) C₆H₅OH
Explanation: This is the commercial method for the preparation of Phenol (C₆H₅OH) from Cumene (Isopropylbenzene).
(vii) The name of metal nanoparticle which acts as highly effective bacterial disinfectant in water purification process is _______.
(b) silver
Explanation: Silver nanoparticles are well-known for their antimicrobial properties and are used in water filters.
(viii) Acid anhydride on reaction with primary amine gives compound having a functional group _______.
(a) amide
Explanation: Acylation of amines with acid anhydride produces amides.
(ix) The standard potential of the cell in the following reaction is _______.
Cd(s) + Cu²⁺(1M) ⟶ Cd²⁺(1M) + Cu(s)
(E°Cd = –0.403V, E°Cu = 0.334 V)
(b) 0.737 V
Explanation: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{Cu} - E^\circ_{Cd} = 0.334 - (-0.403) = 0.737 V\)
(x) The value of [H₃O⁺] in mol lit⁻¹ of 0.001 M acetic acid solution (Ka = 1.8 × 10⁻⁵) is _______.
(d) 1.34 × 10⁻⁴
Explanation: \([H_3O^+] = \sqrt{K_a \cdot C} = \sqrt{1.8 \times 10^{-5} \times 10^{-3}} = \sqrt{1.8 \times 10^{-8}} = 1.34 \times 10^{-4}\)
HSC Chemistry
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(i) Write the product formed when alkyl halide reacts with silver nitrite.
The major product formed is Nitroalkane (\(R-NO_2\)).
(ii) Write the name of product formed, when acetone is treated with 2, 4-dinitrophenyl hydrazine.
The product formed is Acetone 2,4-dinitrophenylhydrazone.
(iii) Write the name of biodegradable polyamide copolymer.
Nylon-2-nylon-6.
(iv) Identify the molecularity of following elementary reaction: NO(g) + O₃(g) ⟶ NO₃(g) + O(g)
The reaction involves the collision of two molecules (one NO and one O₃). Therefore, the molecularity is 2 (Bimolecular).
(v) What is the action of selenium on magnesium metal?
Magnesium reacts with selenium on heating to form Magnesium selenide.
Reaction: \(Mg + Se \xrightarrow{\Delta} MgSe\)
(vi) Write the name of isomerism in the following complexes: [Cu(NH₃)₄] [PtCl₄] and [Pt(NH₃)₄] [CuCl₄]
Coordination Isomerism.
(vii) Write the name of the alloy used in Fischer Tropsch process in the synthesis of gasoline.
Cobalt-Thorium alloy (or Co-Th-Mg). Cobalt based catalysts are typically used.
(viii) Henry’s law constant for CH₃Br(g) is 0.159 mol dm⁻³ bar⁻¹ at 25°C. What is solubility of CH₃Br(g) in water at same temperature and partial pressure of 0.164 bar?
According to Henry's Law: \(S = K_H \times P\)
Given: \(K_H = 0.159 \, \text{mol dm}^{-3} \text{bar}^{-1}\), \(P = 0.164 \, \text{bar}\)
\(S = 0.159 \times 0.164 = 0.026076 \, \text{mol dm}^{-3}\)
Solubility = 0.0261 mol dm⁻³ (approx)
Attempt any EIGHT of the following questions: [16]
Definition: A reaction which has higher order true rate law but behaves as a first order reaction experimentally is called a pseudo-first order reaction. This happens when one of the reactants is present in large excess, so its concentration remains effectively constant.
Example: Hydrolysis of methyl acetate in acidic medium.
\(CH_3COOCH_{3(aq)} + H_2O_{(l)} \xrightarrow{H^+} CH_3COOH_{(aq)} + CH_3OH_{(aq)}\)
Rate Law: \(Rate = k'[CH_3COOCH_3][H_2O]\)
Since water is in excess, \([H_2O]\) is constant. Let \(k = k'[H_2O]\).
Thus, \(Rate = k[CH_3COOCH_3]\), making it a first-order reaction.
- Decrease in Density: As the number of ions decreases due to vacancies (both cations and anions are missing), the mass decreases while volume remains constant, leading to a decrease in the density of the crystal.
- Electrical Conductivity: The crystal becomes slightly electrically conductive due to the movement of ions into the vacancies (holes).
- Stability: The presence of a large number of vacancies lowers the lattice energy, slightly decreasing the stability of the crystal.
(i) Na in dry ether (ii) Mg in dry ether
(i) Action of Na in dry ether (Wurtz Reaction):
When ethyl bromide is treated with sodium metal in the presence of dry ether, it gives butane.
\(2C_2H_5Br + 2Na \xrightarrow{\text{dry ether}} C_2H_5-C_2H_5 + 2NaBr\)
(Ethyl bromide \(\to\) n-Butane)
(ii) Action of Mg in dry ether:
When ethyl bromide reacts with magnesium metal in the presence of dry ether, it forms ethyl magnesium bromide (Grignard Reagent).
\(C_2H_5Br + Mg \xrightarrow{\text{dry ether}} C_2H_5MgBr\)
A peptide linkage (or peptide bond) is an amide linkage \((-CO-NH-)\) formed between the carboxyl group \((-COOH)\) of one amino acid and the amino group \((-NH_2)\) of another amino acid with the elimination of a water molecule.
Example: Formation of Glycylalanine (dipeptide).
When Glycine \((H_2N-CH_2-COOH)\) reacts with Alanine \((H_2N-CH(CH_3)-COOH)\):
\(H_2N-CH_2-COOH + H-NH-CH(CH_3)-COOH \xrightarrow{-H_2O} H_2N-CH_2-CO-NH-CH(CH_3)-COOH\)
The bond \(-CO-NH-\) connects the two amino acid units.
According to Van't Hoff equation for osmotic pressure \((\pi)\) of a dilute solution:
\(\pi = MRT\) ...(1)
Where:
- \(\pi\) = Osmotic pressure
- \(M\) = Molarity of the solution
- \(R\) = Gas constant
- \(T\) = Temperature in Kelvin
Molarity \(M\) is the number of moles of solute \((n_2)\) per volume of solution in litres \((V)\).
\(M = \frac{n_2}{V}\)
Substituting into (1):
\(\pi = \frac{n_2 RT}{V}\)
Also, number of moles \(n_2 = \frac{W_2}{M_2}\), where \(W_2\) is mass of solute and \(M_2\) is molar mass of solute.
\(\pi = \frac{W_2 RT}{M_2 V}\)
Rearranging to find \(M_2\):
\(\mathbf{M_2 = \frac{W_2 RT}{\pi V}}\)
1. Monodentate Ligands:
Ligands that have only one donor atom and can form only one coordinate bond with the central metal ion are called monodentate ligands.
Example: Chloride ion \((Cl^-)\), Ammonia \((NH_3)\), Water \((H_2O)\).
2. Ambidentate Ligands:
Ligands that have two different donor atoms but can coordinate to the central metal ion through only one donor atom at a time are called ambidentate ligands.
Example: Nitrite ion \((NO_2^-)\). It can bond through Nitrogen \((-NO_2)\) or Oxygen \((-ONO)\).
(i) Atomic radii (ii) Ionisation enthalpy (iii) Electronegativity (iv) Electron gain enthalpy
(i) Atomic radii: It increases down the group from Oxygen to Polonium due to the addition of new electron shells.
(ii) Ionisation enthalpy: It decreases down the group due to the increase in atomic size and screening effect, making it easier to remove the outer electron.
(iii) Electronegativity: It decreases down the group as atomic size increases.
(iv) Electron gain enthalpy: Oxygen has a less negative electron gain enthalpy than Sulphur due to its compact size and inter-electronic repulsion. From Sulphur to Polonium, the electron gain enthalpy becomes less negative.
Phenol is prepared from aniline via diazotization.
- Aniline is treated with nitrous acid (generated in situ from \(NaNO_2 + HCl\)) at low temperature \((0-5^\circ C)\) to form Benzene diazonium chloride.
- Benzene diazonium chloride is then hydrolyzed by warming with water or dilute acid to give phenol.
\(C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, \, 273K} C_6H_5N_2^+Cl^- \xrightarrow{H_2O, \, \text{warm}} C_6H_5OH + N_2 + HCl\)
(i) acetonitrile (ii) nitroethane
(i) From Acetonitrile (Mendius Reduction):
Acetonitrile is reduced by sodium and ethanol (or \(LiAlH_4\)) to ethanamine.
\(CH_3-C\equiv N + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3-CH_2-NH_2\)
(ii) From Nitroethane:
Nitroethane is reduced by metals like Sn or Fe in the presence of concentrated HCl to form ethanamine.
\(CH_3-CH_2-NO_2 + 6[H] \xrightarrow{Sn/conc. HCl} CH_3-CH_2-NH_2 + 2H_2O\)
\(2CH_3 – C = O \xrightarrow{Ba(OH)_2} A \xrightarrow{\Delta} B + H_2O\)
(Note: Reactant is Acetaldehyde \(CH_3CHO\))
This is an Aldol Condensation reaction.
Step 1: Two molecules of Acetaldehyde condense in the presence of mild base \(Ba(OH)_2\) to form 3-Hydroxybutanal (Aldol).
\(2CH_3CHO \rightleftharpoons CH_3-CH(OH)-CH_2-CHO\)
A is 3-Hydroxybutanal.
Step 2: On heating \((\Delta)\), A loses a water molecule to form an \(\alpha,\beta\)-unsaturated aldehyde (Crotonaldehyde).
\(CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO + H_2O\)
B is But-2-enal (Crotonaldehyde).
Attempt any EIGHT of the following questions: [24]
Given:
- \(n = 1\) mole
- \(V_1 = 10\) L, \(V_2 = 15\) L
- \(T = 300\) K
- \(R = 8.314 \, J \, K^{-1} \, mol^{-1}\)
Formula: \(W_{max} = -2.303 nRT \log_{10}(\frac{V_2}{V_1})\)
Calculation:
\(W_{max} = -2.303 \times 1 \times 8.314 \times 300 \times \log_{10}(\frac{15}{10})\)
\(W_{max} = -5744.14 \times \log_{10}(1.5)\)
\(\log_{10}(1.5) \approx 0.1761\)
\(W_{max} = -5744.14 \times 0.1761\)
\(W_{max} \approx -1011.5 \, J\)
Work done = -1.01 kJ (Work is done by the system)
Reaction: \(Zn^{2+} + 2e^- \to Zn\)
1 mole of \(Zn^{2+}\) requires 2 moles of electrons.
Therefore, 2 moles of \(Zn^{2+}\) require \(2 \times 2 = \mathbf{4 \, \text{moles of electrons}}\).
Charge of 1 mole of electrons = 1 Faraday (F).
Thus, electricity required = 4 Faradays (4F).
Haematite: The chemical composition is Ferric oxide, \(\mathbf{Fe_2O_3}\).
Group 17 Elements (Halogens):
- Fluorine (F): Atomic No. 9
Electronic Config: \([He] 2s^2 2p^5\) - Chlorine (Cl): Atomic No. 17
Electronic Config: \([Ne] 3s^2 3p^5\)
Polymers are classified into three types based on structure:
- Linear Polymers: These consist of long and straight chains. The chains are stacked over one another to give a well-packed structure. Example: High Density Polyethylene (HDPE), PVC.
- Branched Chain Polymers: These contain linear chains having some branches. These branches prevent tight packing. Example: Low Density Polyethylene (LDPE).
- Cross-linked (or Network) Polymers: These are usually formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains. Example: Bakelite, Melamine.
Green Chemistry: It is the design of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.
Disadvantages of Nanotechnology:
- Health Hazards: Nanoparticles can easily penetrate skin and lungs causing potential toxicity and damage (e.g., lung damage from inhaling nanotubes).
- Environmental Impact: 'Nanopollution' refers to the accumulation of nanoparticles in the environment which may affect aquatic life and enter the food chain.
Commercial Preparation of Glucose:
Glucose is prepared commercially by the hydrolysis of starch with dilute \(H_2SO_4\) at 393K under pressure.
\((C_6H_{10}O_5)_n + nH_2O \xrightarrow{H^+, 393K, 2-3 atm} nC_6H_{12}O_6\)
(Starch \(\to\) Glucose)
Structure of Adipic Acid (Hexanedioic acid):
\(HOOC - CH_2 - CH_2 - CH_2 - CH_2 - COOH\)
Or \(HOOC - (CH_2)_4 - COOH\)
(i) hot HI (ii) PCl₅ (iii) dilute H₂SO₄
Methoxyethane: \(CH_3-O-C_2H_5\)
(i) Action of hot HI: cleavage of both alkyl groups to form alkyl iodides.
\(CH_3-O-C_2H_5 + 2HI \xrightarrow{\Delta} CH_3I + C_2H_5I + H_2O\)
(ii) Action of PCl₅:
\(CH_3-O-C_2H_5 + PCl_5 \xrightarrow{\Delta} CH_3Cl + C_2H_5Cl + POCl_3\)
(iii) Action of dilute H₂SO₄ (Hydrolysis):
\(CH_3-O-C_2H_5 + H_2O \xrightarrow{dil. H_2SO_4, \Delta} CH_3OH + C_2H_5OH\)
- Cationic Sphere Complexes: The coordination sphere (complex ion) carries a net positive charge.
Example: \([Pt(NH_3)_4]^{2+}\), \([Zn(NH_3)_4]^{2+}\). - Anionic Sphere Complexes: The coordination sphere carries a net negative charge.
Example: \([Fe(CN)_6]^{3-}\), \([Ni(CN)_4]^{2-}\). - Neutral Sphere Complexes: The coordination sphere does not carry any net charge.
Example: \([Ni(CO)_4]\), \([PtCl_2(NH_3)_2]\).
Part 1: Magnetic Moment
Atomic Number (Z) = 25 (Manganese, Mn).
Electronic Config of Mn: \([Ar] 3d^5 4s^2\).
Divalent cation \(Mn^{2+}\): \([Ar] 3d^5\).
Number of unpaired electrons \((n) = 5\).
Spin only magnetic moment \(\mu = \sqrt{n(n+2)}\) B.M.
\(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx \mathbf{5.92 \, B.M.}\)
Part 2: Ti⁴⁺ Colour
Ti (Z=22): \([Ar] 3d^2 4s^2\).
\(Ti^{4+}\): \([Ar] 3d^0\).
Since \(Ti^{4+}\) has an empty d-orbital (no unpaired d-electrons), d-d electron transition is not possible. Hence, it does not absorb light in the visible region and appears colourless.
Lanthanoid Contraction: The steady decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number is called lanthanoid contraction. This is due to the poor shielding effect of 4f electrons.
Preparation of Acetic Acid:
(i) From Dry Ice (Solid \(CO_2\)): Reacting Grignard reagent (Methyl magnesium iodide) with dry ice followed by acid hydrolysis.
\(CH_3MgI + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgI \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)I\)
(ii) From Acetyl Chloride: Hydrolysis of acetyl chloride with water.
\(CH_3COCl + H_2O \to CH_3COOH + HCl\)
Classification of Aliphatic Ketones:
- Simple (Symmetrical) Ketones: Both alkyl groups attached to the carbonyl carbon are the same.
Example: Acetone \((CH_3-CO-CH_3)\). - Mixed (Unsymmetrical) Ketones: The alkyl groups attached to the carbonyl carbon are different.
Example: Butanone \((CH_3-CO-C_2H_5)\).
Action of Sodium Hypoiodite on Acetone (Iodoform Test):
Acetone reacts with sodium hypoiodite \((NaOI)\), prepared in situ from \(NaOH + I_2\), to give a yellow precipitate of Iodoform.
\(CH_3COCH_3 + 3NaOI \to CHI_3 \downarrow (\text{Yellow ppt}) + CH_3COONa + 2NaOH\)
Definition: The half-life of a reaction \((t_{1/2})\) is the time required for the reactant concentration to decrease to half of its initial value.
Derivation:
The integrated rate law for a first-order reaction is:
\(k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}\)
At \(t = t_{1/2}\), \([A]_t = \frac{[A]_0}{2}\).
Substituting these values:
\(k = \frac{2.303}{t_{1/2}} \log_{10} \frac{[A]_0}{[A]_0 / 2}\)
\(k = \frac{2.303}{t_{1/2}} \log_{10} (2)\)
Since \(\log_{10}(2) = 0.3010\):
\(k = \frac{2.303 \times 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}}\)
Therefore, \(t_{1/2} = \frac{0.693}{k}\)
(i) \(CH_3OH(l) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(l) \quad \Delta H^\circ = -726 \, kJ \, mol^{-1}\)
(ii) \(C(s) + O_2(g) \to CO_2(g) \quad \Delta_c H^\circ = -393 \, kJ \, mol^{-1}\)
(iii) \(H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l) \quad \Delta_f H^\circ = -286 \, kJ \, mol^{-1}\)
Aim: Find \(\Delta_f H^\circ\) for \(C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \to CH_3OH(l)\).
According to Hess's Law:
\(\Delta_f H^\circ (CH_3OH) = [\Delta H^\circ(C \to CO_2) + 2 \times \Delta H^\circ(H_2 \to H_2O)] - \Delta H^\circ(\text{combustion of } CH_3OH)\)
\(\Delta_f H^\circ = \text{Eq(ii)} + 2 \times \text{Eq(iii)} - \text{Eq(i)}\)
\(\Delta_f H^\circ = (-393) + 2(-286) - (-726)\)
\(\Delta_f H^\circ = -393 - 572 + 726\)
\(\Delta_f H^\circ = -965 + 726\)
\(\mathbf{\Delta_f H^\circ = -239 \, kJ \, mol^{-1}}\)
This is a basic buffer solution.
According to Henderson-Hasselbalch equation:
\(pOH = pK_b + \log_{10} \frac{[\text{Salt}]}{[\text{Base}]}\)
First, calculate \(pK_b\):
\(pK_b = -\log_{10}(K_b) = -\log_{10}(1.8 \times 10^{-5}) = 5 - \log_{10}(1.8) = 5 - 0.2553 = 4.7447\)
Given: \([\text{Salt}] = 0.02 \, M\), \([\text{Base}] = 0.01 \, M\).
\(pOH = 4.7447 + \log_{10} (\frac{0.02}{0.01})\)
\(pOH = 4.7447 + \log_{10} (2)\)
\(pOH = 4.7447 + 0.3010 = 5.0457\)
Now, \(pH + pOH = 14\)
\(pH = 14 - 5.0457\)
\(\mathbf{pH = 8.95}\)
Attempt any THREE of the following questions: [12]
Gold crystallises into face-centred cubic cells. The edge length of unit cell is 4.08 × 10⁻⁸ cm. Calculate the density of gold. [Molar mass of gold = 197 g mol⁻¹]
Definitions:
(i) Isotonic solution: Two or more solutions having the same osmotic pressure at the same temperature are called isotonic solutions.
(ii) Osmosis: The spontaneous flow of solvent molecules from a region of lower solute concentration (pure solvent) to a region of higher solute concentration through a semi-permeable membrane.
Numerical:
Given: FCC structure \(\Rightarrow Z = 4\).
Edge length \(a = 4.08 \times 10^{-8} \, cm\).
Molar mass \(M = 197 \, g \, mol^{-1}\).
Avogadro's No \(N_A = 6.022 \times 10^{23}\).
Formula: \(\rho = \frac{Z \times M}{a^3 \times N_A}\)
\(a^3 = (4.08 \times 10^{-8})^3 = 67.91 \times 10^{-24} \, cm^3\).
\(\rho = \frac{4 \times 197}{67.91 \times 10^{-24} \times 6.022 \times 10^{23}}\)
\(\rho = \frac{788}{67.91 \times 0.6022}\) (Adjusting powers of 10)
Actually: \(67.91 \times 10^{-24} \times 6.022 \times 10^{23} = 67.91 \times 0.6022 = 40.89\)
\(\rho = \frac{788}{40.89} = 19.27 \, g/cm^3\)
Density of Gold = 19.27 g/cm³
First Law Equations (\(\Delta U = q + W\)):
(i) Isothermal Process: Temperature is constant, so internal energy change \(\Delta U = 0\).
Equation: \(0 = q + W \Rightarrow \mathbf{q = -W}\) (Heat absorbed = Work done by system)
(ii) Adiabatic Process: No heat exchange, so \(q = 0\).
Equation: \(\mathbf{\Delta U = W}\) (Internal energy change = Work done on system)
Relationship between pH and pOH:
The ionic product of water is given by:
\(K_w = [H^+][OH^-]\)
At 298 K, \(K_w = 1.0 \times 10^{-14}\).
Taking negative logarithm to the base 10 on both sides:
\(-\log_{10} K_w = -\log_{10} ([H^+][OH^-])\)
\(-\log_{10} K_w = (-\log_{10} [H^+]) + (-\log_{10} [OH^-])\)
By definition, \(pK_w = -\log K_w\), \(pH = -\log [H^+]\), \(pOH = -\log [OH^-]\).
Therefore, \(pH + pOH = pK_w\)
At 25°C, \(pK_w = 14\), so \(pH + pOH = 14\).
Reference Electrode: An electrode whose potential is arbitrarily fixed or exactly known and is used to determine the potential of other electrodes is called a reference electrode.
Functions of Salt Bridge:
- It maintains electrical neutrality in both half-cells by providing ions.
- It completes the electrical circuit by allowing the flow of ions between the two half-cells.
- It prevents the intermixing of the two electrolytic solutions.
Diagram of SHE:
(Diagram description for student to draw): A beaker containing 1M H⁺ solution. An inverted glass tube containing a platinum wire sealed at the top and a platinum foil coated with platinum black at the bottom. Hydrogen gas is bubbled at 1 atm pressure.
Metal Deficiency Defect: This defect arises when a compound contains less metal (cation) than the stoichiometric proportion. It occurs in metals showing variable oxidation states. A cation is missing from its lattice site, and the electrical neutrality is maintained by another nearby cation acquiring a higher positive charge.
Example: \(FeO\), \(FeS\). In FeO, formula is often \(Fe_{0.95}O\).
Preparation of Sulphur Dioxide:
Sulphur dioxide is formed by burning sulphur in air or oxygen.
\(S(s) + O_2(g) \to SO_2(g)\)
Uses of Sulphur:
- Manufacture of sulphuric acid \((H_2SO_4)\).
- Vulcanization of rubber.
- As a fungicide and in skin ointments.
(i) Ethyl bromide to ethyl methyl ether.
(ii) Ethyl bromide to ethene.
(iii) Bromobenzene to toluene.
(iv) Chlorobenzene to biphenyl.
(i) Ethyl bromide to ethyl methyl ether (Williamson Synthesis):
\(C_2H_5Br + NaOCH_3 \xrightarrow{\Delta} C_2H_5-O-CH_3 + NaBr\)
(ii) Ethyl bromide to ethene (Dehydrohalogenation):
Heated with alcoholic KOH.
\(C_2H_5Br + KOH(alc) \xrightarrow{\Delta} CH_2=CH_2 + KBr + H_2O\)
(iii) Bromobenzene to toluene (Wurtz-Fittig Reaction):
Reacted with methyl bromide and sodium in dry ether.
\(C_6H_5Br + CH_3Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5-CH_3 + 2NaBr\)
(iv) Chlorobenzene to biphenyl (Fittig Reaction):
Heated with sodium in dry ether.
\(2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 + 2NaCl\)
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