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International Chemistry Olympiad (IChO) Practice

12 challenging chemistry problems covering organic, inorganic, physical, analytical chemistry, and biochemistry.

Organic Chemistry

1 Diels-Alder Selectivity Mechanisms Hard
A Diels-Alder reaction between 1-methoxy-1,3-butadiene and methyl acrylate can produce two regioisomeric products.
  1. Draw both possible products and predict which is the major product. Justify using frontier molecular orbital (FMO) theory.
  2. Explain the endo rule and predict the stereochemistry of the major product.
  3. If the reaction is run with a Lewis acid catalyst (\(\text{AlCl}_3\)), how does the rate change and why?
Hint
The methoxy group is electron-donating, raising HOMO energy of the diene. The ester is electron-withdrawing, lowering LUMO of the dienophile. Normal electron-demand DA. The "ortho" rule: largest HOMO coefficient on C1 of diene (para to OMe) matches largest LUMO coefficient on C2 of dienophile (beta to COOR). Major product is 1,2-disubstituted (ortho). Endo product favored by secondary orbital interactions. Lewis acid lowers dienophile LUMO further, accelerating the reaction dramatically.
2 Multistep Synthesis Puzzle Mechanisms Olympiad Level
Propose a synthesis of racemic ibuprofen (2-(4-isobutylphenyl)propionic acid) starting from isobutylbenzene. You may use any common reagents.
  1. Outline a 4-5 step synthesis with reagents for each step.
  2. Identify the key carbon-carbon bond-forming step.
  3. How would you modify the synthesis to produce a single enantiomer?
Hint
Classic route: (1) Friedel-Crafts acylation with acetyl chloride/\(\text{AlCl}_3\) to get 4-isobutylacetophenone. (2) Reduction of ketone (\(\text{NaBH}_4\) or Clemmensen) or use the Darzens glycidic ester synthesis. (3) A common approach: acetophenone \(\to\) alpha-methylation \(\to\) oxidation to acid. Or: Arndt-Eistert homologation. For enantiopure: use chiral auxiliary (Evans oxazolidinone) for the alpha-methylation, or enzymatic resolution, or asymmetric hydrogenation of the \(\alpha,\beta\)-unsaturated acid.
3 Aromatic Substitution Challenge Mechanisms Medium
Explain the following observations:
  1. Nitration of toluene gives predominantly ortho and para products, while nitration of nitrobenzene gives predominantly meta product.
  2. Chlorination of anisole (methoxybenzene) is much faster than chlorination of benzene, yet chlorination of acetanilide gives predominantly para product rather than the expected ortho/para mixture.
  3. Predict the major product of bromination of 2-methylanisole and justify your answer.
Hint
(1) \(\text{CH}_3\) is electron-donating (+I), stabilizing the arenium ion at ortho/para. \(\text{NO}_2\) is electron-withdrawing (-M, -I), destabilizing ortho/para more than meta. (2) OMe is strongly activating (+M). In acetanilide, the bulky NHCOMe group sterically blocks ortho positions, giving para selectivity. (3) In 2-methylanisole, OMe directs ortho/para (strong +M) and \(\text{CH}_3\) directs ortho/para. The position para to OMe and ortho to \(\text{CH}_3\) (C-4) is doubly activated. Major product: 4-bromo-2-methylanisole.

Inorganic Chemistry

4 Crystal Field Theory Analysis Coordination Chemistry Hard
Consider the complex \([\text{Co}(\text{NH}_3)_6]^{3+}\).
  1. Determine the oxidation state, d-electron count, and predict whether the complex is high-spin or low-spin using the spectrochemical series.
  2. Draw the crystal field splitting diagram and calculate the CFSE (Crystal Field Stabilization Energy) in terms of \(\Delta_o\) and pairing energy \(P\).
  3. Explain why \([\text{CoF}_6]^{3-}\) is paramagnetic while \([\text{Co}(\text{NH}_3)_6]^{3+}\) is diamagnetic, despite having the same metal center.
  4. Predict the number of unpaired electrons and the magnetic moment for each complex.
Hint
\(\text{Co}^{3+}\) is \(d^6\). \(\text{NH}_3\) is a strong-field ligand (high in spectrochemical series), so \([\text{Co}(\text{NH}_3)_6]^{3+}\) is low-spin: \(t_{2g}^6 e_g^0\), CFSE \(= -24\text{Dq} + 2P\), 0 unpaired electrons, diamagnetic, \(\mu = 0\). \(\text{F}^-\) is a weak-field ligand, so \([\text{CoF}_6]^{3-}\) is high-spin: \(t_{2g}^4 e_g^2\), CFSE \(= -4\text{Dq}\), 4 unpaired electrons, \(\mu = \sqrt{4 \times 6} \approx 4.9\;\text{BM}\). The difference is entirely due to the ligand field strength.
5 Isomerism in Coordination Compounds Coordination Chemistry Medium
The compound with formula \(\text{CrCl}_3 \cdot 6\text{H}_2\text{O}\) exists in three different forms that can be distinguished by their reactions with \(\text{AgNO}_3\) solution.
  1. Write the structural formulas of all three isomers.
  2. Predict how many moles of AgCl precipitate per mole of each isomer when treated with excess \(\text{AgNO}_3\).
  3. How would you distinguish the three isomers using conductivity measurements?
Hint
(1) \([\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3\) (violet), \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}\) (blue-green), \([\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}\) (dark green). (2) 3 mol, 2 mol, and 1 mol AgCl respectively, since only ionic (outer-sphere) chlorides precipitate. (3) Molar conductivity decreases with fewer ions: 3:1 electrolyte > 2:1 electrolyte > 1:1 electrolyte. The violet form has highest conductivity.

Physical Chemistry

6 Complex Reaction Kinetics Kinetics Hard
The decomposition of \(\text{N}_2\text{O}_5\) follows the mechanism:

Step 1: \(\text{N}_2\text{O}_5 \rightleftharpoons \text{NO}_2 + \text{NO}_3\) (fast equilibrium, \(K_{\text{eq}}\))
Step 2: \(\text{NO}_2 + \text{NO}_3 \to \text{NO} + \text{NO}_2 + \text{O}_2\) (slow, \(k_2\))
Step 3: \(\text{NO} + \text{NO}_3 \to 2\text{NO}_2\) (fast, \(k_3\))
  1. Derive the rate law using the steady-state approximation for \([\text{NO}_3]\).
  2. Show that the reaction is first-order in \(\text{N}_2\text{O}_5\).
  3. At \(25°\text{C}\), the half-life is 3.38 hours. At \(35°\text{C}\), it is 1.39 hours. Calculate the activation energy.
Hint
From the fast equilibrium: \([\text{NO}_2][\text{NO}_3] = K_{\text{eq}}[\text{N}_2\text{O}_5]\). Rate-determining step: \(\text{Rate} = k_2[\text{NO}_2][\text{NO}_3] = k_2 K_{\text{eq}}[\text{N}_2\text{O}_5]\). So \(\text{rate} = k_{\text{obs}}[\text{N}_2\text{O}_5]\) where \(k_{\text{obs}} = k_2 K_{\text{eq}}\). For first-order: \(k = \frac{\ln 2}{t_{1/2}}\). At \(25°\text{C}\): \(k_1 = 0.205\;\text{hr}^{-1}\). At \(35°\text{C}\): \(k_2 = 0.499\;\text{hr}^{-1}\). Using Arrhenius: \(\ln\!\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\!\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\). \(E_a \approx 103\;\text{kJ/mol}\).
7 Electrochemistry and Nernst Equation Thermodynamics Medium
A galvanic cell is constructed:

\(\text{Zn}(s) \mid \text{Zn}^{2+}(0.010\;\text{M}) \| \text{Cu}^{2+}(1.0\;\text{M}) \mid \text{Cu}(s)\)

Given: \(E°(\text{Zn}^{2+}/\text{Zn}) = -0.76\;\text{V}\), \(E°(\text{Cu}^{2+}/\text{Cu}) = +0.34\;\text{V}\)
  1. Calculate the standard cell potential and the cell potential under given conditions.
  2. Calculate \(\Delta G°\) and \(K_{\text{eq}}\) for the cell reaction.
  3. At what ratio \([\text{Zn}^{2+}]/[\text{Cu}^{2+}]\) will the cell potential become zero?
Hint
\(E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.34 - (-0.76) = 1.10\;\text{V}\). Nernst: \(E = E° - \frac{RT}{nF}\ln Q = 1.10 - \frac{0.0257}{2}\ln\!\left(\frac{0.010}{1.0}\right) = 1.10 + 0.059 = 1.159\;\text{V}\). \(\Delta G° = -nFE° = -2(96485)(1.10) = -212.3\;\text{kJ/mol}\). \(K = \exp\!\left(\frac{nFE°}{RT}\right) = 10^{37.2}\). Cell dies when \(E = 0\): \(\ln Q = \frac{nFE°}{RT}\), so \([\text{Zn}^{2+}]/[\text{Cu}^{2+}] = e^{85.6} \approx 10^{37.2}\).
8 Chemical Equilibrium Under Pressure Thermodynamics Hard
For the reaction: \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\), \(\Delta H° = -92.4\;\text{kJ/mol}\)

At 500 K and 300 atm, the equilibrium mole fraction of \(\text{NH}_3\) is 0.35 starting from a 1:3 mixture of \(\text{N}_2 : \text{H}_2\).
  1. Calculate \(K_p\) at 500 K under these conditions.
  2. Using van't Hoff equation, estimate \(K_p\) at 700 K.
  3. Explain quantitatively why high pressure and low temperature favor ammonia production, yet industrial processes use 400-500 °C.
Hint
With \(x_{\text{NH}_3} = 0.35\) from 1:3 feed: let \(\alpha\) be conversion. Moles at equilibrium: \(\text{N}_2 = 1-\alpha\), \(\text{H}_2 = 3-3\alpha\), \(\text{NH}_3 = 2\alpha\), total \(= 4-2\alpha\). \(x_{\text{NH}_3} = \frac{2\alpha}{4-2\alpha} = 0.35\), giving \(\alpha \approx 0.412\). \(K_p = \frac{x_{\text{NH}_3}^2}{x_{\text{N}_2} \cdot x_{\text{H}_2}^3} \times P^{-2}\). Van't Hoff: \(\ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H°}{R}\!\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\). At low \(T\), equilibrium favors products, but rate is too slow. The iron catalyst still needs \(400°\text{C}+\) to be effective — a kinetic compromise.

Analytical Chemistry

9 NMR and IR Spectroscopy Puzzle Spectroscopy Hard
An unknown compound has molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) (degree of unsaturation = 1). The IR spectrum shows a strong absorption at \(1740\;\text{cm}^{-1}\) and no broad O-H stretch. The \(^1\text{H}\) NMR shows:
  • Singlet at \(\delta\) 3.67 (3H)
  • Quartet at \(\delta\) 2.34 (2H)
  • Triplet at \(\delta\) 1.14 (3H)
  1. Identify the compound and explain each spectral feature.
  2. If the compound were the isomer ethyl formate, how would the NMR spectrum differ?
  3. Propose how mass spectrometry could distinguish between all possible isomers of \(\text{C}_4\text{H}_8\text{O}_2\).
Hint
\(1740\;\text{cm}^{-1}\) = ester C=O stretch. One degree of unsaturation matches C=O. Singlet 3H at 3.67 = \(\text{OCH}_3\). Quartet 2H at 2.34 = \(\text{CH}_2\) next to one \(\text{CH}_3\) neighbor. Triplet 3H at 1.14 = \(\text{CH}_3\) next to \(\text{CH}_2\). The compound is methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)). Ethyl formate (\(\text{HCOOCH}_2\text{CH}_3\)) would show: singlet ~8.1 ppm (1H, CHO), quartet ~4.2 ppm (2H, \(\text{OCH}_2\)), triplet ~1.3 ppm (3H, \(\text{CH}_3\)). MS: different fragmentation patterns — loss of \(\text{OCH}_3\) (31) vs \(\text{OC}_2\text{H}_5\) (45), McLafferty rearrangement products differ.
10 Complexometric Titration Titrations Medium
A 25.00 mL sample of hard water is titrated with 0.0100 M EDTA solution using Eriochrome Black T indicator at pH 10 (ammoniacal buffer).
  1. The endpoint is reached at 18.40 mL. Calculate the total hardness in mg/L \(\text{CaCO}_3\).
  2. A second 25.00 mL sample is treated with NaOH to precipitate \(\text{Mg(OH)}_2\), filtered, and the filtrate titrated with EDTA requires 11.60 mL. Calculate the individual \(\text{Ca}^{2+}\) and \(\text{Mg}^{2+}\) hardness.
  3. Why must the pH be controlled at 10 for total hardness determination?
Hint
Total moles EDTA \(= 0.01840\;\text{L} \times 0.0100\;\text{M} = 1.84 \times 10^{-4}\;\text{mol}\) = total moles \((\text{Ca}^{2+} + \text{Mg}^{2+})\). As \(\text{CaCO}_3\) equivalent: \(1.84 \times 10^{-4} \times 100.09\;\text{g/mol} = 0.01842\;\text{g}\) in 25 mL. Hardness \(= \frac{0.01842}{0.025} = 0.737\;\text{g/L} = 737\;\text{mg/L}\). \(\text{Ca}^{2+}\) alone: \(0.01160 \times 0.0100 = 1.16 \times 10^{-4}\;\text{mol}\). Ca hardness = 464 mg/L. Mg hardness = 737 - 464 = 273 mg/L. pH 10 needed: EDTA complexes are stable at this pH, Mg-EDTA complex forms fully, and EBT indicator works (color change blue to wine-red).

Biochemistry

11 Enzyme Kinetics — Michaelis-Menten Biochemistry Hard
An enzyme-catalyzed reaction has \(V_{\max} = 120\;\mu\text{mol/min}\) and \(K_m = 4.0\;\text{mM}\).
  1. Calculate the initial rate at substrate concentrations of 1.0, 4.0, 8.0, and 20.0 mM.
  2. A competitive inhibitor is added with \(K_i = 2.0\;\text{mM}\) at concentration \([I] = 6.0\;\text{mM}\). Calculate the apparent \(K_m\) and the new initial rates at the same substrate concentrations.
  3. How would a Lineweaver-Burk plot (\(1/v\) vs \(1/[\text{S}]\)) look for uninhibited vs competitively inhibited reactions? What is the x-intercept and y-intercept for each?
Hint
\(v = \frac{V_{\max}[\text{S}]}{K_m + [\text{S}]}\). At \([\text{S}] = 1.0\): \(v = \frac{120(1)}{4+1} = 24\). At 4.0: \(v = 60\). At 8.0: \(v = 80\). At 20.0: \(v = 100\;\mu\text{mol/min}\). With competitive inhibitor: \(K_{m,\text{app}} = K_m\!\left(1 + \frac{[\text{I}]}{K_i}\right) = 4\!\left(1 + \frac{6}{2}\right) = 16\;\text{mM}\). \(V_{\max}\) unchanged. New rates: at 1.0: 7.06, at 4.0: 24, at 8.0: 40, at 20.0: 66.7. LB plot: uninhibited x-int \(= -1/K_m = -0.25\), y-int \(= 1/V_{\max} = 0.00833\). Inhibited: x-int \(= -1/K_{m,\text{app}} = -0.0625\), same y-int. Both lines converge on y-axis.
12 Amino Acid Chemistry Biochemistry Medium
Glycine (\(\text{H}_2\text{N-CH}_2\text{-COOH}\)) has \(\text{p}K_{a1} = 2.34\) (carboxyl) and \(\text{p}K_{a2} = 9.60\) (amino).
  1. Calculate the isoelectric point (pI) of glycine.
  2. Draw the predominant form of glycine at pH = 1.0, 6.0, and 11.0.
  3. Calculate the fraction of glycine in the zwitterionic form at pH = 6.0.
  4. A mixture of glycine (pI = 5.97), aspartic acid (pI = 2.77), and lysine (pI = 9.74) is subjected to electrophoresis at pH = 6.0. Predict the migration direction of each amino acid.
Hint
\(\text{pI} = \frac{\text{p}K_{a1} + \text{p}K_{a2}}{2} = \frac{2.34 + 9.60}{2} = 5.97\). At pH 1.0: fully protonated cation (\(^+\text{H}_3\text{N-CH}_2\text{-COOH}\)). At pH 6.0: zwitterion (\(^+\text{H}_3\text{N-CH}_2\text{-COO}^-\)). At pH 11.0: fully deprotonated anion (\(\text{H}_2\text{N-CH}_2\text{-COO}^-\)). At pH 6.0, fraction zwitterionic \(\approx \frac{1}{1 + 10^{\text{pH}-\text{p}K_{a2}} + 10^{\text{p}K_{a1}-\text{pH}}} \approx \frac{1}{1 + 10^{-3.6} + 10^{-3.66}} \approx 0.9995\). At pH 6.0: glycine is near pI (barely moves), aspartic acid is anionic (moves to anode +), lysine is cationic (moves to cathode -).