CBSE Class 12 Chemistry: 2023 board paper, set 56/1/1

1. Polymers and proteins have very high molar masses, so their solutions give tiny $\Delta T_f$, $\Delta T_b$ and vapour-pressure lowering, but a measurable osmotic pressure.

2. Osmotic pressure is measured at room temperature, so heat-sensitive biomolecules are not damaged. Answer: osmotic pressure, option (A).

   1 mark

Answer: Osmotic pressure, option (A)

1. By Henry's law, gas solubility is proportional to its partial pressure, $p=K_{H}x$.

2. At high altitude the atmospheric pressure, and so the partial pressure of $O_2$, is low. Less oxygen dissolves in blood (anoxia). Option (C).

   1 mark

Answer: Low atmospheric pressure, option (C)

1. Zn is oxidised, so it is the anode and is written on the left. $A{g}^{+}$ is reduced, so Ag is the cathode and is written on the right.

2. The cell is $Zn|Z{n}^{2+}\Vert A{g}^{+}|Ag$, option (D).

   1 mark

Answer: Option (D)

1. From $\Delta G=-nF{E}_{cell}$, a positive $E_{cell}$ gives a negative $\Delta G$.

2. A negative $\Delta G$ means the reaction is spontaneous. Option (C).

   1 mark

Answer: Negative, positive; option (C)

1. A catalyst provides an alternative pathway with a lower activation energy.

2. $\Delta H$, $\Delta G$ and $\Delta S$ are state functions and do not change. Option (C).

   1 mark

Answer: $E_a$, option (C)

1. The photochemical combination of $H_2$ and $C{l}_2$ is a zero-order reaction: its rate depends on light intensity, not on reactant concentration.

2. Rate $=k$, so the order is 0. Option (C).

   1 mark

Answer: 0, option (C)

1. Lanthanoids most commonly show the +3 oxidation state, formed by losing two 6s electrons and one 4f/5d electron. Option (B).

   1 mark

Answer: +3, option (B)

1. The two compounds exchange the ligand inside the coordination sphere with the counter-ion outside it.

2. In solution they give different ions ($B{r}^{-}$ and $S{O}_4^{2-}$), so they are ionisation isomers. Option (C).

   1 mark

Answer: Ionisation isomerism, option (C)

1. In the Swarts reaction, an alkyl chloride or bromide is heated with a metallic fluoride such as $AgF$, $H{g}_2{F}_2$ or $Sb{F}_3$, giving an alkyl fluoride. Option (B).

   1 mark

Answer: Swarts reaction, option (B)

1. With Lucas reagent the reaction goes through a carbocation, and carbocation stability is $3^{\circ }>2^{\circ }>1^{\circ }$.

2. So the reactivity order is $1^{\circ }<2^{\circ }<3^{\circ }$. Option (A).

   1 mark

Answer: $1^{\circ }<2^{\circ }<3^{\circ }$, option (A)

1. This is the Hoffmann bromamide degradation, which gives an amine with one carbon fewer than the amide.

2. $C{H}_{3}CON{H}_2+B{r}_2+4NaOH\to C{H}_{3}N{H}_2+N{a}_{2}C{O}_3+2NaBr+2H_{2}O$. Option (B).

   1 mark

Answer: $C{H}_{3}N{H}_2$, option (B)

1. In aniline the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept a proton.

2. Aniline is therefore less basic than $N{H}_3$ and the alkylamines. Option (C).

   1 mark

Answer: Aniline, option (C)

1. Amylose is an unbranched chain of α-D-glucose units joined by C1–C4 glycosidic links; the C1–C6 links occur at the branch points of amylopectin. Option (C).

   1 mark

Answer: $C_1$–$C_4$ α linkage, option (C)

1. The α-helix is a secondary structure of proteins (polypeptides), held in shape by hydrogen bonds between –CO and –NH groups of the chain. Option (C).

   1 mark

Answer: Polypeptides, option (C)

1. A is true: $-N{H}_2$ increases electron density at the o- and p-positions through resonance.

2. R is true: aniline forms a salt with the Lewis acid $AlC{l}_3$, so it cannot undergo Friedel-Crafts reaction. However, R does not explain A. Option (B).

   1 mark

Answer: Option (B)

1. A is true: acetylation controls the activity of the ring, so a monosubstituted product is obtained.

2. R is false: the lone pair on N in $-NHCOC{H}_3$ is also delocalised onto the C=O group, so its activating effect is less than that of $-N{H}_2$. Option (C).

   1 mark

Answer: Option (C)

1. Molecularity is the number of reacting species colliding in an elementary step.

2. Taken as an elementary step, one $H_2$ and one $B{r}_2$ molecule collide, so the molecularity appears to be 2. R correctly explains A. Option (A).

   1 mark

Answer: Option (A)

1. For tetrahedral complexes ${\Delta }_t=\frac{4}{9}{\Delta }_o$, which is small.

2. Because ${\Delta }_t$ is less than the pairing energy P, electrons occupy the higher orbitals rather than pair, giving high-spin complexes. So low-spin tetrahedral complexes are rare, and R explains A. Option (A).

   1 mark

Answer: Option (A)

1. Henry's law: At a constant temperature, the partial pressure of a gas in the vapour phase is directly proportional to the mole fraction of that gas in the solution: $p=K_{H}x$. Here $p$ is the partial pressure of the gas, $x$ is its mole fraction in the solution and $K_H$ is Henry's law constant.

   1 mark

2. Application: Soft-drink and soda-water bottles are sealed under high pressure. This increases the solubility of $\mathrm{C}{\mathrm{O}}_2$ in the drink. (Another accepted example: scuba divers breathe air diluted with helium to avoid the bends.)

   1 mark

Answer: $p=K_{H}x$ (partial pressure of a gas ∝ its mole fraction in solution); application: CO₂ dissolved in soft drinks by sealing under high pressure.

1. 'B' is the strong electrolyte. A strong electrolyte is already almost fully ionised, so dilution only slightly increases ${\Lambda }_m$ (here by 1·5 times). Ion–ion attraction decreases on dilution, which raises ${\Lambda }_m$ a little. A weak electrolyte ('A') ionises much more on dilution, so its ${\Lambda }_m$ rises steeply (25 times).

   1 mark

2. **Graph (${\Lambda }_m$ on the y-axis vs $\sqrt{c}$ on the x-axis):**

   • For B (strong electrolyte) the graph is a straight line with a small negative slope. It obeys ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$ and can be extrapolated to $c=0$ to give ${\Lambda }_m^{\circ }$.
   • For A (weak electrolyte) the curve stays low at higher concentrations. It then rises very sharply as $\sqrt{c}\to 0$ and cannot be extrapolated.
   1 mark

Answer: B is the strong electrolyte (Λm rises only slightly and linearly with dilution; A shows a steep rise and is weak).

1. The radius is $r=0\cdot 5$ cm. Area of cross-section: $A=\pi r^2=3\cdot 14\times (0\cdot 5{)}^2=0\cdot 785{\text{cm}}^2$. The length is $l=50$ cm.

   ½ mark

2. Use $\kappa =\frac{l}{R\times A}=\frac{50\text{cm}}{5\cdot 55\times 10^3\Omega \times 0\cdot 785{\text{cm}}^2}$.

   1 mark

3. $\kappa =\frac{63\cdot 69}{5550}=1\cdot 147\times 10^{-2}{\text{S cm}}^{-1}$ (i.e. $11\cdot 47\times 10^{-3}$ S cm⁻¹).

   ½ mark

Answer: κ ≈ 1.147 × 10⁻² S cm⁻¹

1. (a) In acidic medium, permanganate oxidises nitrite to nitrate and is itself reduced to $\mathrm{M}{\mathrm{n}}^{2+}$:
   $2Mn{\mathrm{O}}_4^{-}(\mathrm{aq})+5N{\mathrm{O}}_2^{-}(\mathrm{aq})+6{\mathrm{H}}^{+}(\mathrm{aq})\to 2M{\mathrm{n}}^{2+}(\mathrm{aq})+5N{\mathrm{O}}_3^{-}(\mathrm{aq})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   1 mark

2. (b) Dichromate is reduced to chromium(III) in acidic medium:
   $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}(\mathrm{aq})+14{\mathrm{H}}^{+}(\mathrm{aq})+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}(\mathrm{aq})+7{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   1 mark

Answer: (a) 2Mn²⁺ + 5NO₃⁻ + 3H₂O; (b) 2Cr³⁺ + 7H₂O

1. (a) $\mathrm{PC}{\mathrm{l}}_5$ replaces –OH by –Cl. **A = $\mathrm{C}{\mathrm{H}}_3-\mathrm{CHCl}-\mathrm{C}{\mathrm{H}}_3$ (2-chloropropane).**

   ½ mark

2. AgCN is mainly covalent, so it attacks through the nitrogen atom and gives an isocyanide. **B = $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}(\mathrm{NC})-\mathrm{C}{\mathrm{H}}_3$ (2-propyl isocyanide / isopropyl isocyanide).**

   ½ mark

3. (b) Alcoholic KOH causes dehydrohalogenation. **A = $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ (propene).**

   ½ mark

4. HBr adds by Markovnikov's rule, so Br goes to the carbon with fewer H atoms. **B = $\mathrm{C}{\mathrm{H}}_3-\mathrm{CHBr}-\mathrm{C}{\mathrm{H}}_3$ (2-bromopropane).**

   ½ mark

Answer: (a) A = 2-chloropropane, B = 2-propyl isocyanide (CH₃CH(NC)CH₃); (b) A = propene, B = 2-bromopropane

1. (i) When phenol loses ${\mathrm{H}}^{+}$, it forms the phenoxide ion. The negative charge of phenoxide is delocalised over the benzene ring by resonance, so the ion is stabilised. In the alkoxide ion the charge stays on oxygen, and the +I effect of the alkyl group makes it even less stable. Hence phenol releases ${\mathrm{H}}^{+}$ more easily and is the stronger acid.

   1 mark

2. (ii) Branching makes the molecule more compact, so its surface area decreases. This weakens the van der Waals forces between molecules, so less energy is needed to boil the alcohol. Hence the boiling point falls as branching increases.

   1 mark

Answer: (i) Phenoxide ion is resonance-stabilised, alkoxide is not; (ii) branching lowers surface area and hence van der Waals forces.

1. (i) Step 1 – Protonation of the alcohol:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\ddot{\mathrm{O}}\mathrm{H}+{\mathrm{H}}^{+}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2$ (fast)

   ½ mark

2. Step 2 – Formation of carbocation (slow, rate-determining):
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_3\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$
   Step 3 – Loss of proton:
   $\mathrm{H}-\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}^{+}$

   ½ mark

3. (ii) Reimer–Tiemann reaction: Phenol is heated with chloroform and aqueous NaOH, forming a dichloromethyl (–CHCl₂) intermediate at the ortho position. Hydrolysis then gives sodium salicylaldehyde, and acidification gives salicylaldehyde:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\underset{\text{340 K}}{\overset{\mathrm{CHC}{\mathrm{l}}_3,aq.NaOH}{\to }}\mathrm{o}\text{-}{\mathrm{C}}_6{\mathrm{H}}_4({\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+})(\mathrm{CHO})\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{o}\text{-}{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{OH})\mathrm{CHO}$ (2-hydroxybenzaldehyde)

   1 mark

Answer: (i) Protonation → carbocation (slow) → loss of H⁺ giving ethene; (ii) phenol + CHCl₃/aq NaOH, then H⁺ → salicylaldehyde

1. (a) Carbylamine reaction: A primary amine (aliphatic or aromatic) is heated with chloroform and ethanolic KOH. It forms an isocyanide (carbylamine), which has a very foul smell. Secondary and tertiary amines do not give this test.
   $\mathrm{R}-\mathrm{N}{\mathrm{H}}_2+\mathrm{CHC}{\mathrm{l}}_3+3KOH\stackrel{\mathrm{\Delta }}{\to }\mathrm{R}-\mathrm{NC}+3KCl+3{\mathrm{H}}_2\mathrm{O}$

   1 mark

2. (b) Gabriel phthalimide synthesis:

   • Phthalimide is treated with ethanolic KOH to form potassium phthalimide.
   • Heating this salt with an alkyl halide gives N-alkylphthalimide.
   • Alkaline hydrolysis of N-alkylphthalimide gives a pure primary amine and sodium phthalate.

   ${\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2\mathrm{NH}\stackrel{\mathrm{KOH}(\mathrm{alc})}{\to }{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2{\mathrm{N}}^{-}{\mathrm{K}}^{+}\stackrel{\mathrm{R}-\mathrm{X}}{\to }{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2\mathrm{N}-\mathrm{R}\stackrel{\mathrm{NaOH}(\mathrm{aq})}{\to }\mathrm{R}-\mathrm{N}{\mathrm{H}}_2+{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{COONa}{)}_2$

   1 mark

Answer: (a) RNH₂ + CHCl₃ + 3KOH → RNC (foul smell); (b) phthalimide → K-phthalimide → N-alkylphthalimide → RNH₂ (1° amine)

1. (a) On prolonged heating with HI, glucose forms n-hexane. This shows that all six carbon atoms of glucose are linked in a straight chain:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{\mathrm{\Delta }}{\overset{\mathrm{HI}}{\to }}\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$ (n-hexane)

   1 mark

2. (b) Proteins are formed by peptide linkage ($-\mathrm{CO}-\mathrm{NH}-$). This is an amide bond between the –COOH of one amino acid and the –NH₂ of the next.

   1 mark

Answer: (a) Glucose + HI (heat) → n-hexane; (b) peptide linkage (–CO–NH–)

1. (a) An ideal solution obeys Raoult's law at all concentrations, with ${\Delta }_{mix}H=0$ and ${\Delta }_{mix}V=0$. A non-ideal solution does not obey Raoult's law, with ${\Delta }_{mix}H\ne 0$ and ${\Delta }_{mix}V\ne 0$ (it shows positive or negative deviation).

   1 mark

2. (b) Raoult's law for a non-volatile solute: $\frac{p_A^{\circ }-p_A}{p_A^{\circ }}=x_B=\frac{n_B}{n_A+n_B}$.

   ½ mark

3. Moles of urea: $n_B=\frac{30}{60}=0\cdot 5$ mol. Moles of water: $n_A=\frac{846}{18}=47$ mol.
   So $1-\frac{p_A}{23\cdot 8}=\frac{0\cdot 5}{47+0\cdot 5}=\frac{0\cdot 5}{47\cdot 5}$.

   1 mark

4. $p_A=23\cdot 8\times \frac{47}{47\cdot 5}\approx 23\cdot 5$ mm Hg.

   ½ mark

Answer: (a) Ideal obeys Raoult's law, ΔH_mix = ΔV_mix = 0; non-ideal does not. (b) p ≈ 23.5 mm Hg

1. (a) Finkelstein reaction (halogen exchange): $\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{NaI}\stackrel{\mathrm{acetone}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{I}+\mathrm{NaCl}\downarrow$. The main product is iodomethane (methyl iodide). NaCl precipitates in acetone, which drives the reaction forward.

   1 mark

2. (b) Three –NO₂ groups at the ortho and para positions strongly activate the C–Cl bond, so warm water alone hydrolyses it. The product is 2,4,6-trinitrophenol (picric acid):
   ${\mathrm{C}}_6{\mathrm{H}}_2(\mathrm{N}{\mathrm{O}}_2{)}_3\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{warm}}{\to }{\mathrm{C}}_6{\mathrm{H}}_2(\mathrm{N}{\mathrm{O}}_2{)}_3\mathrm{OH}+\mathrm{HCl}$

   1 mark

3. (c) Alcoholic KOH causes dehydrohalogenation, and only one β-carbon is available:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}+\mathrm{KOH}(\mathrm{alc})\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+\mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}$. The product is but-1-ene.

   1 mark

Answer: (a) CH₃I (iodomethane); (b) picric acid (2,4,6-trinitrophenol); (c) CH₃CH₂CH=CH₂ (but-1-ene)

1. (a) Phenol → picric acid: Phenol is treated with conc. H₂SO₄ to form phenol-2,4-disulphonic acid. This is then treated with conc. HNO₃, which gives 2,4,6-trinitrophenol.
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\stackrel{conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{OH})(\mathrm{S}{\mathrm{O}}_3\mathrm{H}{)}_2\stackrel{conc.\mathrm{HN}{\mathrm{O}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_2(\mathrm{OH})(\mathrm{N}{\mathrm{O}}_2{)}_3$ (picric acid)

   1 mark

2. (b) Propanone → 2-methylpropan-2-ol: Propanone reacts with $\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}$ in dry ether, and the adduct is then hydrolysed.
   $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{O}+\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}\stackrel{\mathrm{dry}\mathrm{ether}}{\to }(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{OMgBr}\stackrel{{\mathrm{H}}_{2}O/{\mathrm{H}}^{+}}{\to }(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{OH}+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

   1 mark

3. (c) Phenol → anisole: Phenol is converted to sodium phenoxide with NaOH. The phenoxide is then heated with methyl iodide (Williamson synthesis).
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{NaOH}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}\stackrel{\mathrm{C}{\mathrm{H}}_3\mathrm{I}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OC}{\mathrm{H}}_3+\mathrm{NaI}$

   1 mark

4. (d) Propene → propan-1-ol: Hydroboration–oxidation gives the anti-Markovnikov alcohol.
   $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2\stackrel{(\mathrm{B}{\mathrm{H}}_3{)}_2}{\to }(\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{B}\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2/O{\mathrm{H}}^{-}}{\to }3C{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$
   (Any three conversions earn the full 3 marks.)

Answer: (a) conc. H₂SO₄ then conc. HNO₃; (b) CH₃MgBr then H₂O/H⁺; (c) NaOH then CH₃I; (d) B₂H₆ then H₂O₂/OH⁻

1. (i) The –COOH group is electron-withdrawing (–R and –I effects). It reduces electron density mainly at the ortho and para positions, so the meta positions are left relatively richer in electrons. Therefore electrophilic substitution occurs at the meta position, and the group also deactivates the ring.

   1 mark

2. (ii) Aldehydes and ketones form crystalline bisulphite addition compounds with $\mathrm{NaHS}{\mathrm{O}}_3$, which can be separated from impurities. These adducts are decomposed by dilute acid or alkali to regenerate the pure aldehyde or ketone.

   1 mark

3. (iii) In –COOH, the lone pair on the –OH oxygen is in resonance with the C=O group. This makes the carboxyl carbon much less electrophilic, so it does not undergo the typical nucleophilic addition reactions of aldehydes and ketones.

   1 mark

Answer: (i) –COOH is electron-withdrawing (–R), deactivating ortho/para; (ii) reversible formation of crystalline bisulphite adducts; (iii) resonance lowers electrophilicity of the carboxyl carbon.

1. Cu at 573 K dehydrogenates alcohols, so A (${\mathrm{C}}_3{\mathrm{H}}_8\mathrm{O}$) is an alcohol. B does not reduce Fehling's solution, so B is a ketone, not an aldehyde. A must therefore be a secondary alcohol: **A = $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{C}{\mathrm{H}}_3$ (propan-2-ol).**

   1 mark

2. $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Cu},573\mathrm{K}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3+{\mathrm{H}}_2$. **B = $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$ (propanone).** It is a methyl ketone, which explains the positive iodoform test.

   1 mark

3. $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3+3{\mathrm{I}}_2+4NaOH\to \mathrm{CH}{\mathrm{I}}_3\downarrow +\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+3NaI+3{\mathrm{H}}_2\mathrm{O}$. **C = $\mathrm{CH}{\mathrm{I}}_3$ (iodoform),** the yellow precipitate.

   1 mark

Answer: A = CH₃CH(OH)CH₃ (propan-2-ol); B = CH₃COCH₃ (propanone); C = CHI₃ (iodoform)

1. (a)(i) Lactose on hydrolysis gives β-D-galactose and β-D-glucose.

   1 mark

2. (a)(ii) Maltose on hydrolysis gives two molecules of α-D-glucose.

   1 mark

3. (b) Starch is a polymer of α-D-glucose: amylose is a linear chain with C1–C4 α-glycosidic links, and amylopectin is branched with C1–C6 links. Cellulose is a linear, unbranched polymer of β-D-glucose joined by C1–C4 β-glycosidic linkages.

   1 mark

Answer: (a)(i) Galactose + glucose; (ii) glucose + glucose; (b) starch = α-D-glucose polymer, cellulose = β-D-glucose polymer

1. (i) Average rate is the change in concentration of a reactant or product divided by the time interval over which the change occurs: $r_{av}=-\frac{\Delta [R]}{\Delta t}=\frac{\Delta [P]}{\Delta t}$.

   1 mark

2. (ii) Any two factors: concentration of reactants, temperature, presence of a catalyst, and surface area of reactants.

   1 mark

3. (iii)(a)(1) For a zero order reaction, the rate is independent of the concentration of reactants. It stays constant: rate = k.

   1 mark

4. (iii)(a)(2) Unit of k for a zero order reaction is $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$.

   1 mark

5. (iii)(b)(1) Order = $\frac{1}{2}+1=\frac{3}{2}$ = 1·5.

6. (iii)(b)(2) A pseudo first order reaction has a higher molecularity or order but behaves as first order. This happens because one reactant is in large excess, so its concentration stays practically constant. Example: acid hydrolysis of ethyl acetate in excess water, $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$, with rate = $k^{\prime }[\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5]$.
   (Alternative (iii)(b) carries the same 2 marks: order 1 mark, definition ½, example ½.)

Answer: (i) Δ(concentration)/Δt; (ii) concentration, temperature (also catalyst, surface area); (iii)(a) rate constant, independent of concentration; mol L⁻¹ s⁻¹; (iii)(b) order 3/2; pseudo first order, e.g. acid hydrolysis of ethyl acetate

1. (i) No precipitate with AgNO₃ means no ionisable Cl⁻, so both Cl are inside the coordination sphere: **$[\mathrm{Pt}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{C}{\mathrm{l}}_2]$**.

   1 mark

2. (ii) Ethylenediamine is bidentate, so three en ligands give 3 × 2 donor atoms. The secondary valency (coordination number) is 6.

   1 mark

3. (iii)(a)(1) Iron(III) hexacyanidoferrate(II): **$\mathrm{F}{\mathrm{e}}_4[\mathrm{Fe}(\mathrm{CN}{)}_6{]}_3$** (Prussian blue).

   1 mark

4. (iii)(a)(2) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{Cl}]\mathrm{C}{\mathrm{l}}_2$: pentaamminechloridocobalt(III) chloride.

   1 mark

5. (iii)(b) In $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$, Ni is +2, so Ni²⁺ is [Ar]3d⁸. CN⁻ is a strong-field ligand, so the eight d electrons pair up in four 3d orbitals. One 3d, one 4s and two 4p orbitals then mix to form dsp² hybrid orbitals (square planar). With no unpaired electrons the complex is diamagnetic.
   (Alternative (iii)(b) carries the same 2 marks: hybridisation 1, magnetic behaviour 1.)

Answer: (i) [Pt(NH₃)₂Cl₂]; (ii) 6; (iii)(a) Fe₄[Fe(CN)₆]₃; pentaamminechloridocobalt(III) chloride; (iii)(b) dsp², diamagnetic

1. (i) Kohlrausch's law: The limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion: ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$.

   1 mark

2. For acetic acid: ${\Lambda }_m^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})={\lambda }_{\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}}^{\circ }+{\lambda }_{{\mathrm{H}}^{+}}^{\circ }$
   (This can also be found as ${\Lambda }_m^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COONa})+{\Lambda }_m^{\circ }(\mathrm{HCl})-{\Lambda }_m^{\circ }(\mathrm{NaCl})$.)

   1 mark

3. (ii) $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0\cdot 80-(-0\cdot 25)=1\cdot 05$ V, with $n=2$.

   ½ mark

4. Maximum work $=-{\Delta }_r{G}^{\circ }=nF{E}_{cell}^{\circ }$.

   ½ mark

5. $=2\times 96500{\text{C mol}}^{-1}\times 1\cdot 05\text{V}=202650{\text{J mol}}^{-1}=202\cdot 65$ kJ mol⁻¹.

   1 mark

6. $\log K_c=\frac{n{E}_{cell}^{\circ }}{0\cdot 059}=\frac{2\times 1\cdot 05}{0\cdot 059}=35\cdot 59$

   1 mark

Answer: (i) Λ°m(CH₃COOH) = λ°(CH₃COO⁻) + λ°(H⁺); (ii) maximum work = 202.65 kJ mol⁻¹, log Kc ≈ 35.59

1. (i) Faraday's first law: The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte: $w=ZIt$.

   1 mark

2. $\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$. One mole of Cu²⁺ needs 2 mol of electrons, so the charge required is 2 F (= 193000 C).

   1 mark

3. (ii) Cell reaction: $\mathrm{Mg}+\mathrm{C}{\mathrm{u}}^{2+}\to \mathrm{M}{\mathrm{g}}^{2+}+\mathrm{Cu}$, $n=2$.
   Nernst equation: $E_{cell}=E_{cell}^{\circ }-\frac{0\cdot 059}{n}\log \frac{[\mathrm{M}{\mathrm{g}}^{2+}]}{[\mathrm{C}{\mathrm{u}}^{2+}]}$

   1 mark

4. $E_{cell}=2\cdot 71-\frac{0\cdot 059}{2}\log \frac{0\cdot 1}{0\cdot 01}=2\cdot 71-0\cdot 0295\log 10$

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5. $E_{cell}=2\cdot 71-0\cdot 0295=2\cdot 68$ V

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Answer: (i) w ∝ Q; 2 F required; (ii) E_cell ≈ 2.68 V

1. (i) Mn has the configuration [Ar]3d⁵4s², with the maximum number of unpaired d electrons in the 3d series. All seven 3d and 4s electrons can take part in bonding, giving the highest oxidation state of +7 (e.g. in MnO₄⁻).

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2. (ii) Transition metals can show variable oxidation states and form unstable intermediate complexes with reactants. They also provide a large surface area for adsorption, which weakens reactant bonds and lowers the activation energy. For example, Fe is used in the Haber process and V₂O₅ in the contact process.

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3. (iii) Cr²⁺ (d⁴) is oxidised to Cr³⁺, which has a stable half-filled $t_{2g}^3$ configuration, so Cr²⁺ acts as a reducing agent. Mn³⁺ (d⁴) is reduced to Mn²⁺, which has the extra-stable half-filled d⁵ configuration, so Mn³⁺ acts as an oxidising agent.

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4. (iv) Zn ([Ar]3d¹⁰4s²) has a completely filled d subshell and no unpaired electrons for metallic bonding. Its interatomic (metallic) bonding is therefore weak, so its enthalpy of atomisation is the lowest.

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5. (v) In water, Cu⁺ disproportionates: $2C{\mathrm{u}}^{+}(\mathrm{aq})\to \mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$. The much more negative hydration enthalpy of the smaller, more highly charged Cu²⁺ ion more than compensates for the second ionisation enthalpy of Cu. Hence Cu²⁺(aq) is more stable than Cu⁺(aq).

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Answer: (i) all 3d and 4s electrons used; (ii) variable oxidation states/complex formation/large surface area; (iii) Cr³⁺ gets t₂g³, Mn²⁺ gets d⁵; (iv) no unpaired electrons, weak metallic bonding; (v) Cu⁺ disproportionates to Cu²⁺ + Cu

1. (i)(1) Ethanal → But-2-en-1-al (aldol condensation)

   Ethanal has $\alpha$-hydrogens, so with dilute NaOH two molecules undergo aldol addition. On heating, the aldol loses water:

   $2\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}\stackrel{\text{dil. NaOH}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ (3-hydroxybutanal)

   $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}\underset{-{\mathrm{H}}_2\mathrm{O}}{\overset{\Delta }{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHCHO}$ (but-2-en-1-al)

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2. (i)(2) Propanoic acid → 2-chloropropanoic acid (Hell–Volhard–Zelinsky reaction)

   A carboxylic acid with an $\alpha$-hydrogen is chlorinated at the $\alpha$-carbon by $\mathrm{C}{\mathrm{l}}_2$ with red phosphorus, followed by hydrolysis:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}\underset{(ii){\mathrm{H}}_2\mathrm{O}}{\overset{(i)\mathrm{C}{\mathrm{l}}_2/red\mathrm{P}}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{Cl})\mathrm{COOH}$

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3. (ii) Identifying B

   B gives a positive Fehling test, so it is an aliphatic aldehyde. It also gives the iodoform test with ${\mathrm{I}}_2$/NaOH, so it has a $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group. The only aldehyde with both features is ethanal.

   **B = $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$ (ethanal)**

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4. Identifying C

   C does not give the Fehling test, so it is a ketone. It forms iodoform, so it is a methyl ketone. A has 5 carbons and B uses 2 of them, so C has 3 carbons.

   **C = $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$ (propanone / acetone)**

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5. Identifying A

   Joining the two carbonyl carbons of B and C by a C=C bond gives the alkene:

   **A = $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CHC}{\mathrm{H}}_3$ (2-methylbut-2-ene)**, formula ${\mathrm{C}}_5{\mathrm{H}}_{10}$.

   Check by ozonolysis: $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CHC}{\mathrm{H}}_3\underset{(ii)Zn/{\mathrm{H}}_2\mathrm{O}}{\overset{(i){\mathrm{O}}_3}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$

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Answer: (i)(1) $2\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}\stackrel{\text{dil. NaOH}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}\stackrel{\Delta ,-{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHCHO}$; (i)(2) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}\stackrel{\mathrm{C}{\mathrm{l}}_2/red\mathrm{P},{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CHClCOOH}$; (ii) A = $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CHC}{\mathrm{H}}_3$ (2-methylbut-2-ene), B = $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$, C = $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$

1. (i)(1) Butanone vs butanal: iodoform test

   Warm each compound with ${\mathrm{I}}_2$ and NaOH solution.

   • Butanone, $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, is a methyl ketone, so it gives a yellow precipitate of iodoform:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3+3{\mathrm{I}}_2+4NaOH\to \mathrm{CH}{\mathrm{I}}_3\downarrow +\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COONa}+3NaI+3{\mathrm{H}}_2\mathrm{O}$

   • Butanal, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$, has no $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group, so no yellow precipitate forms.

   (Tollens' test also works: butanal gives a silver mirror and butanone does not.)

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2. (i)(2) Ethanal vs ethanoic acid: sodium hydrogencarbonate test

   Add $\mathrm{NaHC}{\mathrm{O}}_3$ solution to each.

   • Ethanoic acid gives brisk effervescence of $\mathrm{C}{\mathrm{O}}_2$:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{NaHC}{\mathrm{O}}_3\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2\uparrow$

   • Ethanal gives no effervescence.

   (Alternatively, ethanal gives a silver mirror with Tollens' reagent and ethanoic acid does not.)

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3. (ii) Oxime of acetone

   Acetone condenses with hydroxylamine ($\mathrm{N}{\mathrm{H}}_2\mathrm{OH}$) with loss of water:

   $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{O}+{\mathrm{H}}_2\mathrm{N}-\mathrm{OH}\to (\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{N}-\mathrm{OH}+{\mathrm{H}}_2\mathrm{O}$

   The oxime (acetoxime) is $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{N}-\mathrm{OH}$.

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4. (iii) Identifying A and B

   • $\mathrm{PC}{\mathrm{l}}_5$ converts the acid into its acid chloride. **A = $\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}$** (ethanoyl chloride).
   • ${\mathrm{H}}_2$ over Pd–$\mathrm{BaS}{\mathrm{O}}_4$ (Rosenmund reduction) reduces the acid chloride to an aldehyde. **B = $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$** (ethanal).
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5. Identifying C and D

   • $\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}$ adds to ethanal, and hydrolysis then gives a secondary alcohol:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}\underset{(\mathrm{ii}){\mathrm{H}}_3{\mathrm{O}}^{+}}{\overset{(\mathrm{i})\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}}{\to }}(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHOH}$. C = propan-2-ol.

   • $\mathrm{LiAl}{\mathrm{H}}_4$ reduces the aldehyde to a primary alcohol. **D = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$** (ethanol).
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Answer: (i)(1) Iodoform test: butanone gives a yellow ppt of $\mathrm{CH}{\mathrm{I}}_3$, butanal does not; (i)(2) $\mathrm{NaHC}{\mathrm{O}}_3$ test: ethanoic acid gives $\mathrm{C}{\mathrm{O}}_2$ effervescence, ethanal does not; (ii) $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{N}-\mathrm{OH}$; (iii) A = $\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}$, B = $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$, C = $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHOH}$, D = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$