CBSE Class 12 Chemistry Sample Paper 2: marking scheme and solutions
Every answer with its marking scheme, the way an examiner awards the marks. Chapters in this paper: 1 Solutions; 2 Electrochemistry; 3 Chemical Kinetics; 4 The d- and f-Block Elements; 5 Coordination Compounds; 6 Haloalkanes and Haloarenes; 7 Alcohols, Phenols and Ethers; 8 Aldehydes, Ketones and Carboxylic Acids; 9 Amines; 10 Biomolecules.
Section A (16 × 1 mark)
Answer: (b) For a polyhydric alcohol the 'e' of the alkane is kept and 'di' goes before 'ol'; the chain is numbered to give the –OH groups the lowest locants: butane-1,2-diol. (a) drops the 'e'; (c) numbers from the wrong end; (d) treats –OH as a prefix.
Answer: (b) D gives the configuration relative to glyceraldehyde, judged at the lowest asymmetric carbon (C-5) with –CHO at the top; the sign of rotation is shown separately by (+). In D-glucose the –OH on C-3 is on the left, so not every –OH is on the right.
Answer: (b) Two Na⁺ counter ions make the entity [Fe(CN)₅(NH₃)]²⁻. NH₃ is neutral and each CN⁻ is −1: x + 5(−1) + 0 = −2, so x = +3.
Answer: (b) It is a secondary alcohol (–OH on C-2, which carries one H); secondary alcohols are oxidised to ketones: CH₃COCH(CH₃)₂, 3-methylbutan-2-one.
Answer: (d) Δsol H > 0, so dissolution absorbs heat. By Le Chatelier's principle, raising the temperature favours the endothermic change, dissolution, and solubility rises. Incompressibility is the reason pressure, not temperature, has little effect.
Answer: (d) The longer chain on nitrogen, propyl, gives the parent propan-1-amine; the ethyl group on nitrogen takes the locant N. Ethylpropylamine is the common name, not the IUPAC name; (a) takes the shorter chain as parent; (b) puts the nitrogen on C-2.
Answer: (b) Both chlorine atoms sit on the same (terminal) carbon, so the compound is geminal (an alkylidene halide, butylidene chloride); its IUPAC name is 1,1-dichlorobutane. Vicinal would need the two halogens on adjacent carbons.
Answer: (d) The cation is named first. The entity [Zn(CN)₄]²⁻ is an anion, so the metal name ends in -ate; x + 4(−1) = −2 gives Zn(II).
Answer: (a) The slowest step is the rate determining step; the overall reaction cannot go faster than it, as a relay team is held to its slowest runner.
Answer: (a) The longest chain has five carbons and is numbered from the end nearest –OH: OH on C-1, CH₃ on C-3; the 'e' of pentane becomes 'ol': 3-methylpentan-1-ol. Numbering from the other end (b) gives the –OH a higher locant; (c) uses a four-carbon chain; (d) names –OH as a prefix.
Answer: (c) A single line separates a metal from its electrolyte; the double line is the salt bridge joining the two half-cells.
Answer: (c) Arylamines are usually colourless but become coloured on storage through atmospheric oxidation.
Answer: (a) Dissolving a gas evolves heat, like condensation; by Le Chatelier's principle a rise in temperature shifts the equilibrium back, so solubility falls. R explains A.
Answer: (b) Both true. A lower aliphatic amine dissolves because it hydrogen bonds with water; the carbylamine reaction shows only that it is primary and does not explain its solubility.
Answer: (a) Both true. Because phenol is a stronger acid than water, it gives up its proton to OH⁻: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. R explains A.
Answer: (a) Both intervals are one half-life; t½ = 0.693/k is independent of [R]₀, which is exactly why the two times are equal.
Section B (5 × 2 marks)
CH₃CH₂CH₂Br + Mg → CH₃CH₂CH₂MgBr (dry ether). CH₃CH₂CH₂MgBr + H₂O → CH₃CH₂CH₃ + Mg(OH)Br. In the C–Mg bond the carbon is δ–: it pulls electrons from the electropositive magnesium.
- Grignard formation in dry ether (1)
- hydrolysis to propane, Mg(OH)Br; carbon δ– (1)
C₆H₁₁Br + Mg → C₆H₁₁MgBr (cyclohexylmagnesium bromide, dry ether). C₆H₁₁MgBr + CH₃OH → C₆H₁₂ + Mg(OCH₃)Br. Methanol, like water, is a proton source, so the Grignard reagent is converted to the hydrocarbon cyclohexane.
- Grignard formation in dry ether (1)
- protonation by methanol gives cyclohexane (1)
CH₃CH₂CH₂COOH + Cl₂ → CH₃CH₂CH(Cl)COOH + HCl, with a little red phosphorus. Hell–Volhard–Zelinsky reaction: halogenation at the α-carbon of an acid that has an α-hydrogen.
- correct equation with Cl₂ and red phosphorus, Cl on the α-carbon (1)
- names the Hell–Volhard–Zelinsky reaction (1)
Order is the sum of the powers of concentration terms in the experimental rate law; it can be 0, a fraction or a whole number. Molecularity is the number of species colliding simultaneously in an elementary step; it is a whole number (1, 2 or 3), never 0 or a fraction. Order applies to elementary and complex reactions; molecularity has meaning only for an elementary reaction.
- one difference, correctly stated for both quantities (e.g. 0 or fraction allowed for order only) (1)
- a second difference (e.g. experimental against theoretical, or complex against elementary) (1)
P lacks vitamin E, whose deficiency causes increased fragility of red blood cells and a loss of muscle strength; E is fat soluble. Q lacks vitamin B₂ (riboflavin), whose deficiency causes cheilosis, digestive disorders and a burning sensation of the skin; B₂ is water soluble.
- P: vitamin E, fat soluble (1)
- Q: vitamin B₂ (riboflavin), water soluble (1)
Denticity is the number of donor atoms through which a ligand binds the central atom (the number of its ligating groups). (i) Oxalate binds through two O atoms: didentate. (ii) N(CH₂CH₂NH₂)₃ binds through its central N and three NH₂ nitrogens: tetradentate (polydentate).
- denticity: the number of donor atoms of a ligand bound to the metal (1)
- oxalate 2 (didentate); N(CH₂CH₂NH₂)₃ 4 (polydentate) (1)
Section C (7 × 3 marks)
Step 1 (slow, reversible): (CH₃)₂CBrCH₂CH₃ ⇌ (CH₃)₂C⁺CH₂CH₃ + Br⁻; the energy to break C–Br comes from solvation of the ions by water. Step 2 (fast): (CH₃)₂C⁺CH₂CH₃ + H₂O → (CH₃)₂C(OH)CH₂CH₃ + H⁺, giving 2-methylbutan-2-ol. Rate = k[(CH₃)₂CBrCH₂CH₃], first order: only the halide is in the slow step. The reaction is fast because the carbocation is tertiary and stabilised by three alkyl groups.
- slow ionisation to the tertiary carbocation (1)
- fast attack, 2-methylbutan-2-ol (1)
- first-order rate law, tertiary cation stability (1)
Step 1 (slow, reversible): (C₂H₅)₃C–Cl ⇌ (C₂H₅)₃C⁺ + Cl⁻. Step 2 (fast): (C₂H₅)₃C⁺ + H₂O → (C₂H₅)₃C–OH + H⁺, giving 3-ethylpentan-3-ol. Rate = k[(C₂H₅)₃CCl], first order. A polar protic solvent such as water or alcohol solvates the ions formed in step 1, supplying the energy needed to break the C–Cl bond.
- slow ionisation to the tertiary carbocation (1)
- fast attack, 3-ethylpentan-3-ol (1)
- first-order rate law, solvation of the ions (1)
(i) Anode: Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻. (ii) Cathode: PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l). (iii) Overall Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O: sulphuric acid is used up and water formed, so the acid becomes more dilute and its density falls as the battery discharges.
- anode reaction (1)
- cathode reaction (1)
- H₂SO₄ consumed and water formed, so density falls (1)
1. The alcohol is protonated: (CH₃)₃C–OH + H⁺ → (CH₃)₃C–OH₂⁺. 2. Water leaves, giving the carbocation (CH₃)₃C⁺; this is the slowest, rate-determining step. 3. A proton is lost from a CH₃ group: (CH₃)₂C=CH₂ + H⁺, and the acid is regenerated. Because step 2 decides the rate and the tertiary carbocation is the most stable, it forms easily; ethanol would have to form a much less stable primary carbocation, so it needs harsher conditions.
- step 1: protonation of the alcohol (1)
- step 2: (CH₃)₃C⁺ formed, rate determining (1)
- step 3: loss of H⁺ to 2-methylpropene; tertiary carbocation most stable, so easier (1)
(i) Hypo (sodium thiosulphate) dissolves the unexposed AgBr as the complex [Ag(S₂O₃)₂]³⁻: AgBr + 2Na₂S₂O₃ → Na₃[Ag(S₂O₃)₂] + NaBr. (ii) Silver plates more smoothly and evenly from a solution of [Ag(CN)₂]⁻ than from simple Ag⁺ ions. (iii) Wilkinson catalyst, [(Ph₃P)₃RhCl], a rhodium complex, catalyses the hydrogenation of alkenes.
- hypo dissolves AgBr as [Ag(S₂O₃)₂]³⁻ (1)
- smooth plating from [Ag(CN)₂]⁻ (1)
- Wilkinson catalyst [(Ph₃P)₃RhCl] (1)
(i) V₂O₃ is basic, V₂O₄ less basic, V₂O₅ amphoteric though mainly acidic (it gives VO₄³⁻ and VO₂⁺ salts). (ii) VO²⁺ salts. (iii) The oxidation number rises from +3 to +5; ionic character falls as the oxidation number rises, and the higher, more covalent oxides are acidic.
- nature of all three (1)
- VO²⁺ salts (1)
- rising oxidation number, falling ionic character (1)
(i) k = ([R]₀ − [R])/t = (0.50 − 0.38)/30 = 4.0 × 10⁻³ mol L⁻¹ min⁻¹. (ii) [R] = 0 when t = [R]₀/k = 0.50/0.004 = 125 min. (iii) t½ = [R]₀/2k = 0.50/0.008 = 62.5 min.
- k = 4.0 × 10⁻³ mol L⁻¹ min⁻¹ (1)
- 125 min to completion (1)
- t½ = 62.5 min (1)
(a) Gabriel synthesis: phthalimide + ethanolic KOH → potassium phthalimide; with CH₃CH₂CH₂Br on heating → N-propylphthalimide; hydrolysis with NaOH(aq) → CH₃CH₂CH₂NH₂ + the disodium salt of phthalic acid. Only a primary amine can form. (b) A mixture of propan-1-amine, N-propylpropan-1-amine, tripropylamine and the quaternary salt (CH₃CH₂CH₂)₄N⁺Br⁻, because each amine formed is itself a nucleophile and reacts further with the halide. (c) 1-Iodopropane: the reactivity of halides with amines is RI > RBr > RCl, so the iodide reacts faster and the chloride slower.
- (a) Gabriel route with the three stages (1)
- (b) mixture of 1°, 2°, 3° amines and quaternary salt, with reason (1)
- (c) 1-iodopropane, RI > RBr > RCl (1)
Section D (2 × 4 marks)
| Sample | Tollens' reagent | Products of complete hydrolysis |
|---|---|---|
| P | silver mirror | not hydrolysed |
| Q | no change | glucose and fructose, equimolar |
| R | silver mirror | glucose only, two units per molecule |
| S | silver mirror | galactose and glucose |
| T | no change | glucose only, several hundred units in an unbranched chain |
Answer: (c) Q is a non-reducing sugar giving equimolar glucose and fructose: sucrose. Maltose gives only glucose, lactose galactose and glucose, cellulose many glucose units.
Answer: (c) R is maltose: two α-D-glucose units, C1 of one to C4 of the other. (a) is lactose (sample S), (b) sucrose (Q), (d) a branch point of amylopectin.
In Q, sucrose, the glycosidic linkage joins C1 of α-D-glucose to C2 of β-D-fructose, so the reducing groups of both units are bonded and none is free. In R, maltose, C1 of the first glucose is joined to C4 of the second; C1 of the second unit is free and a free aldehyde group can form there in solution, which reduces Tollens' reagent.
- Q: both reducing groups used in the C1–C2 linkage (1)
- R: free aldehyde group can form at C1 of the second glucose unit (1)
T is amylose, the water-soluble component of starch (about 15–20%), an unbranched chain of α-D-glucose units joined by C1–C4 glycosidic linkages. The insoluble component is amylopectin (about 80–85%), a branched polymer of α-D-glucose: its chain is formed by C1–C4 linkages and its branches by C1–C6 linkages.
- T amylose, unbranched α C1–C4 chain (1)
- amylopectin, branched, with C1–C6 linkages at the branches (1)
Answer: (d) Lu³⁺ is 4f¹⁴: with a filled f subshell no f–f transition is possible, so it is colourless. Pr³⁺, Nd³⁺ and Sm³⁺ have partly filled 4f sets and are coloured.
Answer: (b) The lanthanoid contraction: the decrease is regular for the M³⁺ ions (not quite regular for the atoms), because 4f electrons shield the rising nuclear charge poorly.
Eu (Z = 63) is [Xe]4f⁷6s²; Eu²⁺ (4f⁷) forms by losing the two s electrons, and it readily loses one more electron to return to the common +3 state, so it reduces other species. Tb⁴⁺ has a half-filled 4f⁷ set, which favours its formation, but it readily gains an electron to return to +3, so it oxidises other species.
- Eu²⁺ 4f⁷, gives up an electron to reach +3: reductant (1)
- Tb⁴⁺ 4f⁷, takes an electron to reach +3: oxidant (1)
Ce (Z = 58) in the +4 state has a noble gas configuration ([Xe], 4f⁰), which favours Ce⁴⁺; it is a strong oxidant that reverts to the common +3 state. Its reaction with water is very slow, so Ce(IV) solutions are kinetically stable and serve as a good analytical oxidant.
- Ce⁴⁺ has the noble gas configuration (4f⁰) (1)
- oxidation of water is very slow, so the reagent is usable (1)
Section E (3 × 5 marks)
(a) The measured vapour pressure exceeds the Raoult value: positive deviation. (b) P–Q attractions are less strong than P–P and Q–Q, so molecules escape more easily; energy is absorbed on mixing, so Δmix H is usually > 0. (c) A minimum boiling azeotrope: at its composition the vapour pressure is highest, so it boils at a lower temperature than mixtures of nearby composition. (d) Ethanol and acetone: acetone molecules get between ethanol molecules and break some of their hydrogen bonds, weakening the interactions.
- positive deviation (1)
- P–Q less strong than P–P and Q–Q (1)
- Δmix H usually positive (1)
- minimum boiling azeotrope, lowest boiling point (1)
- ethanol + acetone, with hydrogen bonds of ethanol broken (1)
(a) The measured vapour pressure is below the Raoult value: negative deviation. (b) L–M attractions are stronger than L–L and M–M, so the escaping tendency falls; heat is evolved on mixing, so Δmix H is usually < 0. (c) A maximum boiling azeotrope, for example nitric acid–water at about 68% HNO₃ by mass, boiling at 393.5 K. (d) Chloroform and acetone: the hydrogen of chloroform forms a hydrogen bond with the oxygen of acetone, (CH₃)₂C=O ···· H–CCl₃.
- negative deviation (1)
- L–M stronger than L–L and M–M (1)
- Δmix H usually negative (1)
- maximum boiling azeotrope, e.g. HNO₃–water about 68% (1)
- chloroform + acetone with the hydrogen bond (1)
(a) Cell constant G* = l/A, the distance between the electrodes divided by their area of cross-section (unit cm⁻¹). l and A cannot be measured reliably, so G* = Rκ is found from the resistance of a KCl solution of known κ. (b) G* = 1000 × 1.41 × 10⁻³ = 1.41 cm⁻¹. κ(HX) = G*/R = 1.41/6000 = 2.35 × 10⁻⁴ S cm⁻¹. Λm = κ × 1000/c = 2.35 × 10⁻⁴ × 1000/0.0050 = 47.0 S cm² mol⁻¹. α = 47.0/390 = 0.121; Ka = cα²/(1 − α) = 0.0050 × 0.01452/0.879 = 8.3 × 10⁻⁵ mol L⁻¹.
- (a) G* = l/A and why KCl calibration (G* = Rκ) (1)
- G* = 1.41 cm⁻¹ (1)
- κ = 2.35 × 10⁻⁴ S cm⁻¹ (1)
- Λm = 47.0 S cm² mol⁻¹ (1)
- α = 0.121 and Ka = 8.3 × 10⁻⁵ (1)
(a) The bridge is balanced when no current flows through the detector; then the cell's resistance R₂ = R₁R₄/R₃, and κ = G*/R₂. (b) G* = Rκ = 200 × 0.0129 = 2.58 cm⁻¹. κ(HY) = 2.58/5160 = 5.00 × 10⁻⁴ S cm⁻¹. Λm = 5.00 × 10⁻⁴ × 1000/0.0020 = 250 S cm² mol⁻¹. α = 250/400 = 0.625; Ka = cα²/(1 − α) = 0.0020 × 0.3906/0.375 = 2.08 × 10⁻³ mol L⁻¹.
- (a) no detector current, R₂ = R₁R₄/R₃, κ = G*/R (1)
- G* = 2.58 cm⁻¹ (1)
- κ = 5.00 × 10⁻⁴ S cm⁻¹ (1)
- Λm = 250 S cm² mol⁻¹ (1)
- α = 0.625 and Ka = 2.08 × 10⁻³ (1)
(a) C₆H₅COOH + NH₃ → C₆H₅COONH₄, ammonium benzoate. C₆H₅COONH₄ → (strong heating) C₆H₅CONH₂ + H₂O, benzamide. (b) m-O₂N–C₆H₄–CHO, m-nitrobenzaldehyde (3-nitrobenzaldehyde). The –CHO group withdraws electrons from the ring, deactivating it, and directs the incoming electrophile to the meta position. (c) The –COOH group deactivates the ring towards electrophilic substitution, and the Lewis acid AlCl₃ bonds to the –COOH group, so the Friedel–Crafts reaction does not occur.
- (a) ammonium benzoate equation (1)
- (a) heating to benzamide, water lost (1)
- (b) m-nitrobenzaldehyde, structure and name (1)
- (b) –CHO deactivates the ring and directs meta (1)
- (c) ring deactivated, and AlCl₃ bonds to –COOH (1)
(a) C₆H₄(COOH)₂ + 2NH₃ → C₆H₄(COONH₄)₂; on heating, 2H₂O is lost: P = C₆H₄(CONH₂)₂, phthalamide. On strong heating P loses NH₃ and closes a ring: Q = C₆H₄(CO)₂NH, phthalimide. (b) m-nitrobenzaldehyde and m-nitrobenzoic acid. Both –CHO and –COOH withdraw electrons from the ring, deactivate it and direct the electrophile meta. (c) m-bromobenzoic acid. The Friedel–Crafts alkylation fails because –COOH deactivates the ring and the AlCl₃ bonds to the –COOH group.
- (a) P phthalamide (1)
- (a) Q phthalimide, NH₃ lost (1)
- (b) m-nitrobenzaldehyde and m-nitrobenzoic acid (1)
- (b) both groups deactivate and direct meta (1)
- (c) m-bromobenzoic acid; Friedel–Crafts fails, deactivation and AlCl₃ bonding (1)