CBSE Class 12 Chemistry: sample paper 2024-25

1. Ammonolysis: ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}+\mathrm{N}{\mathrm{H}}_3\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$ (primary amine). With 1 mol $\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}$: ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{NHC}{\mathrm{H}}_3$ (secondary amine).

2. A secondary amine reacts with benzenesulphonyl chloride (Hinsberg reagent) to give an N,N-disubstituted sulphonamide. It has no H on N, so it is not acidic and is insoluble in alkali. Option (C).

   1 mark

Answer: Option (C)

1. Dipole moment $\mu =q\times d$. F is the most electronegative, but the C–F bond is very short. The C–Cl bond has a large charge separation and a longer bond, so the product is the largest for $\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}$.

   1 mark

2. Order: $\mathrm{C}{\mathrm{H}}_3\mathrm{I}<\mathrm{C}{\mathrm{H}}_3\mathrm{Br}<\mathrm{C}{\mathrm{H}}_3\mathrm{F}<\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}$. Option (B).

Answer: $\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}$, option (B)

1. [Rn]$5f^{7}6d^{1}7s^2$ is Cm (Z = 96). Lu${}^{3+}$ is [Xe]$4f^{14}$. Ce shows +4 because it attains the stable $4f^0$ configuration.

   1 mark

2. So (i)-(C), (ii)-(B), (iii)-(A). Option (A).

Answer: Option (A)

1. Bromomethane has the smallest molar mass, so it has the lowest b.p. (277 K).

2. Among the isomeric bromobutanes, b.p. falls as branching increases because surface area and van der Waals forces decrease. So 1-bromobutane (375 K) > 2-bromobutane (364 K) > 2-bromo-2-methylpropane (346 K). Option (D).

   1 mark

Answer: Option (D)

1. The unit of k is mol L${}^{-1}$ s${}^{-1}$, so the reaction is zero order.

2. $t_{1/2}=\frac{[R{]}_0}{2k}=\frac{4.62\times 10^{-2}}{2\times 2.31\times 10^{-2}}=1$ s. Option (C).

   1 mark

Answer: 1 s, option (C)

1. The compound is a mixed anhydride. Hydrolysis breaks the C–O–C link: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOCOC}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$.

   1 mark

2. The products are benzoic acid and ethanoic acid. Option (B).

Answer: Option (B)

1. $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6]\mathrm{C}{\mathrm{l}}_3$ gives 1 complex cation and 3 Cl${}^{-}$ ions, so Y = 1:3 electrolyte.

2. A 1:1 electrolyte has only one ionisable Cl${}^{-}$, so X = $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{C}{\mathrm{l}}_2{]}^{+}\mathrm{C}{\mathrm{l}}^{-}$. Option (B).

   1 mark

Answer: Option (B)

1. Sucrose contains α-D-glucose and β-D-fructose. Starch and maltose contain α-D-glucose.

2. Cellulose is a linear polymer of β-D-glucose only. Option (B).

   1 mark

Answer: Cellulose, option (B)

1. Unpaired electrons: Ti${}^{3+}$ ($3d^1$) = 1, Cr${}^{3+}$ ($3d^3$) = 3, Fe${}^{2+}$ ($3d^6$) = 4, Mn${}^{2+}$ ($3d^5$) = 5.

2. Paramagnetism increases with the number of unpaired electrons. Option (D).

   1 mark

Answer: Option (D)

1. For a first order reaction, $[R]=[R{]}_0{e}^{-kt}$.

2. [R] = 0 needs $e^{-kt}=0$, which happens only as $t\to \infty$. So the reaction never goes to completion. Option (D).

   1 mark

Answer: Option (D)

1. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{O}}_2+3{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+2{\mathrm{H}}_2\mathrm{O}$.

   1 mark

2. Benzonitrile would give benzylamine, and the other two give aliphatic amines. Option (A).

Answer: Nitrobenzene, option (A)

1. Oximes form when hydroxylamine adds to the carbonyl of an aldehyde or ketone and water is then lost.

2. Acetone gives $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{NOH}$. Acids, anhydrides and acyl chlorides undergo substitution instead. Option (A).

   1 mark

Answer: $C{H}_{3}COC{H}_3$, option (A)

1. Mn${}^{3+}$ is $3d^4$. CN${}^{-}$ is a strong field ligand, so the electrons pair as $t_{2g}^4$, leaving 2 unpaired. Two inner d orbitals are free, giving $d^{2}s{p}^3$.

2. Cl${}^{-}$ is a weak field ligand, so there is no pairing and 4 electrons stay unpaired. Outer d orbitals are used, giving $s{p}^3{d}^2$. A and R are both true and R explains A. Option (A).

   1 mark

Answer: Option (A)

1. The Debye–Hückel–Onsager equation is ${\Lambda }_m={\Lambda }_m^{\circ }-A{c}^{1/2}$. The equation given in A is wrong, so A is false.

2. A depends on the type of electrolyte (1:1, 2:1, 2:2), so it differs for NaCl, CaCl${}_2$ and MgSO${}_4$. R is true. Option (D).

   1 mark

Answer: Option (D)

1. Glucose has no free –CHO group, so it does not give a bisulphite addition product. A is true.

2. Glucose does exist as a six-membered pyranose ring, so R is true. As per the scheme, R is not taken as the correct explanation. Option (B).

   1 mark

Answer: Option (B)

1. For zero order, $t_{1/2}=[R{]}_0/2k$, so the half-life depends on the initial concentration. A is false.

2. Rate = $k[R{]}^0$ = k, so R is true. Option (D).

   1 mark

Answer: Option (D)

1. (a) By Henry's law, $p=K_{H}x$, so the mole fraction (solubility) of the gas is $x=p/K_H$. Solubility is therefore inversely proportional to $K_H$.

   ½ mark

2. Above 293 K, $K_H$ rises from 76.48 kbar to 88.8 kbar. At the same partial pressure, $x$ falls, so the solubility of nitrogen decreases.

   ½ mark

3. (b) The boiling point of the azeotrope will be (ii) 64.5 °C.

   ½ mark

4. Reason: H-bonding between chloroform and acetone ($C{l}_{3}C–H\cdots O=C(C{H}_3{)}_2$) is stronger than the attractions in the pure liquids. The mixture shows negative deviation from Raoult's law and forms a maximum-boiling azeotrope. Its boiling point is higher than both components (61.2 °C and 56 °C), so only 64.5 °C fits.

   ½ mark

Answer: (a) Solubility of N₂ decreases; (b) (ii) 64.5 °C (maximum-boiling azeotrope, negative deviation)

1. (a) Yes, the statement is correct. Srinagar is at a high altitude, so atmospheric pressure (and the partial pressure of $C{O}_2$ above the opened drink) is lower than in Delhi.

   ½ mark

2. By Henry's law, the solubility of a gas is directly proportional to its partial pressure above the solution. Less $C{O}_2$ stays dissolved at lower pressure, so the gas escapes faster and the soda goes flat sooner in Srinagar.

   ½ mark

3. (b) A high concentration of sugar makes the surroundings hypertonic to any microbe on the food.

   ½ mark

4. By osmosis, water moves out of the bacterial or fungal cells. They shrink (plasmolyse) and die, so the product does not spoil and its shelf life increases.

   ½ mark

Answer: (a) Correct: lower pressure at Srinagar means lower CO₂ solubility (Henry's law); (b) sugar draws water out of microbes by osmosis, killing them

1. (a) Find the oxidation state of Cr: $x+2(0)+2(-2)=-1$, so $x=+3$. Name the ligands alphabetically (aqua before oxalato). Name the anionic complex with the suffix '-ate' and add the water of crystallisation.

2. IUPAC name: Potassium diaquadioxalatochromate(III) hydrate (monohydrate).

   1 mark

3. (b) (i) Haemoglobin contains iron (Fe).

   ½ mark

4. (ii) Vitamin B-12 (cyanocobalamin) contains cobalt (Co).

   ½ mark

Answer: (a) Potassium diaquadioxalatochromate(III) hydrate; (b) (i) Iron (ii) Cobalt

1. (a) Y is oxidised (anode) and X⁺ is reduced (cathode). Writing the anode on the left:
   $Y(s)|Y^{2+}(aq)||X^{+}(aq)|X(s)$

   1 mark

2. (b) The current in the salt bridge is carried by ions: the cations and anions of the inert electrolyte in it (e.g. $K^{+}$ and $C{l}^{-}$ or $N{O}_3^{-}$ from KCl/KNO₃). They migrate to keep both solutions electrically neutral.

   ½ mark

3. (c) Anode (oxidation): $Y(s)\to Y^{2+}(aq)+2e^{-}$

   ½ mark

Answer: (a) Y(s)|Y²⁺(aq)||X⁺(aq)|X(s); (b) ions of the salt-bridge electrolyte; (c) Y(s) → Y²⁺(aq) + 2e⁻

1. (a) $C{H}_{3}C{H}_{2}Br+KCN\to C{H}_{3}C{H}_{2}CN$ (major, propanenitrile) $+C{H}_{3}C{H}_{2}NC$ (minor, ethyl isocyanide).
   KCN is ionic, so $C{N}^{-}$ attacks mainly through carbon, forming the stronger C–C bond.

   1 mark

2. (b) $C{H}_{3}C{H}_{2}CH=C{H}_2+HBr\to C{H}_{3}C{H}_{2}CHBrC{H}_3$ (major, 2-bromobutane) $+C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}Br$ (minor, 1-bromobutane).
   By Markovnikov's rule, Br goes to the carbon that gives the more stable 2° carbocation.

   1 mark

3. (c) (Alternative to (a) or (b); 1 mark if attempted instead.)
   $(C{H}_3{)}_{2}CHCHClC{H}_3\stackrel{alc.KOH}{\to }(C{H}_3{)}_{2}C=CHC{H}_3$ (major, 2-methylbut-2-ene) $+(C{H}_3{)}_{2}CHCH=C{H}_2$ (minor, 3-methylbut-1-ene).
   By Saytzeff's rule, the more substituted alkene is the major product.

Answer: (a) CH₃CH₂CN (major), CH₃CH₂NC (minor); (b) CH₃CH₂CHBrCH₃ (major), CH₃CH₂CH₂CH₂Br (minor); (c) (CH₃)₂C=CHCH₃ (major), (CH₃)₂CHCH=CH₂ (minor)

1. The carbonyl group in glucose is an aldehyde (–CHO) group.

   ½ mark

2. It is present at the C-1 (terminal) carbon of the six-carbon chain.

   ½ mark

3. Supporting reaction: a mild oxidising agent, bromine water, oxidises glucose to gluconic acid, a six-carbon acid with –COOH at C-1. Only an aldehyde is oxidised so easily without breaking the chain, which confirms –CHO at C-1.
   $CHO(CHOH{)}_{4}C{H}_{2}OH\stackrel{B{r}_{2}water}{\to }COOH(CHOH{)}_{4}C{H}_{2}OH$ (gluconic acid)

   1 mark

Answer: Aldehydic group at C-1; confirmed by oxidation with bromine water to gluconic acid

1. (a) Electrolysis of aqueous $CuC{l}_2$ with inert (Pt) electrodes:
   Cathode (–): $C{u}^{2+}(aq)+2e^{-}\to Cu(s)$ (copper is deposited)
   Anode (+): $2C{l}^{-}(aq)\to C{l}_2(g)+2e^{-}$ (chlorine gas evolves)

   1 mark

2. With concentrated $CuS{O}_4$ instead: copper is still deposited at the cathode, $C{u}^{2+}+2e^{-}\to Cu(s)$.
   There is no $C{l}^{-}$, so $C{l}_2$ is no longer formed. In concentrated solution, sulphate ions are oxidised at the anode to peroxodisulphate: $2S{O}_4^{2-}(aq)\to S_2{O}_8^{2-}(aq)+2e^{-}$. In dilute solution, water is oxidised to $O_2$ instead.

   1 mark

3. (b) By Kohlrausch's law, each formula unit gives 2 $A{l}^{3+}$ and 3 $S{O}_4^{2-}$ ions:
   ${\Lambda }_m^{\circ }[A{l}_2(S{O}_4{)}_3]=2{\lambda }^{\circ }(A{l}^{3+})+3{\lambda }^{\circ }(S{O}_4^{2-})$

   1 mark

Answer: (a) CuCl₂: Cu at cathode, Cl₂ at anode; conc. CuSO₄: Cu at cathode, SO₄²⁻ → S₂O₈²⁻ at anode (no Cl₂); (b) Λ°m = 2λ°(Al³⁺) + 3λ°(SO₄²⁻)

1. (a) In the lowest oxide (e.g. MnO), the metal is in a low oxidation state and keeps valence electrons not used in bonding. It can donate electrons and the oxide is basic. In the highest oxide (e.g. $M{n}_2{O}_7$), the metal is in a high oxidation state with no spare valence electrons, so it accepts electrons and the oxide is acidic (covalent, acid-forming).

   1 mark

2. (b) Chromium ($3d^{5}4s^1$) has many unpaired d-electrons. These take part in strong metallic bonding, so Cr is hard. Mercury ($5d^{10}6s^2$) has completely filled orbitals with no unpaired electrons, so its metallic bonding is very weak and it is a liquid at room temperature.

   1 mark

3. (c) Across the 3d series, nuclear charge rises by one unit at each step, but the added electron goes into the inner (n–1)d subshell. These d-electrons shield the outer 4s electrons, largely cancelling the rise in nuclear charge. The effective nuclear charge on the 4s electrons rises only slightly, so ionisation enthalpy increases only slowly.

   1 mark

Answer: (a) Low oxidation state metal can donate electrons (basic); high oxidation state accepts electrons (acidic); (b) unpaired d-electrons give strong metallic bonding in Cr, none in Hg; (c) increase in nuclear charge is offset by shielding by (n–1)d electrons

1. (a) Étard reaction: chromyl chloride in $C{S}_2$ oxidises the methyl group to a chromium complex, which is hydrolysed to the aldehyde.
   $p\text{-}{O}_{2}N–C_6{H}_4–C{H}_3\stackrel{Cr{O}_{2}C{l}_2,C{S}_2}{\to }p\text{-}{O}_{2}N–C_6{H}_4–CH(OCrOHC{l}_2{)}_2\stackrel{H_3{O}^{+}}{\to }p\text{-}{O}_{2}N–C_6{H}_4–CHO$ (p-nitrobenzaldehyde)

   1 mark

2. (b) In an aprotic solvent, there is no H-bonding with the solvent. Two benzoic acid molecules therefore form two intermolecular hydrogen bonds between their –COOH groups ($C=O\cdots H–O$). This gives a cyclic dimer.

   1 mark

3. (c) No Friedel–Crafts methylation takes place; toluic acid is not formed. The –COOH group is strongly electron-withdrawing and deactivates the ring. Also, the Lewis acid catalyst $AlC{l}_3$ gets bonded to the carboxyl group, so it is no longer available as a catalyst.

   1 mark

Answer: (a) p-nitrotoluene → (CrO₂Cl₂/CS₂, then H₃O⁺) p-nitrobenzaldehyde; (b) intermolecular H-bonding forms a cyclic dimer; (c) no reaction: –COOH deactivates the ring and binds AlCl₃

1. (a) X does not undergo aldol condensation, so it has no α-hydrogen: **X = benzaldehyde ($C_6{H}_{5}CHO$)**.

   ½ mark

2. The product $C_6{H}_{5}CH=CH–CO–C_6{H}_5$ needs a $C_6{H}_{5}CO–C{H}_2$ part, so **Y = acetophenone ($C_6{H}_{5}COC{H}_3$)**.

   ½ mark

3. (b) Cross aldol (Claisen–Schmidt) condensation:
   $C_6{H}_{5}CHO+C{H}_{3}CO{C}_6{H}_5\stackrel{dil.NaOH,\Delta }{\to }{C}_6{H}_{5}CH=CH–CO–C_6{H}_5+H_{2}O$ (1,3-diphenylprop-2-en-1-one)

   1 mark

4. (c) Tollens' test: warm each compound with ammoniacal silver nitrate. Benzaldehyde (X) gives a silver mirror, $C_6{H}_{5}CHO+2[Ag(N{H}_3{)}_2{]}^{+}+3O{H}^{-}\to C_6{H}_{5}CO{O}^{-}+2Ag\downarrow +2H_{2}O+4N{H}_3$. Acetophenone (Y) does not react. (Alternatively, Y gives a yellow iodoform precipitate with $I_2$/NaOH and X does not.)

   1 mark

Answer: X = benzaldehyde, Y = acetophenone; Claisen–Schmidt condensation; Tollens' test (silver mirror with X only)

1. (a) Two dipeptides are possible, depending on which amino acid gives its –COOH to the peptide bond (–CO–NH–).
   Ala–Gly (alanylglycine): $H_{2}N–CH(C{H}_3)–CO–NH–C{H}_2–COOH$

   1 mark

2. Gly–Ala (glycylalanine): $H_{2}N–C{H}_2–CO–NH–CH(C{H}_3)–COOH$

   1 mark

3. (b) Keratin is a fibrous protein. Its polypeptide chains lie side by side, held by H-bonds and disulphide links, giving a fibre-like structure. It is generally insoluble in water.

   ½ mark

4. Insulin is a globular protein. Its chains coil to give a spherical shape, and it is generally soluble in water.

   ½ mark

Answer: (a) Ala–Gly: H₂N–CH(CH₃)–CO–NH–CH₂–COOH and Gly–Ala: H₂N–CH₂–CO–NH–CH(CH₃)–COOH; (b) keratin fibrous (water-insoluble), insulin globular (water-soluble)

1. (a) At 140 °C with conc. $H_{2}S{O}_4$ (excess alcohol), intermolecular dehydration gives an ether. The major product is dipropyl ether (1-propoxypropane):
   $2C{H}_{3}C{H}_{2}C{H}_{2}OH\stackrel{H_{2}S{O}_4,413K}{\to }C{H}_{3}C{H}_{2}C{H}_2–O–C{H}_{2}C{H}_{2}C{H}_3+H_{2}O$

   1 mark

2. Mechanism: **bimolecular nucleophilic substitution ($S_{N}2$)**. A second propanol molecule attacks the protonated alcohol ($C{H}_{3}C{H}_{2}C{H}_2–O{H}_2^{+}$), displacing water.

   ½ mark

3. (b) At about 180 °C, intramolecular dehydration gives the alkene. The major product is propene:
   $C{H}_{3}C{H}_{2}C{H}_{2}OH\stackrel{conc.H_{2}S{O}_4,443K}{\to }C{H}_{3}CH=C{H}_2+H_{2}O$

   1 mark

4. Mechanism: protonation of –OH and loss of water (unimolecular step), then loss of $H^{+}$. This is an elimination (E1-type, unimolecular) pathway rather than a substitution.

   ½ mark

Answer: (a) Dipropyl ether, S_N2; (b) Propene, unimolecular elimination (E1-type)

1. The isomers of $C_4{H}_{9}Cl$ are 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane and 2-chloro-2-methylpropane. **A = 2-chloro-2-methylpropane, $(C{H}_3{)}_{3}CCl$** (tertiary butyl chloride).

   ½ mark

2. Reason: $S_{N}1$ goes through a carbocation. A gives a tertiary carbocation, $(C{H}_3{)}_3{C}^{+}$, which is the most stable because of the +I effect and hyperconjugation of three methyl groups. So A reacts fastest of all the isomers.

   ½ mark

3. Step 1 (slow, rate-determining): ionisation of the C–Cl bond
   $(C{H}_3{)}_{3}C–Cl\to (C{H}_3{)}_3{C}^{+}+C{l}^{-}$

   1 mark

4. Step 2 (fast): attack of the nucleophile on the planar carbocation
   $(C{H}_3{)}_3{C}^{+}+O{H}^{-}\to (C{H}_3{)}_{3}C–OH$ (2-methylpropan-2-ol)
   Rate = k[(CH₃)₃CCl], i.e. first order.

   1 mark

Answer: A = (CH₃)₃CCl (2-chloro-2-methylpropane); forms the most stable 3° carbocation; S_N1: slow ionisation then fast OH⁻ attack giving (CH₃)₃COH

1. (a) Balanced cell reaction: $3S{n}^{4+}+2Al\to 2A{l}^{3+}+3S{n}^{2+}$, so $n=6$.

   ½ mark

2. At equilibrium $E_{cell}=0$, so $E_{cell}^{\circ }=\frac{0.059}{n}\log K_c$ at 298 K.

   ½ mark

3. $E_{cell}^{\circ }=\frac{0.059}{6}\times \log (4.617\times 10^{184})=0.00983\times 184.6644$

   ½ mark

4. $E_{cell}^{\circ }=1.815V$

   ½ mark

5. (b) $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=E_{S{n}^{4+}/S{n}^{2+}}^{\circ }-E_{A{l}^{3+}/Al}^{\circ }$
   $1.815=0.15-E_{A{l}^{3+}/Al}^{\circ }$

   ½ mark

6. $E_{A{l}^{3+}/Al}^{\circ }=0.15-1.815=-1.66V$

   ½ mark

Answer: (a) E°cell ≈ 1.815 V; (b) E°(Al³⁺/Al) = −1.66 V

1. a. From the passage, rate ∝ [H₂] and ∝ [Br₂]^½:
   Rate $=k[H_2][B{r}_2{]}^{1/2}$; order $=1+\frac{1}{2}=\frac{3}{2}$

   1 mark

2. a (cont.) $k=\frac{\text{Rate}}{[H_2][B{r}_2{]}^{1/2}}=\frac{mol{L}^{-1}{s}^{-1}}{(mol{L}^{-1}{)}^{3/2}}$
   Units of k: $mo{l}^{-1/2}{L}^{1/2}{s}^{-1}$

   1 mark

3. b (first alternative). Rate′ $=k[H_2](3[B{r}_2]{)}^{1/2}=\sqrt{3}k[H_2][B{r}_2{]}^{1/2}$, so the rate becomes $\sqrt{3}\approx 1.73$ times.

   1 mark

4. b (OR alternative). $3\text{Rate}=k(x[H_2])[B{r}_2{]}^{1/2}$ gives $x=3$. Since the rate is first order in H₂, tripling [H₂] triples the rate.

5. c. The slow step is rate-determining. In Mechanism 1 the slow step involves only A, giving Rate = k[A]. In Mechanism 2 the slow step is A + B → C + D, giving Rate = k[A][B]. So Mechanism 2 is consistent with the experimental finding.

   1 mark

Answer: a. Rate = k[H₂][Br₂]^½, order 3/2; units mol^−½ L^½ s^−1; b. rate increases √3 (≈1.73) times / [H₂] must be tripled; c. Mechanism 2

1. a. Use pKb = 14 − pKa for the substituted propylamines:

   X	EN of X	pKa	pKb
   CH₂	2.55	10.67	3.33
   NH	3.12	10.08	3.92
   O	3.44	9.45	4.55
   CH₃CON	3.6	9.28	4.72

   Plot EN on the x-axis and pKb on the y-axis. The points rise steadily and lie close to a straight line of best fit with positive slope.

   1½ marks

2. Relation: pKb increases as the electronegativity of X increases. A more electronegative substituent withdraws electron density from N (–I effect), so basic strength decreases with increasing electronegativity.

   ½ mark

3. b. pKa falls as EN rises (9.45 at 3.44; 9.28 at 3.6). Extrapolating to EN 3.7 gives a value just below 9.28, about 9.1–9.2. Answer: (iv) 9.1.

   1 mark

4. c. For piperidines, pKa 8.36 (EN 3.44) and 7.94 (EN 3.6). A pKa of 8.28 lies just past O, so EN ≈ 3.44 + (0.08/0.42)(0.16) ≈ 3.47. Answer: (i) 3.5.

   1 mark

5. c (OR). For propylamines, EN 2.55 → 10.67 and 3.12 → 10.08. At EN 3.0, pKa ≈ 10.67 − (0.45/0.57)(0.59) ≈ 10.2. Answer: (iii) 10.15.

Answer: a. pKb rises (basic strength falls) with increasing electronegativity of X; b. (iv) 9.1; c. (i) 3.5 / OR (iii) 10.15

1. A is $KMn{O}_4$ (purple, strong oxidant, used to bleach textiles).
   a(i) Violet vapours ($I_2$) form in test tube C (KI).

   ½ mark

2. $10I^{-}+2Mn{O}_4^{-}+16H^{+}\to 5I_2+2M{n}^{2+}+8H_{2}O$

   ½ mark

3. a(ii) The gas that puts out a burning matchstick is $C{O}_2$, formed in test tube A (oxalic acid).

   ½ mark

4. $5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 10C{O}_2+2M{n}^{2+}+8H_{2}O$

   ½ mark

5. b(i) The complex is octahedral $[M(AA{)}_3]$ and high spin, since ${\Delta }_0<P$. The diagram shows the five degenerate 3d orbitals of the free ion split into: a lower $t_{2g}$ set ($d_{xy},d_{yz},d_{xz}$) at $-0.4{\Delta }_0$ with one electron each, and an upper $e_g$ set ($d_{x^2-y^2},d_{z^2}$) at $+0.6{\Delta }_0$ with one electron each. The gap between them is ${\Delta }_0$. Configuration: $t_{2g}^3{e}_g^2$, 5 unpaired electrons.

   1 mark

6. b(ii) Hybridisation: **$s{p}^3{d}^2$** (outer-orbital complex). Because ${\Delta }_0<P$, the five 3d electrons do not pair and occupy all five 3d orbitals. No inner 3d orbitals are free, so the metal uses $4s,4p$ and $4d$ orbitals for the six donor atoms.

   1 mark

7. b(iii) Optical isomerism. $[M(AA{)}_3]$ exists as two non-superimposable mirror images (d and l forms).

   1 mark

Answer: a(i) Test tube C: 10I⁻ + 2MnO₄⁻ + 16H⁺ → 5I₂ + 2Mn²⁺ + 8H₂O; a(ii) Test tube A: 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O; b(i) t₂g³e_g² high-spin splitting; b(ii) sp³d² outer orbital; b(iii) optical isomerism

1. a. In $[Co{F}_4{]}^{2-}$: $x+4(-1)=-2$, so $x=+2$. **A = $C{o}^{2+}$** ($3d^7$).

   ½ mark

2. F⁻ is a weak-field ligand, so there is no pairing and $3d^7$ has B = 3 unpaired electrons.

   ½ mark

3. $C{r}^{3+}$ ($3d^3$) leaves two empty inner 3d orbitals for the six ligands, so the hybridisation is **C = $d^{2}s{p}^3$** (inner orbital, octahedral).

   ½ mark

4. It has 3 unpaired electrons, so it is D = paramagnetic.

   ½ mark

5. In $[Ni(CO{)}_4]$, strong-field CO makes the $4s^2$ electrons pair into 3d, giving $3d^{10}$. The empty 4s and three 4p orbitals hybridise: **E = $s{p}^3$**.

   ½ mark

6. $s{p}^3$ hybridisation gives a F = tetrahedral geometry.

   ½ mark

7. b(i) $C{r}_2{O}_7^{2-}+8H^{+}+3H_{2}S\to 2C{r}^{3+}+3S\downarrow +7H_{2}O$

   1 mark

8. b(ii) $C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$

   1 mark

Answer: A = Co²⁺, B = 3, C = d²sp³, D = paramagnetic, E = sp³, F = tetrahedral; (i) Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O; (ii) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

1. a(i) $C_2{H}_{5}OH+C{H}_{3}COCl\stackrel{pyridine}{\to }C{H}_{3}COO{C}_2{H}_5+HCl$. Pyridine, an organic base, neutralises the HCl as it forms. This prevents the reverse reaction and shifts the equilibrium forward, increasing the yield of ester.

   1 mark

2. a(ii) In cresols, the –CH₃ group is electron-releasing (+I effect and hyperconjugation). It increases electron density on the phenoxide oxygen and destabilises the cresoxide ion relative to phenoxide. Cresols therefore lose $H^{+}$ less readily and are less acidic than phenol.

   1 mark

3. b. The target is $C{H}_3–CH(O{C}_2{H}_5)–CH(C{H}_3)–C{H}_{2}C{H}_3$. Williamson synthesis needs a primary alkyl halide to avoid elimination, so use **bromoethane, $C_2{H}_{5}Br$, and sodium 3-methylpentan-2-olate, $C{H}_{3}CH(ONa)CH(C{H}_3)C{H}_{2}C{H}_3$**:
   $C{H}_{3}CH(ONa)CH(C{H}_3)C{H}_{2}C{H}_3+C_2{H}_{5}Br\to C{H}_{3}CH(O{C}_2{H}_5)CH(C{H}_3)C{H}_{2}C{H}_3+NaBr$

   1 mark

4. c(i) Toluene $\stackrel{KMn{O}_4/KOH,\Delta ;then{H}_3{O}^{+}}{\to }$ benzoic acid $\stackrel{conc.HN{O}_3/conc.H_{2}S{O}_4}{\to }$ 3-nitrobenzoic acid. Oxidation is done first because –COOH is meta-directing.

   1 mark

5. c(ii) Benzene $\stackrel{C{H}_{3}COCl,anhyd.AlC{l}_3}{\to }$ acetophenone $\stackrel{conc.HN{O}_3/conc.H_{2}S{O}_4}{\to }$ m-nitroacetophenone. Acylation is done first because –COCH₃ is meta-directing; nitrobenzene would not undergo Friedel–Crafts acylation.

   1 mark

Answer: (i) Pyridine removes HCl and shifts equilibrium forward; (ii) +I/hyperconjugation of –CH₃ destabilises the phenoxide ion; b. C₂H₅Br + sodium 3-methylpentan-2-olate; c(i) KMnO₄ oxidation then nitration; c(ii) Friedel–Crafts acetylation then nitration

1. a. Acetic acid gives the HVZ reaction. The HVZ reaction replaces an α-hydrogen of a carboxylic acid by halogen, using $C{l}_2$ or $B{r}_2$ with a little red phosphorus. Acetic acid ($C{H}_{3}COOH$) has three α-hydrogens. Formic acid ($HCOOH$) has no α-carbon, so it has no α-hydrogen and cannot react.

   1 mark

2. Reaction:
   $$C{H}_{3}COOH\underset{\text{(ii) }{H}_{2}O}{\overset{\text{(i) }B{r}_2/\text{red P}}{\to }}BrC{H}_{2}COOH(\text{bromoacetic acid})+HBr$$

   1 mark

3. b. Increasing order of acidic strength of the butanol isomers:
   $$\underset{3^{\circ }}{\text{2-methylpropan-2-ol}}<\underset{2^{\circ }}{\text{butan-2-ol}}<\underset{1^{\circ },\text{branched}}{\text{2-methylpropan-1-ol}}<\underset{1^{\circ }}{\text{butan-1-ol}}$$

   ½ mark

4. Reason: the acidity of an alcohol comes from the polar O–H bond. Alkyl groups are electron-releasing (+I). More alkyl groups on, or close to, the carbinol carbon push more electron density onto oxygen. This makes the O–H bond less polar and destabilises the alkoxide ion. So acidity falls in the order $1^{\circ }>2^{\circ }>3^{\circ }$.

   ½ mark

5. c. The product is 2-methylbutan-2-ol, $C{H}_{3}C{H}_{2}C(OH)(C{H}_3{)}_2$, a tertiary alcohol. A Grignard reagent gives a tertiary alcohol only with a ketone, so B is a ketone.

   Disconnect one methyl group from the carbinol carbon. This gives **A = $C{H}_{3}MgBr$ (methylmagnesium bromide) and B = $C{H}_{3}COC{H}_{2}C{H}_3$** (butan-2-one).

   1 mark

6. Equation:
   $$C{H}_{3}MgBr+C{H}_{3}COC{H}_{2}C{H}_3\to C{H}_{3}C{H}_{2}C(C{H}_3{)}_{2}OMgBr\stackrel{H_{2}O/H^{+}}{\to }C{H}_{3}C{H}_{2}C(OH)(C{H}_3{)}_2+Mg(OH)Br$$

   (Another valid pair is $C_2{H}_{5}MgBr$ with propanone, which gives the same alcohol.)

   1 mark

Answer: a. Acetic acid (it has α-H; formic acid has none): CH₃COOH + Br₂/red P → BrCH₂COOH + HBr. b. 2-methylpropan-2-ol < butan-2-ol < 2-methylpropan-1-ol < butan-1-ol (+I effect of alkyl groups). c. A = CH₃MgBr, B = butan-2-one.

1. a. In the graph, Solution 2 has the lowest vapour-pressure curve. It therefore meets the solid-solvent curve at the lowest temperature, giving the larger depression in freezing point. So **Solution 2 = 1 M $Al(N{O}_3{)}_3$ and Solution 1 = 1 M glucose**.

   1 mark

2. Reason: depression in freezing point is a colligative property, $\Delta T_f=i{K}_{f}m$.

   • $Al(N{O}_3{)}_3\to A{l}^{3+}+3N{O}_3^{-}$, so $i=4$ for complete dissociation.
   • Glucose does not dissociate, so $i=1$.

   At the same molality, $Al(N{O}_3{)}_3$ gives more solute particles and hence the larger $\Delta T_f$.

   1 mark

3. b. Use $\pi =\frac{n_2}{V}RT$. At constant $T$ and fixed $n_2$, $\pi \propto \frac{1}{V}$, so $\frac{{\pi }_1}{{\pi }_2}=\frac{V_2}{V_1}$.

   ½ mark

4. Diluting five times gives $V_2=5V_1$. Then
   $$\frac{2.46}{{\pi }_2}=\frac{5V_1}{V_1}=5$$

   ½ mark

5. $${\pi }_2=\frac{2.46}{5}=0.492\text{atm}$$

   1 mark

6. To lower the osmotic pressure without changing the volume, decrease the temperature. Since $\pi =CRT$, at fixed concentration $\pi$ is directly proportional to the absolute temperature $T$.

   1 mark

Answer: a. Solution 1 = glucose, Solution 2 = Al(NO₃)₃ (larger i, larger ΔTf). b. π = 0.492 atm; lower it by decreasing the temperature, since π ∝ T.

1. a. An intravenous fluid must be isotonic with blood, i.e. have the same osmotic pressure as the fluid inside the blood cells (about 0.9% saline). Only then is there no net movement of water across the cell membranes, and the cells keep their normal shape.

   1 mark
2. • If the solution is more concentrated (hypertonic), water flows out of the red blood cells by osmosis, and the cells shrink.
   • If it is less concentrated (hypotonic), water flows into the cells, which swell and may burst.

   Both situations can be life-threatening, so the concentration must be checked.

   1 mark

3. b. Phenol dimerises in benzene: $2C_6{H}_{5}OH\rightleftharpoons (C_6{H}_{5}OH{)}_2$.

   For association into n-mers, $i=1-\alpha \left(1-\frac{1}{n}\right)$, with $n=2$ and $\alpha =0.73$.

   ½ mark

4. $$i=1-\frac{0.73}{2}=1-0.365=0.635$$

   ½ mark

5. Use $\Delta T_f=i{K}_{f}m=i{K}_f\frac{w_B}{M_B\times w_A(\text{kg})}$.

   Data: $K_f$ (benzene) $=5.12$ K kg mol⁻¹, $w_B=2\times 10^{-2}$ kg $=20$ g, $M_B$ (phenol) $=94$ g mol⁻¹, $w_A=1$ kg.

   ½ mark

6. $$\Delta T_f=0.635\times 5.12\times \frac{20}{94\times 1}$$

   ½ mark

7. $$\Delta T_f=3.2512\times 0.2128\approx 0.69\text{K}$$

   1 mark

Answer: a. The IV solution must be isotonic with blood; hypertonic fluid shrinks the cells, hypotonic fluid swells or bursts them. b. i = 0.635, ΔTf ≈ 0.69 K.