CBSE Class 12 Chemistry: 2023 board paper, set 56/5/1

1. A chiral carbon carries four different groups. In (B) the starred carbon has $\mathrm{C}{\mathrm{H}}_3$, H, Cl and $\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$, which are all different, so it is chiral.

   1 mark

2. In (A) and (C) the starred carbon carries two identical groups ($\mathrm{C}{\mathrm{H}}_3$ or $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$), and in (D) it carries two Br atoms, so none of these is chiral.

Answer: (B)

1. 2-Methylbutan-2-ol is a tertiary alcohol, so the carbon bearing –OH has no α-hydrogen. It therefore resists oxidation under normal conditions.

   1 mark

Answer: (C) 2-Methylbutan-2-ol

1. The couple with the lower reduction potential acts as the anode. So Sn is oxidised to $\mathrm{S}{\mathrm{n}}^{2+}$, and $\mathrm{F}{\mathrm{e}}^{3+}$ is reduced to $\mathrm{F}{\mathrm{e}}^{2+}$.

2. $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.77-(-0.14)=0.91$ V

   1 mark

Answer: (B)

1. In concentrated brine, $\mathrm{C}{\mathrm{l}}^{-}$ is oxidised at the anode in preference to $\mathrm{O}{\mathrm{H}}^{-}$ (because of overpotential), giving $\mathrm{C}{\mathrm{l}}_2$.

2. At the cathode, ${\mathrm{H}}^{+}$ (from water) is reduced to ${\mathrm{H}}_2$ rather than $\mathrm{N}{\mathrm{a}}^{+}$ to Na. So reactions I and IV occur.

   1 mark

Answer: (B) I and IV

1. Transition metals can easily change their oxidation state. This lets them form unstable intermediates with reactants, which lowers the activation energy, so they act as catalysts.

   1 mark

Answer: (D) Variable oxidation states

1. $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}$ is made of two $\mathrm{Cr}{\mathrm{O}}_4$ tetrahedra that share one bridging oxygen.

2. The six terminal Cr–O bonds are equivalent in length. The two bridging Cr–O–Cr bonds are longer.

   1 mark

Answer: (B)

1. Mole fractions: $x_A=\frac{1}{1+2}=\frac{1}{3}$, $x_B=\frac{2}{3}$.

2. Ideal (Raoult's law) vapour pressure: $p=p_A^0{x}_A+p_B^0{x}_B=45\times \frac{1}{3}+30\times \frac{2}{3}=15+20=35$ torr.

3. The observed vapour pressure (40 torr) is greater than the ideal value (35 torr), so the solution shows positive deviation from Raoult's law. Answer: (b).

   1 mark

Answer: (b) shows positive deviation

1. $\mathrm{LiAl}{\mathrm{H}}_4$ reduces aromatic nitro compounds to azo compounds, not to aniline. So it is not a good choice.

   1 mark

Answer: (A) LiAlH₄

1. $K_b$ is a characteristic of the solvent alone. Changing the molality does not change it.

   1 mark

Answer: (D) unchanged

1. Sucrose is dextrorotatory, but its hydrolysis mixture of glucose and fructose is laevorotatory. Because the sign of rotation reverses, the hydrolysis is called inversion.

   1 mark

Answer: (A) inversion

1. $-\mathrm{N}{\mathrm{O}}_2$ has the strongest –I effect of the substituents given. It best stabilises the carboxylate ion, making ${\mathrm{O}}_2\mathrm{NC}{\mathrm{H}}_2\mathrm{COOH}$ the strongest acid, with the lowest $p{K}_a$.

   1 mark

Answer: (B)

1. The Apollo space programme used ${\mathrm{H}}_2–{\mathrm{O}}_2$ fuel cells for electrical power. The water they produced was used for drinking.

   1 mark

Answer: (C) H₂–O₂ fuel cell

1. Experiments 1 and 2: doubling [B] leaves the rate unchanged, so the order w.r.t. B is 0.

2. Experiments 2 and 3: doubling [A] increases the rate 4 times, so the order w.r.t. A is 2.

   1 mark

Answer: (C) A second, B zero

1. $\mathrm{N}{\mathrm{i}}^{2+}$ is $3d^8$. $\mathrm{C}{\mathrm{l}}^{-}$ is a weak-field ligand, so the complex is tetrahedral ($s{p}^3$) with 2 unpaired electrons.

2. $\mu =\sqrt{n(n+2)}=\sqrt{8}=2.82$ BM

   1 mark

Answer: (B) 2.82 BM

1. The Assertion is true: proteins are polymers of α-amino acids joined by peptide bonds.

2. The Reason is false: a tetrapeptide has 4 amino acids linked by 3 peptide bonds, not 4.

   1 mark

Answer: (C)

1. For a zero order reaction, rate $=k[R{]}^0=k$, so the rate is independent of concentration. Hence k has the same unit as rate, mol L⁻¹ s⁻¹.

   1 mark

2. Both statements are true and R explains A. Option (B) is also accepted by the scheme.

Answer: (A) (scheme also accepts B)

1. HVZ halogenation needs an α-hydrogen. Acetic acid has α-hydrogens, but formic acid has none, so the Assertion is true.

2. Acetic acid is indeed weaker than formic acid (because of the +I effect of $\mathrm{C}{\mathrm{H}}_3$), so the Reason is true. However, this is not why HVZ halogenation occurs.

   1 mark

Answer: (B)

1. The trans isomer has a mirror plane containing the two ox ligands, so it is achiral. The Assertion is false.

2. Octahedral complexes with didentate ligands (for example cis isomers) commonly show optical isomerism, so the Reason is true.

   1 mark

Answer: (D)

1. (i) Using $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$: a reaction is spontaneous only when $\Delta G^{\circ }$ is negative. Since $n$ and $F$ are positive, this requires $E_{cell}^{\circ }$ to be positive.

   1 mark

2. (ii) Faraday's first law: the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte, i.e. $w=ZQ=ZIt$, where $Z$ is the electrochemical equivalent.

   1 mark

Answer: (i) $E_{cell}^{\circ }$ positive, $\Delta G^{\circ }$ negative; (ii) $w\propto Q$, i.e. $w=ZIt$

1. Cell reaction: $\mathrm{Fe}(\mathrm{s})+2{\mathrm{H}}^{+}(\mathrm{aq})\to \mathrm{F}{\mathrm{e}}^{2+}(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$, so $n=2$.

   Nernst equation at 298 K:
   $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}\log \frac{[\mathrm{F}{\mathrm{e}}^{2+}]}{[{\mathrm{H}}^{+}{]}^2}$

   ½ mark

2. Substitute the values:
   $E_{cell}=0.44-\frac{0.059}{2}\log \frac{0.01}{(1{)}^2}=0.44-0.0295\log 10^{-2}$

   1 mark

3. $E_{cell}=0.44-0.0295\times (-2)=0.44+0.059=0.499\text{V}$

   ½ mark

Answer: $E_{cell}=0.499$ V

1. **Rate constant $k$ increases** with rise in temperature.

   ½ mark

2. Justification: by the Arrhenius equation $k=A{e}^{-E_a/RT}$, increasing $T$ makes $-E_a/RT$ less negative, so $k$ increases; more molecules acquire energy ≥ $E_a$, so the rate (and $k$) rises.

   ½ mark

3. **Activation energy $E_a$ remains unchanged** (no effect of temperature).

   ½ mark

4. Justification: $E_a$ is the energy difference between the activated complex and the reactants; it depends only on the nature of the reactants (and catalyst), not on temperature.

   ½ mark

Answer: $k$ increases; $E_a$ remains unchanged

1. $\mathrm{N}{\mathrm{H}}_4^{+}$ cannot act as a ligand.

   1 mark

2. Reason: a ligand must donate a lone pair to the metal (Lewis base). In $\mathrm{N}{\mathrm{H}}_4^{+}$ the lone pair of N is already used in forming the fourth N–H bond, so no lone pair is available for donation. $\mathrm{O}{\mathrm{H}}^{-}$, $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ and ${\mathrm{H}}_2\mathrm{O}$ all have lone pairs.

   1 mark

Answer: $\mathrm{N}{\mathrm{H}}_4^{+}$ — it has no lone pair of electrons

1. $\mathrm{N}{\mathrm{O}}_2^{-}$ is an ambidentate ligand: it can bind through N (nitrito-N) or through O (nitrito-O, –ONO). In the given formula it is written as N-bonded $\mathrm{N}{\mathrm{O}}_2$, so the linkage isomer has the ligand bonded through oxygen: $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{ONO})]\mathrm{C}{\mathrm{l}}_2$.

   1 mark

2. IUPAC name of the linkage isomer: Pentaamminenitrito-O-cobalt(III) chloride.

   1 mark

Answer: Pentaamminenitrito-O-cobalt(III) chloride, $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{ONO})]\mathrm{C}{\mathrm{l}}_2$

1. Boiling point: o-dichlorobenzene is polar (the two C–Cl dipoles do not cancel), while in p-dichlorobenzene they cancel (μ = 0). The stronger dipole–dipole forces in the ortho isomer give it the higher boiling point.

   1 mark

2. Melting point: p-dichlorobenzene is more symmetrical, so its molecules pack more closely and fit better in the crystal lattice; more energy is needed to break the lattice, so it has the higher melting point.

   1 mark

Answer: o-isomer: higher dipole moment → higher b.p.; p-isomer: symmetrical, better packing → higher m.p.

1. (a) On losing ${\mathrm{H}}^{+}$, phenol forms phenoxide ion whose negative charge is delocalised over the benzene ring by resonance, so it is stabilised. Cyclohexoxide ion has no resonance and its charge is localised on O (further destabilised by the +I effect of the alkyl ring). Hence phenol ionises more readily and is more acidic.

   1 mark

2. (b) **Neutral $\mathrm{FeC}{\mathrm{l}}_3$ test:** phenol gives a violet colour with neutral ferric chloride; cyclohexanol gives no such colour. (Alternatively, bromine water gives a white precipitate of 2,4,6-tribromophenol with phenol but not with cyclohexanol.)

   1 mark

Answer: (a) Phenoxide ion is resonance-stabilised; (b) neutral FeCl₃ — violet colour with phenol only

1. (i) Sulphanilic acid is p-aminobenzenesulphonic acid. The acidic $-\mathrm{S}{\mathrm{O}}_3\mathrm{H}$ protonates the basic $-\mathrm{N}{\mathrm{H}}_2$, giving the zwitter ion:
   ${}^{+}{\mathrm{H}}_3\mathrm{N}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{S}{\mathrm{O}}_3^{-}$ (benzene ring with $-\mathrm{N}{\mathrm{H}}_3^{+}$ and $-\mathrm{S}{\mathrm{O}}_3^{-}$ para to each other).

   1 mark

2. (ii) By acetylation of aniline with acetic anhydride (or acetyl chloride) in pyridine to form acetanilide, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{)}_2\mathrm{O}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NHCOC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$. The N lone pair is now also delocalised onto the C=O group, so the ring is less activated; after the desired substitution, hydrolysis regenerates the $-\mathrm{N}{\mathrm{H}}_2$ group.

   1 mark

Answer: (i) ${}^{+}{\mathrm{H}}_3\mathrm{N}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{S}{\mathrm{O}}_3^{-}$ (para); (ii) protect –NH₂ by acetylation (acetanilide)

1. (i) Ethanol reduces the diazonium salt to benzene:
   ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to {\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{N}}_2+\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}+\mathrm{HCl}$
   Main product: benzene.

   1 mark

2. (ii) Step-up by one carbon via the nitrile:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Br}\stackrel{\mathrm{KCN}(alc.)}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CN}\stackrel{\mathrm{LiAl}{\mathrm{H}}_4\text{or}{\mathrm{H}}_2/Ni}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$
   (bromoethane → propanenitrile → propan-1-amine)

   1 mark

Answer: (i) Benzene (with N₂, CH₃CHO, HCl); (ii) CH₃CH₂Br → (KCN) CH₃CH₂CN → (LiAlH₄) CH₃CH₂CH₂NH₂

1. Glucose adds HCN across its C=O group to form a cyanohydrin:
   $\mathrm{HOC}{\mathrm{H}}_2(\mathrm{CHOH}{)}_4\mathrm{CHO}+\mathrm{HCN}\to \mathrm{HOC}{\mathrm{H}}_2(\mathrm{CHOH}{)}_4\mathrm{CH}(\mathrm{OH})\mathrm{CN}$ (glucose cyanohydrin)

   1 mark

2. This addition reaction confirms the presence of a carbonyl group (>C=O), which in glucose is the aldehydic (–CHO) group.

   1 mark

Answer: Glucose cyanohydrin formed; confirms carbonyl (aldehyde) group

1. (a) Rate of reaction $=-\frac{1}{2}\frac{d[{\mathrm{N}}_2{\mathrm{O}}_5]}{dt}$ (coefficient of ${\mathrm{N}}_2{\mathrm{O}}_5$ is 2).

   ½ mark

2. $=\frac{1}{2}\times 1.4\times 10^{-3}=0.7\times 10^{-3}=7\times 10^{-4}{\text{mol L}}^{-1}{\text{s}}^{-1}$

   ½ mark

3. (b) For a first order reaction: $t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}$

   ½ mark

4. For 99% completion, $[R]=\frac{[R{]}_0}{100}$: $t_{99\%}=\frac{2.303}{k}\log 100=\frac{2.303}{k}\times 2$

   ½ mark

5. For 90% completion, $[R]=\frac{[R{]}_0}{10}$: $t_{90\%}=\frac{2.303}{k}\log 10=\frac{2.303}{k}\times 1$

   ½ mark

6. $\frac{t_{99\%}}{t_{90\%}}=\frac{\log 100}{\log 10}=\frac{2}{1}$, hence $t_{99\%}=2t_{90\%}$.

   ½ mark

Answer: (a) $7\times 10^{-4}$ mol L⁻¹ s⁻¹; (b) $t_{99\%}=2t_{90\%}$ proved

1. (a) With a strong field ligand (${\Delta }_0>P$), pairing in the lower $t_{2g}$ set is preferred, so the $d^5$ configuration is $t_{2g}^5{e}_g^0$ (low spin, one unpaired electron).

   1 mark

2. (b) $[\mathrm{Ni}(\mathrm{CO}{)}_4]$: Ni is in 0 state, $3d^{8}4s^2$. CO is a strong field ligand, so the 4s electrons are pushed into 3d, giving $3d^{10}$. The empty 4s and three 4p orbitals form **$s{p}^3$** hybrids → tetrahedral.

   ½ mark

3. $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$: Ni is $\mathrm{N}{\mathrm{i}}^{2+}$, $3d^8$. Strong field $\mathrm{C}{\mathrm{N}}^{-}$ pairs the two unpaired 3d electrons, leaving one 3d orbital empty; one 3d, the 4s and two 4p orbitals form **$ds{p}^2$** hybrids → square planar.

   ½ mark

4. In both complexes all electrons are paired (no unpaired electrons), so both are diamagnetic despite their different geometries.

   1 mark

Answer: (a) $t_{2g}^5{e}_g^0$; (b) [Ni(CO)₄] sp³ (tetrahedral), [Ni(CN)₄]²⁻ dsp² (square planar); both have all electrons paired → diamagnetic

1. (a) Sandmeyer reaction: the diazonium group is replaced by Cl, Br or CN using the corresponding cuprous salt:
   ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}\stackrel{\mathrm{C}{\mathrm{u}}_2\mathrm{C}{\mathrm{l}}_2/HCl}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Cl}+{\mathrm{N}}_2$
   (similarly $\mathrm{C}{\mathrm{u}}_2\mathrm{B}{\mathrm{r}}_2/HBr$ gives ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}$; $CuCN/KCN$ gives ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CN}$).

   1 mark

2. (b) In aqueous solution basicity depends on the +I effect of alkyl groups and on solvation (H-bonding) of the protonated amine with water. +I effect alone would make the tertiary amine strongest.

   1 mark

3. $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{N}{\mathrm{H}}_2^{+}$ has two N–H hydrogens to form H-bonds with water and is better hydrated (stabilised) than $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}{\mathrm{H}}^{+}$, which has only one N–H and is crowded by three methyl groups (steric hindrance to solvation). The combined effect makes $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{NH}$ more basic.

   1 mark

Answer: (a) C₆H₅N₂⁺Cl⁻ + Cu₂Cl₂/HCl → C₆H₅Cl + N₂; (b) greater solvation (H-bonding) of (CH₃)₂NH₂⁺ with +I effect makes (CH₃)₂NH more basic

1. (a) In glucose pentaacetate the –OH at C-1 is acetylated, so the ring cannot open to give a free –CHO group. Without a free aldehyde group, it cannot form an oxime with hydroxylamine.

   1 mark

2. (b) Amino acids contain both an acidic (–COOH) and a basic (–NH₂) group, which react internally to form a dipolar zwitter ion ${\mathrm{H}}_3{\mathrm{N}}^{+}-\mathrm{CHR}-\mathrm{CO}{\mathrm{O}}^{-}$. Being ionic, they show salt-like properties (high m.p., water solubility).

   1 mark

3. (c) Water-soluble vitamins (B-complex, C) are readily excreted in urine and cannot be stored in the body (except B₁₂), so they must be supplied regularly through diet.

   1 mark

4. (d) The bases pair specifically through hydrogen bonds — adenine with thymine (2 H-bonds) and guanine with cytosine (3 H-bonds). Hence the base sequence of one strand fixes the sequence of the other, making them complementary. (Any three of (a)–(d) earn full 3 marks.)

Answer: (a) no free –CHO; (b) zwitter ion formation; (c) excreted in urine, not stored; (d) specific base pairing A–T, G–C

1. (i) In phenol the lone pair on O is in conjugation with the benzene ring, giving the C–O bond partial double bond character by resonance; also the carbon is $s{p}^2$ hybridised (more s-character) compared with $s{p}^3$ carbon in methanol. Hence C–O bond in phenol (≈136 pm) is shorter than in methanol (≈142 pm).

   1 mark

2. (ii) n-Butane < Ethoxyethane < Butanal < Butanol
   (only van der Waals forces < weak dipole forces < stronger dipole–dipole of C=O < intermolecular H-bonding in butanol)

   1 mark

3. (iii) Anisole is cleaved by hot conc. HI (or HBr):
   ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{O}-\mathrm{C}{\mathrm{H}}_3+\mathrm{HI}\stackrel{\mathrm{\Delta }}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{I}$
   The O–CH₃ bond breaks (not the O–aryl bond, which has partial double bond character), giving phenol.

   1 mark

Answer: (i) resonance/partial double bond and sp² carbon in phenol; (ii) n-butane < ethoxyethane < butanal < butanol; (iii) C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I

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

   ½ mark

2. Step 2 – Nucleophilic attack ($S_{N}2$) by a second ethanol molecule on the protonated alcohol, with loss of water:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\ddot{\mathrm{O}}\mathrm{H}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}(\mathrm{H})-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}$

   1 mark

3. Step 3 – Deprotonation:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}(\mathrm{H})-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3+{\mathrm{H}}^{+}$

   ½ mark

4. (ii) Hydroboration–oxidation: diborane adds to an alkene, and the trialkylborane is oxidised by alkaline ${\mathrm{H}}_2{\mathrm{O}}_2$ to an alcohol (anti-Markovnikov product):
   $3C{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+\mathrm{B}{\mathrm{H}}_3\to (\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{B}\stackrel{3{\mathrm{H}}_2{\mathrm{O}}_2/O{\mathrm{H}}^{-}}{\to }3C{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{B}(\mathrm{OH}{)}_3$
   (propene → propan-1-ol)

   1 mark

Answer: (i) protonation → SN2 attack by ethanol → deprotonation gives ethoxyethane; (ii) propene + B₂H₆, then H₂O₂/OH⁻ → propan-1-ol

1. (a) In $S_{N}1$ the intermediate carbocation is $s{p}^2$ hybridised and planar, so the nucleophile can attack equally from both faces, giving equal amounts of the two enantiomers (retention and inversion) — a racemic mixture.

   1 mark

2. (b) In ethanol the ethyl group exerts a +I (electron-releasing) effect, which reduces the polarity of the O–H bond; the large non-polar hydrocarbon part also lowers overall polarity. Hence ethanol (dielectric constant 24) is less polar than water (79).

   1 mark

3. (c)(i) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\mathrm{I}$ is more reactive: the C–I bond is weaker and ${\mathrm{I}}^{-}$ is a better leaving group than $\mathrm{C}{\mathrm{l}}^{-}$.

   1 mark

4. (c)(ii) (Chloromethyl)cyclohexane is more reactive: it is a primary halide with less steric hindrance at the reacting carbon than the secondary chlorocyclohexane, so backside attack is easier.

   1 mark

5. OR (c) [alternative, 2 marks]
   (i) 1-Bromopentane < 2-Bromopentane < 2-Bromo-2-methylbutane (1°< 2°< 3°, carbocation stability) [1 mark]
   (ii) 1-Bromo-3-methylbutane < 2-Bromo-3-methylbutane < 2-Bromo-2-methylbutane (1°< 2°< 3°) [1 mark]

Answer: (a) planar carbocation attacked from both sides; (b) +I effect of ethyl group; (c)(i) CH₃CH₂I, (ii) (chloromethyl)cyclohexane; OR (c)(i) 1-bromopentane < 2-bromopentane < 2-bromo-2-methylbutane, (ii) 1-bromo-3-methylbutane < 2-bromo-3-methylbutane < 2-bromo-2-methylbutane

1. (a) Conductivity is the conductance of 1 cm³ of solution; on dilution the number of ions per unit volume decreases, so conductivity decreases.

   1 mark

2. (b) $\alpha =\frac{{\wedge }_m^c}{{\wedge }_m^o}=\frac{141.0}{150.0}=0.94$

   1 mark

3. (c) ${\wedge }_m$ values for HCl would be more than those for KCl at the same concentration.

   1 mark

4. Justification: ${\mathrm{H}}^{+}$ has a much higher ionic mobility (limiting molar conductivity ≈ 349.6 S cm² mol⁻¹) than ${\mathrm{K}}^{+}$ (≈ 73.5 S cm² mol⁻¹), because it moves by proton-hopping through H-bonded water molecules; the anion ($\mathrm{C}{\mathrm{l}}^{-}$) is the same.

   1 mark

5. OR (c) [alternative, 2 marks]
   Similarity: for both KCl and $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$, ${\wedge }_m$ increases (and κ decreases) on dilution. [1 mark]
   Difference: for KCl (strong electrolyte) ${\wedge }_m$ increases only slightly and linearly with $\sqrt{c}$, so ${\wedge }_m^o$ can be found by extrapolation; for $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ (weak electrolyte) ${\wedge }_m$ is much lower and rises steeply at high dilution, so ${\wedge }_m^o$ cannot be obtained by extrapolation. [1 mark]

Answer: (a) fewer ions per unit volume; (b) α = 0.94; (c) more, H⁺ more mobile than K⁺; OR (c) similarity: Λm increases on dilution; difference: steep rise for CH₃COOH, gradual for KCl

1. (i) Elevation of boiling point is a colligative property depending on the number of solute particles. NaCl dissociates into $\mathrm{N}{\mathrm{a}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$ ($i\approx 2$), while glucose does not dissociate ($i=1$). So 1 M NaCl has about twice as many particles, giving larger $\Delta T_b$ and a higher boiling point.

   1 mark

2. (ii) Raoult's law: $\frac{p^{\circ }-p}{p^{\circ }}=x_2=\frac{n_2}{n_1+n_2}$

   ½ mark

3. $p=0.9p^{\circ }\Rightarrow \frac{p^{\circ }-0.9p^{\circ }}{p^{\circ }}=0.1$; $n_1=\frac{78}{78}=1$ mol (benzene, M = 78 g mol⁻¹), $n_2=\frac{w}{50}$
   $0.1=\frac{w/50}{1+w/50}$

   1 mark

4. $0.1+\frac{0.1w}{50}=\frac{w}{50}\Rightarrow 0.9\frac{w}{50}=0.1\Rightarrow w=\frac{5}{0.9}=5.55\text{g}$ (≈ 5 g if the dilute-solution approximation $x_2\approx n_2/n_1$ is used)

   ½ mark

5. (iii) $\mathrm{MgC}{\mathrm{l}}_2\to \mathrm{M}{\mathrm{g}}^{2+}+2C{\mathrm{l}}^{-}$, so $i=3$.

   ½ mark

6. $\Delta T_b=i{K}_{b}m=i{K}_b\times \frac{w_2\times 1000}{M_2\times w_1}$

   ½ mark

7. $\Delta T_b=3\times 0.512\times \frac{10\times 1000}{95\times 200}$

   ½ mark

8. $\Delta T_b=3\times 0.512\times 0.526=0.81\text{K}$

   ½ mark

Answer: (i) NaCl gives more particles (i ≈ 2); (ii) mass of X = 5.55 g; (iii) ΔTb = 0.81 K

1. (i) In benzene (non-polar), ethanoic acid molecules associate through hydrogen bonding to form dimers, $2C{\mathrm{H}}_3\mathrm{COOH}\rightleftharpoons (\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}{)}_2$. The number of particles is nearly halved, so $i\approx 0.5$.

   1 mark

2. (ii) ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4\to 2{\mathrm{K}}^{+}+\mathrm{S}{\mathrm{O}}_4^{2-}$, so $i=3$.

   ½ mark

3. $\pi =iCRT=i\frac{n}{V}RT$

   ½ mark

4. $\pi =3\times \frac{2.32\times 10^{-2}}{174\times 2}\times 0.082\times 298$

   ½ mark

5. $\pi =3\times 6.67\times 10^{-5}\times 24.44=4.89\times 10^{-3}\text{atm}$

   ½ mark

6. (iii) $\Delta T_f=K_f\times \frac{w_2\times 1000}{M_2\times w_1}$

   ½ mark

7. $0.512=5.12\times \frac{25.6\times 1000}{M_2\times 1000}$

   ½ mark

8. $M_2=\frac{5.12\times 25.6}{0.512}=256{\text{g mol}}^{-1}$

   ½ mark

9. $32x=256\Rightarrow x=8$; formula of sulphur is ${\mathrm{S}}_8$.

   ½ mark

Answer: (i) dimerisation of ethanoic acid; (ii) π = 4.88–4.89 × 10⁻³ atm; (iii) S₈

1. (i) Cannizzaro reaction: an aldehyde with no α-hydrogen, heated with concentrated alkali, disproportionates. One molecule is oxidised to the carboxylate salt and the other is reduced to the alcohol:

   $2\mathrm{HCHO}+\text{conc. }\mathrm{KOH}\stackrel{\Delta }{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{OH}+\mathrm{HCOOK}$

   Other examples are also accepted, e.g. $2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}+\text{conc. NaOH}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}$.

   1 mark

2. (ii) Carboxylic acid molecules are held together by strong intermolecular hydrogen bonds and exist as dimers. Aldehydes and ketones cannot form H-bonds among themselves; they have only weaker dipole–dipole attractions. Less energy is therefore needed to separate their molecules, so their boiling points are lower.

   1 mark

3. **(iii) Identifying A ($C_5{H}_8{O}_2$):**

   • Degree of unsaturation $=\frac{(2\times 5+2)-8}{2}=2$, and A forms a dioxime, so A has two C=O groups.
   • Wolff–Kishner reduction gives n-pentane, so the carbon chain is straight (unbranched).
   • A positive Tollens' test shows a –CHO group.
   • A positive iodoform test shows a $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group.

   Hence **A = $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ (4-oxopentanal)**.

   Check by Wolff–Kishner: $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}\underset{\text{(ii) KOH/NaOH, ethylene glycol, }\Delta }{\overset{\text{(i) }\mathrm{N}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$

   1 mark

4. Iodoform test (the methyl ketone end reacts):

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}+3{\mathrm{I}}_2+4\mathrm{NaOH}\to \mathrm{CH}{\mathrm{I}}_3\downarrow \text{(yellow)}+\mathrm{NaOOCC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}+3\mathrm{NaI}+3{\mathrm{H}}_2\mathrm{O}$

   1 mark

5. Tollens' test (the aldehyde end is oxidised):

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}+2[\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{)}_2{]}^{+}+3\mathrm{O}{\mathrm{H}}^{-}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CO}{\mathrm{O}}^{-}+2\mathrm{Ag}\downarrow \text{(silver mirror)}+4\mathrm{N}{\mathrm{H}}_3+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) 2HCHO + conc. KOH → CH3OH + HCOOK; (ii) acids have strong intermolecular H-bonding, aldehydes/ketones only dipole–dipole attraction; (iii) A = CH3COCH2CH2CHO (4-oxopentanal), with iodoform and Tollens' equations as shown

1. (i) Taking the intended pair as ethanal and ethanoic acid: 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, while ethanoic acid does not.

   1 mark

2. (ii) The C=O group is strongly electron-withdrawing (–I effect), which weakens the α C–H bond. The carbanion (enolate ion) formed after losing the α-hydrogen is stabilised by resonance, because the negative charge is delocalised onto the oxygen:

   ${}^{-}\mathrm{C}{\mathrm{H}}_2-\mathrm{CHO}\leftrightarrow \mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-{\mathrm{O}}^{-}$

   Hence α-hydrogens are acidic.

   1 mark

3. (iii)(1) The sodium salt of B gives methane on heating with soda lime (decarboxylation), so B is ethanoic acid:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+\mathrm{NaOH}\stackrel{\mathrm{CaO},\mathrm{\Delta }}{\to }\mathrm{C}{\mathrm{H}}_4+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$

   C on oxidation also gives B, so C is ethanol. A ($C_4{H}_8{O}_2$) is the ester formed from these two.

   **A = $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5$ (ethyl ethanoate)**

   1 mark

4. Acid hydrolysis of A gives B and C:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}(\mathrm{B})+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(\mathrm{C})$

   **B = $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ (ethanoic acid)**

   ½ mark

5. Oxidation of C gives B:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{KMn}{\mathrm{O}}_4/{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   **C = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ (ethanol)**

   ½ mark

6. (iii)(2) B (ethanoic acid) has the higher boiling point.

   ½ mark

7. Reason: carboxylic acid molecules are associated more extensively through stronger intermolecular hydrogen bonding, forming dimers. This hydrogen bonding is stronger than that in ethanol.

   ½ mark

Answer: (i) NaHCO3 test: ethanoic acid gives CO2 effervescence, ethanal does not; (ii) –I effect of C=O and resonance-stabilised enolate; (iii) A = CH3COOC2H5, B = CH3COOH, C = C2H5OH; B has the higher b.p. due to stronger H-bonding (dimer formation)

1. (a) Actinoids show a much wider range of oxidation states than lanthanoids, because the 5f, 6d and 7s levels are close in energy. Most actinoids are also radioactive, and many have short half-lives, so their study is difficult. Hence their chemistry is more complicated.

   1 mark

2. (b) Completed and balanced equation:

   $3\mathrm{Mn}{\mathrm{O}}_4^{2-}+4{\mathrm{H}}^{+}\to 2\mathrm{Mn}{\mathrm{O}}_4^{-}+\mathrm{Mn}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

3. Justification: the oxidation state of Mn in $\mathrm{Mn}{\mathrm{O}}_4^{2-}$ is +6.

   • It rises to +7 in $\mathrm{Mn}{\mathrm{O}}_4^{-}$ (oxidation).
   • It falls to +4 in $\mathrm{Mn}{\mathrm{O}}_2$ (reduction).

   The same species is simultaneously oxidised and reduced, so this is a disproportionation reaction.

   1 mark

4. (c) Cr ($3d^{5}4s^1$) has six unpaired electrons, the maximum possible in the series. All of these can take part in interatomic metallic bonding, giving very strong metal–metal bonds and hence the highest melting point in the first series.

   1 mark

5. Mn ($3d^{5}4s^2$) has a stable half-filled $3d^5$ configuration. Its d-electrons are held tightly and take little part in metallic bonding, so the metallic bonds are weaker and the melting point drops, as shown by the dip in the graph.

   1 mark

Answer: (a) Wide range of oxidation states and radioactivity of actinoids; (b) 3MnO4²⁻ + 4H⁺ → 2MnO4⁻ + MnO2 + 2H2O, Mn goes from +6 to +7 and +4; (c) Cr has the maximum unpaired electrons and strongest metallic bonding; Mn's stable half-filled d⁵ gives weaker metallic bonding