CBSE Class 12 Chemistry: sample paper 2025-26

1. Ozonolysis cleaves the C=C: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)=\mathrm{CHC}{\mathrm{H}}_3\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$. The ketone formed is butan-2-one (ethyl methyl ketone).

   1 mark

2. (A) gives an aldehyde/acid, (B) a 2° alcohol over Cu gives acetone, (D) acetylene gives acetaldehyde, so only (C) is correct.

Answer: Option (C)

1. Alcoholic NaOH causes dehydrohalogenation: A = but-1-ene, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$.

2. Acid-catalysed hydration follows Markovnikov's rule: B = butan-2-ol. Hydroboration–oxidation gives the anti-Markovnikov product: C = butan-1-ol.

   1 mark

Answer: B = Butan-2-ol, C = Butanol; option (C)

1. Ions written outside the square bracket are counter ions; here they are the two $\mathrm{C}{\mathrm{l}}^{-}$ ions.

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Answer: Chloride, option (C)

1. Hinsberg test: A's product dissolves in NaOH, so A is a 1° amine. B's product is insoluble, so B is a 2° amine. C does not react, so C is a 3° amine.

2. Hydrogen bonding falls as the number of N–H bonds falls (1° > 2° > 3°), so the boiling points are A > B > C.

   1 mark

Answer: A > B > C, option (A)

1. Molality $m=\frac{70/M}{700}\times 1000$. Volume of solution $V=770/1.5$ mL, so molarity $=\frac{70/M}{770}\times 1.5\times 1000$.

2. Ratio $\frac{m}{M}=\frac{770}{700\times 1.5}=0.73$.

   1 mark

Answer: 0.73, option (C)

1. A: alcohol + HCl on heating goes via a carbocation, so it is $S_{N}1$. B: inversion of configuration means $S_{N}2$. C: an alkene is formed, so it is elimination. D: HI adds across C=C, so it is addition.

   1 mark

Answer: A-(iv), B-(iii), C-(ii), D-(i); option (D)

1. In o-cresol the –OH group is bonded directly to a benzene ring carbon. All ring carbons are $s{p}^2$ hybridised.

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Answer: $s{p}^2$, option (B)

1. Glucose (both anomers) and sucrose are dextrorotatory. D-fructose is laevorotatory, so β-D-fructose is the answer.

   1 mark

Answer: β-D-fructose, option (C)

1. The f-block elements (lanthanoids and actinoids) are called inner transition metals.

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Answer: Option (C)

1. By Kohlrausch's law, set 1 gives ${\Lambda }_{C{H}_{3}COOH}^o={\Lambda }_{HCl}^o+{\Lambda }_{C{H}_{3}COOK}^o-{\Lambda }_{KCl}^o$.

2. Set 3 gives ${\Lambda }_{C{H}_{3}COOH}^o=\frac{1}{2}{\Lambda }_{H_{2}S{O}_4}^o+{\Lambda }_{C{H}_{3}COONa}^o-\frac{1}{2}{\Lambda }_{N{a}_{2}S{O}_4}^o$. Set 2 contains no ${\mathrm{H}}^{+}$ source, so it cannot be used.

   1 mark

Answer: Either 1 or 3, option (C)

1. 2,4-DNP reacts with the carbonyl group of aldehydes and ketones to give a coloured hydrazone. Propanal and propanone react. Propanoic acid's C=O is part of a carboxyl group and does not react.

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Answer: (i) and (ii), option (A)

1. (iv): Cl is on the carbon attached to the ring, and that carbon carries two carbon groups, so it is a secondary benzylic halide.

2. (ii): Cl is on a terminal $\mathrm{C}{\mathrm{H}}_2$ away from the ring, so it is a primary alkyl halide. The answer is (iv) and (ii).

   1 mark

Answer: (iv) and (ii), option (B)

1. It is primary (not secondary) aliphatic amines that form unstable diazonium salts and release ${\mathrm{N}}_2$, so A is false.

2. The quantitative evolution of ${\mathrm{N}}_2$ is used to estimate amino acids and proteins, so R is true.

   1 mark

Answer: Option (D)

1. A is true: solutions in the blood must be isotonic, otherwise endosmosis or exosmosis damages the cells.

2. R is also true (Na⁺/K⁺ maintain osmotic balance), but it does not explain why intravenous solutions must be isotonic.

   1 mark

Answer: Option (B)

1. Starch is about 80–85% amylopectin, which is insoluble in water. Starch therefore does not dissolve completely but forms a colloidal solution, so R correctly explains A.

   1 mark

Answer: Option (A)

1. A is true: inverters use rechargeable (secondary) cells such as lead storage batteries.

2. R is false: a primary cell cannot be recharged once used; only secondary cells can be.

   1 mark

Answer: Option (C)

1. I. The volume will be less than 100 mL. A phenol–aniline mixture shows negative deviation from Raoult's law. Strong intermolecular hydrogen bonding forms between the –OH of phenol and the –NH${}_2$ of aniline. These attractions are stronger than phenol–phenol and aniline–aniline attractions, so the molecules pack more closely and $\Delta V_{mix}<0$.

   1 mark

2. II. Salt dissolves in the water from the melting ice and causes depression in freezing point ($\Delta T_f=i{K}_{f}m$). The ice–brine mixture therefore stays in equilibrium at a temperature below 0 °C (a freezing mixture). The bottles stay colder for a longer time.

   1 mark

Answer: I. Less than 100 mL (negative deviation; stronger phenol–aniline H-bonding). II. Salt depresses the freezing point, so the ice–salt mixture stays below 0 °C and keeps the drinks cold longer.

1. I. No, no white precipitate of BaSO${}_4$ will appear. A semi-permeable membrane lets only solvent (water) molecules pass during osmosis. The solute ions Ba${}^{2+}$ and SO${}_4^{2-}$ cannot cross the membrane, so they never meet.

   1 mark

2. II. Sugar is a non-volatile solute, so it lowers the vapour pressure of water. The vapour pressure of the solution at 100 °C falls below atmospheric pressure, so boiling stops. The boiling point is elevated ($\Delta T_b=K_{b}m$), and the solution boils again only on further heating.

   1 mark

Answer: I. No; only solvent passes through the SPM, so Ba²⁺ and SO₄²⁻ never meet. II. Sugar lowers the vapour pressure, so the boiling point rises and boiling stops until the solution is heated further.

1. I. Reading from the graph: reactants (H${}_2$ + I${}_2$) at 110 kJ mol${}^{-1}$, activated complex at 140 kJ mol${}^{-1}$, products (2HI) at 100 kJ mol${}^{-1}$.

   $\Delta H=E_{products}-E_{reactants}=100-110=-10$ kJ mol${}^{-1}$ (magnitude 10 kJ mol${}^{-1}$; the reaction is exothermic).

   $E_a(\text{backward})=E_{activatedcomplex}-E_{products}=140-100=40$ kJ mol${}^{-1}$.

   1 mark

2. II. A catalyst increases the rate of reaction. It provides an alternative pathway with a lower activation energy, forming an intermediate complex with the reactants. More molecules then have energy ≥ $E_a$. It lowers $E_a$ for the forward and backward reactions equally and does not change $\Delta H$.

   1 mark

3. (For visually challenged learners)

   I. Activation energy is the minimum extra energy the reactants must absorb to form the activated complex. Raising the temperature does not change $E_a$; it only increases the fraction of molecules with energy ≥ $E_a$.

   II. A catalyst increases the rate of an endothermic reaction too, by providing a path of lower activation energy.

Answer: I. ΔH = −10 kJ mol⁻¹ (magnitude 10 kJ mol⁻¹); Ea(backward) = 40 kJ mol⁻¹. II. A catalyst increases the rate by lowering the activation energy.

1. I. Nitrobenzene $\stackrel{\text{Sn/HCl}}{\to }$ aniline $\stackrel{({\text{CH}}_3\text{CO}{)}_2\text{O, pyridine}}{\to }$ acetanilide $\stackrel{{\text{Br}}_2/{\text{CH}}_3\text{COOH}}{\to }$ 4-bromoacetanilide $\stackrel{{\text{H}}^{+}\text{ or }{\text{OH}}^{-},{\text{H}}_2\text{O}}{\to }$ 4-bromobenzenamine.

   Acetylation lowers the activating power of –NH${}_2$, so mainly the mono para-bromo product forms.

   1 mark

2. II. ${\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{Cl}\stackrel{\text{KCN (alc)}}{\to }{\text{C}}_6{\text{H}}_5{\text{CH}}_2\text{CN}\stackrel{{\text{LiAlH}}_4{\text{ (or H}}_2\text{/Ni)}}{\to }{\text{C}}_6{\text{H}}_5{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2$ (2-phenylethanamine).

   1 mark

Answer: I. PhNO₂ → (Sn/HCl) PhNH₂ → (Ac₂O) PhNHCOCH₃ → (Br₂) p-BrC₆H₄NHCOCH₃ → (hydrolysis) p-BrC₆H₄NH₂. II. PhCH₂Cl → (KCN) PhCH₂CN → (LiAlH₄) PhCH₂CH₂NH₂.

1. I. The cation is [Ag(H${}_2$O)${}_2$]${}^{+}$ and the anion is [AgCl${}_2$]${}^{-}$, so the formula is $[{\text{Ag(H}}_2{\text{O)}}_2][{\text{AgCl}}_2]$.

   1 mark

2. II. Ni${}^{2+}$ with two OH${}^{-}$ and two PPh${}_3$ ligands gives a neutral complex: $[{\text{Ni(OH)}}_2({\text{PPh}}_3{)}_2]$.

   1 mark

Answer: I. [Ag(H₂O)₂][AgCl₂]; II. [Ni(OH)₂(PPh₃)₂]

1. **Step 1 – Protonation of the alkene (electrophilic attack by H${}_3$O${}^{+}$):** $>\text{C=C}<+{\text{H}}_3{\text{O}}^{+}\rightleftharpoons >\text{CH}-\stackrel{+}{C}<+{\text{H}}_2\text{O}$. The curved arrow starts at the π bond and points to H of H${}_3$O${}^{+}$. A second arrow goes from the H–O bond to O${}^{+}$, which becomes neutral H${}_2$O. The carbocation is formed on the carbon that gives the more stable carbocation (Markovnikov).

   ½ mark

2. Step 2 – Nucleophilic attack of water on the carbocation: $>\text{CH}-\stackrel{+}{C}<+{\text{H}}_2\ddot{\text{O}}\rightleftharpoons >\text{CH}-\text{C}<(\stackrel{+}{O}{\text{H}}_2)$. The arrow goes from a lone pair on the O of water to the positive carbon, giving the protonated alcohol (oxonium ion).

   ½ mark

3. Step 3 – Deprotonation: $>\text{CH}-\text{C}<(\stackrel{+}{O}{\text{H}}_2)+{\text{H}}_2\text{O}\rightleftharpoons >\text{CH}-\text{C}<(\text{OH})+{\text{H}}_3{\text{O}}^{+}$. The arrow goes from a lone pair of a water molecule to an H on O${}^{+}$, and the O–H bond electrons move back onto oxygen. This step is also reversible.

   ½ mark

4. Corrections: every step is an equilibrium; each curved arrow starts from an electron source (π bond or lone pair) and ends at the electron-deficient atom. The positive charge sits on carbon in step 1 and on oxygen in the protonated alcohol, and H${}_3$O${}^{+}$ is regenerated, so it acts only as a catalyst.

   ½ mark

Answer: Corrected mechanism: (1) alkene + H₃O⁺ ⇌ carbocation + H₂O; (2) carbocation + H₂O ⇌ protonated alcohol; (3) protonated alcohol + H₂O ⇌ alcohol + H₃O⁺, with arrows from electron-rich to electron-poor sites and all steps reversible.

1. 1 molal means 1 mol MgCl${}_2$ in 1000 g water: $n_B=1$ mol and $n_A=\frac{1000}{18}=55.5$ mol.

   ½ mark

2. Dissociation: ${\text{MgCl}}_2\to {\text{Mg}}^{2+}+2{\text{Cl}}^{-}$, so $n=3$ ions and $\alpha =0.7$.

   ½ mark

3. $\alpha =\frac{i-1}{n-1}\Rightarrow i=1+\alpha (n-1)=1+0.7\times 2=2.4$

   ½ mark

4. Raoult's law with van't Hoff factor: $\frac{P_A^o-P_s}{P_A^o}=i\frac{n_B}{n_A+n_B}$

   ½ mark

5. $\frac{23.8-P_s}{23.8}=\frac{2.4\times 1}{55.5+1}=\frac{2.4}{56.5}=0.0425$

   ½ mark

6. $23.8-P_s=23.8\times 0.0425=1.01\Rightarrow P_s=22.79\approx 22.8$ mm Hg

   ½ mark

Answer: Vapour pressure of the solution ≈ 22.8 mm Hg

1. I. Here $n=2$: $E_{cell}=E_{cell}^o-\frac{2.303RT}{2F}\log \frac{[{\text{Ni}}^{2+}]}{[{\text{Cu}}^{2+}]}$. At 298 K this is $E_{cell}^o-\frac{0.059}{2}\log \frac{[{\text{Ni}}^{2+}]}{[{\text{Cu}}^{2+}]}$.

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2. II. Balanced ionic equation: $2\text{Al(s)}+3{\text{Fe}}^{2+}\text{(aq)}\to 2{\text{Al}}^{3+}\text{(aq)}+3\text{Fe(s)}$, so $n=6$.

   $E_{cell}=E_{cell}^o-\frac{2.303RT}{6F}\log \frac{[{\text{Al}}^{3+}{]}^2}{[{\text{Fe}}^{2+}{]}^3}$

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3. III. Cell reaction: $\text{Mg(s)}+2{\text{Ag}}^{+}\text{(aq)}\to {\text{Mg}}^{2+}\text{(aq)}+2\text{Ag(s)}$, so $n=2$.

   $E_{cell}=E_{cell}^o-\frac{2.303RT}{2F}\log \frac{[{\text{Mg}}^{2+}]}{[{\text{Ag}}^{+}{]}^2}$

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Answer: I. E = E° − (2.303RT/2F) log([Ni²⁺]/[Cu²⁺]); II. E = E° − (2.303RT/6F) log([Al³⁺]²/[Fe²⁺]³); III. E = E° − (2.303RT/2F) log([Mg²⁺]/[Ag⁺]²)

1. I. In sunlight Cl${}_2$ undergoes homolysis to Cl${}^{\bullet }$ free radicals. These attack the side-chain CH${}_3$ (free-radical substitution), giving benzyl chloride, C${}_6$H${}_5$CH${}_2$Cl.

   In the dark with a Lewis acid (FeCl${}_3$), Cl${}^{+}$ (an electrophile) forms and attacks the ring (electrophilic substitution). Since –CH${}_3$ is ortho/para directing, o- and p-chlorotoluene are formed.

   1 mark

2. II. In the Finkelstein reaction, $\text{R–Cl/R–Br}+\text{NaI}\stackrel{\text{dry acetone}}{\to }\text{R–I}+\text{NaCl/NaBr}$. NaI dissolves in dry acetone but NaCl/NaBr do not, so they precipitate out. This shifts the equilibrium forward (Le Chatelier's principle). Water would dissolve NaCl and allow the reverse reaction.

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3. III. Neopentyl chloride is more branched than isopentyl chloride, so it is more compact (nearly spherical) with a smaller surface area. Its van der Waals forces are weaker, so its boiling point is lower.

   1 mark

Answer: I. Sunlight: free-radical side-chain substitution gives benzyl chloride; dark with FeCl₃: electrophilic ring substitution gives o/p-chlorotoluene. II. NaCl/NaBr precipitate in dry acetone, driving the reaction forward. III. Greater branching means smaller surface area, weaker van der Waals forces and a lower boiling point.

1. I. Chromium and Molybdenum show similar magnetic behaviour.

   ½ mark

2. Cr: [Ar]3d${}^5$4s${}^1$ and Mo: [Kr]4d${}^5$5s${}^1$ have similar configurations, each with 6 unpaired electrons. So both are strongly paramagnetic with similar magnetic moments. (Mg has no unpaired electrons; Fe [Ar]3d${}^6$4s${}^2$ has 4.)

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3. II. Zinc and Scandium form white salts.

   ½ mark

4. Zn${}^{2+}$ is [Ar]3d${}^{10}$ (completely filled) and Sc${}^{3+}$ is [Ar]3d${}^0$ (empty), so neither has unpaired d electrons. No d–d transition is possible, so their salts are white. Ni${}^{2+}$ (3d${}^8$) and V ions (e.g. V${}^{3+}$, 3d${}^2$) have unpaired electrons, so their salts are coloured.

   1 mark

Answer: I. Cr and Mo (similar ns¹(n−1)d⁵ configuration, 6 unpaired electrons each). II. Zn and Sc (Zn²⁺ 3d¹⁰, Sc³⁺ 3d⁰: no unpaired electrons, no d–d transition).

1. A. ${\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{K}}_2{\text{Cr}}_2{\text{O}}_7/{\text{H}}^{+}}{\to }{\text{CH}}_3\text{COOH}$ (ethanoic acid)

   ½ mark

2. B. ${\text{CH}}_3{\text{CH}}_2\text{COOH}\stackrel{{\text{Br}}_2/\text{red P}}{\to }{\text{CH}}_3\text{CHBrCOOH}$ (2-bromopropanoic acid; HVZ reaction)

   ½ mark

3. C. $({\text{CH}}_3{)}_2\text{CHMgBr}\stackrel{{\text{(i) CO}}_2{\text{(ii) H}}_3{\text{O}}^{+}}{\to }({\text{CH}}_3{)}_2\text{CHCOOH}$ (2-methylpropanoic acid)

   ½ mark

4. D. ${\text{CH}}_3{\text{CH}}_2\text{COOH}\stackrel{{\text{Cl}}_2/\text{red P}}{\to }{\text{CH}}_3\text{CHClCOOH}$ (2-chloropropanoic acid)

   ½ mark

5. Halogen at the α-carbon withdraws electrons (–I), stabilising the carboxylate ion; Cl (more electronegative) does this more than Br. Alkyl groups (+I) destabilise the anion. Acid strength: D > B > A > C, so the increasing order of pKa is:

   ${\text{CH}}_3\text{CHClCOOH}<{\text{CH}}_3\text{CHBrCOOH}<{\text{CH}}_3\text{COOH}<({\text{CH}}_3{)}_2\text{CHCOOH}$

   1 mark

Answer: pKa: CH₃CHClCOOH (D) < CH₃CHBrCOOH (B) < CH₃COOH (A) < (CH₃)₂CHCOOH (C)

1. I. ${\text{CH}}_3{\text{COCH}}_2{\text{CH}}_3\stackrel{{\text{C}}_2{\text{H}}_5\text{MgBr, dry ether}}{\to }({\text{C}}_2{\text{H}}_5{)}_2{\text{C(CH}}_3)\text{OMgBr}\stackrel{{\text{H}}_3{\text{O}}^{+}}{\to }({\text{C}}_2{\text{H}}_5{)}_2{\text{C(OH)CH}}_3$ (3-methylpentan-3-ol)

   1 mark

2. II. ${\text{C}}_6{\text{H}}_5{\text{OCH}}_3+{\text{CH}}_3\text{Cl}\stackrel{{\text{anhyd. AlCl}}_3}{\to }p{\text{-CH}}_3{\text{C}}_6{\text{H}}_4{\text{OCH}}_3$ (4-methoxytoluene, major) + some ortho isomer (Friedel–Crafts alkylation; –OCH${}_3$ is o/p directing).

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3. III. ${\text{C}}_6{\text{H}}_5\text{OH}+\text{Zn}\stackrel{\Delta }{\to }{\text{C}}_6{\text{H}}_6+\text{ZnO}$ (distillation with zinc dust)

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4. IV. Williamson synthesis: ${\text{C}}_2{\text{H}}_5\text{OH}+\text{Na}\to {\text{C}}_2{\text{H}}_5\text{ONa}$; then ${\text{C}}_2{\text{H}}_5\text{Cl}+{\text{C}}_2{\text{H}}_5\text{ONa}\to {\text{C}}_2{\text{H}}_5{\text{OC}}_2{\text{H}}_5+\text{NaCl}$ (alternatively, heat C${}_2$H${}_5$Cl with dry Ag${}_2$O).

Answer: I. Butan-2-one + C₂H₅MgBr, then H₃O⁺; II. CH₃Cl/anhyd. AlCl₃; III. Zn dust, heat; IV. C₂H₅ONa (Williamson synthesis). Any three carry 1 mark each.

1. I. No, the values are interchanged. Zinc ([Ar]3d${}^{10}$4s${}^2$) has no unpaired d electrons, so its metallic bonding is weak and its enthalpy of atomisation is low, **130 kJ mol${}^{-1}$. Copper has stronger metallic bonding, so its value is 339 kJ mol${}^{-1}$**.

   1 mark

2. II. Sulphuric acid is preferred. HCl is oxidised by KMnO${}_4$ to Cl${}_2$, so part of the permanganate is used up and the titre is wrong. Dilute H${}_2$SO${}_4$ is not oxidised by MnO${}_4^{-}$.

   1 mark

3. III. $5{\text{NO}}_2^{-}+2{\text{MnO}}_4^{-}+6{\text{H}}^{+}\to 2{\text{Mn}}^{2+}+5{\text{NO}}_3^{-}+3{\text{H}}_2\text{O}$

   1 mark

Answer: I. No; Zn = 130 and Cu = 339 kJ/mol (Zn has no unpaired d electrons, so weaker metallic bonding). II. H₂SO₄, because HCl is oxidised to Cl₂ by KMnO₄. III. 2Mn²⁺ + 5NO₃⁻ + 3H₂O

1. I. No. 1 M NaOH (strong base) and 1 M HCl (strong acid) are both strong electrolytes that ionise completely. Both give very high conductivity (red LED very bright, green LED medium, scale 4). The meter only shows the amount of conductivity, so it cannot tell the two apart.

   1 mark

2. II. Red very bright and green medium means scale 4, very high conductivity, so the solution is a strong acid or strong base. Possible pH values are 1 and 13, i.e. option (a) (i) and (ii).

   OR (distilled water): both LEDs stay off (scale 0). Distilled water has very few ions, so its conductivity is negligible.

   1 mark

3. III. Solution A (red bright, green dim, high conductivity) is a strong electrolyte; B (red dim, green off, low conductivity) is a weak electrolyte.

   Curve for A: ${\Lambda }_m$ is high and decreases only slightly and almost linearly as concentration ($\sqrt{c}$) increases (${\Lambda }_m={\Lambda }_m^o-A\sqrt{c}$), so the line can be extrapolated to ${\Lambda }_m^o$.

   1 mark

4. Curve for B: ${\Lambda }_m$ is low at higher concentration but rises very steeply as concentration approaches zero, because the degree of dissociation increases on dilution. The curve is not linear and cannot be extrapolated. (Axes: ${\Lambda }_m$ on y-axis, $\sqrt{c}$ on x-axis.)

   1 mark

Answer: I. No, both are strong electrolytes giving the same high reading. II. (a) (i) and (ii), pH 1 and 13 [OR distilled water: both LEDs off, scale 0]. III. A is a strong electrolyte (Λm falls slightly, linearly with √c); B is a weak electrolyte (Λm rises steeply on dilution).

1. I. EDTA is a hexadentate chelating ligand. It binds Fe(III) through six donor atoms, forming several five-membered chelate rings, and the chelate effect makes the complex very stable. It does not release Fe to precipitate in tea, unlike simple ferrous sulphate.

   OR: When hard water is titrated with Na${}_2$EDTA, the Ca${}^{2+}$ and Mg${}^{2+}$ ions form stable complexes with EDTA. The volume of EDTA used gives the hardness of water.

   1 mark

2. II. Yes (with caution). The passage says 40% of people in African countries are anaemic. Fortification is described as safe, cost-effective and proven, and Fe(III)-EDTA is well absorbed and not precipitated in tea for 24 h. The 60 mg of EDTA needed for 10 mg of iron is far below the 3 g/day safe limit. I would advise patients to stay within the recommended limits and not to take EDTA continuously for more than 5–7 days.

   (Alternative 'No': prescribe it only to patients whose tests show iron deficiency, since excess EDTA or iron can harm the kidneys and cause cramps, nausea and low BP.)

   1 mark

3. III. (a) EDTA is hexadentate, so its denticity is 6. It binds through 2 nitrogen atoms and 4 oxygen atoms (of the carboxylate groups).

   2 marks

4. OR (b) Structure of EDTA (ethylenediaminetetraacetate ion): ${(}^{-}{\text{OOCCH}}_2{)}_2{\text{N–CH}}_2{\text{–CH}}_2{\text{–N(CH}}_2{\text{COO}}^{-}{)}_2$. The parent acid is $({\text{HOOCCH}}_2{)}_2{\text{NCH}}_2{\text{CH}}_2{\text{N(CH}}_2{\text{COOH)}}_2$.

Answer: I. EDTA is a hexadentate chelating ligand, so the chelate effect makes the complex stable [OR Ca²⁺/Mg²⁺ form stable EDTA complexes, used to estimate hardness]. II. Yes, within safe limits (40% anaemia; 60 mg EDTA ≪ 3 g limit). III. (a) Denticity 6, via 2 N and 4 O atoms; (b) (⁻OOCCH₂)₂N–CH₂CH₂–N(CH₂COO⁻)₂.

1. I. $2(2{\text{-ClC}}_6{\text{H}}_4\text{CHO})\stackrel{\text{conc. NaOH}}{\to }$ 2-chlorobenzyl alcohol ($2{\text{-ClC}}_6{\text{H}}_4{\text{CH}}_2\text{OH}$) + sodium 2-chlorobenzoate ($2{\text{-ClC}}_6{\text{H}}_4{\text{COO}}^{-}{\text{Na}}^{+}$). Acidifying the salt gives 2-chlorobenzoic acid.

   1 mark

2. II. –SO${}_3$H is a strong electron-withdrawing group (–I, –R). It lowers the electron density on the nitrogen of –NH${}_2$, so the lone pair is less available for donation. The basic strength of aniline decreases.

   1 mark

3. III. ${\text{CH}}_3\text{COOH}\stackrel{{\text{LiAlH}}_4}{\to }{\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{SOCl}}_2}{\to }{\text{CH}}_3{\text{CH}}_2\text{Cl}\stackrel{{\text{excess NH}}_3}{\to }{\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$

   (Alternatively: CH${}_3$COOH → SOCl${}_2$ → CH${}_3$COCl → NH${}_3$ → CH${}_3$CONH${}_2$ → LiAlH${}_4$ → CH${}_3$CH${}_2$NH${}_2$.)

   1 mark

4. IV. ${\text{C}}_6{\text{H}}_5\text{COCl}\stackrel{{\text{H}}_2,{\text{Pd/BaSO}}_4}{\to }$ C${}_6$H${}_5$CHO (Rosenmund reduction) $\stackrel{{\text{alk. KMnO}}_4{\text{ then H}}^{+}}{\to }$ C${}_6$H${}_5$COOH. (Benzoyl chloride also gives benzoic acid directly by hydrolysis: C${}_6$H${}_5$COCl + H${}_2$O → C${}_6$H${}_5$COOH + HCl.)

   1 mark

5. V. Tollens' test: warm each compound with ammoniacal silver nitrate on a water bath. Propanal gives a silver mirror: CH${}_3$CH${}_2$CHO + 2[Ag(NH${}_3$)${}_2$]${}^{+}$ + 3OH${}^{-}$ → CH${}_3$CH${}_2$COO${}^{-}$ + 2Ag↓ + 2H${}_2$O + 4NH${}_3$. Propanone gives no silver mirror.

   1 mark

Answer: I. 2-chlorobenzyl alcohol + sodium 2-chlorobenzoate. II. Basicity decreases (electron-withdrawing –SO₃H). III. CH₃COOH → LiAlH₄ → C₂H₅OH → SOCl₂ → C₂H₅Cl → NH₃ → C₂H₅NH₂. IV. Rosenmund reduction then oxidation (or direct hydrolysis). V. Tollens' test: propanal gives a silver mirror, propanone does not.

1. I. ${\text{CH}}_3{\text{CH}}_2{\text{CH(CH}}_3)\text{CHO}\stackrel{{\text{(i) NH}}_2{\text{NH}}_2\text{(ii) KOH/ethylene glycol, heat}}{\to }{\text{CH}}_3{\text{CH}}_2{\text{CH(CH}}_3){\text{CH}}_3$ (2-methylbutane); the >C=O becomes >CH${}_2$.

   1 mark

2. II. –SO${}_3$H is electron-withdrawing. It disperses the negative charge of the benzoate ion and stabilises the conjugate base, which makes proton release easier. The acidic strength of benzoic acid increases.

   1 mark

3. III. ${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2\stackrel{{\text{NaNO}}_2+{\text{HCl (HNO}}_2)}{\to }{\text{CH}}_3{\text{CH}}_2\text{OH}+{\text{N}}_2+{\text{H}}_2\text{O}$; then ${\text{CH}}_3{\text{CH}}_2\text{OH}\stackrel{{\text{acidified KMnO}}_4{\text{ or K}}_2{\text{Cr}}_2{\text{O}}_7}{\to }{\text{CH}}_3\text{COOH}$

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4. IV. ${\text{C}}_6{\text{H}}_5{\text{NH}}_2\stackrel{{\text{NaNO}}_2/\text{HCl, 273–278 K}}{\to }{\text{C}}_6{\text{H}}_5{\text{N}}_2^{+}{\text{Cl}}^{-}\stackrel{\text{CuCN/KCN}}{\to }{\text{C}}_6{\text{H}}_5\text{CN}\stackrel{{\text{H}}_3{\text{O}}^{+},\Delta }{\to }{\text{C}}_6{\text{H}}_5\text{COOH}$

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5. V. Iodoform test: warm each compound with I${}_2$ and NaOH. Ethanal (has a CH${}_3$CO– group) gives a yellow precipitate of iodoform, CHI${}_3$: CH${}_3$CHO + 3I${}_2$ + 4NaOH → CHI${}_3$↓ + HCOONa + 3NaI + 3H${}_2$O. Propanal gives no yellow precipitate.

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Answer: I. 2-methylbutane, CH₃CH₂CH(CH₃)CH₃. II. Acidity increases. III. Ethanamine → HNO₂ → ethanol → acidified KMnO₄ → acetic acid. IV. Diazotisation → CuCN → benzonitrile → hydrolysis → benzoic acid. V. Iodoform test: ethanal gives a yellow ppt, propanal does not.

1. I. Hair contains keratin, a fibrous protein; egg albumin is a globular protein. Fibrous proteins have long, thread-like parallel chains held by H-bonds and disulphide bonds and are insoluble in water. Globular proteins are folded into spherical shapes and are generally soluble in water.

   1 mark

2. II. In glucose the aldehydic group is free (open chain in equilibrium with the cyclic hemiacetal), so it reduces Fehling's solution. In sucrose, C1 of α-glucose is joined to C2 of β-fructose by a glycosidic bond, so both reducing groups are used up. No free –CHO (or >C=O) is left, so sucrose is a non-reducing sugar.

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3. III. An α-amino acid has both an acidic –COOH group and a basic –NH${}_2$ group. The proton moves from –COOH to –NH${}_2$, forming a dipolar ion (zwitterion), H${}_3\stackrel{+}{N}$–CHR–COO${}^{-}$. Because of this ionic nature, amino acids have high melting points and are water-soluble, like salts.

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4. IV. Bacteria (e.g. Lactobacillus) ferment the lactose in milk into lactic acid. The acid lowers the pH, which denatures and coagulates the milk protein (casein), forming curd.

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5. V. Sunlight, fish and egg yolk are sources of Vitamin D, so the woman has Vitamin D deficiency. The likely disease is osteomalacia/osteoporosis (soft, weak bones) in adults.

   1 mark

Answer: I. Keratin (fibrous, insoluble) vs albumin (globular, soluble). II. Sucrose has no free aldehyde group (glycosidic linkage involves C1 of glucose and C2 of fructose). III. Zwitterion formation. IV. Lactose → lactic acid, which coagulates casein. V. Vitamin D deficiency: osteomalacia/osteoporosis.

1. I. Cane sugar contains sucrose, while milk contains lactose. Both are disaccharides. Difference: sucrose is made of $\alpha$-D-glucose and $\beta$-D-fructose, whereas lactose is made of $\beta$-D-galactose and $\beta$-D-glucose. (Another valid difference: sucrose is a non-reducing sugar, while lactose is a reducing sugar.)

   1 mark

2. II. Use bromine water. It is a mild oxidising agent that oxidises only the $-\mathrm{CHO}$ group of glucose to $-\mathrm{COOH}$, giving gluconic acid: ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\stackrel{\mathrm{B}{\mathrm{r}}_2\mathrm{water}}{\to }\mathrm{HOOC}(\mathrm{CHOH}{)}_4\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$.

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3. III. The molecule has two $-\mathrm{COOH}$ groups but only one $-\mathrm{N}{\mathrm{H}}_2$ group. It is therefore an acidic amino acid, and its aqueous solution has pH < 7.

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4. IV. The two major shapes are the **$\alpha$-helix and the $\beta$-pleated sheet**. Both are held in place by hydrogen bonds between $>\mathrm{C}=\mathrm{O}$ and $-\mathrm{NH}-$ groups of the peptide chain.

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5. V. Scurvy is caused by a deficiency of vitamin C (ascorbic acid). Suitable food sources are citrus fruits (orange, lemon), amla and green leafy vegetables.

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Answer: I. Sucrose (glucose + fructose) vs lactose (galactose + glucose); II. Bromine water; III. pH < 7 (acidic amino acid); IV. α-helix and β-pleated sheet; V. Vitamin C sources: citrus fruits, amla (or green leafy vegetables)

1. I. Here $T_1=298$ K, $T_2=318$ K and $\frac{k_2}{k_1}=3$. Using the Arrhenius equation: $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

   ½ mark

2. Substituting: $0.4771=\frac{E_a}{2.303\times 8.314}\left(\frac{318-298}{298\times 318}\right)=\frac{E_a}{19.147}\times \frac{20}{94764}$

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3. Rearranging: $E_a=\frac{0.4771\times 19.147\times 94764}{20}=9.135\times 4738.2$

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4. $E_a\approx 43284\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}=43.28\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

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5. II. The graph of [R] against t is a straight line with a negative slope. This means concentration falls at a constant rate, independent of [R], so the reaction is of zero order.

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6. Integrated rate equation: $[R]=[R{]}_0-kt$ (or $k=\frac{[R{]}_0-[R]}{t}$). Here $[R{]}_0$ is the initial concentration of the reactant, $[R]$ is its concentration at time $t$, and $k$ is the rate constant (equal to $-$slope of the line).

   1 mark

Answer: I. Ea ≈ 43.28 kJ mol⁻¹; II. Zero order, [R] = [R]₀ − kt

1. I. Let $P_i$ be the initial pressure of $\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2$ and let $x$ torr of it decompose in 10 s.

   	$\mathrm{S}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2$	$\mathrm{S}{\mathrm{O}}_2$	$\mathrm{C}{\mathrm{l}}_2$
   $t=0$	$P_i$	0	0
   $t=10$ s	$P_i-x$	$x$	$x$
   completion	0	$P_i$	$P_i$

   ½ mark

2. On complete decomposition, $P_{total}=2P_i=300$ torr, so $P_i=150$ torr.

   ½ mark

3. After 10 s, $P_{total}=P_i+x=200$ torr, so $x=200-150=50$ torr and $P_i-x=100$ torr.

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4. First-order rate equation in terms of pressure: $k=\frac{2.303}{t}\log \frac{P_i}{P_i-x}$

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5. Substituting: $k=\frac{2.303}{10}\log \frac{150}{100}=0.2303(\log 3-\log 2)=0.2303\times (0.4771-0.3010)=0.2303\times 0.1761$

   ½ mark

6. $k\approx 0.0406{\mathrm{s}}^{-1}$ (about $4.06\times 10^{-2}{\mathrm{s}}^{-1}$)

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7. II. By collision theory, for $A+B\to$ Products: $\text{Rate}=Z_{AB}{e}^{-E_a/RT}$

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8. In this expression:

   • $Z_{AB}$ is the collision frequency, the number of collisions between A and B per second per unit volume.
   • $e^{-E_a/RT}$ is the fraction of molecules with energy equal to or greater than $E_a$.
   • $E_a$ is the activation energy, $R$ is the gas constant and $T$ is the absolute temperature.
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Answer: I. k ≈ 0.040 s⁻¹; II. Rate = Z_AB e^(−Ea/RT)