CBSE Class 12 Chemistry Question Paper 2025 Solved Code 56-1-2

SECTION A

1. Assertion (A): All naturally occurring \( \alpha \)-amino acids except glycine are optically active.
Reason (R): Most naturally occurring amino acids have L-configuration. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) Optical activity comes from a chiral \( \alpha \)-carbon; glycine has two H atoms on it, so it is not chiral.
b) L-configuration only tells the arrangement of groups, so it does not explain optical activity.

2. Assertion (A): The boiling point of ethanol is higher than that of methoxymethane.
Reason (R): There is intramolecular hydrogen bonding in ethanol. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Ethanol has intermolecular (not intramolecular) hydrogen bonding, which raises its boiling point.
b) Watch the words "inter" and "intra" carefully in Assertion-Reason questions.

3. Assertion (A): The boiling points of alkyl halides decrease in the order: RI \( \gt \) RBr \( \gt \) RCl \( \gt \) RF.
Reason (R): The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) The order RI \( \gt \) RBr \( \gt \) RCl \( \gt \) RF is due to increasing size and van der Waals forces of the halogen.
b) The Reason compares alkyl halides with hydrocarbons, so it does not explain the order among halides.

4. Assertion (A): [Cr(H₂O)₆]Cl₂ and [Fe(H₂O)₆]Cl₂ are examples of homoleptic complexes.
Reason (R): All the ligands attached to the metal are the same. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) Homoleptic complexes have only one kind of ligand bonded to the metal (here only H₂O).
b) Cl^- ions outside the square bracket are counter ions, not ligands.

5. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because ________. [1 Mark]
(A) it gains water due to osmosis
(B) it loses water due to reverse osmosis
(C) it gains water due to reverse osmosis
(D) it loses water due to osmosis

Answer: (D) it loses water due to osmosis

Teacher's Note:
a) The salt solution is hypertonic, so water moves out of the mango cells by osmosis.
b) Reverse osmosis needs external pressure greater than osmotic pressure, which is not applied here.

6. Which of the following statements is not true about glucose? [1 Mark]
(A) It is an aldohexose.
(B) On heating with HI it forms n-hexane.
(C) It exists in furanose form.
(D) It does not give Schiff's test.

Answer: (C) It exists in furanose form.

Teacher's Note:
a) Glucose exists in the six-membered pyranose form; fructose exists in the furanose form.
b) Not giving Schiff's test is one of the anomalies of the open-chain structure of glucose.

7. The best reagent for converting propanamide into propanamine is ________. [1 Mark]
(A) excess H₂
(B) Br₂ in aqueous NaOH
(C) iodine in the presence of red phosphorus
(D) LiAlH₄ in ether

Answer: (D) \( LiAlH_4 \) in ether

Teacher's Note:
a) LiAlH₄ reduces the amide CH₃CH₂CONH₂ to CH₃CH₂CH₂NH₂ with the same number of carbon atoms.
b) Br₂/NaOH (Hoffmann bromamide reaction) would give ethanamine, one carbon less.

8. The acid formed when propyl magnesium bromide is treated with CO₂ followed by acid hydrolysis is: [1 Mark]
(A) C₃H₇COOH
(B) C₂H₅COOH
(C) CH₃COOH
(D) C₃H₇OH

Answer: (A) \( C_3H_7COOH \)

Teacher's Note:
a) A Grignard reagent RMgX adds to CO₂ and gives RCOOH after hydrolysis, with one carbon more.
b) C₃H₇MgBr therefore gives butanoic acid, C₃H₇COOH.

9. Which is the correct order of acid strength from the following? [1 Mark]
(A) C₆H₅OH \( \gt \) H₂O \( \gt \) ROH
(B) C₆H₅OH \( \gt \) ROH \( \gt \) H₂O
(C) ROH \( \gt \) C₆H₅OH \( \gt \) H₂O
(D) H₂O \( \gt \) C₆H₅OH \( \gt \) ROH

Answer: (A) \( C_6H_5OH \gt H_2O \gt ROH \)

Teacher's Note:
a) The phenoxide ion is stabilised by resonance, so phenol is the strongest acid here.
b) The alkyl group in ROH has +I effect, which makes alcohols weaker acids than water.

10. Alkyl halides undergoing nucleophilic bimolecular substitution reaction involve [1 Mark]
(A) retention of configuration
(B) formation of racemic mixture
(C) inversion of configuration
(D) formation of carbocation

Answer: (C) inversion of configuration

Teacher's Note:
a) In S_N2 the nucleophile attacks from the back side, opposite to the leaving group.
b) Racemisation and carbocation formation belong to the S_N1 mechanism.

11. Arrange the following compounds in increasing order of their boiling points:
(i) (CH₃)₂CH - CH₂Br (ii) CH₃CH₂CH₂CH₂Br (iii) (CH₃)₃C - Br
The correct order is [1 Mark]
(A) (ii) \( \lt \) (i) \( \lt \) (iii)
(B) (i) \( \lt \) (ii) \( \lt \) (iii)
(C) (iii) \( \lt \) (i) \( \lt \) (ii)
(D) (iii) \( \lt \) (ii) \( \lt \) (i)

[Figure: Structures of the three compounds: (i) a CH carbon bearing two CH₃ groups and joined to CH₂Br; (ii) straight chain CH₃CH₂CH₂CH₂Br; (iii) a central carbon bonded to three CH₃ groups (one on each side and one above) and to Br below.]

Answer: (C) (iii) \( \lt \) (i) \( \lt \) (ii)

Teacher's Note:
a) All three are isomers of C₄H₉Br; boiling point falls as branching increases.
b) More branching gives a more spherical shape, less surface area and weaker van der Waals forces.

12. The correct IUPAC name of [Pt(NH₃)₂Cl₂]²⁺ is [1 Mark]
(A) Diamminedichloridoplatinum (II)
(B) Diamminedichloridoplatinum (IV)
(C) Diamminedichloridoplatinum (O)
(D) Diamminedichloridoplatinate (IV)

Answer: (B) Diamminedichloridoplatinum (IV)

Teacher's Note:
a) Oxidation state of Pt: \( x + 2(0) + 2(-1) = +2 \), so \( x = +4 \).
b) The suffix "-ate" is used only for anionic complexes; this complex is a cation.

13. Acidified KMnO₄ oxidises sulphite to [1 Mark]
(A) S₂O₃^2-
(B) S₂O₈^2-
(C) SO₂(g)
(D) SO₄^2-

Answer: (D) \( SO_4^{2-} \)

Teacher's Note:
a) In acidic medium, 2MnO₄^- + 5SO₃^2- + 6H⁺ → 2Mn²⁺ + 5SO₄^2- + 3H₂O.
b) Sulphur changes from +4 in sulphite to +6 in sulphate.

14. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr³⁺ ion (Atomic no.: Cr = 24) is ________. [1 Mark]
(A) 2.87 B.M.
(B) 3.87 B.M.
(C) 3.47 B.M.
(D) 3.57 B.M.

Answer: (B) 3.87 B.M.

Teacher's Note:
a) Cr³⁺ is 3d³, so it has \( n = 3 \) unpaired electrons.
b) \( \mu = \sqrt{n(n+2)} = \sqrt{15} \approx 3.87 \) B.M.

15. Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be [1 Mark]
(A) +1.19 V
(B) +0.89 V
(C) +0.18 V
(D) +1.83 V

Answer: (B) +0.89 V

Teacher's Note:
a) The couple with higher E° (Sn⁴⁺/Sn²⁺) is the cathode.
b) \( E^{\circ}_{cell} = 0.15 - (-0.74) = +0.89\ V \); do not multiply E° by stoichiometric coefficients.

16. In case of association, abnormal molar mass of solute will [1 Mark]
(A) increase
(B) decrease
(C) remain same
(D) first increase and then decrease

Answer: (A) increase

Teacher's Note:
a) Association reduces the number of particles, so the colligative property decreases.
b) Molar mass is inversely related to the colligative property, so the calculated molar mass increases (\( i \lt 1 \)).

SECTION B

17. Identify A and B in each of the following reaction sequence:
(a) CH₃CH₂Cl →(NaCN) A →(H₂/Ni) B
(b) C₆H₅NH₂ →(NaNO₂/HCl, 0 - 5 \( ^{\circ} \)C) A →(C₆H₅NH₂, H⁺) B [2 Marks]

Answer:
(a) A = CH₃CH₂CN (propanenitrile); B = CH₃CH₂CH₂NH₂ (propan-1-amine).
(b) A = C₆H₅N₂⁺Cl^- (benzenediazonium chloride); B = C₆H₅-N=N-C₆H₄-NH₂ (p-aminoazobenzene, the -NH₂ group is para to the azo group).

Teacher's Note:
a) Each of A and B carries half a mark, so write both formula and name clearly.
b) Coupling with aniline occurs at the para position of aniline in mildly acidic medium.

18. When FeCr₂O₄ is fused with Na₂CO₃ in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured (C). An acidified solution of compound (C) oxidises iodide to (D). Identify (A), (B), (C) and (D). [2 Marks]

Answer:
(A) Na₂CrO₄ - sodium chromate (yellow).
(B) Na₂Cr₂O₇ - sodium dichromate.
(C) K₂Cr₂O₇ - potassium dichromate (orange).
(D) I₂ - iodine.

Teacher's Note:
a) 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ is the first step.
b) Colour clues help: chromate is yellow and dichromate is orange.
c) Cr₂O₇^2- + 14H⁺ + 6I^- → 2Cr³⁺ + 3I₂ + 7H₂O.

19. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Justify your answer. [2 Marks]

Answer:
1. Benzaldehyde is less reactive than propanal in nucleophilic addition reactions.
2. In benzaldehyde, the benzene ring is in resonance with the C=O group. This reduces the polarity of the carbonyl group, so its carbon is less electrophilic than the carbonyl carbon of propanal.

Teacher's Note:
a) One mark is for "less reactive" and one mark is for the resonance reason.
b) Key words: resonance, reduced polarity, less electrophilic carbonyl carbon.

20. (A) Give reasons:
(a) Cooking is faster in pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? [2 Marks]

Answer:
(a) Pressure inside the pressure cooker is high. Higher pressure raises the boiling point of water, so food cooks at a higher temperature and faster.
(b) A decrease in volume shows negative deviation from Raoult's law. The temperature of the solution increases on mixing (heat is released).

Teacher's Note:
a) For negative deviation, \( \Delta_{mix}V \lt 0 \) and \( \Delta_{mix}H \lt 0 \), so the mixture warms up.
b) Link pressure to boiling point clearly in part (a).

OR

(B) Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example. [2 Marks]

Answer:
1. Azeotropes are binary mixtures that have the same composition in the liquid and vapour phase and boil at a constant temperature.
2. Solutions showing negative deviation form maximum boiling azeotropes.
3. Example: a mixture of about 68% HNO₃ and 32% water by mass.

Teacher's Note:
a) The definition carries 1 mark; the type and the example carry half a mark each.
b) Do not mix up: positive deviation gives minimum boiling azeotropes (for example ethanol-water).

21. Give reasons for the following:
(a) The melting points of \( \alpha \)-amino acids are generally higher than that of the corresponding carboxylic acids.
(b) Amino acids show amphoteric behaviour. [2 Marks]

Answer:
(a) Amino acids exist as dipolar ions (zwitter ions), H₃N⁺-CHR-COO^-. Strong electrostatic forces act between these ions, so their melting points are high.
(b) Amino acids contain both an acidic carboxyl group and a basic amino group (zwitter ion form), so they can react with both acids and bases.

Teacher's Note:
a) The key word for both parts is "zwitter ion" (dipolar ion).
b) Amphoteric means able to react with both acids and bases.

SECTION C

22. A solution containing 15 g urea (molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. [3 Marks]

Answer:
Given: \( w_U = 15\ g \), \( M_U = 60\ g\ mol^{-1} \), \( M_G = 180\ g\ mol^{-1} \), volume of each solution = 1 L.
Formula: \( \pi = CRT = \dfrac{w_B RT}{M_B V} \)
For isotonic solutions: \( \pi_{glucose} = \pi_{urea} \), so \( C_G = C_U \) (same T and V).
\( \dfrac{w_G}{M_G} = \dfrac{w_U}{M_U} \)
\( \dfrac{w_G}{180} = \dfrac{15}{60} \)
\( w_G = \dfrac{15 \times 180}{60} = 45\ g \)
Mass of glucose in one litre of solution = 45 g.

Teacher's Note:
a) Isotonic solutions have equal molar concentration, which is the key step for marks.
b) Quick check: glucose is 3 times heavier than urea per mole, so it needs 3 times the mass.

23. Calculate \( \Delta_r G^{\circ} \) and log K_C of the reaction:
Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Given \( E^{\circ}_{Ag^{+}/Ag} = 0.80\ V \), \( E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77\ V \)
[R = 8.314 J K^-1 mol^-1, F = 96500 C mol^-1] [3 Marks]

Answer:
Given: \( E^{\circ}_{cathode} = E^{\circ}_{Ag^{+}/Ag} = 0.80\ V \), \( E^{\circ}_{anode} = E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77\ V \), \( n = 1 \), T = 298 K.
\( E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - 0.77 = 0.03\ V \)
\( \Delta_r G^{\circ} = -nFE^{\circ}_{cell} = -1 \times 96500 \times 0.03 \)
\( \Delta_r G^{\circ} = -2895\ J\ mol^{-1} = -2.895\ kJ\ mol^{-1} \)
\( \Delta_r G^{\circ} = -2.303RT \log K_C \)
\( \log K_C = \dfrac{-\Delta_r G^{\circ}}{2.303RT} = \dfrac{2895}{2.303 \times 8.314 \times 298} = \dfrac{2895}{5705.8} \)
\( \log K_C \approx 0.507 \) (about 0.51)

Teacher's Note:
a) Only one electron is transferred (Fe²⁺ → Fe³⁺), so n = 1.
b) Keep \( \Delta_r G^{\circ} \) in joules while calculating log K_C, and give units with \( \Delta_r G^{\circ} \).
c) A negative \( \Delta_r G^{\circ} \) and positive log K_C confirm the reaction is spontaneous.

24. (a) Arrange the following in decreasing order of pK_b:
Aniline, p-nitroaniline, p-methylaniline
(b) Account for the following:
(i) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(ii) Methylamine in water reacts with FeCl₃ to precipitate hydrated ferric oxide. [3 Marks]

Answer:
(a) p-nitroaniline \( \gt \) Aniline \( \gt \) p-methylaniline (decreasing pK_b).
(b) (i) In arenediazonium salts, the positive charge of the -N₂⁺ group is spread over the benzene ring by resonance. This resonance stabilisation is absent in aliphatic diazonium salts.
(ii) Methylamine is a base. In water it releases OH^- ions: CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH^-. These OH^- ions react with FeCl₃: 2FeCl₃ + 6OH^- → Fe₂O₃.3H₂O (hydrated ferric oxide) + 6Cl^-.

Teacher's Note:
a) Higher pK_b means weaker base; the -NO₂ group withdraws electrons and -CH₃ donates them.
b) Write "resonance stabilisation" as the key word in (b)(i).

25. (A) Draw the structure of the major monohalo product for each of the following reaction:
(a) →(Br₂, Heat) ?
(b) + HBr → ?
(c) →(HCl, Heat) ? [3 Marks]

[Figure: (a) A benzene ring with a CH₂ - CH₃ group at one position and Cl at the para position, reacting with Br₂, Heat; (b) a cyclohexene ring with a CH₃ group on one carbon of the double bond, reacting with HBr; (c) a benzene ring with an OH group at one position and an HO - H₂C group at the para position, reacting with HCl, Heat.]

Answer:
(a) 1-(1-Bromoethyl)-4-chlorobenzene: Cl-C₆H₄-CH(Br)-CH₃ (Br replaces an H on the benzylic carbon; Cl stays para).
(b) 1-Bromo-1-methylcyclohexane: Br and CH₃ are both on the same ring carbon (Markovnikov addition).
(c) 4-(Chloromethyl)phenol: HO-C₆H₄-CH₂Cl (only the benzylic -OH is replaced by Cl; the phenolic -OH stays).

Teacher's Note:
a) Br₂ with heat gives free radical substitution at the benzylic position, not on the ring.
b) The phenolic C-O bond has partial double bond character, so phenol -OH is not replaced by HCl.

OR

(B) How do you convert:
(a) Chlorobenzene to biphenyl
(b) Propene to 1-Iodopropane
(c) 2-bromobutane to but-2-ene. [3 Marks]

Answer:
(a) Fittig reaction: 2C₆H₅Cl + 2Na →(dry ether) C₆H₅-C₆H₅ (biphenyl) + 2NaCl
(b) CH₃CH=CH₂ + HBr →(peroxide) CH₃CH₂CH₂Br; then CH₃CH₂CH₂Br + NaI →(acetone) CH₃CH₂CH₂I + NaBr
(c) CH₃CH(Br)CH₂CH₃ →(ethanolic KOH, heat) CH₃CH=CHCH₃ + HBr (dehydrohalogenation)

Teacher's Note:
a) In (b), the peroxide effect (anti-Markovnikov addition) puts Br on the terminal carbon; the Finkelstein reaction then swaps Br for I.
b) In (c), Saytzeff rule makes the more substituted alkene, but-2-ene, the major product.

26. The elements of 3d transition series are given as:
Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
Answer the following:
(a) Copper has exceptionally positive \( E^{\circ}_{M^{2+}/M} \) value, why?
(b) Which element is a strong reducing agent in +2 oxidation state and why?
(c) Zn²⁺ salts are colourless. Why? [3 Marks]

Answer:
(a) Copper has a high enthalpy of atomisation (\( \Delta_a H^{\circ} \)) and a low hydration enthalpy (\( \Delta_{hyd} H^{\circ} \)) of Cu²⁺. The energy needed to form Cu²⁺(aq) is not balanced by its hydration energy, so \( E^{\circ} \) is positive.
(b) Chromium (Cr). Cr²⁺ (d⁴) easily loses an electron to become Cr³⁺ (d³), which has a stable half-filled t_2g level.
(c) Zn²⁺ has a completely filled 3d¹⁰ configuration with no unpaired electron, so no d-d transition is possible and the salts are colourless.

Teacher's Note:
a) In (a) mention both terms: high atomisation enthalpy and low hydration enthalpy.
b) In (b) the name "Cr" and the reason (d⁴ to d³) carry half a mark each.

27. A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.
[R = 8.314 J K^-1 mol^-1; log 4 = 0.602] [3 Marks]

Answer:
Given: \( t_{1/2} = 20 \) min at \( T_1 = 300 \) K; \( t_{1/2} = 5 \) min at \( T_2 = 350 \) K.
For a first order reaction, \( t_{1/2} = \dfrac{0.693}{k} \)
\( k_1 = \dfrac{0.693}{20} = 0.03465\ min^{-1} \)
\( k_2 = \dfrac{0.693}{5} = 0.1386\ min^{-1} \)
Arrhenius equation: \( \log \dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left[\dfrac{T_2 - T_1}{T_1 T_2}\right] \)
\( \log \dfrac{0.1386}{0.03465} = \dfrac{E_a}{2.303 \times 8.314}\left[\dfrac{350 - 300}{350 \times 300}\right] \)
\( \log 4 = \dfrac{E_a}{19.15} \times \dfrac{50}{105000} \)
\( E_a = \dfrac{0.602 \times 19.15 \times 105000}{50} \approx 24209\ J\ mol^{-1} \)
\( E_a \approx 24.2\ kJ\ mol^{-1} \)

Teacher's Note:
a) Note that \( \dfrac{k_2}{k_1} = \dfrac{20}{5} = 4 \), which is why log 4 is given.
b) Half a mark is deducted for a missing or wrong unit of E_a.

28. (a) Write the reaction when D-glucose reacts with the following:
(i) NH₂OH
(ii) Acetic anhydride
(b) Why vitamin C cannot be stored in our body? [3 Marks]

Answer:
(a) (i) CHO-(CHOH)₄-CH₂OH + NH₂OH → HO-N=CH-(CHOH)₄-CH₂OH (glucose oxime) + H₂O
(ii) CHO-(CHOH)₄-CH₂OH + 5(CH₃CO)₂O → CHO-(CH-O-CO-CH₃)₄-CH₂-O-CO-CH₃ (glucose pentaacetate) + 5CH₃COOH
(b) Vitamin C is water soluble. It is regularly excreted in urine, so it cannot be stored in the body.

Teacher's Note:
a) Oxime formation shows the presence of a carbonyl (-CHO) group in glucose.
b) Pentaacetate formation shows five -OH groups in glucose.
c) Fat-soluble vitamins (A, D, E, K) can be stored; water-soluble ones (B group, C) cannot.

SECTION D

29. Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more to o- and p- positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols.
Answer the following questions:

(a) What happens when phenol reacts with
(i) Br₂/CS₂
(ii) Conc. HNO₃ [2 Marks]

Answer:
(i) With bromine in CS₂ at low temperature (273 K), monobromination occurs. A mixture of 2-bromophenol (o-bromophenol) and 4-bromophenol (p-bromophenol) is formed.
C₆H₅OH + Br₂ →(CS₂, 273 K) o-BrC₆H₄OH + p-BrC₆H₄OH
(ii) With conc. HNO₃, 2,4,6-trinitrophenol (picric acid) is formed.
C₆H₅OH →(conc. HNO₃) C₆H₂(NO₂)₃OH

Teacher's Note:
a) A non-polar solvent (CS₂) and low temperature give only monobromo products; bromine water gives 2,4,6-tribromophenol.
b) Name the products clearly: o- and p-bromophenol, and picric acid.

(b) Why phenol does not undergo protonation readily? [1 Mark]

Answer: Due to resonance, the lone pair of electrons on the oxygen atom is delocalised into the benzene ring, so it is not easily available for protonation.

Teacher's Note:
a) The key idea is that the lone pair on oxygen takes part in resonance with the ring.
b) This is also why the C-O bond in phenol has partial double bond character.

(c) Which is a stronger acid - phenol or cresol? Give reason. [1 Mark]

Answer: Phenol is the stronger acid. The methyl group in cresol is electron releasing (+I effect), so the phenoxide ion formed from cresol is less stable.

Teacher's Note:
a) Electron releasing groups decrease acidity; electron withdrawing groups increase it.
b) Half a mark is for "phenol" and half a mark for the reason.

OR

(c) Write the IUPAC name of the product formed in the Reimer-Tiemann reaction. [1 Mark]

Answer: 2-Hydroxybenzaldehyde (2-hydroxybenzenecarbaldehyde), commonly called salicylaldehyde.

Teacher's Note:
a) Reimer-Tiemann reaction uses CHCl₃ and aqueous NaOH, followed by acid hydrolysis.
b) The -CHO group enters ortho to the -OH group, so write "2-" in the name.

30. The rate of a chemical reaction is expressed either in terms of decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reaction is directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the Rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction.
Answer the following questions:

(a) (i) What is a rate determining step?
(ii) Define complex reaction. [2 Marks]

Answer:
(i) The slowest step in the mechanism of a reaction is called the rate determining step.
(ii) A complex reaction is one that takes place through a series of elementary reactions, that is, in two or more steps.

Teacher's Note:
a) The key word in (i) is "slowest step".
b) The key words in (ii) are "series of elementary steps".

(b) What is the effect of temperature on the rate constant of a reaction? [1 Mark]

Answer: The rate constant increases with increase in temperature (it nearly doubles for every 10 K rise).

Teacher's Note:
a) This follows the Arrhenius equation, \( k = Ae^{-E_a/RT} \).
b) A higher temperature gives more molecules with energy equal to or more than E_a.

OR

(b) Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions? [1 Mark]

Answer: Molecularity is the number of species colliding in a single step, so it is defined only for elementary reactions. In a complex reaction each step may have a different molecularity, so it has no meaning for the overall reaction. Order is determined experimentally (by the slowest step), so it applies to both.

Teacher's Note:
a) Molecularity is theoretical; order is experimental.
b) Order can be zero or fractional, but molecularity is always a whole number.

(c) The conversion of molecule X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y? [1 Mark]

Answer: Rate = k[X]². When [X] becomes 3[X], rate = k(3[X])² = 9k[X]². So the rate of formation of Y increases 9 times.

Teacher's Note:
a) For order n, the rate changes by (factor)ⁿ; here \( 3^2 = 9 \).
b) Write the rate law first; it makes the working clear.

SECTION E

31. (A) (a) Carry out the following conversions:
(i) Ethanal to But-2-enal
(ii) Propanoic acid to ethane
(b) An alkene A with molecular formula C₅H₁₀ on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C. [5 Marks]

Answer:
(a) (i) Aldol condensation: 2CH₃CHO →(dil. NaOH / OH^-) CH₃-CH(OH)-CH₂-CHO (aldol) →(heat) CH₃-CH=CH-CHO (but-2-enal) + H₂O
(ii) Decarboxylation: CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O; then CH₃CH₂COONa + NaOH →(CaO, heat) CH₃-CH₃ + Na₂CO₃
(b) A = (CH₃)₂C=CHCH₃, 2-methylbut-2-ene.
B = CH₃CHO, ethanal (an aldehyde, so it gives Fehling test; it has CH₃CO- group, so it gives the iodoform test).
C = CH₃COCH₃, propanone (acetone) (a ketone, so no Fehling test, but it forms iodoform).
Ozonolysis: (CH₃)₂C=CHCH₃ →(O₃, then Zn/H₂O) CH₃COCH₃ + CH₃CHO

Teacher's Note:
a) Soda lime decarboxylation removes one carbon, so propanoic acid (3 C) gives ethane (2 C).
b) Check the carbon count: 3 C in acetone + 2 C in ethanal = 5 C in the alkene C₅H₁₀.
c) Each of A, B and C carries 1 mark; give formula and name.

OR

(B) An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to get a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives But-l-ene. Identify (A), (B) and (C) and write chemical equations for the reactions involved. [5 Marks]

Answer:
(A) = C₃H₇COOC₄H₉ (CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃), butyl butanoate.
(B) = C₃H₇COOH (CH₃CH₂CH₂COOH), butanoic acid.
(C) = C₄H₉OH (CH₃CH₂CH₂CH₂OH), butan-1-ol.
Equations:
1. Hydrolysis: C₃H₇COOC₄H₉ + H₂O →(dil. H₂SO₄) C₃H₇COOH + C₄H₉OH
2. Dehydration: CH₃CH₂CH₂CH₂OH →(conc. H₂SO₄, heat) CH₃CH₂CH=CH₂ + H₂O
3. Oxidation: CH₃CH₂CH₂CH₂OH →(CrO₃ / chromic acid) CH₃CH₂CH₂COOH

Teacher's Note:
a) Since oxidation of C gives B, both B and C have the same carbon chain (4 C each), which fits C₈H₁₆O₂.
b) But-1-ene on dehydration tells us C is a straight chain primary alcohol, butan-1-ol.
c) Marks: A = 1, B and C = half each, and 1 mark for each equation.

32. (A) For the complex [Fe(en)₂Cl₂] Cl, identify:
(a) the oxidation number of iron.
(b) the hybridization and the shape of the complex.
(c) the magnetic behaviour of the complex
(d) whether there is an optical isomer of the complex? If so draw its structure.
(e) IUPAC name of the complex.
(At. no. of Fe = 26) [5 Marks]

Answer:
(a) Oxidation number of Fe: \( x + 2(0) + 2(-1) = +1 \), so \( x = +3 \).
(b) Hybridisation is d²sp³ and the shape is octahedral.
(c) Fe³⁺ is 3d⁵. Even after pairing by the strong field en ligand, one electron remains unpaired, so the complex is paramagnetic.
(d) Yes. The cis isomer is optically active. In cis-[Fe(en)₂Cl₂]⁺ the two Cl atoms are on adjacent positions of the octahedron and the two en rings span the remaining positions; this arrangement is not superimposable on its mirror image, so it exists as a pair of enantiomers (d and l forms). The trans form is optically inactive.
(e) Dichloridobis(ethane-1,2-diamine)iron(III) chloride.

Teacher's Note:
a) "en" is a neutral bidentate ligand, so the two en ligands occupy four positions.
b) Ligands are named in alphabetical order (chlorido before ethane-1,2-diamine), and "bis" is used for a ligand with a number in its name.
c) Only the cis form has an optical isomer; say "cis" clearly.

OR

(B) (a) Using IUPAC norms write the names of the following:
(i) [Co(NH₃)₄ Cl(NO₂)]Cl
(ii) K₃[Fe(CN)₆]
(iii) [Cr(C₂O₄)₃]^3-
(b) What is crystal field splitting energy? Why low spin tetrahedral complexes are not formed? [5 Marks]

Answer:
(a) (i) Tetraamminechloridonitrito-N-cobalt(III) chloride
(ii) Potassium hexacyanidoferrate(III)
(iii) Trioxalatochromate(III) ion
(b) Crystal field splitting energy is the energy difference between the two sets of d-orbitals (t_2g and e_g) formed when the degenerate d-orbitals split in the presence of ligands in a definite geometry.
In tetrahedral complexes the splitting energy is small (\( \Delta_t = \frac{4}{9}\Delta_o \)). It is not large enough to force the pairing of electrons, so low spin tetrahedral complexes are not formed.

Teacher's Note:
a) Anionic complexes end in "-ate" (ferrate, chromate); cationic and neutral complexes keep the metal name (cobalt).
b) In (b), always compare \( \Delta_t \) with the pairing energy: \( \Delta_t \lt P \) gives high spin.

33. (A) (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:
Sn(s) | Sn²⁺ (0.004 M) || H⁺ (0.02 M) | H₂(g) (1 Bar) | Pt (s)
(Given: \( E^{\circ}_{Sn^{2+}/Sn} = -0.14\ V \), \( E^{\circ}_{H^{+}|H_2(g),\ Pt} = 0.00\ V \))
(b) Account for the following;
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH₃COOH decreases on dilution. [5 Marks]

Answer:
(a) Cell reaction: Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g); n = 2
Given: [Sn²⁺] = 0.004 M, [H⁺] = 0.02 M, \( E^{\circ}_{cathode} = 0.00\ V \), \( E^{\circ}_{anode} = -0.14\ V \).
Nernst equation: \( E_{cell} = (E^{\circ}_{c} - E^{\circ}_{a}) - \dfrac{0.059}{2}\log \dfrac{[Sn^{2+}]}{[H^{+}]^{2}} \)
\( E_{cell} = [0 - (-0.14)] - \dfrac{0.059}{2}\log \dfrac{0.004}{(0.02)^{2}} \)
\( E_{cell} = 0.14 - 0.0295 \log \dfrac{0.004}{0.0004} = 0.14 - 0.0295 \log 10 \)
\( E_{cell} = 0.14 - 0.0295 = 0.1105\ V \)
(b) (i) Because of the overpotential of oxygen. The oxidation of water to O₂ is kinetically slow and needs extra voltage, so Cl^- is oxidised to Cl₂ at the anode.
(ii) CH₃COOH is a weak electrolyte. Conductivity is the conductance of unit volume of solution; on dilution the number of ions carrying current per unit volume decreases, so conductivity decreases.

Teacher's Note:
a) H₂ is at 1 bar, so it does not appear in the log term; square [H⁺] because 2H⁺ are involved.
b) Do not confuse conductivity (decreases on dilution) with molar conductivity (increases on dilution).
c) The key word for (b)(i) is "overpotential".

OR

(B) (a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use.
(b) Calculate the potential for half-cell containing 0.01 M K₂Cr₂O₇(aq), 0.01 M Cr³⁺ (aq) and 1.0 \( \times \) 10^-4 M H⁺(aq).
The half cell reaction is
Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- → 2Cr³⁺(aq) + 7H₂O(l)
and the standard electrode potential is given as E° = 1.33 V.
[Given: log 10 = 1] [5 Marks]

Answer:
(a) At anode: Pb(s) + SO₄^2-(aq) → PbSO₄(s) + 2e^-
At cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^- → PbSO₄(s) + 2H₂O(l)
Overall: Pb(s) + PbO₂(s) + 2SO₄^2-(aq) + 4H⁺(aq) → 2PbSO₄(s) + 2H₂O(l)
(b) Given: [Cr₂O₇^2-] = \( 10^{-2} \) M, [Cr³⁺] = \( 10^{-2} \) M, [H⁺] = \( 10^{-4} \) M, E° = 1.33 V, n = 6.
\( E = E^{\circ} - \dfrac{0.059}{n}\log \dfrac{[Cr^{3+}]^{2}}{[Cr_2O_7^{2-}][H^{+}]^{14}} \)
\( E = 1.33 - \dfrac{0.059}{6}\log \dfrac{(10^{-2})^{2}}{(10^{-2})(10^{-4})^{14}} \)
\( E = 1.33 - \dfrac{0.059}{6}\log \dfrac{10^{-4}}{10^{-58}} = 1.33 - \dfrac{0.059}{6}(54)\log 10 \)
\( E = 1.33 - 0.059 \times 9 = 1.33 - 0.531 \)
\( E = 0.799\ V \)

Teacher's Note:
a) Both electrodes change to PbSO₄ during discharge, and sulphuric acid is used up.
b) Raise [H⁺] to the power 14 and [Cr³⁺] to the power 2; this is where most errors happen.
c) Write the Nernst equation first, as it carries 1 mark on its own.

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