ISC Class 12 Chemistry Sample Paper 2027 with Solutions

SECTION A - 14 MARKS

Question 1

(A) Fill in the blanks by choosing the appropriate word(s) from those given in the brackets. [4x1]
[methyl alcohol, ethyl alcohol, paired, one, Cannizzaro, d-d, salicylaldehyde, Perkin's, unpaired, f-f, salicylic acid, Kolbe's, two, 193000, 96500, Reimer-Tiemann]

(i) Formaldehyde reacts with conc. alkali to give sodium formate and __________ due to the absence of alpha - hydrogen atom. The reaction is known as __________ reaction. [1 Mark]

Answer: Methyl alcohol, Cannizzaro

Teacher's Note:
a) Aldehydes without alpha-hydrogen atoms undergo Cannizzaro reaction in the presence of concentrated alkali via intermolecular redox (disproportionation).
b) Students must remember that one molecule is oxidized to a salt of carboxylic acid and another is reduced to an alcohol.

(ii) Electrochemical equivalent (Z) is the quantity of substance deposited or discharged by a charge of __________ coulomb, whereas the equivalent conductance (E) is the amount of substance deposited or discharged by a charge of __________ coulombs. [1 Mark]

Answer: One, 96,500

Teacher's Note:
a) Electrochemical equivalent is mass deposited per coulomb of charge, hence for a charge of 1 coulomb it equals Z.
b) Equivalent conductance relates to the conductivity of an electrolyte solution containing one gram equivalent of electrolyte, but the second blank refers to the definition of equivalent mass deposition requiring 1 Faraday (96,500 coulombs).

(iii) The colour of transition metal ions is due to __________ electron(s) in d-subshell and __________ transition. [1 Mark]

Answer: Unpaired, d-d

Teacher's Note:
a) Presence of unpaired d-electrons and the subsequent absorption of visible light causing d-d transition gives colour to transition metal ions.
b) Ions with completely filled (d¹⁰) or empty (d⁰) subshells are colorless due to the absence of d-d transitions.

(iv) The product formed when sodium phenoxide is treated with CO₂ at 400K, under 4-7 atm pressure and then acidified with dilute HCl is __________. The reaction is known as __________ reaction. [1 Mark]

Answer: Salicylic acid, Kolbe's

Teacher's Note:
a) Kolbe's reaction involves the carboxylation of sodium phenoxide using carbon dioxide under pressure followed by acidification to form 2-hydroxybenzoic acid (salicylic acid).
b) Do not confuse this with Reimer-Tiemann reaction, which uses chloroform and NaOH to form salicylaldehyde.

(B) Select and write the correct alternative from the choices given below. [7x1]

(i) The correct order of freezing point of 0.5M solution of urea, AlCl₃, KCl and MgCl₂ is: [1 Mark]
(a) AlCl₃ > MgCl₂ > KCl > urea
(b) Urea > KCl > MgCl₂ > AlCl₃
(c) MgCl₂ > KCl > AlCl₃ > urea
(d) KCl > MgCl₂ > AlCl₃ > urea

Answer: (b) Urea > KCl > MgCl₂ > AlCl₃

Depression in freezing point is directly proportional to the van't Hoff factor (i). Higher particles mean lower freezing point.

Teacher's Note:
a) Urea (i=1), KCl (i=2), MgCl₂ (i=3), AlCl₃ (i=4). Therefore, freezing point depression is maximum for AlCl₃ and minimum for urea.
b) Since freezing point decreases with an increase in particle concentration, the order of freezing points is exactly reversed: Urea > KCl > MgCl₂ > AlCl₃.

(ii) Glucose molecules react with (x) number of molecules of phenyl hydrazine to yield glucosazone. The melting point of glucosazone is 206⁰C. The value of (x) is: [1 Mark]
(a) one
(b) two
(c) three
(d) four

Answer: (c) three

Three molecules of phenylhydrazine are consumed per molecule of glucose to form osazone.

Teacher's Note:
a) One molecule reacts with the aldehyde group at C-1, and two molecules react at C-2 with oxidation of C-3, consuming 3 moles of phenylhydrazine total.
b) Students often mistakenly write one or two by looking only at the terminal carbonyl group.

(iii) Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will NOT give a test for chloride ions with silver nitrate solution at 25⁰C? [1 Mark]
(a) CoCl₃.6NH₃
(b) CoCl₃.5NH₃
(c) CoCl₃.4NH₃
(d) CoCl₃.3NH3

Answer: (d) CoCl₃.3NH₃

Formulated as [Co(NH₃)₃Cl₃], where all chloride ions are inside the coordination sphere.

Teacher's Note:
a) Only chloride ions present outside the coordination sphere (ionizable chlorides) give a white precipitate with silver nitrate.
b) CoCl₃.3NH₃ has no ionizable chloride ions, so it yields no precipitate.

(iv) The following data was obtained for a hypothetical reaction 2A + B → 2AB
S. No. | [A] (mol L^-1) | [B] (mol L^-1) | Initial rate (mol L^-1 sec^-1)
1 | 0.5 | 0.5 | 1.8 × 10^-4
2 | 1.0 | 0.5 | 3.6 × 10^-4
3 | 1.0 | 1.0 | 3.6 × 10^-4
The rate law expression for this reaction is: [1 Mark]
(a) rate = k[A]¹[B]¹
(b) rate = k[A]¹[B]⁰
(c) rate = k[A]²[B]¹
(d) rate = k[A]⁰[B]¹

Answer: (b) rate = k[A]¹[B]⁰

Comparing experiments 1 and 2, doubling [A] doubles the rate, making order with respect to [A] equal to 1. Comparing 2 and 3, doubling [B] does not change the rate, making order with respect to [B] equal to 0.

Teacher's Note:
a) Always use initial rate comparisons where concentrations of one reactant are kept constant while the other changes.
b) Zero order with respect to [B] signifies that the rate is independent of the concentration of [B] under these conditions.

(v) Which of the following reactions will NOT give primary amine? [1 Mark]
(a) C₂H₅NC →[LiAlH₄] __________
(b) C₂H₅CN →[LiAlH₄] __________
(c) C₂H₅CONH₂ →[Br₂/KOH] __________
(d) CH₃CONH₂ →[Br₂/KOH] __________

Answer: (a) C₂H₅NC →[LiAlH₄] __________

Reduction of isocyanides with LiAlH₄ yields secondary amines (N-methylpropanamine / ethylmethylamine), not primary amines.

Teacher's Note:
a) Isocyanides (RNC) on reduction give secondary amines (R-NH-CH₃), whereas cyanides (RCN) give primary amines (R-CH₂-NH₂).
b) Hoffmann bromamide degradation (options c and d) always produces primary amines with one carbon less.

(vi) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: When an ether containing a primary alkyl group and tertiary alkyl group is treated with hydrogen iodide, tertiary alkyl iodide is formed.
Reason: The reaction takes place via S_N1 mechanism, and the more stable tertiary carbocation is formed easily which combines with iodide ion to form tertiary alkyl iodide. [1 Mark]
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.

Cleavage of mixed ethers with tertiary and primary groups via HI proceeds through an S_N1 path favored by tertiary carbocation stability.

Teacher's Note:
a) The attack of iodide ion occurs at the tertiary carbon because it forms a stable tertiary carbocation intermediate.
b) Primary alkyl groups react via S_N2 pathway due to steric factors, but here the tertiary path dominates.

(vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: Magnesium metal can liberate H₂ gas from an aqueous solution of HCl.
Reason: Magnesium has positive value of standard reduction electrode potential. [1 Mark]
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (c) Assertion is true and Reason is false.

Magnesium has a negative standard reduction electrode potential (-2.37V), making it a strong reducing agent capable of displacing hydrogen from acids.

Teacher's Note:
a) Metals with negative standard reduction potentials can liberate hydrogen from acids.
b) Since the reduction potential of Mg is highly negative, the reason statement is factually incorrect.

(C) Read the passage carefully and answer the questions that follow. [3x1]

The hydroxy derivatives of aliphatic hydrocarbons are called alcohols. Ethanol is perhaps the most important aliphatic alcohol. During COVID-19 pandemic alcohol-based sanitizers became an essential protective tool worldwide, reducing the spread of the virus. Nowadays, it is also used as a substitute for petrol in internal combustion engines.
The boiling points of alcohols are considerably higher than those of the corresponding ethers due to the presence of intermolecular hydrogen bonding in alcohols and it increases with increase in molecular mass.
For isomeric alcohols the boiling point follows the order: Primary (1^o) alcohol > Secondary (2^o) alcohol > Tertiary (3^o) alcohol, which is evident from the following data:
Aliphatic alcohols behave as weak acids and ionize to a small extent. Their acidic strength is even less than that of both water and phenol. Phenols are hydroxy derivatives of arenes and have varied application in manufacturing of dyes, plastics, drugs, explosives, etc. At different concentrations, phenols are used both as a disinfectant and an antiseptic.
Compound | Propan-1-ol | Butan-1-ol | Butan-2-ol | 2-methyl-propan-2-ol
Boiling point | 370K | 391K | 371K | 356K

(i) The boiling point of 2-methyl-propan-2-ol is much lower than that of butan-1-ol, although both have the same molecular formula. Give a reason to explain such observation. [1 Mark]

Answer: The boiling point of 2-methyl-propan-2-ol is much lower than that of butan-1-ol. This is due to the fact that with branching the surface area decreases. Consequently, boiling point also decreases.

Teacher's Note:
a) Branched chain isomers form more compact spherical shapes with smaller surface areas, reducing van der Waals forces.
b) Straight chain isomers like butan-1-ol have larger surface contact areas, resulting in higher boiling points.

(ii) Although alcohols are weaker acids than water, phenol is more acidic than alcohols. Justify. [1 Mark]

Answer: The greater acidic character of phenol as compared to alcohols can be explained on the basis of resonance. The oxygen attracts the electron pair of O-H bond strongly towards itself, weakens the O-H bond and therefore facilitates the release of H⁺ more easily.

Teacher's Note:
a) Phenoxide ion formed after the loss of a proton is resonance stabilized, driving the equilibrium forward.
b) Alkoxide ions from alcohols lack such resonance stabilization, making alcohols less acidic.

(iii) In what form (i.e. concentration), does phenol act as a disinfectant or an antiseptic? [1 Mark]

Answer: A 0.2% solution of phenol acts as an antiseptic, whereas its 1% solution is used as a disinfectant.

Teacher's Note:
a) Antiseptics are safe for living tissues at lower concentrations (0.2%).
b) Disinfectants are applied to inanimate objects at higher concentrations (1%).

SECTION B - 20 MARKS

Question 2. The standard reduction electrode potential (E^o) of some metals is given below:
Metal | E^o
Ag⁺/Ag | +0.80V
Mg²⁺/Mg | -2.37V
Li⁺/Li | -3.05V
Cu²⁺/Cu | +0.34V
Fe²⁺/Fe | -0.44V
Arrange these metals in the increasing order of their reducing power. Justify the order. [2 Marks]

Answer:
Increasing order of reducing power of metals is Ag < Cu < Fe < Mg < Li.
Reducing power = 1 / Reduction potential.
Therefore, as the reduction potential value of a metal decreases, its reducing power increases.

Teacher's Note:
a) Reducing power is the tendency to lose electrons, which is inversely related to standard reduction potential.
b) Lithium with the most negative reduction potential (-3.05V) is the strongest reducing agent among them.

Question 3. An alkene [A] with molecular formula, C₅H₁₀ on ozonolysis and subsequent hydrolysis gives a mixture of two compounds, [B] and [C]. Compound [B] gives positive Fehling's test and also reacts with iodine and NaOH solution. Compound [C] does not give Fehling's test but forms iodoform. Identify compounds [A], [B] and [C]. Write the iodoform reaction either with compound [B] or [C]. [2 Marks]

Answer:
[A] CH₃-CH=C(CH₃)₂ or 2-methylbut-2-ene
[B] CH₃CHO (acetaldehyde)
[C] CH₃-CO-CH₃ (propanone or acetone)
Iodoform reaction with compound [B]:
CH₃CHO + 3I₂ + 4NaOH →[Δ] CHI₃ + 3NaI + HCOONa + 3H₂O
Or with compound [C]:
CH₃-CO-CH₃ + 3I₂ + 4NaOH →[Δ] CHI₃ + 3NaI + CH₃COONa + 3H₂O

Teacher's Note:
a) Reverse ozonolysis by joining the carbonyl oxygens of [B] (CH₃CHO) and [C] ((CH₃)₂CO) yields alkene [A].
b) Both acetaldehyde and acetone respond to the iodoform test because they contain the CH₃-CO- group.

Question 4

(i) Answer the following questions with respect to the coordination complex ion [Cr(NH₃)₆]³⁺
(a) What is the hybridisation state of the central metal atom? [1 Mark]
(b) Comment on the magnetic nature of the complex ion. [1 Mark]

Answer:
(a) d²sp³
(b) Paramagnetic. Reason: There are three unpaired electrons in Cr³⁺ (d³ configuration).

Teacher's Note:
a) Chromium in +3 oxidation state has 3 electrons in the d-subshell (d³).
b) With inner d-orbitals available, it forms an inner orbital octahedral complex with d²sp³ hybridization.

OR

(ii) Observe the diagram below of splitting of d-orbital in octahedral field and answer the following questions. [2 Marks]

[Figure: Energy level diagram showing crystal field splitting of d-orbitals in an octahedral field. The lower degenerate set t_2g contains 3 electrons with energy -0.40Δ_o, and the upper degenerate set e_g has an energy gap of Δ_o or 10Dq with +0.60Δ_o relative to the barycenter.]

(a) Identify the electronic configuration of d⁵ ion, if Δ_o < P. [1 Mark]
(b) Comment on the relationship between crystal field stabilisation energy (Δ_o) and the strength of the ligand. [1 Mark]

Answer:
(a) t_2g³ e_g²
(b) If the value of CFSE (Δ_o) is more, the ligand is strong (i.e., low spin complex).

Teacher's Note:
a) When Δ_o < P (weak field ligand), pairing energy is greater than crystal field splitting energy, leading to high spin configurations.
b) Strong field ligands cause greater splitting, resulting in high CFSE values.

Question 5. Rohan, a patient of thyroid disorder, was given certain amount of radioactive Iodine-131 as a part of his treatment. What amount of Iodine-131 will be left in Rohan's body after 48 days, if it is known that the half-life period of radioactive iodine-131 is 12 days? Assume that no amount of the isotope was eliminated through biological processes. [2 Marks]

Answer:
Given: t_1/2 = 12 days, t = 48 days.
Number of half-lives (n) = Total time / Half-life period = 48 / 12 = 4.
Amount left after 4 half-lives:
A = A_o × (1/2)⁴ = A_o × (1/16).
i.e., 1/16 of the initial amount is left.

Teacher's Note:
a) Radioactive decay follows first-order kinetics, allowing the use of half-life formulas directly.
b) Always calculate the number of half-lives first before applying the fraction remaining formula.

Question 6. How can the following conversions be carried out? (Write the chemical equations):
(i) Aniline to chlorobenzene [1 Mark]
(ii) Chloroethane to n-butane [1 Mark]

Answer:
(i) C₆H₅N₂Cl →[Cu₂Cl₂/HCl] C₆H₅Cl + N₂
(or C₆H₅N₂Cl →[Cu/HCl] C₆H₅Cl + N₂)
(ii) 2C₂H₅Cl + 2Na →[ether] C₂H₅-C₂H₅ (n-butane) + 2NaCl

Teacher's Note:
a) Conversion (i) is Sandmeyer's reaction or Gattermann reaction via diazonium salt formation.
b) Conversion (ii) is the Wurtz reaction used for ascending the carbon chain symmetrically.

Question 7. Substantiate the following statements with reasons.
(i) Cobalt is attracted in a magnetic field whereas Zinc is not. (Atomic number of Co = 27 and Zn = 30) [1 Mark]
(ii) The melting points and boiling points of Zn, Cd and Hg are low. (Atomic number of Zn = 30, Cd = 48 and Hg = 80) [1 Mark]

Answer:
(i) Cobalt has d⁷ electrons in the outer shell and therefore has three unpaired electrons, making it attracted in a magnetic field. Zinc has a completely filled d¹⁰ configuration and therefore will not be attracted by the magnetic field.
(ii) In Zn, Cd and Hg, all electrons in the d-subshell are paired. Hence, the metallic bonds present in them are weak, resulting in low melting and boiling points.

Teacher's Note:
a) Paramagnetism arises due to the presence of unpaired electrons, which is a characteristic feature of many transition metals.
b) Group 12 elements (Zn, Cd, Hg) lack d-electron involvement in metallic bonding due to completely filled d-orbitals.

Question 8. Write the Nernst equation and calculate the emf of the following cell at 298K.
Mg(s) / Mg²⁺ (1 × 10^-3M) // Cu²⁺ (1 × 10^-4M) / Cu(s)
Given: E^o(Mg²⁺/Mg) = -2.373V, E^o(Cu²⁺/Cu) = +0.337V [2 Marks]

Answer:
Nernst Equation:
E_cell = E^o_cell - (0.059 / n) log ([Mg²⁺] / [Cu²⁺])
E^o_cell = E^o_cathode - E^o_anode = 0.337V - (-2.373V) = +2.71V.
E_cell = 2.71 - (0.059 / 2) log ((1 × 10^-3) / (1 × 10^-4))
= 2.71 - 0.0295 log(10) = 2.71 - 0.0295(1) = 2.6805 V.

Teacher's Note:
a) Always write the standard cell potential first by subtracting anode potential from cathode potential.
b) Ensure the stoichiometry of electrons transferred (n = 2) is applied correctly in the Nernst log term divisor.

Question 9

(i) Illustrate the following reactions with an example in each case:
(a) Coupling reaction [1 Mark]
(b) Balz-Schiemann reaction [1 Mark]

Answer:
(a) Coupling reaction:
C₆H₅N₂⁺Cl^- + C₆H₅OH →[alkali, 273-278K] C₆H₅-N=N-C₆H₄-OH (p-hydroxyazobenzene) + HCl.
(b) Balz-Schiemann reaction:
C₆H₅N₂⁺Cl^- + HBF₄ → C₆H₅N₂⁺BF₄^- →[heat] C₆H₅F (Fluorobenzene) + BF₃ + N₂.

Teacher's Note:
a) Coupling reactions involve diazonium salts reacting with activated aromatic rings like phenols or amines in mild alkaline conditions.
b) Balz-Schiemann reaction is a standard method for preparing aryl fluorides from diazonium tetrafluoroborates.

OR

(ii) An organic compound [A] with molecular formula C₃H₇NO forms compound [B] on heating with Br₂ and KOH. Compound [B] on heating with CHCl₃ and alcoholic KOH produces a pungent smelling compound [C]. Compound [B] on reacting with C₆H₅SO₂Cl forms compound [D] which is soluble in alkali. Write the structures of compounds [A], [B], [C] and [D]. [2 Marks]

Answer:
[A] C₂H₅CONH₂ (propanamide)
[B] C₂H₅NH₂ (ethanamine)
[C] C₂H₅N≡C (ethyl isocyanide)
[D] C₆H₅SO₂-NH-C₂H₅ (N-ethylbenzenesulfonamide)

Teacher's Note:
a) Compound [B] is a primary amine because it gives carbylamine reaction and forms a sulfonamide soluble in alkali.
b) Working backwards from [B] via Hoffmann bromamide degradation identifies [A] as propanamide.

Question 10. In general, it is observed that the rate of a reaction becomes double with every 10^oC rise in temperature. If this generalisation holds true for a reaction, calculate the value of activation energy when temperature changes from 295 K to 305 K. [2 Marks]

Answer:
Given: T₁ = 295K, T₂ = 305K, k₂/k₁ = 2.
log(k₂/k₁) = (E_a / 2.303 R) × [(T₂ - T₁) / (T₁T₂)]
log(2) = (E_a / (2.303 × 8.314)) × [(305 - 295) / (295 × 305)]
0.3010 = (E_a / 19.147) × (10 / 89975)
E_a = (0.3010 × 19.147 × 295 × 305) / 10 = 151854.8 J K^-1 mol^-1 = 151.854 kJ mol^-1.

Teacher's Note:
a) Use the Arrhenius integrated equation for two different temperatures to compute activation energy.
b) Ensure units for gas constant R and calculated activation energy (Joules to kiloJoules) match standard reporting conventions.

Question 11. The students of Class XII were on a study trip. They visited a nearby lake and took some water samples which were rich in sodium chloride salt. The boiling point of lake water was found to be 373.032 K. If 500 g of lake water sample was used which contained 0.45 g of NaCl, what is the observed molecular weight of NaCl in the lake sample? Assume that NaCl ionises completely in lake water. (Given: k_b of water = 0.52 K kg mol^-1, boiling point of pure water = 373 K) [2 Marks]

Answer:
Given: ΔT_b = 373.032 - 373 = 0.032 K, w = 0.45 g, W = 500 g, k_b = 0.52 K kg mol^-1.
NaCl → Na⁺ + Cl^-, so i = 2.
M_(obs) = (i × 1000 × k_b × w) / (ΔT_b × W)
M_(obs) = (2 × 1000 × 0.52 × 0.45) / (0.032 × 500)
M_(obs) = 29.25 g mol^-1.

Teacher's Note:
a) Elevation in boiling point is a colligative property modified by van't Hoff factor (i) for electrolytes.
b) The observed molecular weight is lower than the normal molecular weight due to dissociation in aqueous solution.

SECTION C - 21 MARKS

Question 12. Two students Rehan and Ananya were given the same compound, alkyl halide (C₂H₅Cl) by their chemistry teacher and asked to perform different reactions.
- Rehan was instructed to use alcoholic KCN as a reagent and Ananya was instructed to use alcoholic AgCN.
- Rehan observed a compound was formed that could be converted to a carboxylic acid on hydrolysis but Ananya could not obtain the same acid.
- The teacher explained that the difference was due to the bonding behaviour of cyanide ions in the reagents.
Write the chemical equations of the reactions and the IUPAC names of the compounds formed by Rehan and Ananya. [3 Marks]

Answer:
Rehan's Reaction:
C₂H₅Cl + KCN (alc) → C₂H₅CN (propanenitrile) + KCl.
Ananya's Reaction:
C₂H₅Cl + AgCN (alc) → C₂H₅NC (ethyl isocyanide / ethyl carbylamine) + AgCl.

Teacher's Note:
a) Potassium cyanide is ionic and provides cyanide ions, leading to nucleophilic attack through carbon to form nitriles.
b) Silver cyanide is predominantly covalent, making nitrogen the more reactive nucleophile, thus yielding isocyanides.

Question 13

(i) For a given first order reaction, it takes 5 minutes for the initial concentration of 0.6 moles litre^-1 to become 0.2 moles litre^-1. Calculate the rate constant for this reaction. [2 Marks]
(ii) Find the overall order of the reactions which have the following rate law expression: [1 Mark]
(a) Rate = k[A]^1/2[B]^3/2
(b) Rate = k[A]^3/2[B]^-1
(c) Rate = k[A]¹[B]²

Answer:
(i) Given: t = 5 minutes, a = 0.6 moles litre^-1, (a - x) = 0.2 moles litre^-1.
k = (2.303 / t) log₁₀ (a / (a - x))
= (2.303 / 5) log₁₀ (0.6 / 0.2) = (2.303 / 5) log(3)
= (2.303 / 5) × 0.4771 = 0.2197 min^-1.
(ii) Overall order of reactions:
(a) 1/2 + 3/2 = 2
(b) 3/2 - 1 = 1/2
(c) 1 + 2 = 3

Teacher's Note:
a) First-order rate constant formula uses logarithms of initial to remaining concentrations.
b) Overall order is the simple algebraic sum of all individual exponents present in the rate law expression.

Question 14

(i) Compounds [X] and [Y] are functional isomers of each other with molecular formula C₂H₆O.
(a) Draw the structural formula of both the isomers. [1 Mark]
(b) Which of the compounds will have a lower boiling point and why? [1 Mark]
(ii) Methanol does not give yellow precipitate of iodoform when heated with iodine and alkali, but ethanol gives iodoform test with the same reagents. Explain. [1 Mark]

Answer:
(i) (a) [X] = CH₃CH₂OH (ethanol), [Y] = CH₃-O-CH₃ (methoxymethane / dimethyl ether).
(b) Compound [Y] will have a lower boiling point as it is an ether which cannot form associated molecules by intermolecular hydrogen bonding.
(ii) Methanol does not have the CH₃-CH(OH)- group and therefore does not give the iodoform test:
CH₃OH + I₂ + NaOH →[Δ] No reaction.
Ethanol contains the CH₃-CH(OH)- group and gives the iodoform test:
CH₃CH₂OH + 4I₂ + 6NaOH →[Δ] CHI₃ (yellow ppt) + 5NaI + HCOONa + 5H₂O.

Teacher's Note:
a) Alcohols exhibit hydrogen bonding leading to higher boiling points compared to their isomeric ethers.
b) The iodoform test specifically requires a secondary alcohol or ethanol containing the CH₃CH(OH)- structural unit.

OR

(iii) How will the following be converted? (Give chemical equations) [3 Marks]
(a) Diethyl ether from sodium ethoxide
(b) Propan-2-ol from Grignard's reagent
(c) Phenol from Cumene

Answer:
(a) C₂H₅ONa + BrC₂H₅ →[330K] C₂H₅-O-C₂H₅ (diethyl ether) + NaBr.
(b) CH₃CHO + CH₃MgBr →[dry ether] CH₃-CH(OMgBr)-CH₃ →[H₂O/H⁺] CH₃-CH(OH)-CH₃ (propan-2-ol) + Mg(OH)Br.
(c) Cumene + O₂ → Cumene hydroperoxide →[H⁺/H₂O] Phenol + Acetone (CH₃COCH₃).

Teacher's Note:
a) Conversion (a) is Williamson ether synthesis involving haloalkanes and sodium alkoxides.
b) Conversion (b) uses acetaldehyde with methylmagnesium bromide to yield a secondary alcohol upon hydrolysis.

Question 15

(i) Complete and balance the following chemical equation.
KMnO₄ + H₂SO₄ + H₂S → _____ + _____ + _____ + _____ [1 Mark]
(ii) Why do transition metals form alloys? [1 Mark]
(iii) Potassium dichromate (K₂Cr₂O₇) acts as a powerful oxidising agent in acidic medium. Explain. [1 Mark]

Answer:
(i) 2KMnO₄ + 3H₂SO₄ + 5H₂S → K₂SO₄ + 2MnSO₄ + 8H₂O + 5S.
(ii) Atoms of transition metals have similar atomic radii and crystal structures, allowing them to easily replace each other in a crystal lattice in the molten state and form solid solutions (alloys).
(iii) In the presence of dilute sulfuric acid, K₂Cr₂O₇ liberates nascent oxygen or undergoes reduction from Cr(VI) to Cr(III), thereby acting as an oxidising agent.
Equation: K₂Cr₂O₇ + 4H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O + 3[O].

Teacher's Note:
a) Balance redox reactions by equating oxidation number changes or ion-electron half reactions.
b) Transition metals readily form alloys due to close atomic size similarities.

Question 16

(i) Identify the non-reducing sugar which on hydrolysis gives two reducing monosaccharides. [1 Mark]
(ii) Which structure of protein is normally unaffected during denaturation of protein? [1 Mark]
(iii) Name the vitamins whose deficiency cause the following diseases. [1 Mark]
(a) Rickets
(b) Night blindness

Answer:
(i) Sucrose.
(ii) Primary structure.
(iii) (a) Rickets: Vitamin D.
(b) Night blindness: Vitamin A.

Teacher's Note:
a) Sucrose is non-reducing because its glycosidic linkage involves the reducing centers of both glucose and fructose.
b) Denaturation disrupts secondary and tertiary structures while leaving the primary peptide backbone intact.

Question 17

(i) Three electrolytic cells (X), (Y) and (Z) containing solutions of AgNO₃, CuSO₄ and ZnSO₄ respectively are connected in series. A steady current of 1.5 ampere is passed through these electrolytic cells until 1.45g of silver is deposited at the cathode of cell (X). (Atomic weight of Ag = 108, Cu = 63.5 and Zn = 65.3)
(a) How much of charge is given to the electrolyte solution? [1 Mark]
(b) What mass of copper and zinc is deposited at the respective cathode? [2 Marks]

Answer:
(a) Cell (X) contains AgNO₃. Reaction: Ag⁺ + e^- ⇄ Ag.
1 mole or 108 g of Ag is deposited by 96500 coulombs.
Charge for 1.45 g of Ag = (96500 × 1.45) / 108 = 1295.6 coulombs.
(b) In cell (Y), reaction: Cu²⁺ + 2e^- → Cu.
2 × 96500 coulombs deposit 63.5 g of Cu.
Mass of Cu deposited by 1295.6 coulombs = (63.5 × 1295.6) / (2 × 96500) = 0.426 g of Cu.
In cell (Z), reaction: Zn²⁺ + 2e^- → Zn.
2 × 96500 coulombs deposit 65.3 g of Zn.
Mass of Zn deposited by 1295.6 coulombs = (65.3 × 1295.6) / (2 × 96500) = 0.438 g of Zn.

Teacher's Note:
a) Faraday's second law of electrolysis states that masses of different substances deposited by the same quantity of electricity are proportional to their equivalent weights.
b) Always divide the molar mass by the valence factor (n-factor) to find the correct equivalent mass.

OR

(ii) The molar conductivities at infinite dilution for NaI, CH₃COONa and (CH₃COO)₂Mg are 126.9, 91.0 and 187.8 ohm^-1 cm² mol^-1 respectively at 25⁰C. Find out the molar conductivity of MgI₂ at infinite dilution by using Kohlrausch's law of independent migration of ions. [3 Marks]

Answer:
Given: ∧^o_m(NaI) = 126.9, ∧^o_m(CH₃COONa) = 91.0, ∧^o_m((CH₃COO)₂Mg) = 187.8 ohm^-1 cm² mol^-1.
∧^o_m(MgI₂) = ∧^o_m((CH₃COO)₂Mg) + 2∧^o_m(NaI) - 2∧^o_m(CH₃COONa)
= 187.8 + 2(126.9) - 2(91.0)
= 187.8 + 253.8 - 182.0 = 259.6 ohm^-1 cm² mol^-1.

Teacher's Note:
a) Express target molar conductivity in terms of constituent ionic molar conductivities.
b) Combine given electrolyte values with proper stoichiometric coefficients to cancel out unwanted ions.

Question 18. Write the chemical equations for the following conversions: [3 Marks]
(i) Nitrobenzene to benzene
(ii) Aniline to bromobenzene
(iii) Methyl cyanide to ethylamine

Answer:
(i) C₆H₅NO₂ + 6[H] →[Sn/HCl] C₆H₅NH₂ →[HNO₂ + HCl, 273-278K] C₆H₅N₂⁺Cl^- →[H₃PO₂ + H₂O] C₆H₆ + N₂ + H₃PO₃ + HCl.
(ii) C₆H₅NH₂ →[HNO₂ + HCl, 273-278K] C₆H₅N₂⁺Cl^- →[Cu₂Br₂ + HBr] C₆H₅Br (bromobenzene) + N₂.
(iii) CH₃CN + 4[H] →[Na/alcohol or H₂/Ni] CH₃CH₂NH2 (ethylamine).

Teacher's Note:
a) Conversion (i) requires reduction to aniline, diazotization, and subsequent deamination using hypophosphorous acid.
b) Conversion (ii) utilizes Sandmeyer bromination on benzene diazonium chloride.

SECTION D - 15 MARKS

Question 19

(i) An organic compound [A] with molecular formula C₂Cl₃O₂H is obtained when compound [B] reacts with red phosphorous and Cl₂. The organic compound [B] can be obtained on the reaction of methyl magnesium bromide with carbon dioxide followed by acid hydrolysis. Upon partial reduction, compound [B] forms compound [C] with molecular formula C₂H₄O, which gives yellow precipitate of iodoform on heating with iodine in presence of sodium hydroxide. Compound [C] also reacts with dilute NaOH to form compound [D].
(a) Identify the compounds [A], [B], [C] and [D]. [1 Mark]
(b) Write down the reaction for the formation of [A] from [B]. What is the reaction called? [1 Mark]
(c) Write the chemical test to distinguish between compound [C] and acetone. [2 Marks]
(d) Which will be more acidic, compound [A] or [B]? Why? [1 Mark]

Answer:
(a) [A] Cl₃C-COOH (trichloroacetic acid), [B] CH₃COOH (acetic acid), [C] CH₃CHO (acetaldehyde), [D] CH₃-CH(OH)-CH₂-CHO (3-hydroxybutanal / aldol).
(b) CH₃COOH + 3Cl₂ →[Red P] Cl₃C-COOH + 3HCl.
Reaction name: Hell-Volhard-Zelinsky (HVZ) reaction.
(c) Compound [C] (acetaldehyde) gives Tollens' test (silver mirror) and Fehling's test (red precipitate), whereas acetone does not respond to these tests.
(d) Compound [A] (trichloroacetic acid) will be more acidic due to the electron-withdrawing inductive effect (-I effect) of three chlorine atoms, which stabilize the carboxylate anion.

Teacher's Note:
a) Establish identities step-by-step: Grignard reaction with CO₂ gives acetic acid ([B]), whose reduction gives acetaldehyde ([C]).
b) HVZ reaction selectively halogenates alpha-hydrogens of carboxylic acids in the presence of red phosphorus.

OR

(ii) Identify the compounds [A], [B] and [C] in the following reactions: [3 Marks]
(a) CH₃C≡N →[SnCl₂/HCl / H₂O] [A] →[CH₃MgBr / H₂O] [B] →[Acidified K₂Cr₂O₇] [C]
(b) H-C≡C-H →[H₂O / Hg²⁺/H₂SO₄] [A] →[[O] / K₂Cr₂O₇ + H₂SO₄] [B] →[PCl₅] [C]

Answer:
(a) [A] CH₃CHO (acetaldehyde), [B] CH₃CH(OH)CH₃ (propan-2-ol), [C] CH₃COCH₃ (propanone / acetone).
(b) [A] CH₃CHO (acetaldehyde), [B] CH₃COOH (acetic acid), [C] CH₃COCl (acetyl chloride).

Teacher's Note:
a) Stephen reduction of acetonitrile yields acetaldehyde ([A]), which with Grignard reagent forms a secondary alcohol ([B]), oxidized to acetone ([C]).
b) Hydration of acetylene gives acetaldehyde, oxidation gives acetic acid, and reaction with PCl₅ gives acetyl chloride.

(iii) Write chemical equations to convert the following: [2 Marks]
(a) Acetone to propene
(b) Benzaldehyde to cinnamic acid

Answer:
(a) Acetone →[LiAlH₄] propan-2-ol →[H₂SO₄, 443K] propene + H₂O.
(b) C₆H₅CHO + (CH₃CO)₂O →[(CH₃COONa, boil)] C₆H₅CH=CHCOOH (cinnamic acid) + CH₃COOH (Perkin reaction).

Teacher's Note:
a) Reduction of ketone yields secondary alcohol followed by acid-catalyzed dehydration to alkene.
b) Benzaldehyde undergoes Perkin condensation with acetic anhydride in the presence of sodium acetate to form cinnamic acid.

Question 20

(i) The compound [PtCl₂(NH₃)₂] also known as cis-platin, is used in chemotherapy. It also demonstrates how coordination geometry impacts medical science.
(a) Write its IUPAC name and the type of isomerism it exhibits. [1 Mark]
(b) What is the geometry of the compound? [1 Mark]
(ii) What type of structural isomers are [Pt(NH₃)₃Br]NO₃ and [Pt(NH3)₃(NO3)]Br? How can they be distinguished by using a chemical test? [2 Marks]
(iii) A coordination compound CoCl₃.4H₂O precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write the structural formula of the compound and name it. [1 Mark]

Answer:
(i) (a) Diamminedichloridoplatinum(II); Geometrical isomerism.
(b) Square planar geometry.
(ii) Ionisation isomers.
Distinction test: Aqueous solution of [Pt(NH₃)₃Br]NO₃ gives a white/pale yellow precipitate with AgNO₃ (or gives brown ring test with FeSO₄ and conc. H₂SO₄ confirming NO₃^-), whereas [Pt(NH₃)₃(NO3)]Br gives a yellow precipitate of AgBr with AgNO₃ solution which is sparingly soluble in excess NH₄OH.
(iii) Structural formula: [Co(H2O)₄Cl₂]Cl.
IUPAC name: Tetraaquadichlorocobalt(III) chloride.

Teacher's Note:
a) Square planar complexes of Pt(II) commonly exhibit cis-trans isomerism.
b) Ionization isomers yield different ions in solution, identifiable by specific reagent tests like silver nitrate.

Question 21

(i) Answer the following:
(a) 300 ml of aqueous solution of protein contains 1.85g of protein, the osmotic pressure of the solution at 25^oC is found to be 3.05 × 10^-3 atm. Calculate the molar mass of the protein. [2 Marks]
(b) Phenol (C₆H₅OH) associates in benzene to form a dimer. A solution of 2.5g of phenol in 120g of benzene lowers its freezing point by 0.85K. Calculate the degree of association of phenol. (k_f for benzene = 5.12 K kg mol^-1, molecular mass of phenol = 94 g mol^-1) [2 Marks]
(c) The molecular weight of potassium chloride and sucrose is determined by the depression of freezing point method. Compared to their theoretical molecular weight, what will be their observed molecular weights when determined by the above method? Justify your answer. [1 Mark]

Answer:
(a) Given: π = 3.05 × 10^-3 atm, T = 298K, V = 300 / 1000 = 0.3 L, w = 1.85 g, R = 0.0821 L atm K^-1 mol^-1.
π = (w × R × T) / (m × V)
m = (1.85 × 0.0821 × 298) / (3.05 × 10^-3 × 0.3) = 49466 g mol^-1.
(b) Given: k_f = 5.12, ΔT_f = 0.85, M_normal = 94, w = 2.5, W = 120 g, n = 2.
M_(observed) = (1000 × k_f × w) / (ΔT_f × W) = (1000 × 5.12 × 2.5) / (0.85 × 120) = 125.49 g mol^-1.
van't Hoff factor i = M_normal / M_observed = 94 / 125.49 = 0.749.
Degree of association α = (1 - i) / (1 - (1/n)) = (1 - 0.749) / (1 - 0.5) = 0.251 / 0.5 = 0.502 or 50.2%.
(c) The observed molecular mass of potassium chloride will be half of its theoretical molecular mass because KCl dissociates into two ions (K⁺ and Cl^-). The observed molecular mass of sucrose will be the same as its theoretical molecular mass because sucrose is a non-electrolyte and does not dissociate.

Teacher's Note:
a) Osmotic pressure equation relates molar mass directly with solution volume and concentration.
b) For association, van't Hoff factor is less than 1, and degree of association formulas must account for dimer formation (n=2).

OR

(ii) (a) Arrange the following aqueous solutions in increasing order of osmotic pressure. Give reasons in support of your answer. [2 Marks]
(1) 6.0g per litre of urea solution (molecular weight of urea = 60g mol^-1)
(2) 72.0g per litre of glucose solution (molecular weight of glucose = 180g mol^-1)
(3) 5.85g per litre of sodium chloride solution (molecular weight of sodium chloride = 58.5g mol^-1)
(b) Calculate the mole fraction (x) of ethanol and water if 92g of ethanol is dissolved in 540g of water. (Atomic mass of C = 12, O = 16 and H = 1) [2 Marks]
(c) What type of azeotropic mixture will be formed by a solution of chloroform and acetone? Justify your answer based on strength of intermolecular interactions that develops in the solution. [1 Mark]

Answer:
(a) Moles in 1 L:
(1) Urea = 6 / 60 = 0.1 mol.
(2) Glucose = 72 / 180 = 0.4 mol.
(3) NaCl = 5.85 / 58.5 = 0.1 mol × 2 (i=2) = 0.2 mol of particles.
Since osmotic pressure is proportional to total particle concentration: Urea (0.1) < NaCl (0.2) < glucose (0.4).
Increasing order: Urea < NaCl < glucose.
(b) Moles of ethanol (n) = 92 / 46 = 2 moles.
Moles of water (N) = 540 / 18 = 30 moles.
Mole fraction of ethanol X_(C2H5OH) = 2 / (2 + 30) = 2 / 32 = 0.0625.
Mole fraction of water X_(H2O) = 30 / (2 + 30) = 30 / 32 = 0.9375.
(c) A mixture of chloroform and acetone forms a maximum boiling azeotrope because it shows a negative deviation from Raoult's law due to hydrogen bonding forming between chloroform and acetone molecules, strengthening intermolecular attractive forces.

Teacher's Note:
a) Always convert given mass concentrations into total particle molarities by factoring in dissociation constants before comparing osmotic pressures.
b) Negative deviations from Raoult's law result in maximum boiling azeotropes due to enhanced intermolecular attraction.