ISC Class 12 Chemistry Board Exam Question Paper 2017 with Solutions

ISC Class 12 Chemistry Board Exam Question Paper with Solutions 2017

Part - I (20 Marks)

Question 1.

(a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets: [5]
(iodoform, acetaldehyde, positive, greater, acidic, acetone, disaccharide, negative, increases, glucose, decreases, chloroform, polysaccharide, lactose, lesser, basic, cationic hydrolysis, anionic hydrolysis)
(i) Calcium acetate on heating gives .......... which gives ......... on heating with iodine and sodium hydroxide solution.
(ii) On dilution of a solution, its specific conductance .......... while its equivalent conductance ..........
(iii) Sucrose is a .......... and yields upon hydrolysis, a mixture of .......... and fructose.
(iv) More .......... the standard reduction potential of a substance, the .......... is its ability to displace hydrogen from acids.
(v) An aqueous solution of CH₃COONa is .......... due to ..........

Answer:
(i) acetone, iodoform
(ii) decreases, increases
(iii) disaccharide, glucose
(iv) negative, greater
(v) basic, anionic hydrolysis

Teacher's Note:
a) Dry distillation of calcium acetate yields acetone \((CH_3)_2CO\), which undergoes the iodoform test when heated with \(I_2\) and \(NaOH\).
b) Specific conductance decreases on dilution due to fewer ions per unit volume, whereas equivalent conductance increases due to increased degree of dissociation.

(b) Complete the following statements by selecting the correct alternative from the choices given: [5]
(i) In a face-centred cubic lattice, atom (A) occupies the corner positions and atom (B) occupies the face centre positions. If one atom of (B) is missing from one of the face-centred points, the formula of the compound is:
(1) A₂B₅
(2) A₂B₃
(3) AB₂
(4) A₂B

Answer: (1) A₂B₅

Number of atoms A per unit cell = \(8 \times \frac{1}{8} = 1\). Number of atoms B per unit cell = \(5 \times \frac{1}{2} = \frac{5}{2}\). Ratio A : B = \(1 : \frac{5}{2} = 2 : 5\).

Teacher's Note:
a) Calculate contributions from corners (\(\frac{1}{8}\)) and face centres (\(\frac{1}{2}\)).
b) Account for the missing face-centred atom by subtracting \(\frac{1}{2}\) from the total face contribution before finding the simple whole-number ratio.

(ii) The half-life period of a first-order reaction is 20 minutes. The time required for the concentration of the reactant to change from 0.16 M to 0.02 M is:
(1) 80 minutes
(2) 60 minutes
(3) 40 minutes
(4) 20 minutes

Answer: (2) 60 minutes

Concentration changes from \(0.16 \text{ M} \xrightarrow{t_{1/2}} 0.08 \text{ M} \xrightarrow{t_{1/2}} 0.04 \text{ M} \xrightarrow{t_{1/2}} 0.02 \text{ M}\). Total time = \(3 \times 20 = 60\) minutes.

Teacher's Note:
a) For a first-order reaction, time taken can be found by successive half-life reductions or using the integrated rate equation \(t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}\).
b) Remember that the half-life of a first-order reaction is independent of the initial concentration.

(iii) For a spontaneous reaction \(\Delta G^{\circ}\) and \(E^{\circ}_{\text{cell}}\) will be respectively:
(1) -ve and +ve
(2) +ve and -ve
(3) +ve and +ve
(4) -ve and -ve

Answer: (1) -ve and +ve

For a spontaneous process, \(\Delta G^{\circ}\) must be negative and \(E^{\circ}_{\text{cell}}\) must be positive according to the relation \(\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}\).

Teacher's Note:
a) Relate spontaneity to Gibbs free energy change (\(\Delta G\) negative) and standard cell potential (\(E^{\circ}\) positive).
b) Do not confuse standard cell potential signs with non-standard conditions.

(iv) The conjugate acid of \(\text{HPO}_4^{2-}\) is:
(1) \(\text{H}_3\text{PO}_3\)
(2) \(\text{H}_3\text{PO}_4\)
(3) \(\text{H}_2\text{PO}_4^-\)
(4) \(\text{PO}_4^{3-}\)

Answer: (3) \(\text{H}_2\text{PO}_4^-\)

A conjugate acid is formed by adding a proton (\(\text{H}^+\)) to the given species: \(\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^-\).

Teacher's Note:
a) Add one \(\text{H}^+\) ion to find the conjugate acid and remove one \(\text{H}^+\) ion to find the conjugate base.
b) Pay close attention to the net charges on the ions.

(v) The polymer formed by the condensation of hexamethylenediamine and adipic acid is:
(1) Teflon
(2) Bakelite
(3) Dacron
(4) Nylon-66

Answer: (4) Nylon-66

Hexamethylenediamine (6 carbon atoms) and adipic acid (6 carbon atoms) undergo condensation polymerization to form Nylon-6,6.

Teacher's Note:
a) Learn monomers for common polymers like Nylon-6,6, Dacron, and Bakelite.
b) The '6,6' in Nylon-6,6 indicates that both monomers contain 6 carbon atoms each.

(c) Answer the following questions: [5]
(i) Why the freezing point depression (\(\Delta T_f\)) of 0.4 M NaCl solution is nearly twice than that of 0.4 M glucose solution?

Answer:
This is because Van't Hoff factor for NaCl is 2 and for glucose, it is 1.

Teacher's Note:
a) Freezing point depression is a colligative property dependent on the number of particles in solution.
b) Strong electrolytes like NaCl dissociate completely into two ions, doubling the effective particle concentration.

(ii) Identify the order of reaction from each of the following units of rate constant (\(k\)):
(a) \(\text{mol L}^{-1}\text{ sec}^{-1}\)
(b) \(\text{mol}^{-1}\text{ L sec}^{-1}\)

Answer:
(a) Zero
(b) Two

Teacher's Note:
a) General unit of rate constant is \(\text{mol}^{1-n}\text{ L}^{n-1}\text{ s}^{-1}\), where \(n\) is the order of the reaction.
b) Memorize or derive units quickly by substituting \(n = 0, 1, 2\) into the general formula.

(iii) Specific conductivity of 0.20 M solution of KCl at 298 K is \(0.025 \text{ S cm}^{-1}\). Calculate its molar conductivity.

Answer:
\(\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{0.025 \times 1000}{0.20} = 125 \text{ S cm}^2 \text{ mol}^{-1}\).

Teacher's Note:
a) State the formula \(\Lambda_m = \frac{\kappa \times 1000}{M}\) clearly before substituting values.
b) Ensure proper unit conversion between \(\text{cm}\) and \(\text{L}\) when calculating molar conductivity.

(iv) Name the order of reaction which proceeds with a uniform rate throughout.

Answer:
Zero

Teacher's Note:
a) Zero-order reactions have rates independent of reactant concentrations.
b) Photochemical reactions or surface-catalyzed decompositions often exhibit zero-order kinetics.

(v) What are the products formed when phenol and nitrobenzene are treated separately with a mixture of concentrated sulphuric acid and concentrated nitric acid?

Answer:
Phenol gives 2, 4, 6 - Trinitrophenol (Picric acid), and nitrobenzene gives m - Dinitrobenzene.

Teacher's Note:
a) Phenol's hydroxyl group is strongly activating and directs incoming groups to ortho and para positions.
b) Nitro group is strongly deactivating and meta-directing.

(d) Match the following: [5]

Answer:
(i) - (c)
(ii) - (e)
(iii) - (a)
(iv) - (b)
(v) - (d)

Teacher's Note:
a) Match each term with its direct application or classification from physical, organic, and inorganic chemistry.
b) Read all options carefully before finalizing pairings to avoid cross-matching errors.

Part - II (50 Marks)

Section - A

Answer any two questions.

Question 2.
(a) (i) Determine the freezing point of a solution containing 0.625 g of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) dissolved in 102.8 g of water. [2]
(Freezing point of water = 273 K, \(K_f\) for water = \(1.87 \text{ K kg mol}^{-1}\), at. wt. C = 12, H = 1, O = 16)

Answer:
Molar mass of glucose (\(M_B\)) = \(6(12) + 12(1) + 6(16) = 180 \text{ g mol}^{-1}\)
\(\Delta T_f = \frac{K_f \times W_B \times 1000}{M_B \times W_A} = \frac{1.87 \times 0.625 \times 1000}{180 \times 102.8} = 0.063 \text{ K}\)
Freezing point of solution (\(T_f\)) = \(273 - 0.063 = 272.937 \text{ K}\)

Teacher's Note:
a) Apply the formula \(\Delta T_f = \frac{K_f \times w \times 1000}{M \times W}\) correctly.
b) Subtract the depression from the pure solvent freezing point to get the solution freezing point.

(ii) A 0.15 M aqueous solution of KCl exerts an osmotic pressure of 6.8 atm at 310 K. Calculate the degree of dissociation of KCl. (\(R = 0.0821 \text{ Lit. atm K}^{-1}\text{ mol}^{-1}\)). [2]

Answer:
\(\pi = iCRT \implies 6.8 = i \times 0.15 \times 0.0821 \times 310\)
\(i = \frac{6.8}{3.81765} = 1.78\)
\(\alpha = \frac{i - 1}{n - 1}\). For KCl, \(n = 2\).
\(\alpha = \frac{1.78 - 1}{2 - 1} = 0.78\) or \(78\%\).

Teacher's Note:
a) Calculate the Van't Hoff factor \(i\) using the experimental osmotic pressure formula \(\pi = iCRT\).
b) Use the relation between degree of dissociation and \(i\) for strong electrolytes.

(iii) A solution containing 8.44 g of sucrose in 100 g of water has a vapour pressure 4.56 mm of Hg at 273 K. If the vapour pressure of pure water is 4.58 mm of Hg at the same temperature, calculate the molecular weight of sucrose. [1]

Answer:
\(\frac{P^{\circ} - P}{P^{\circ}} = \frac{w_B / M_B}{w_A / M_A + w_B / M_B}\)
\(\frac{4.58 - 4.56}{4.58} = \frac{8.44 / M_B}{100 / 18 + 8.44 / M_B}\)
\(\frac{0.02}{4.58} = \frac{8.44 \times 18}{100 M_B} \implies M_B = 346.38 \text{ g mol}^{-1}\).

Teacher's Note:
a) Apply Raoult's law for relative lowering of vapour pressure for dilute solutions.
b) Neglect solute moles in the denominator for very dilute solutions if approximation is valid, though exact calculation gives higher precision.

(b) (i) When ammonium chloride and ammonium hydroxide are added to a solution containing both \(\text{Al}^{3+}\) and \(\text{Ca}^{2+}\) ions, which ion is precipitated first and why? [2]

Answer:
\(\text{Al}^{3+}\) ions are precipitated first as \(\text{Al(OH)}_3\) because the solubility product (\(K_{sp}\)) of \(\text{Al(OH)}_3\) is lower than that of \(\text{Ca(OH)}_2\).

Teacher's Note:
a) Precipitation occurs when ionic product exceeds solubility product.
b) Lower \(K_{sp}\) means lesser concentration of hydroxyl ions is required to precipitate the metal hydroxide.

(ii) A solution of potassium chloride has no effect on litmus whereas, a solution of zinc chloride turns the blue litmus red. Give a reason. [2]

Answer:
This is because KCl is a salt of a strong acid and strong base, so it does not undergo hydrolysis and remains neutral. \(\text{ZnCl}_2\) is a salt of a weak base (\(\text{Zn(OH)}_2\)) and strong acid (\(\text{HCl}\)), undergoing hydrolysis to give an acidic solution.

Teacher's Note:
a) Explain salt hydrolysis based on the strengths of parent acids and bases.
b) Strong acid-weak base salts produce acidic solutions due to cationic hydrolysis.

(c) How many sodium ions and chloride ions are present in a unit cell of sodium chloride crystal? [1]

Answer:
Number of \(\text{Na}^+\) ions = 4
Number of \(\text{Cl}^-\) ions = 4

Teacher's Note:
a) In an fcc unit cell of NaCl, \(\text{Cl}^-\) ions occupy corners and face centres, while \(\text{Na}^+\) ions occupy edge centres and the body centre.
b) Total effective number of formula units per unit cell (\(Z\)) is 4.

Question 3.
(a) (i) Lead sulphide has a face-centred cubic crystal structure. If the edge length of the unit cell of lead sulphide is 495 pm, calculate the density of the crystal. [1]
(at. wt. of Pb = 207, S = 32)

Answer:
\(\rho = \frac{Z \times M}{N_A \times a^3} = \frac{4 \times (207 + 32)}{6.022 \times 10^{23} \times (495 \times 10^{-10})^3} = 10.35 \text{ g cm}^{-3}\).

Teacher's Note:
a) Convert edge length from picometers to centimeters (\(1 \text{ pm} = 10^{-10} \text{ cm}\)).
b) Use \(Z = 4\) for fcc lattice structure.

(ii) For the reaction: \(2\text{H}_2 + 2\text{NO} \rightleftharpoons 2\text{H}_2\text{O} + \text{N}_2\), the following rate data was obtained: [3]

Calculate the following:
(1) The overall order of a reaction.
(2) The rate law.
(3) The value of rate constant (\(k\)).

Answer:
(1) Overall order = 3
(2) Rate law = \(k[\text{H}_2]^1[\text{NO}]^2$
(3) \(k = 0.07 \text{ mol}^{-2}\text{ L}^2\text{ sec}^{-1}\)

Teacher's Note:
a) Compare rate equations from different trials to determine reaction orders with respect to each reactant.
b) Sum up individual orders to get the overall order and substitute values back into any trial to compute \(k\).

(b) (i) The following electrochemical cell is set up at 298 K: [2]
(1) Write the cell reaction.
(2) Calculate the emf and free energy change at 298 K.
(a) Answer the following: [2]
(1) What is the effect of temperature on the ionic product of water (\(K_w\))?
(2) What happens to the ionic product of water (\(K_w\)) if some acid is added to it?
(c) Frenkel defect does not change the density of the ionic crystal whereas, Schottky defect lowers the density of ionic crystal. Give a reason; [2]

Answer:
(b) (i) (1) Cell reaction: \(\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}\)
(2) \(E_{\text{cell}} = 0.339 + 0.761 = 1.10 \text{ V}\); \(\Delta G^{\circ} = -nFE_{\text{cell}} = -2 \times 96500 \times 1.10 = -212.3 \text{ kJ}\).
(a) (1) \(K_w\) increases with an increase in temperature due to increased ionization.
(2) \(K_w\) remains unchanged because it depends only on temperature.
(c) In Frenkel defect, ions just shift to interstitial sites without leaving the crystal, whereas in Schottky defect, equal numbers of cations and anions are missing from the lattice.

Teacher's Note:
a) Remember that equilibrium constants like \(K_w\) and \(K_c\) vary exclusively with temperature.
b) Crystal defects alter physical properties like density based on whether particles leave the lattice entirely.

Question 4.
(a) (i) Name the law or principle to which the following observations confirm: [3]
(1) When water is added to a 1.0 M aqueous solution of acetic acid, the number of hydrogen ion (\(\text{H}^+\)) increases.
(2) When 9650 coulombs of electricity is passed through a solution of copper sulphate, 3.175 g of copper is deposited on the cathode (at. wt. of Cu = 63.5).
(3) When ammonium chloride is added to a solution of ammonium hydroxide, the concentration of hydroxyl ion decreases.
(ii) What is the difference between the order of a reaction and its molecularity? [2]

Answer:
(a) (i) (1) Ostwald's Dilution Law
(2) Faraday's First Law of Electrolysis
(3) Common Ion Effect
(ii)

Teacher's Note:
a) Distinguish clearly between experimental properties (order) and theoretical properties (molecularity).
b) Associate quantitative laws correctly with electrochemical and equilibrium observations.

(b) (i) Explain why high pressure is required in the manufacture of sulphur trioxide by the contact process. State the law or principle used. [2]
(ii) Calculate the equilibrium constant (\(K_c\)) for the formation of \(\text{NH}_3\) in the following reaction: [1]
\(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\)
At equilibrium, the concentration of \(\text{NH}_3\), \(\text{H}_2\) and \(\text{N}_2\) are \(1.2 \times 10^{-2}\), \(3.0 \times 10^{-2}\) and \(1.5 \times 10^{-2}\text{ M}\) respectively.

Answer:
(b) (i) High pressure favors the forward reaction because it proceeds with a decrease in volume (from 4 volumes to 2 volumes). The principle used is Le Chatelier's Principle.
(ii) \(K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2}) \times (3.0 \times 10^{-2})^3} = 400\).

Teacher's Note:
a) Apply Le Chatelier's principle by analyzing volume changes in gaseous equilibria.
b) Substitute equilibrium concentrations carefully into the expression for \(K_c\).

(c) Explain the following: [2]
(i) Hydrolysis of ester (ethyl acetate) begins slowly but becomes fast after some time.
(ii) The pH value of acetic acid increases on the addition of a few drops of sodium acetate.

Answer:
(c) (i) This is because the acetic acid produced during the hydrolysis acts as an auto-catalyst, increasing the reaction rate.
(ii) This is due to the common ion effect caused by acetate ions, which suppresses the dissociation of acetic acid and decreases hydrogen ion concentration, thus increasing pH.

Teacher's Note:
a) Recognize auto-catalytic reactions where a product speeds up the process.
b) Explain pH shifts using equilibrium suppression via common ions.

Section - B

Answer any two questions.

Question 5.
(a) Write the formula of the following compounds: [2]
(i) Potassium trioxalatoaluminate (III).
(ii) Hexaaquairon (II) sulphate.
(b) Name the types of isomerism shown by the following pairs of compounds: [1]
(i) \([\text{Cu(NH}_3)_4][\text{PtCl}_4]\) and \([\text{Pt(NH}_3)_4][\text{CuCl}_4]\)
(ii) \([\text{Co(Pn)}_2\text{Cl}_2]^+\) and \([\text{Co(en)}_2\text{Cl}_2]^+\)
(c) For the coordination complex ion \([\text{Co(NH}_3)_6]^{3+}\) [2]
(i) Give the IUPAC name of the complex ion.
(ii) What is the oxidation number of cobalt in the complex ion?
(iii) State the type of hybridisation of the complex ion.
(iv) State the magnetic behaviour of the complex ion.

Answer:
(a) (i) \(\text{K}_3[\text{Al(C}_2\text{O}_4)_3]\)
(ii) \([\text{Fe(H}_2\text{O)}_6]\text{SO}_4$
(b) (i) Coordination isomerism
(ii) Ligand isomerism (or structural isomerism due to ligand variation)
(c) (i) hexaamminecobalt (III) ion
(ii) +3
(iii) \(d^2sp^3$
(iv) Diamagnetic

Teacher's Note:
a) Follow IUPAC nomenclature rules for writing coordination formulas and names.
b) Determine hybridization and magnetic properties using Valence Bond Theory by checking for unpaired electrons.

Question 6.
(a) Give balanced equations for the following reactions: [3]
(i) Potassium permanganate is heated with concentrated hydrochloric acid.
(ii) Lead sulphide is heated with hydrogen peroxide.
(iii) Ozone is treated with potassium iodide solution.
(b) Discuss the theory involved in the manufacture of sulphuric acid by the contact process. [2]

Answer:
(a) (i) \(2\text{KMnO}_4 + 16\text{HCl} \rightarrow 2\text{KCl} + 2\text{MnCl}_2 + 8\text{H}_2\text{O} + 5\text{Cl}_2\
(ii) \(\text{PbS} + 4\text{H}_2\text{O}_2 \rightarrow \text{PbSO}_4 + 4\text{H}_2\text{O}\)
(iii) \(2\text{KI} + \text{H}_2\text{O} + \text{O}_3 \rightarrow 2\text{KOH} + \text{I}_2 + \text{O}_2\)
(b) The Contact process involves three main steps: (1) Production of \(\text{SO}_2\) by burning sulphur or roasting sulphide ores; (2) Catalytic oxidation of \(\text{SO}_2\) to \(\text{SO}_3\) using \(\text{V}_2\text{O}_5\) at 1075 K and 2 atm; (3) Absorption of \(\text{SO}_3\) in concentrated \(\text{H}_2\text{SO}_4\) to form oleum, which is then diluted with water to get \(\text{H}_2\text{SO}_4\).

Teacher's Note:
a) Ensure all equations for oxidizing properties of reagents like \(\text{KMnO}_4\) and \(\text{O}_3\) are fully balanced.
b) Clearly outline all three chemical stages of the Contact process.

Question 7.
(a) (i) What are the types of hybridisation of iodine in interhalogen compounds \(\text{IF}_3\), \(\text{IF}_5\) and \(\text{IF}_7\), respectively? [3]
(ii) Draw the structure of xenon hexafluoride (\(\text{XeF}_6\)) molecule and state the hybridisation of the central atom.
(b) Give the balanced equations for the conversion of argentite (\(\text{Ag}_2\text{S}\)) to metallic silver. [2]

Answer:
(a) (i) \(\text{IF}_3\): \(sp^3d\); \(\text{IF}_5\): \(sp^3d^2\); \(\text{IF}_7\): \(sp^3d^3$
(ii) Hybridisation: \(sp^3d^3\).
[Figure: Distorted octahedral geometry of \(\text{XeF}_6\) showing Xe bonded to six F atoms with one lone pair]
(b) Conversion of argentite:
\(\text{Ag}_2\text{S} + 4\text{NaCN} \rightleftharpoons 2\text{Na}[\text{Ag(CN)}_2] + \text{Na}_2\text{S}\)
\(2\text{Na}_2\text{S} + 2\text{H}_2\text{O} + 2\text{O}_2 \rightarrow 2\text{Na}_2\text{SO}_4 + 4\text{NaOH} + 2\text{S}\)
\(2\text{Na}[\text{Ag(CN)}_2] + \text{Zn} \rightarrow \text{Na}_2[\text{Zn(CN)}_4] + 2\text{Ag}\downarrow\)

Teacher's Note:
a) Use VSEPR theory to predict hybridisation and geometry of interhalogen and noble gas compounds.
b) Describe MacArthur-Forrest cyanidation process equations accurately for silver extraction.

Section - C

Answer any two questions.

Question 8.
(a) How can the following conversions be brought about:
(i) Acetaldehyde to propan-2-ol. [1]
(ii) Nitrobenzene to p-aminoazobenzene. [1]
(iii) Acetic acid to methylamine. [2]
(iv) Aniline to benzene. [1]
(b) (i) How will you distinguish between primary, secondary and tertiary amines by Hinsberg's test? [1]
(ii) Why do alcohols possess higher boiling points as compared to those of corresponding alkanes? [1]
(c) Identify the compounds A, B and C: [3]
(i) \(\text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{PCl}_5} \text{A} \xrightarrow{\text{H}_2 - \text{Pd/BaSO}_4} \text{B} \xrightarrow{\text{KCN alc. distil.}} \text{C}\)
(ii) \(\text{H} - \text{C}\equiv\text{C} - \text{H} \xrightarrow[\text{dil. }\text{H}_2\text{SO}_4 + \text{HgSO}_4]{\text{H}_2\text{O}} \text{A} \xrightarrow[\text{[Ni]}]{\text{H}_2} \text{B} \xrightarrow[\text{conc. }\text{H}_2\text{SO}_4]{140^{\circ}\text{C}} \text{C}\)

Answer:
(a) (i) \(\text{CH}_3\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow \text{Adduct} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{propan-2-ol}\)
(ii) Nitrobenzene reduced to aniline, then coupled with benzene diazonium chloride in acidic medium.
(iii) \(\text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{NH}_2\)
(iv) \(\text{Aniline} \xrightarrow{\text{NaNO}_2 + \text{HCl}, 0-5^{\circ}\text{C}} \text{Benzenediazonium chloride} \xrightarrow{\text{H}_3\text{PO}_2/\text{H}_2\text{O}} \text{Benzene}\)
(b) (i) Primary amines give N-alkylbenzenesulphonamide soluble in alkali; secondary amines give insoluble derivatives; tertiary amines do not react.
(ii) Alcohols form intermolecular hydrogen bonds, requiring more energy to break than weak van der Waals forces in alkanes.
(c) (i) A: \(\text{C}_6\text{H}_5\text{COCl}\) (Benzoyl chloride), B: \(\text{C}_6\text{H}_5\text{CHO}\) (Benzaldehyde), C: Benzoin.
(ii) A: \(\text{CH}_3\text{CHO}\) (Acetaldehyde), B: \(\text{CH}_3\text{CH}_2\text{OH}\) (Ethyl alcohol), C: \(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5\) (Diethyl ether).

Teacher's Note:
a) Master name reactions and functional group interconversions for organic synthesis sequences.
b) Pay close attention to reagent specifications in step-by-step reaction chains.

Question 9.
(a) Give balanced equations for the following name reactions: [3]
(i) Friedel-Crafts reaction (alkylation)
(ii) Williamson's synthesis
(iii) Aldol condensation
(b) Give the chemical test to distinguish: [3]
(i) Ethyl alcohol and sec-propyl alcohol
(ii) Acetaldehyde and acetic acid
(c) (i) Deficiency of which vitamin causes the following diseases: [4]
(1) Scurvy
(2) Night blindness
(ii) Write two differences between globular and fibrous proteins.

Answer:
(a) (i) \(\text{Benzene} + \text{CH}_3\text{Cl} \xrightarrow{\text{anhydrous }\text{AlCl}_3} \text{Toluene} + \text{HCl}\)
(ii) \(\text{C}_2\text{H}_5\text{Br} + \text{NaOC}_2\text{H}_5 \rightarrow \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + \text{NaBr}\)
(iii) \(2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CHO}\)
(b) (i) Victor Meyer's test: Ethyl alcohol gives blood red colour, sec-propyl alcohol gives deep blue colour.
(ii) Acetic acid gives brisk effervescence with \(\text{NaHCO}_3\) solution due to \(\text{CO}_2\) release, whereas acetaldehyde does not.
(c) (i) (1) Vitamin C
(2) Vitamin A
(ii)

Teacher's Note:
a) Memorize standard name reactions along with their specific catalysts and conditions.
b) Distinguish biomolecules clearly by their solubility, structural shapes, and deficiency symptoms.

Question 10.
(a) An aliphatic unsaturated hydrocarbon (A) when treated with \(\text{HgSO}_4/\text{H}_2\text{SO}_4\) yields a compound (B) having molecular formula \(\text{C}_3\text{H}_6\text{O}\). (B) on oxidation with concentrated \(\text{HNO}_3\) gives two compounds (C) and (D). Compound (C), when treated with \(\text{PCl}_5\), gives compound (E). (E) when reacts with ethanol gives a sweet-smelling liquid (F). Compound (F) is also formed when (C) reacts with ethanol in the presence of concentrated \(\text{H}_2\text{SO}_4\). [4]
(i) Identify the compounds A, B, C, D, E and F.
(ii) Give the chemical equation for the reaction of (C) with chlorine in the presence of red phosphorus and name the reaction.
(b) Answer the following: [3]
(i) What is the common name of the polymer obtained by the polymerisation of caprolactam? Is it addition polymer or condensation polymer?
(c) Give balanced equations for the following reactions: [3]
(i) Methyl magnesium bromide with ethyl alcohol.
(ii) Acetic anhydride with phosphorus pentachloride.
(iii) Acetaldehyde with hydroxylamine.

Answer:
(a) (i) A: Propyne (\(\text{CH}_3-\text{C}\equiv\text{CH}\)); B: Acetone (\(\text{CH}_3\text{COCH}_3\)); C: Acetic acid (\(\text{CH}_3\text{COOH}\)); D: Formic acid (\(\text{HCOOH}\)); E: Acetyl chloride (\(\text{CH}_3\text{COCl}\)); F: Ethyl acetate (\(\text{CH}_3\text{COOC}_2\text{H}_5\)).
(ii) \(\text{CH}_3\text{COOH} + \text{Cl}_2 \xrightarrow{\text{Red P}} \text{ClCH}_2\text{COOH} + \text{HCl}\). Name of reaction: Hell-Volhard-Zelinsky (HVZ) reaction.
(b) (i) Nylon-6; Condensation polymer (specifically ring-opening polymerization acting via condensation mechanism).
(c) (i) \(\text{CH}_3\text{MgBr} + \text{C}_2\text{H}_5\text{OH} \rightarrow \text{CH}_4 + \text{Mg(Br)(OC}_2\text{H}_5)\)
(ii) \((\text{CH}_3\text{CO})_2\text{O} + \text{PCl}_5 \rightarrow 2\text{CH}_3\text{COCl} + \text{POCl}_3\)
(iii) \(\text{CH}_3\text{CHO} + \text{NH}_2\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}=\text{N}-\text{OH} + \text{H}_2\text{O}\)

Teacher's Note:
a) Decode organic structure identification problems step-by-step using molecular formulas and functional group reactions.
b) Ensure all substitution and addition reactions are balanced with appropriate co-products.