CBSE Class 12 Chemistry: 2024 board paper, set 56/5/1

1. The primary structure of a protein is the specific order in which amino acids are linked by peptide bonds. Secondary, tertiary and quaternary structures describe folding and the association of subunits. Option (A).

   1 mark

Answer: Primary structure, option (A)

1. When a diazonium group is replaced by –Cl, –Br or –CN using the corresponding Cu(I) salt, the reaction is called Sandmeyer's reaction. Option (A).

   1 mark

Answer: Sandmeyer's reaction, option (A)

1. $(C{H}_3{)}_{2}C=C{H}_2+H^{+}\to (C{H}_3{)}_3{C}^{+}$, a tertiary carbocation (Markovnikov addition).

2. Water then attacks this carbocation and loses $H^{+}$ to give $(C{H}_3{)}_{3}COH$, tert-butyl alcohol. Option (D).

   1 mark

Answer: $(C{H}_3{)}_{2}C=C{H}_2$, option (D)

1. In air and light, chloroform is oxidised to phosgene: $2CHC{l}_3+O_2\to 2COC{l}_2+2HCl$. Option (C).

   1 mark

Answer: Phosgene, option (C)

1. Small atoms such as H, C and N trapped in the voids of the metal lattice make the metal harder and less ductile, and also raise its melting point. Option (A).

   1 mark

Answer: More hard, option (A)

1. Isotonic solutions are solutions that have the same osmotic pressure at a given temperature. Option (C).

   1 mark

Answer: Osmotic pressure, option (C)

1. The $H_2$–$O_2$ fuel cell was used to supply electrical power in the Apollo space programme. Option (B).

   1 mark

Answer: $H_2$–$O_2$ fuel cell, option (B)

1. Rate $=k[A{]}^n$, so $16=4^n$, which gives $n=2$. Option (B).

   1 mark

Answer: Order = 2, option (B)

1. In I (H, CH3, Br, Cl), II (D, I, Br, Cl) and III (D, OH, C2H5, CH3), the starred carbon carries four different groups, so each is chiral.

2. In IV the starred carbon carries two CH3 groups, so it is not chiral. The answer is I, II, III, option (A).

   1 mark

Answer: I, II, III, option (A)

1. The acid strengths are 3-nitrophenol > phenol > water > alcohol. The weakest acid has the strongest conjugate base.

2. The alcohol is the weakest acid, so $R{O}^{-}$ is the strongest base. Its alkyl group (+I effect) also intensifies the negative charge. Option (C).

   1 mark

Answer: $R{O}^{-}$, option (C)

1. On dilution there are fewer ions per unit volume, so conductivity ($\kappa$) decreases.

2. Molar conductivity, ${\Lambda }_m=\kappa \times 1000/c$, increases because the total volume containing 1 mol of electrolyte rises. Option (D).

   1 mark

Answer: Option (D)

1. $N{a}_{2}S{O}_4\cdot 10H_{2}O\to 2N{a}^{+}+S{O}_4^{2-}+10H_{2}O$. The water of crystallisation simply becomes part of the solvent.

2. Three ions are formed, so $i=3$. Option (B).

   1 mark

Answer: i = 3, option (B)

1. Zr (160 pm) and Hf (159 pm) have nearly equal atomic radii.

2. This is because the poorly shielding 4f electrons filled before Hf cause the lanthanoid contraction. Both statements are true and R explains A, so the answer is option (A).

   1 mark

Answer: Option (A)

1. For a zero order reaction, rate $=k[A{]}^0=k$, so both have units of mol L${}^{-1}$ s${}^{-1}$. Assertion is true.

2. Because the rate does not depend on concentration, rate equals $k$, which is why their units are the same. Reason is true and explains the assertion. Option (A).

   1 mark

Answer: Option (A) (scheme also accepts B)

1. $S_{N}2$ is a one-step reaction with backside attack through a transition state, so inversion of configuration occurs. Assertion is true.

2. Carbocations form in $S_{N}1$, not $S_{N}2$, so the Reason is false. Option (C).

   1 mark

Answer: Option (C)

1. The electron-withdrawing $N{O}_2$ group stabilises the phenoxide ion, while the electron-donating $OC{H}_3$ group destabilises it. So p-nitrophenol is the stronger acid and the Assertion is false.

2. The Reason is taken as true (methoxy donating, nitro withdrawing). Option (D).

   1 mark

Answer: Option (D)

1. (a) Kohlrausch's law: The limiting molar conductivity of an electrolyte equals the sum of the individual contributions of its cation and anion: ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$. Here ${\lambda }_{+}^{\circ }$ and ${\lambda }_{-}^{\circ }$ are the limiting molar conductivities of the cation and anion, and ${\nu }_{+}$, ${\nu }_{-}$ are the numbers of each ion per formula unit.

   1 mark

2. (b) Faraday's first law: The mass of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed: $m=ZIt$. Here $m$ is the mass deposited, $Z$ the electrochemical equivalent, $I$ the current and $t$ the time.

   1 mark

Answer: (a) ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$; (b) $m=ZIt$ (mass deposited ∝ charge passed)

1. Monosaccharides (cannot be hydrolysed further): Galactose and Glucose, each $C_6{H}_{12}{O}_6$.

   1 mark

2. Disaccharides (each gives two monosaccharide units on hydrolysis): Lactose (gives galactose + glucose) and Maltose (gives two glucose units).

   1 mark

Answer: Monosaccharides: Galactose, Glucose; Disaccharides: Lactose, Maltose

1. (i) Reduce nitrobenzene with Sn/HCl (or Fe/HCl):
   $C_6{H}_{5}N{O}_2+6[H]\stackrel{Sn/HCl}{\to }{C}_6{H}_{5}N{H}_2+2H_{2}O$

   1 mark

2. (ii) Diazotise aniline with $NaN{O}_2+HCl$ at 273–278 K, then warm the diazonium salt with water:
   $C_6{H}_{5}N{H}_2\stackrel{NaN{O}_2/HCl,273-278K}{\to }{C}_6{H}_5{N}_2^{+}C{l}^{-}\stackrel{H_{2}O,warm}{\to }{C}_6{H}_{5}OH+N_2+HCl$

   1 mark

Answer: (i) Sn/HCl reduction; (ii) diazotisation followed by warming with water

1. (i) Carbylamine test: Warm each amine with chloroform and alcoholic KOH. Ethanamine, a primary amine, gives the foul-smelling ethyl isocyanide; dimethylamine, a secondary amine, gives no such smell.
   $C_2{H}_{5}N{H}_2+CHC{l}_3+3KOH\to C_2{H}_{5}NC+3KCl+3H_{2}O$

   1 mark

2. (ii) Iodobenzene is formed:
   $C_6{H}_5{N}_2^{+}C{l}^{-}+KI\to C_6{H}_{5}I+KCl+N_2$

   1 mark

Answer: (i) Carbylamine test – ethanamine gives foul-smelling isocyanide, dimethylamine does not; (ii) Iodobenzene ($C_6{H}_{5}I$)

1. For a first order reaction, $t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R]}$.

   For 99% completion, $[R]=0.01[R{]}_0$:
   $t_{99\%}=\frac{2.303}{k}\log 100=\frac{2.303}{k}\times 2$

   1 mark

2. For 90% completion, $[R]=0.1[R{]}_0$:
   $t_{90\%}=\frac{2.303}{k}\log 10=\frac{2.303}{k}$

   Dividing, $\frac{t_{99\%}}{t_{90\%}}=2$, so $t_{99\%}=2t_{90\%}$.

   1 mark

Answer: $t_{99\%}=2t_{90\%}$

1. Find the cell constant from the 0.2 M solution:
   $G^{\ast }=\kappa \times R=0.0248{\text{S cm}}^{-1}\times 200\Omega =4.96{\text{cm}}^{-1}$

   ½ mark

2. Conductivity of the 0.05 M solution:
   $\kappa =\frac{G^{\ast }}{R}=\frac{4.96}{620}=0.008{\text{S cm}}^{-1}$

   ½ mark

3. Molar conductivity formula:
   ${\Lambda }_m=\frac{\kappa \times 1000}{c}$

   ½ mark

4. Substitute:
   ${\Lambda }_m=\frac{0.008\times 1000}{0.05}=160{\text{S cm}}^2{\text{ mol}}^{-1}$

   ½ mark

Answer: $\kappa =0.008$ S cm${}^{-1}$; ${\Lambda }_m=160$ S cm${}^2$ mol${}^{-1}$

1. (a) $[Co(N{H}_3{)}_{4}Cl(N{O}_2)]Cl$: Tetraamminechloridonitrito-N-cobalt(III) chloride

   1 mark

2. (b) $[Ni(N{H}_3{)}_6]C{l}_2$: Hexaamminenickel(II) chloride

   1 mark

3. (c) $K_3[Cr(C_2{O}_4{)}_3]$: Potassium trioxalatochromate(III)

   1 mark

4. (d) $[Co(en{)}_{2}B{r}_2{]}^{+}$: Dibromidobis(ethane-1,2-diamine)cobalt(III) ion. This is an alternative part; any three correct names earn the full 3 marks.

Answer: (a) Tetraamminechloridonitrito-N-cobalt(III) chloride; (b) Hexaamminenickel(II) chloride; (c) Potassium trioxalatochromate(III); (d) Dibromidobis(ethane-1,2-diamine)cobalt(III) ion

1. (a) $C{H}_{3}MgBr$ adds to the C=O of ethanal and hydrolysis then gives a secondary alcohol:
   $C{H}_{3}CHO\stackrel{(i)C{H}_{3}MgBr,(ii)H_{2}O}{\to }C{H}_{3}CH(OH)C{H}_3$ (propan-2-ol)

   1 mark

2. (b) Acid-catalysed hydration follows Markovnikov's rule:
   $C{H}_2=CHC{H}_{2}C{H}_3+H_{2}O\stackrel{dil.H_{2}S{O}_4}{\to }C{H}_{3}CH(OH)C{H}_{2}C{H}_3$ (butan-2-ol)

   1 mark

3. (c) Bromine water brominates all the ortho and para positions, giving a white precipitate of 2,4,6-tribromophenol (benzene ring with –OH at C-1 and Br at C-2, C-4, C-6):
   $C_6{H}_{5}OH+3B{r}_2(aq)\to C_6{H}_{2}B{r}_{3}OH+3HBr$

   1 mark

Answer: (a) $C{H}_{3}CH(OH)C{H}_3$; (b) $C{H}_{3}CH(OH)C{H}_{2}C{H}_3$; (c) 2,4,6-tribromophenol

1. (a) Tollens' reagent oxidises only the –CHO group; the ring ketone is unaffected. The product is 4-oxocyclohexane-1-carboxylic acid (formed as its carboxylate, acid after acidification): cyclohexanone ring with –COOH at C-4.

   1 mark

2. (b) A methyl ketone undergoes the haloform (iodoform) reaction with NaOI. The $-COC{H}_3$ group becomes $-CO{O}^{-}N{a}^{+}$, the C=C is untouched, and yellow iodoform separates:
   $C{H}_{3}CH=C(C{H}_3)COC{H}_3\stackrel{NaOI}{\to }C{H}_{3}CH=C(C{H}_3)COONa+CH{I}_3$
   (the acid is 2-methylbut-2-enoic acid)

   1 mark

3. (c) HCN (from NaCN/HCl) adds across the aldehyde C=O to form a cyanohydrin: benzene ring with –COOH and an ortho –CH(OH)CN group, i.e. 2-[cyano(hydroxy)methyl]benzoic acid.

   1 mark

Answer: (a) 4-oxocyclohexane-1-carboxylic acid; (b) $C{H}_{3}CH=C(C{H}_3)COONa$ + $CH{I}_3$; (c) 2-[cyano(hydroxy)methyl]benzoic acid (cyanohydrin)

1. Nernst equation for the cell ($n=2$):
   $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[N{i}^{2+}]}{[A{g}^{+}{]}^2}$

   1 mark

2. Substitute:
   $E_{cell}=1.05-\frac{0.059}{2}\log \frac{0.1}{(0.01{)}^2}=1.05-0.0295\log 10^3$

   1 mark

3. $E_{cell}=1.05-0.0295\times 3=1.05-0.0885=0.9615$ V

   1 mark

Answer: $E_{cell}=0.9615$ V

1. (a) AgCN is mainly covalent, so the carbon of cyanide is tied up in the Ag–C bond. Only the nitrogen lone pair is free to attack the alkyl carbon, so the alkyl isocyanide R–NC is the main product.

   1 mark

2. (b) Allyl chloride ionises to the allyl carbocation $C{H}_2=CH-C{H}_2^{+}$. This ion is resonance-stabilised ($C{H}_2=CH-C{H}_2^{+}\leftrightarrow {}^{+}C{H}_2-CH=C{H}_2$), so it forms easily and the $S_{N}1$ reaction is fast.

   1 mark

3. (c) In haloarenes the halogen lone pair is in resonance with the ring, giving the C–X bond partial double-bond character. The carbon is also $s{p}^2$ hybridised, so the bond is shorter and stronger. In addition, the phenyl cation is unstable and the electron-rich ring repels the nucleophile, so nucleophilic substitution is very difficult.

   1 mark

Answer: (a) Covalent AgCN – only N donates; (b) resonance-stabilised allyl carbocation; (c) partial double-bond character of C–X (resonance, sp² carbon)

1. Arrhenius equation in two-temperature form:
   $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

   1 mark

2. With $k_2/k_1=4$, $T_1=300$ K and $T_2=320$ K:
   $0.60=\frac{E_a}{19.15}\times \frac{20}{300\times 320}$

   1 mark

3. $E_a=\frac{0.60\times 19.15\times 96000}{20}=55152{\text{J mol}}^{-1}\approx 55.15{\text{kJ mol}}^{-1}$

   1 mark

Answer: $E_a\approx 55.15$ kJ mol${}^{-1}$ (55152 J mol${}^{-1}$)

1. A product with Hinsberg's reagent ($C_6{H}_{5}S{O}_{2}Cl$) that is insoluble in alkali shows that X is a secondary amine; the sulphonamide has no acidic N–H. The only 2° amine with formula $C_3{H}_{9}N$ is $C{H}_{3}C{H}_{2}NHC{H}_3$, N-methylethanamine, so X = N-methylethanamine.

   1 mark

2. Reaction:
   $C_6{H}_{5}S{O}_{2}Cl+C{H}_{3}C{H}_{2}NHC{H}_3\to C_6{H}_{5}S{O}_{2}N(C{H}_3)C{H}_{2}C{H}_3+HCl$

   1 mark

3. IUPAC name of the product: N-ethyl-N-methylbenzenesulphonamide

   1 mark

Answer: X = N-methylethanamine ($C{H}_{3}C{H}_{2}NHC{H}_3$); product: N-ethyl-N-methylbenzenesulphonamide

1. (a) An ambidentate ligand has two different donor atoms and can bind to the metal through either one of them. Example: $N{O}_2^{-}$ binds through N (nitrito-N) or through O (nitrito-O); $SC{N}^{-}$ binds through S or N.

   1 mark

2. (b) The two compounds exchange the ion inside and outside the coordination sphere ($C{l}^{-}$/$S{O}_4^{2-}$), so they give different ions in solution. This is ionisation isomerism.

   1 mark

3. (c) Chelate effect: a bidentate or polydentate ligand binds a metal ion through two or more donor atoms, forming a ring (chelate).

   1 mark

4. (c) Chelate complexes are more stable than comparable complexes of monodentate ligands; the greater the chelation, the greater the stability.

   1 mark

Answer: (a) Ligand with two different donor atoms, e.g. $N{O}_2^{-}$; (b) Ionisation isomerism; (c) Ring formation by polydentate ligand; chelation increases stability

1. (a) An ambidentate ligand has two different donor atoms and coordinates through either one of them, e.g. $N{O}_2^{-}$ (N or O donor) or $SC{N}^{-}$ (S or N donor).

   1 mark

2. (b) The two compounds give different ions in solution ($S{O}_4^{2-}$ vs $C{l}^{-}$), so they show ionisation isomerism.

   1 mark

3. (c) Coordination number: each oxalate $C_2{O}_4^{2-}$ is bidentate, so 3 × 2 = 6.

   1 mark

4. (c) Oxidation state: the complex ion is $[Cr(C_2{O}_4{)}_3{]}^{3-}$, so $x+3(-2)=-3$, giving $x=$ +3.

   1 mark

Answer: (a) e.g. $N{O}_2^{-}$; (b) Ionisation isomerism; (c) CN = 6, oxidation state = +3

1. (a) Vitamin $B_6$ is pyridoxine.

   1 mark

2. (b) Deficiency of vitamin K increases blood clotting time.

   1 mark

3. (c) Xerophthalmia (hardening of the cornea) is caused by deficiency of vitamin A.

   1 mark

4. (c) Sources: fish liver oil, carrots (also butter, milk).

   1 mark

Answer: (a) Pyridoxine; (b) Vitamin K; (c) Vitamin A – fish liver oil, carrots

1. (a) Vitamin $B_6$ is pyridoxine.

   1 mark

2. (b) Vitamin K: its deficiency increases blood clotting time.

   1 mark

3. (c) Vitamin C is water soluble, so it is readily excreted in urine and cannot be stored in the body. It must be taken regularly in the diet.

   1 mark

4. (c) Deficiency of vitamin C causes scurvy (bleeding gums).

   1 mark

Answer: (a) Pyridoxine; (b) Vitamin K; (c) Water soluble – excreted in urine; Scurvy

1. (a) Ce(III) is $4f^1$, while Ce(IV) is $4f^0$, the stable noble gas configuration. Ce(III) therefore readily loses one more electron to become Ce(IV).

   1 mark

2. (b) $M{n}^{2+}$ has a half-filled $3d^5$ configuration with extra stability, so reducing $M{n}^{2+}$ to Mn is difficult. Hence $E^{\circ }$ is much more negative than for its neighbours.

   1 mark

3. (c) Zinc has the lowest enthalpy of atomisation. Its $3d^{10}$ orbitals are completely filled, so no unpaired d electrons take part in metallic bonding and the metallic bonding is weak.

   1 mark

4. (d) Acidified sodium chromate changes to sodium dichromate, and the colour changes from yellow to orange:
   $2Cr{O}_4^{2-}+2H^{+}\to C{r}_2{O}_7^{2-}+H_{2}O$

   1 mark

5. (e) Zn, Cd and Hg have completely filled $(n-1)d^{10}$ orbitals with no unpaired electrons, so metallic bonding is weak and the metals are soft.

   1 mark

6. (f) $KMn{O}_4$ oxidises HCl to $C{l}_2$, so part of the oxidant is used up by the acid and the titration gives wrong results:
   $2Mn{O}_4^{-}+10C{l}^{-}+16H^{+}\to 2M{n}^{2+}+5C{l}_2+8H_{2}O$
   (This is an additional part; any five of (a)–(g) earn 5 marks.)

7. (g) In lower oxidation states the metal oxides are ionic, so they are basic (e.g. MnO). As the oxidation state rises, ionic character falls and covalent character increases, so the highest oxides are amphoteric or acidic (e.g. $M{n}_2{O}_7$ is acidic). (Additional part.)

Answer: (a) Ce⁴⁺ has stable 4f⁰; (b) stable d⁵ Mn²⁺; (c) Zn – filled d¹⁰, weak metallic bonding; (d) forms dichromate, yellow → orange; (e) filled d orbitals, weak metallic bonding; (f) HCl oxidised to Cl₂; (g) covalent character increases with oxidation state

1. (i) Depression in freezing point: $\Delta T_f=0-(-2.8)=2.8$ K.
   Formula: $\Delta T_f=K_f\times \frac{w_B}{M_B\times W_A(\text{kg})}$

   1 mark

2. Substitute: $2.8=1.86\times \frac{w_B}{62\times 1.0}$

   1 mark

3. $w_B=\frac{2.8\times 62}{1.86}=93.33$ g of ethylene glycol

   1 mark

4. (ii) An ethanol–acetone mixture shows positive deviation from Raoult's law.

   1 mark

5. Reason: pure ethanol molecules are held by hydrogen bonds. Acetone molecules come between them and break some of these bonds, so ethanol–acetone attractions are weaker than ethanol–ethanol attractions. Molecules escape more easily, and the vapour pressure is higher than Raoult's law predicts (${\Delta }_{mix}H>0$).

   1 mark

Answer: (i) 93.33 g ethylene glycol; (ii) Positive deviation – ethanol–acetone interactions weaker than ethanol–ethanol (H-bonds broken)

1. (i) Rise in boiling point needed: $\Delta T_b=100-99.68=0.32\text{°C}=0.32\text{K}$

   ½ mark

2. Formula: $\Delta T_b=K_b\times \frac{w_B\times 1000}{M_B\times w_A}$, so $w_B=\frac{\Delta T_b\times M_B\times w_A}{K_b\times 1000}$

   1 mark

3. Substitution: $w_B=\frac{0.32\times 342\times 500}{0.52\times 1000}=\frac{0.32\times 342}{0.52\times 2}=\frac{109.44}{1.04}$

   1 mark

4. $w_B=105.23\text{g}$ of sucrose must be added.

   ½ mark

5. (ii) Henry's law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. In mole-fraction form, the partial pressure of the gas in the vapour phase is proportional to its mole fraction in the solution: $p=K_{H}x$.

   1 mark

6. Application: Bottles of soft drinks and soda water are sealed under high pressure so that more CO${}_2$ dissolves in them. Other acceptable applications are adding helium to scuba-divers' tanks to avoid the bends, and the low O${}_2$ in the blood of climbers at high altitude (anoxia).

   1 mark

Answer: (i) 105.23 g sucrose; (ii) $p=K_{H}x$ at constant temperature; application: soft-drink bottles are sealed under high CO${}_2$ pressure.

1. (i) Deduction:

   • It forms a 2,4-DNP derivative, so it has a C=O group.
   • It reduces Fehling's solution, so the C=O is an aldehyde (–CHO).
   • It undergoes the Cannizzaro reaction, so the –CHO has no α-hydrogen and is attached directly to the ring.
   • Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid), so the ring carries two side chains in ortho positions.

   C${}_9$H${}_{10}$O minus C${}_6$H${}_4$ and –CHO leaves –C${}_2$H${}_5$. Hence (A) is $o$-ethylbenzaldehyde, ${\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CHO})({\mathrm{C}}_2{\mathrm{H}}_5)$ with the two groups at positions 1 and 2.

   1 mark

2. IUPAC name of (A): 2-Ethylbenzaldehyde

   1 mark

3. (ii)(1) With 2,4-dinitrophenylhydrazine:
   $${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}+{\mathrm{H}}_2\mathrm{N}-\mathrm{NH}-{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{N}{\mathrm{O}}_2{)}_2\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CH}=\mathrm{N}-\mathrm{NH}-{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{N}{\mathrm{O}}_2{)}_2+{\mathrm{H}}_2\mathrm{O}$$
   The product is the orange-yellow 2-ethylbenzaldehyde 2,4-dinitrophenylhydrazone.

   1 mark

4. (ii)(2) With Fehling's solution:
   $${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}+2C{\mathrm{u}}^{2+}+5O{\mathrm{H}}^{-}\to {\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CO}{\mathrm{O}}^{-}+\mathrm{C}{\mathrm{u}}_2\mathrm{O}\downarrow (\text{red-brown})+3{\mathrm{H}}_2\mathrm{O}$$
   Note: aromatic aldehydes usually do not respond to Fehling's test. The scheme therefore awards this mark for any reasonable attempt.

   1 mark

5. (iii) Cannizzaro reaction (conc. NaOH):
   $$2{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}+\mathrm{NaOH}(conc.)\stackrel{\mathrm{\Delta }}{\to }{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COONa}$$
   The products are 2-ethylbenzyl alcohol, i.e. (2-ethylphenyl)methanol, and sodium 2-ethylbenzoate.

   1 mark

Answer: (A) = 2-ethylbenzaldehyde (o-C${}_2$H${}_5$C${}_6$H${}_4$CHO); it gives the 2,4-DNP hydrazone, reduces Fehling's solution to Cu${}_2$O, and in the Cannizzaro reaction gives 2-ethylbenzyl alcohol and sodium 2-ethylbenzoate.

1. (i)(1) The carbonyl group is strongly electron-withdrawing, so it weakens the α C–H bond. When the α-H leaves as H${}^{+}$, the carbanion formed (enolate ion) is resonance-stabilised, with the negative charge delocalised onto the oxygen. This makes the α-hydrogens acidic.

   1 mark

2. (i)(2) Aldehydes have a hydrogen on the carbonyl carbon, so oxidation only needs a C–H bond to break, which happens easily even with mild oxidants. Ketones can be oxidised only by breaking a C–C bond, which needs strong oxidising agents. So aldehydes are oxidised more easily.

   1 mark

3. (ii)(1) Decreasing reactivity towards nucleophilic addition: Propanal > Benzaldehyde > Acetone.

   Propanal has only one alkyl group, so it has the least steric hindrance and an electrophilic carbonyl carbon. In benzaldehyde, resonance with the ring lowers the electrophilicity of the carbonyl carbon. Acetone has two electron-releasing methyl groups, which give it the most steric hindrance and the least positive carbonyl carbon.

   1 mark

4. (ii)(2) Increasing order of boiling point: Propane < Dimethyl ether < Propanal < Ethanol.

   Propane is non-polar and has only weak van der Waals forces. The ether and the aldehyde have dipole–dipole forces, and these are stronger in the more polar propanal. Ethanol has intermolecular hydrogen bonding, so it boils highest.

   1 mark

5. (iii) Add sodium bicarbonate solution to each compound. Benzoic acid gives brisk effervescence of CO${}_2$: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{NaHC}{\mathrm{O}}_3\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+\mathrm{C}{\mathrm{O}}_2\uparrow +{\mathrm{H}}_2\mathrm{O}$. Benzaldehyde gives no effervescence.

   Alternatively, benzaldehyde gives a silver mirror with Tollens' reagent, while benzoic acid does not.

   1 mark

Answer: (i) Electron-withdrawing C=O and resonance-stabilised enolate; aldehyde oxidation breaks only a C–H bond, not a C–C bond. (ii)(1) Propanal > Benzaldehyde > Acetone; (2) Propane < Dimethyl ether < Propanal < Ethanol. (iii) NaHCO${}_3$ test: benzoic acid effervesces, benzaldehyde does not.