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Butadiene π-electron system

Problem 2 Localization and delocalization in benzene 13

Problem 3 Study of liquid benzene hydrogenation 15

Problem 4 Use of dihydrogen: fuel cells 16

Problem 6 Deacidification and desulfurization of natural gas 22

Problem 8 Which wine is it? Blind tasting challenge 26

Problem 9 Nitrophenols: synthesis and physical properties 27

Problem 11 The mineral of winners 33

Problem 14 Synthesis of block copolymers followed by size-exclusion chromatography 39

Problem 18 A kinetic study of the Maillard reaction 50

Problem 20 Fluoro-deoxyglucose and PET imaging 56

Problem 21 Catalysis and stereoselective synthesis of cobalt glycocomplexes 61

Problem 22 Structural study of copper (II) complexes 62

Problem 23 Synthesis and study of a molecular motor 64

Problem 24 Some steps of a synthesis of cantharidin 67

Problem 25 Study of ricinoleic acid 68

Problem 27 Formal synthesis of testosterone 71

Back to 1990: Aqueous solutions of copper salts 73

Problem P2: Oxidation of (‒)-borneol to (‒)-camphor 77

We are excited to offer Preparatory Problems for the 51st International Chemistry Olympiad, providing students with a valuable opportunity to train and explore various topics in both modern and traditional chemistry These problems, designed to be solved using high school curriculum concepts along with some advanced topics, include six theoretical challenges and two practical ones.

This booklet features 27 theoretical problems and 6 practical problems, designed to mirror the format and spirit of the final challenges The first section concludes with an additional theoretical task titled "Back to 1990," which is an excerpt from the 1990 Olympiad in France and does not require extensive study Official solutions will be provided to Head Mentors by the end of February 2019 and will be available on the IChO 2019 website starting June 1, 2019.

We will be happy to read and reply to your comments, corrections and questions about the problems Please send them to contact-icho2019@laligue.org

Looking forward to seeing you in Paris to enjoy chemistry and to make science together!

The members of the Scientific Committee in charge of the preparatory problems

Didier Bourissou, CNRS, Toulouse Aurélien Moncomble, Université de Lille Élise Duboué-Dijon, CNRS, Paris Clément Guibert, Sorbonne Université, Paris Baptiste Haddou, Lycée Darius Milhaud, Le Kremlin-Bicêtre

Hakim Lakmini, Lycée Saint Louis, Paris

We extend our gratitude to all the authors for their dedication in crafting these problems over several months, culminating in this booklet aimed at benefiting young chemists participating in the Olympiad Additionally, we appreciate the reviewers, including steering committee members, whose meticulous attention to detail has greatly enhanced the quality of these problems.

Pierre Aubertin, Lycée Léonard de Vinci, Calais

Tahar Ayad, Chimie ParisTech, Paris

Jean-Marc Campagne, ENSCM, Montpellier

Baptiste Chappaz, Collège Les Pyramides, Évry

Sylvain Clède, Lycée Stanislas, Paris Éric Clot, CNRS, Montpellier

Bénédicte Colnet, Mines ParisTech, Paris

Antton Curutchet, ENS Lyon Élise Duboué-Dijon, CNRS, Paris

Ludivine Garcia, Lycée Jean Moulin, Béziers

Catherine Gautier, Lycée Algoud Laffemas, Valence

Laetitia Guerret, ENS Paris-Saclay, Cachan

Clément Guibert, Sorbonne Université, Paris

Dayana Gulevich, Moscow State University

Baptiste Haddou, Lycée D Milhaud, Kremlin-Bicêtre

Laurent Heinrich, Lycée Pierre Corneille, Rouen Lucas Henry, ENS, Paris

Claire Kammerer, Univ Paul Sabatier, Toulouse Dmytro Kandaskalov, Aix Marseille Université Iuliia Karpenko, Université de Strasbourg Maxime Lacuve, ENSAM, Paris

Hakim Lakmini from Lycée Saint Louis in Paris, Julien Lalande and Alix Lenormand from Lycée Henri IV and Lycée Henri Poincaré respectively, along with Étienne Mangaud and Jean-Daniel Marty from Université Paul Sabatier in Toulouse, and Olivier Maury from CNRS in Lyon, represent a notable group of scholars in France.

Bastien Mettra, IUT-Lyon1, Villeurbanne Aurélien Moncomble, Université de Lille Pierre-Adrien Payard, ENS, Paris

The article features a diverse group of esteemed individuals from various prestigious institutions, including Daniel Pla from CNRS in Toulouse, Romain Ramozzi from Lycée Henri Poincaré in Nancy, Clémence Richaud from Lycée Jules Verne in Limours, Vincent Robert from Université de Strasbourg, Jean-Marie Swiecicki from MIT in Cambridge, USA, Guillaume Vives from Sorbonne Université in Paris, and Hanna Zhdanova from Université de Strasbourg.

Olivier Durupthy Matthieu Emond Johanna Foret Mickặl Four Emma Gendre Isabelle Girard Antoine Hoste Damien Lavergne Quentin Lefebvre Christina Legendre Anne Leleu Marion Livecchi

Guillaume Mériguet Mathilde Niocel Artem Osypenko Valentin Quint Emelyne Renaglia Tristan Ribeyre-Stecki Clément Robert Nell Saunders Laura Scalfi Freddy Szymczak Julien Valentin Dominique Vichard Vincent Wieczny

1 Thermodynamics: relation between equilibrium constants and standard reaction Gibbs free energy, relation between thermodynamic and electrochemical data

2 Kinetics: orders of reaction, half-life, rates defined as time derivatives of concentrations, use of integrated rate laws, classic approximations

3 Basic quantum chemistry: notion of wavefunction, expression of simple molecular orbitals, electronic energy levels, crystal field theory

4 Spectroscopy: simple IR spectroscopy (identification of chemical groups only), 1 H NMR spectroscopy (chemical shifts, integrals, couplings and multiplicity)

5 Polymers: block copolymers, polymerization, polydispersity, simple size exclusion chromatography (SEC)

6 Stereochemistry: stereoisomers in organic and inorganic chemistry, stereoselectivity in organic synthesis

1 Techniques in organic synthesis (drying of a precipitate, recrystallization, TLC)

2 Use of a spectrophotometer (mono-wavelength measurements)

Theoretical: the following advanced skills or knowledge WILL NOT appear in the exam set:

 Specific carbohydrates chemistry (reactivity at the anomeric position, nomenclature, representation);

 Stereochemical aspects associated with the Diels-Alder reaction (supra-supra and endo approaches);

Practical: the following techniques WILL NOT be required during the competition:

 Use of a separatory funnel and extraction using immiscible solvents;

 Use of a melting point apparatus;

This booklet operates under the assumption that the activities of all aqueous species can be effectively represented by their concentrations in mol L⁻¹ To streamline the formulas and expressions, the standard concentration of c° = 1 mol L⁻¹ is excluded from consideration.

Atmospheric pressure: P atm = 1 atm = 1.013 bar = 1.013∙10 5 Pa

Zero of the Celsius scale: 273.15 K

Ideal gas equation: pV = nRT

Gibbs free energy: G = H – TS Δr G° = –RT lnK° Δr G° = –n F E cell ° Δr G = Δr G° + RT lnQ

Reaction quotient Q for a reaction a A(aq) + b B(aq) = c C(aq) + d D(aq): Q = [C] c [D] d

[A] a [B] b Henderson–Hasselbalch equation: pH = pK a + log [A – ]

Nernst–Peterson equation: E = E o – RT zF ln𝑄 where Q is the reaction quotient of the reduction half-reaction at T = 298 K, RT

Rate laws in integrated form:

First order: ln[A] = ln[A]0 – kt

[A] 0 + kt Half-life for a first order process: t 1/2 = ln2 k

The above constants and formulas will be given to the students for the theoretical exam

Chemical shifts of hydrogen (in ppm /TMS) phenols alcohols alkenes aromatics alkynes carboxylic acids CH 3 —NR 2 CH 3 —SiR 3 aldehydes CH 3 —OR ketones

2-12 if free rotation: 6-8 ax-ax (cyclohexane): 8-12 ax-eq or eq-eq (cyclohexane): 2-5

R 2 H a C—CR 2 —CR 2 H b if free rotation: < 0.1 otherwise (rigid): 1-8

Vibrational mode σ (cm −1 ) Intensity alcohol O—H (stretching) carboxylic acid O—H (stretching)

=C—H (stretching) C—H (stretching) –(CO)—H (stretching) C≡N (stretching)

2250 strong strong strong strong weak weak weak strong

C≡C (stretching) aldehyde C=O (stretching) anhydride C=O (stretching) ester C=O (stretching) ketone C=O (stretching) amide C=O (stretching) alkene C=C (stretching) aromatic C=C (stretching)

1250-1050 (several) 1200-1020 1600-1500; 1400-1300 variable strong weak; strong strong strong strong weak weak medium medium strong strong strong

Buta-1,3-diene, with the chemical formula C4H6, was first isolated in 1863 by French chemist E Caventou and later identified in 1886 by English chemist H E Armstrong This important diene serves as a key reagent in the production of synthetic rubber, with annual production exceeding 12.7 million tons This article will explore the properties of its π-electron system and compare them to those of the hypothetical cyclobutadiene, which has never been isolated in its free form.

1 Give the number of π-electrons of butadiene

The Molecular Orbitals (MO) Ψ i of the π-electron system can be written as a weighted sum (linear combination) of the 2pz atomic orbitals of each carbon atom, φ j:

We present an approximate expression for molecular orbitals (MOs) along with their corresponding energy levels The energy of each MO is determined by two parameters, α and β, which are both negative real numbers Here, α signifies the energy of an electron within an isolated 2pz orbital, while β denotes the interaction energy between adjacent 2pz orbitals.

2 Draw and fill in the MO diagram of butadiene Draw schematically each MO and identify its nature (bonding or anti-bonding)

We consider the formation of the butadiene π-electron system, starting from four carbon atoms, each bringing an electron in a 2pz orbital of energy α

3 Calculate the formation energy ΔE f associated with this transformation

The conjugation energy is defined as the difference between the total π-energy of the studied compound and that of two non-interacting ethylene molecules, with the π-energy of ethylene calculated as 2(α + β).

4 Calculate the conjugation energy ΔE c of butadiene Give its sign Which system is the most stable? Choose the correct answer

The net charge \( q_j \) on each carbon atom, representing the charge gained or lost compared to its neutral state, can be determined for butadiene and cyclobutadiene.

The sum runs over the occupied molecular orbitals (MOs), where \( n_i \) represents the number of electrons in the \( i \)th MO, and \( c_{ij} \) denotes the coefficient of the \( j \)th carbon atom within the \( i \)th MO.

5 Calculate the net charges q 1 and q 2 of the butadiene carbon atoms 1 and 2 Deduce the values of q 3 and q 4

Bond order I serves as an estimate of the number of π chemical bonds between two atoms, with a pure single bond represented by I = 0 and a pure double bond indicated by I = 1 The bond order I between neighboring atoms r and s can be determined using molecular orbitals (MOs).

The quantity \( c \) is defined as the sum of the occupied molecular orbitals (MOs), calculated by multiplying the number of electrons in each MO by the coefficients associated with the two atoms \( r \) and \( s \) within that MO.

6 Calculate for each bond the associated bond order: I 12, I 23, and I 34 Identify the bond(s) that has (have) the strongest double-bond character

7 Draw alternative Lewis structures of butadiene to reflect the previously obtained results

The molecular orbital (MO) diagram for the hypothetical cyclobutadiene illustrates the relationship between atomic orbitals and their contributions to the MO, with the size of each orbital indicating its coefficient Additionally, the color of the orbitals, either grey or white, signifies the sign of the wavefunction.

8 Fill in the MO diagram of cyclobutadiene

9 Using the provided diagram and considering the symmetry of the molecule, determine the missing coefficients (c ij) in the following MO expressions

10 Calculate the formation and conjugation energies, ΔE f’ and ΔE c’, for cyclobutadiene Which system is the most stable? Choose the correct answer

11 Compare the formation energy of cyclobutadiene and that of butadiene Which compound is the most stable? Choose the correct answer

We now consider a rectangular deformation of cyclobutadiene, with localization and shortening of the double bonds and elongation of the simple bonds compared to the square geometry

12 Choose the correct statement(s) among the following:

 This deformation stabilizes C=C double bonds

 This deformation weakens C=C double bonds

 This deformation does not affect the stability of C=C double bonds

 This deformation increases the stability due to electronic conjugation

 This deformation diminishes the stability due to electronic conjugation

 This deformation does not affect the stability due to electronic conjugation

13 Using your previous answers, choose the correct statement among the following The π-system after deformation is:

 More stable than the square cyclobutadiene

 Less stable than the square cyclobutadiene

 As stable as the square cyclobutadiene.

Localization and delocalization in benzene

Benzene, a key aromatic compound, was first isolated from benjoin, the essence of "Papier d’Arménie." In the mid-19th century, French chemist M Berthelot synthesized benzene through acetylene trimerization This article aims to explore the electronic properties of benzene, highlighting its significance as a representative of aromatic molecules.

1 Write the reaction from acetylene C2H2 generating benzene

2 Draw a structure of benzene using three single bonds and three double bonds between carbon atoms It is referred to as Kekulé’s benzene

3 Draw a structure of benzene holding five single and two double bonds This structure is called Dewar’s benzene

The Kekulé structure K1 features a double bond between the C1 and C2 atoms, which can be effectively modeled by representing the delocalization of a single electron with an energy value of t < 0.

4 Give the energy E π of the -system of this bond as a function of t

5 In K1, double bonds are supposed to be fixed For this structure K1, calculate the energy of the π-system E K1 as a function of t

6 Write an analog to K1 It will be called K2

7 Express the energy E K2 of this structure K2.

Mathematically, the benzene molecule is expressed as a mix between K1 and K2,

K = c 1 K1 + c 2 K2, where c 1 and c 2 are real numbers with c 1 2 + c 2 2 = 1 and c 1 > 0 and c 2 > 0 This expression stresses that a proper description of benzene cannot be restricted to K1 or K2

8 On a scheme, show the displacement of the double bond localized between C1 and C2 and the movement of the other double bonds These formulae are the resonance structures of benzene

Starting from a localized view K1 or K2, the electronic delocalization over all the carbon atoms can be accounted for by the introduction of a supplementary energetic term The energy E K of

E K = c 1 2 E K1 + c 2 2 E K2 + 2 c 1 c 2 H 12 where H 12 varies between t and 0, with t < 0 Therefore, E K is a function of c 1 and c 2

9 Express E K as a function of c 1 only

It can be shown that E K is minimal for c 1 = 1 / √2 From now on, we assume that c 1 = 1 / √2

10 If H 12 = 0, what is the expression of E K? The resonance energy is defined as the difference

11 Specify the sign of E 1 Choose the correct statement between:

 electronic delocalization contributes to stabilize the benzene molecule

 electronic delocalization contributes to destabilize the benzene molecule

The π energy of an n-carbon atom system can be determined by analyzing the occupations of its molecular orbitals (MOs) C A Coulson demonstrated in 1939 that the energies of the MOs (εk) in a cyclic n-carbon atom system can be expressed in a specific manner, regardless of their energy order.

12 Draw the MOs diagram of the -system of benzene (n = 6) and calculate the corresponding energies for each MO

14 Evaluate the -system energy of benzene, E MO, from the filling of the MOs in ascending order Then, calculate the resonance energy E 2 = E MO –E K(H 12 = 0)

16 From the previous results, choose one expression for the relation between the standard hydrogenation enthalpy of cyclohexene (Δr H c°) and that of benzene (Δr H b°)

Study of liquid benzene hydrogenation

Determination of the standard enthalpy of formation of liquid benzene

1 Write down the balanced chemical equation for the formation of liquid benzene from its constituent elements in their standard states

2 Calculate the standard enthalpy of formation of liquid benzene ∆f H°(C6H6(l)) using standard bond enthalpies, standard enthalpies of dissociation, and the standard enthalpies of vaporization of benzene and sublimation of graphite

3 Calculate the standard enthalpy of formation of liquid benzene ∆f H°(C6H6(l)) using Hess law and the standard combustion enthalpies

4 Calculate the difference between the ∆f H°(C6H6(l)) values obtained in the two previous questions Choose the correct explanation for this difference

 The difference is due to experimental errors on the values of standard enthalpies of combustion reactions

 The method used at question 2 does not take into account the nature of bonds in benzene

 The Hess law is only rigorously applicable with standard enthalpies of formation

 The method used in question 3 does not take into account the electronic delocalization

Successive hydrogenation reactions of liquid benzene study

5 Calculate the enthalpy of reaction for the full hydrogenation of liquid benzene into liquid cyclohexane

The different steps of benzene hydrogenation into cyclohexane are given in the scheme 1

6 Complete this scheme by calculating the standard enthalpy of hydrogenation of benzene into cyclohexa-1,3-diene

The sign of the standard enthalpy of this reaction differs from the sign of the other standard enthalpies of hydrogenation in scheme 1

7 What is the main reason for such a difference?

 All the double bounds are not equivalent in benzene: one is stronger than the others

 The breaking of benzene aromaticity

 The formation of a reaction intermediate (cyclohexa-1,3-diene) with a constrained geometry

8 Using only the values given in scheme 1, calculate the resonance energy of cyclohexa-1,3- diene and the resonance energy of benzene

Standard combustion enthalpies ∆ comb H ° at 298 K in kJ mol ‒1

Standard enthalpy of formation of cyclohexane at 298 K

Standard bond enthalpies ∆ d H ° at 298 K in kJ mol ‒1

Standard enthalpies of dissociation D ° at 298 K in kJ mol ‒1

Standard latent heat at 298 K in kJ mol ‒1

Use of dihydrogen: fuel cells

To generate electricity, fuel combustion (such as dihydrogen or methanol) heats liquid water to produce steam, which drives a turbine connected to a generator This process converts chemical energy into thermal energy, then mechanical energy, and finally electrical energy However, energy losses occur at each conversion stage, primarily due to heat dissipation, reducing overall efficiency In contrast, fuel cells offer a more efficient solution by directly converting chemical energy into electrical energy.

The balanced chemical equation for the combustion of one equivalent of fuel A is:

∆comb H°(A) and ∆comb G°(A) are respectively the standard enthalpy of reaction and the standard

Gibbs free energy of reaction associated with reaction (1)

The global response in hydrogen fuel cells mirrors that of hydrogen combustion This discussion will focus on the compounds associated with hydrogen fuel cells in their standard state.

1 Write down the redox half-reactions occurring at the anode and the cathode Write down the balanced chemical equation for the global reaction, for one equivalent of dihydrogen

2 Compute the open circuit voltage of such a cell

3 Compute the theoretical maximum electrical energy recoverable by mole of dihydrogen consumed

4 Electric cars consume between 10 and 20 kWh / 100 km Compute the volume of dihydrogen necessary to produce an electrical energy of 20 kWh at 1.0 bar

The thermodynamic efficiency of a cell is defined as:

∆ r 𝐻° where ∆r G° and ∆r H° are respectively the standard Gibbs free energy of reaction and the standard enthalpy of reaction associated with the global reaction of the running cell

5 Calculate the standard enthalpy of the combustion reaction of gaseous dihydrogen

∆comb H°298K(H2(g)) at 298 K Deduce the thermodynamic efficiency of the dihydrogen fuel cell

The thermodynamic efficiency is smaller than 1 because there is a variation of the entropy of the system

6 Calculate the standard entropy of the dihydrogen combustion reaction ∆comb S°298K(H2(g)) at 298 K

7 Determine if the sign of this standard entropy is consistent with the balanced chemical equation for the reaction (Yes/No) Justify it by a short calculation using the stoichiometric coefficients

The low energy density of dihydrogen and the requirement for high pressure storage have led to the innovation of batteries utilizing alternative fuels In a fuel cell employing liquid methanol, the overall reaction mirrors the combustion process of liquid methanol.

8 Determine the oxidation state of the carbon atom in methanol and in carbon dioxide

In a fuel cell utilizing liquid methanol, the redox half-reactions at the anode involve the oxidation of methanol, while the cathode experiences the reduction of oxygen The balanced chemical equation for the overall reaction of the cell, representing one equivalent of liquid methanol, can be expressed as: CH3OH + O2 → CO2 + 2H2O This equation highlights the conversion of methanol and oxygen into carbon dioxide and water, demonstrating the fundamental electrochemical processes at play in the fuel cell.

10 Calculate the associated thermodynamic efficiency Compare it to the efficiency of the dihydrogen fuel cell

11 Calculate the volume of liquid methanol required to produce 20 kWh Compare this value to the previously calculated volume of gaseous dihydrogen

12 Assuming H2 is an ideal gas, determine the pressure to store the dihydrogen necessary to produce 20 kWh in the same volume as methanol (question 11)

Standard enthalpies of formation ∆ f H ° at 298 K in kJ mol ‒1

Molar heat capacities at constant pressure C ° P in J mol ‒1 K ‒1 They are supposed to be independent of the temperature

Standard latent heat of water at 373 K

Standard potentials at 25 °C related to the standard hydrogen electrode (SHE)

Liquid methanol density ρ methanol = 0.79 g cm ‒3

Hydrogen storage

Dihydrogen is a promising fuel for the future, notably for power production or mobility purposes

Hydrogen presents a compelling alternative to fossil fuels, as it does not emit carbon dioxide during combustion, thereby helping to mitigate global warming However, efficiently storing substantial quantities of hydrogen poses significant challenges due to its low energy density at room temperature, high flammability, and the need for technological advancements to compete with fossil fuels This article explores the pros and cons of various hydrogen storage methods.

Compressing dihydrogen is one of the methods commonly used to store it The gas is stored in containers at a pressure kept between 350 and 700 bars

1 Calculate the density of an ideal dihydrogen gas at a pressure of 500 bar and at room temperature (293 K)

Dihydrogen gas is stored in a Dewar flask, a thermally insulated container, under low pressure (1 to 4 bar) and must be maintained at extremely low temperatures The melting point of hydrogen at 1 atm pressure is -259.2 °C, while its boiling point is -252.78 °C Additionally, the critical point of hydrogen is a crucial factor in its storage and handling.

2 At which temperatures can liquid hydrogen be observed?

3 Using the Clausius-Clapeyron relation, calculate the pressure needed to liquefy ideal gaseous dihydrogen at 27.15 K

In 1984, using measurements obtained from neutron diffraction, G J Kubas and his collaborators (G J Kubas et al., J Am Chem Soc., 1984) identified a tungsten complex

The complex W(CO)3(P(iPr)3)2(η2-H2) features a hydrogen-hydrogen bond measuring 0.82 Å, which is comparable to the bond length of an isolated H2 molecule at 0.74 Å This compound readily dissociates in a partial vacuum or an argon atmosphere but can be reformed when exposed to dihydrogen.

To determine the mass of the dehydrogenated complex required to store 1 kg of dihydrogen, it is essential to calculate the density of hydrogen in the complex, denoted as ρ H This density is defined as the mass of hydrogen atoms per unit volume of the complex By performing these calculations, one can establish the necessary mass of the dehydrogenated complex for effective hydrogen storage.

This section will explore the interaction of an H2 molecule with a dehydrogenated complex, which is modeled as a square-based pyramid, in the presence of various other ligands The study focuses on how the dihydrogen molecule integrates into this complex structure.

5 Give the electronic configuration of atomic tungsten Specify the number of valence electrons

6 Fill in the table with the name of each depicted atomic orbital (s, d yz, d z 2 , d (x 2 – y 2 ), d xz, d xy).

7 Draw and fill the molecular orbital diagram of dihydrogen

The complex, modeled as a square-based pyramid with the addition of an H2 molecule, requires consideration of the effects of other ligands, as illustrated in the accompanying diagram of the resulting splitting.

Figure 1: Simplified diagram of molecular orbitals of the Kubas complex

To construct the molecular orbital diagram of the Kubas complex [W(CO)3(P(iPr)3)2], we analyze the interaction between the d orbitals of the central metal atom and the molecular orbitals of the H2 molecule.

Figure 2: Kubas complex and reference axes

8 Give the two planes of symmetry of the Kubas complex (using the axes of figure 2)

9 Indicate for each orbital d of the metallic central atom if they are symmetric or antisymmetric with respect to each of the symmetry planes (using the axes of figure 2)

Two conformations have been proposed: (1) where H2 is parallel to the phosphine ligands P(iPr)3, and (2) where H2 is parallel to the CO ligands Even if steric effects favor conformation

(2), conformation (1) is actually more stable

10 Fill in the diagram in figure 1 with electrons

11 Knowing that only orbitals with the same symmetry interact, enumerate the possible interactions for each conformation Which conformation is the most stable one?

Storing hydrogen in form of formic acid

In 2006, a research team from EPFL in Switzerland, led by C Fellay, proposed an innovative method for hydrogen storage using formic acid Their approach involves utilizing formic acid as a fuel, which can be decomposed over a ruthenium catalyst to generate dihydrogen and carbon dioxide This process highlights the potential of formic acid as a sustainable energy source.

To calculate the density of hydrogen (ρ H) at 25 °C, determine the mass of hydrogen atoms per unit volume of formic acid Once calculated, compare this density to the values obtained for gaseous dihydrogen at 500 bar and for liquid dihydrogen to understand the differences in hydrogen density under varying conditions.

13 Calculate the standard enthalpy and entropy of reaction at 20 °C for reaction (R1)

14 Using the Ellingham approximation (that supposes enthalpy and entropy independent of temperature), calculate the equilibrium constant at 20 °C for reaction (R1)

Formic acid (2.3 g) is added to a 1 L container with 0.1 g of ruthenium catalyst, under constant atmospheric pressure and at an initial temperature of 25 °C The container initially contains dinitrogen

15 Determine the final composition of the mixture

Storing hydrogen in metal hydrides

Metal hydrides are being explored as an effective means to store dihydrogen, utilizing compounds with the formula XxYyHn to achieve high hydrogen storage density in a compact form The hydrogen adsorption-desorption characteristics can be optimized by selecting element X from light elements like lithium, magnesium, or boron, or from electropositive lanthanides, which exhibit strong affinity for hydride ligands, while element Y is chosen from transition metals that show low affinity for these ligands This article focuses on the operational conditions of two specific metal hydrides: LaNi5H6 at 300 K and 2 bar, and Mg2NiH4 at 550 K and 4 bar.

16 Determine ρ H (the density of hydrogen, which is defined as the mass of hydrogen atoms per volume for these two compounds in their operating conditions)

The adsorption-desorption equilibrium is represented by the phase change A(g) → A(ads), where dihydrogen behaves as an ideal gas The Clausius-Clapeyron relation effectively approximates the phase transformation of an ideal gas, allowing the latent heat to be equated with the adsorption enthalpy The accompanying tables illustrate the relationship between pressure (MPa) and temperature (K).

Table 1: Van’t Hoff plot data (pressure (MPa) as a function of the temperature (K)) of several metal hydrides (A Züttel, Naturwissenschaften, 2004)

17 Using table 1, determine the adsorption enthalpies of LaNi5H6 and Mg2NiH4.

Van der Waals gas equation: (𝑝 + 𝑛²𝑎

𝑉²) (𝑉 − 𝑛𝑏) = 𝑛R𝑇 Van der Waals coefficients for dihydrogen: a = 0.2476 L 2 bar mol ‒2 b = 0.02661 L mol ‒1

Specific latent heat of fusion (at standard pressure): Δfus H°m = 58.089 kJ kg ‒1

Specific latent heat of vaporization (at standard pressure): Δvap H°m = 448.69 kJ kg ‒1

Deacidification and desulfurization of natural gas

Liquid dihydrogen, ‒252.78 °C: 70.849 g L ‒1 compound Kubas cplx formic acid LaNi5H6 Mg2NiH4 conditions 25 °C 300 K 550 K ρ 1.94 g cm ‒3 1.22 kg L ‒1 8620 kg m ‒3 2643 kg m ‒3

Thermodynamic data at normal conditions of temperature and pressure (20 °C, 1 atm) compound HCOOH(g) HCOOH(l) CO2(g) H2(g) N2(g) Δf H° kJ mol ‒1 ‒378.60 ‒425.09 ‒393.51 0.00 0.00

Problem 6 Deacidification and desulfurization of natural gas

Approximately 95% of dihydrogen is generated through steam reforming of natural gas, a process similar to the reaction involving methane, which occurs at around 900 °C with the aid of a catalyst.

Approximately 35% to 40% of the produced dihydrogen is utilized for ammonia synthesis However, even a small concentration of sulfur—just one atom per 1000 nickel atoms—can significantly poison nickel-based catalysts Additionally, the acidic gases present in natural gas, such as H2S and CO2, can harm pipelines, necessitating the processes of deacidification and desulfurization for natural gas.

Steam reforming from natural gas

1 Give the chemical reaction of steam reforming for an alkane CnH2n+2

2 Calculate the equilibrium constant K° of reaction (1) at 900 °C

A prevalent technique for removing acidic gases from natural gas involves the use of an amine solution, with some solutions capable of solubilizing all acidic gases while others selectively target H2S or CO2 due to kinetic differences This article explores the deacidification process utilizing two specific amines: monoethanolamine (MEA) and methyl-diethanolamine (MDEA), employing the experimental apparatus illustrated below.

Flask F1 initially contains 100 mL of a 0.5 mol L ‒1 amine solution (n 0 = 50 mmol: large excess)

Flask F2 initially contains 100 mL of a 0.5 mol L ‒1 NaOH solution (large excess too)

Step 1: a gas sample (gas 1) is driven by N2 into a flask containing an amine solution; the outgoing gas (gas 2) bubbles in a second flask containing a NaOH solution; the final gas (gas

3) no longer contains acidic gas

In Step 2, the liquid contents of each flask are titrated using a hydrochloric acid solution with a concentration of 1.0 mol L ‒1 During the titration process, both pH and conductivity are monitored, resulting in the generation of two distinct curves for each experiment.

The gas sample 1 consists of n1 mmol of CO2, n2 mmol of H2S, and n3 mmol of CH3SH The initial experiment utilizes the primary amine MEA, while the subsequent experiment employs the tertiary amine MDEA.

3 Write down the thermodynamic quantitative (K° >> 1) reactions between the different gases and (i) the amine solution and (ii) the NaOH solution

We first study the experiment with MEA There is no kinetic blockage with this amine

4 Determine the amount of each species in the solution (as a function of n 1, n 2 and n 3) in the flask F1 before titration

5 Which chemical species is/are present in gas 2?

6 Using the curves A1F1 and A1F2, determine (i) n 3 and (ii) a relation between n 1 and n 2

MDEA reacts with only one of the acid species, the other reaction being kinetically blocked

7 Determine the amount of the reacting species using curve A2F1

8 Using curve A2F2, determine if MDEA selectively reacts with CO2 or with H2S Calculate the two remaining n 1 and n 2

CO2(g) H2O(g) CH4(g) C5H12(l) CO(g) CH3CH3(g) H2(g) Δf H° (kJ mol ‒1 ) ‒393.5 ‒241.8 ‒74.6 ‒178.4 ‒110.5 ‒84.0 0.0

Amines: MEAH + /MEA; MDEAH + /MDEA pK a = 9.5

Lavoisier’s experiment

In 1775, the French chemist A L de Lavoisier, father of modern chemistry, showed by an experiment that oxygen is one of the constituents of air

Lavoisier’s experiment (Bussard and Dubois, Leỗons ộlộmentaires de chimie, 1897)

The experiment he performed can be summarized as follows:

 he first introduced 122 g of mercury into a retort, the end of which was inside a cloche (see illustration above) containing 0.80 L of air and placed upside down on a tank containing mercury,

 he then heated the retort in such a way as to keep the mercury boiling for several days,

 after two days, the surface of the mercury began to get covered with red particles,

 after twelve days, the calcination of mercury seemed to have finished because the thickness of the particle layer was no longer increasing, he then stopped heating,

 after cooling, he observed the following:

 only 0.66 L of “air” subsisted under the cloche,

 this remaining “air” could extinguish a candle or kill a mouse,

 2.3 g of red particles had been formed He called them “rust of mercury” The table below shows thermodynamic data at 298 K of some mercury-based compounds and dioxygen

1 Standard molar entropy S m° of mercurous oxide Hg2O has not been experimentally determined Choose the value that seems closest to reality:

2 Write down equations for the formation of HgO(s) and Hg2O(s)

The reaction of mercury is believed to occur solely in its liquid state, leading to the potential formation of either red or yellow mercury oxide (HgO) By utilizing the value selected in the previous section, one can calculate the equilibrium constants (K°) at 298 K for red HgO, yellow HgO, and mercury(I) oxide (Hg2O).

The red and yellow forms of mercury (II) oxide exhibit similar standard potentials and comparable magnetic susceptibilities, yet the yellow variant has more significant structural defects Red mercury oxide is produced through the slow pyrolysis of Hg(NO3)2, while the yellow oxide is synthesized by precipitating aqueous mercury (II) ions in an alkaline environment.

4 Write the chemical equations of these processes

Lavoisier's experiment parallels pyrolysis due to the application of heat and the lack of an aqueous environment, which likely accounts for the production of red oxide For our analysis, we will focus solely on this as the resultant product of the reaction.

5 Calculate the theoretical amount of each species in the final state of Lavoisier’s reaction

6 Calculate the theoretical mass of mercury (II) oxide in the final state

7 Choose an explanation to the difference from the mass obtained by Lavoisier

 Other oxides of the type HgOx (x < 1) are obtained

 The yield is not maximum

 Mercury rust also contains nitride HgxNy.

Which wine is it? Blind tasting challenge

The fermentation of grape juice is an essential phase in wine production, where naturally occurring microorganisms transform the sugars in grapes, primarily glucose, into ethanol This biochemical process relies on the microorganisms found on the surface of fruits and in the surrounding environment.

1 Write a balanced equation for the transformation of solid glucose (C6H12O6(s)) into liquid ethanol (C2H6O(l)) and gaseous carbon dioxide Does this reaction require the presence of dioxygen? (Yes/No)

2 Calculate the standard enthalpy, the standard entropy and the standard Gibbs free energy associated with this reaction at 298 K Does this reaction generate heat? (Yes/No)

Conversion of glucose into carbon dioxide and water is called cellular respiration

3 Write a balanced equation for the transformation of glucose into carbon dioxide and water

Does this reaction require the presence of dioxygen? (Yes/No)

Ethanol concentration in wine can significantly differ across varieties, with German Riesling wines labeled as "kabinett" typically having an alcohol content of only 7-8% vol, while Châteauneuf du Pape wines from the Rhône Valley in France generally contain around 14% vol of ethanol.

The "percent of alcohol by volume" (ABV) is defined as the ratio of ethanol volume to the total wine volume, multiplied by 100 at 298 K Controlling ethanol concentration in grape juice during fermentation is crucial To measure ethanol concentration in wine, a protocol involves diluting wine X 50 times with distilled water, followed by the dropwise addition of the diluted wine solution.

100 mL aqueous solution of potassium dichromate (5.0ã10 ‒3 mol L ‒1 ) containing sulfuric acid (0.1 mol L ‒1 ) The volume at the equivalence point 𝑉 𝑒 is 15 mL

4 Write a balanced equation for the oxidation reaction of ethanol by dichromate anions

5 Calculate the equilibrium constant of this reaction Can it be used to determine the concentration of ethanol in wine? (Yes/No)

6 Calculate the pH of the solution of potassium dichromate and sulfuric acid before starting the titration Here, sulfuric acid can be treated as a strong monoacid

To calculate the pH of a potassium dichromate and sulfuric acid solution at the equivalence point, it is essential to recognize that sulfuric acid is a strong monoacid At this stage, the pH can indeed be determined by observing the changes in the solution's pH, confirming that monitoring pH shifts is a viable method for identifying the equivalence point.

8 Calculate the concentration (in % vol) of ethanol contained in wine X Is this wine a

German Riesling or a French Châteauneuf du Pape?

CO2(g) Glucose(s) Ethanol(l) Δf H° (kJ mol ‒1 ) ‒393.5 ‒1274 ‒277.0

Cr2O7 2−/Cr 3+ CH3COOH/CH3CH2OH

Density of ethanol at 293 K: 0.79 g cm ‒3

Nitrophenols: synthesis and physical properties

A multicomponent reaction involves three or more reactants combining to produce a product that incorporates all the reactants An example is the Ugi-Smiles coupling, which was researched by French scientists L El-Kạm and L Grimaud in 2005 Over the past decade, this coupling has been utilized for synthesizing various heterocyclic compounds through different post-condensations The reaction specifically includes an aldehyde, an amine, an isocyanide, and activated phenols, such as nitrophenols.

In this problem, the synthesis of nitrophenols is examined and some physical properties of the 4- nitrophenol are studied

In a three-neck reaction flask, 20.0 g (235 mmol) of sodium nitrate is dissolved in 50.0 mL of water The solution is cooled using an ice bath, after which 14.5 mL of concentrated sulfuric acid (H2SO4) is gradually added Additionally, a separate solution of 12.5 g of phenol in 5.00 mL of water is prepared for the reaction.

In a controlled reaction, 133 mmol of a reagent is gradually added while maintaining a temperature below 20 °C and stirring vigorously for 2 hours Following this, the mixture undergoes distillation to yield 46.5 mmol of the yellow compound 2-nitrophenol The remaining residue is cooled, and a 2.00 mol L ‒1 sodium hydroxide (NaOH) solution is introduced to adjust the pH to between 8 and 9, along with the addition of 2.00 g of charcoal The mixture is then refluxed for 5 minutes before being filtered After distilling 30 mL of water, the concentrated solution is cooled in an ice bath, and the resulting crystals are dissolved in 50.0 mL of 3.7% hydrochloric acid (HCl) and filtered, yielding 20.0 mmol of 4-nitrophenol.

1 A partial scheme for the formation of the 2-nitrophenol starting from the nitronium ion

NO2 + is proposed below Draw the missing intermediates and products

2 Give at least two products other than 2-nitrophenol and 4-nitrophenol that could explain the low yield

Various characterizations of the 2-nitrophenol and 4-nitrophenol were performed: 1 H NMR, and measurement of their melting point, boiling point and solubility The results were attributed anonymously with two labels: A and B

A (, ppm in CDCl3): 10.6 (large s, 1H), 8.1 (d, J = 8.4 Hz, 1H), 7.6 (dd, J = 8.5, 8.4 Hz, 1H), 7.2 (d, J = 8.4 Hz, 1H), 7.0 (dd, J = 8.5, 8.4 Hz, 1H)

B (, ppm in DMSO-d 6 ): 11.1 (large s, 1H), 8.1 (d, J = 9.1 Hz, 2H), 7.0 (d, J = 9.1 Hz, 2H)

3 Using the NMR data, determine which product (2-nitrophenol or 4-nitrophenol) corresponds to A and B To justify your answer, interpret the NMR chemical shifts of the products

4 Which interaction(s) between B and water can explain the higher solubility in comparison to A? Choose the correct answer(s)

A Thin Layer Chromatography (TLC) analysis was conducted on silica to assess the purity of compounds A and B The eluent used for the separation was a 7:3 volume mixture of pentane and diethylether Following the visualization of the TLC plate under UV light, the retention factors for the two observed spots were determined to be 0.4 and 0.9.

A has a ☐ lower ☐ higher retardation factor (R f) than B on the TLC because:

 A develops intermolecular hydrogen bonds with the silica

 A develops an intramolecular hydrogen bond

 B develops intermolecular hydrogen bonds with the silica

 B develops an intramolecular hydrogen bond

Absorbance The absorbance (A) versus the wavelength at various pH is given in the figure below Absorbance beyond 450 nm is negligible The two maxima of the absorbance are at

6 Which is the color of a solution of 4-nitrophenol in neutral water? Choose the correct answer

☐ Blue ☐ Green ☐ Pink ☐ Purple ☐ Red ☐ Yellow

 4-nitrophenol has a longer absorption wavelength than its conjugated base because its conjugation is more important

 4-nitrophenol has a longer absorption wavelength than its conjugated base because its conjugation is less important

 4-nitrophenol has a shorter absorption wavelength than its conjugated base because its conjugation is more important

 4-nitrophenol has a shorter absorption wavelength than its conjugated base because its conjugation is less important

In a study to determine the pKa of 4-nitrophenol, a 10 mL solution at a concentration of 1.00 × 10⁻⁴ mol L⁻¹ was titrated with a 1.00 × 10⁻² mol L⁻¹ sodium hydroxide (NaOH) solution The resulting pH changes were plotted against the volume of NaOH added, with dashed curves illustrating the percentage fractions of 4-nitrophenol and its conjugate base, 4-nitrophenolate The pH values are represented by a solid line on the graph, providing a clear visual representation of the titration process.

8 Assign each curve to 4-nitrophenol or 4-nitrophenolate

9 Estimate the pK a of the 4-nitrophenol

According to the theoretical curve, the pH jump is expected to be small, which makes the experimental titration data difficult to analyze

10 Which alternative method(s) can be used for the titration of 4-nitrophenol? Choose the correct answer(s)

French stone flower

Laumontite is a natural zeolite, a hydrated calcium aluminosilicate of formula

(CaO) x (A) y (B) z ∙yH 2 O, where A and B are oxides

In dry air, this mineral dehydrates and becomes extremely brittle, which is why it was originally referred to as zéolithe efflorescente, or "stone flower." Eventually, it was named in honor of the French mineralogist who studied it.

F Gillet de Laumont who discovered it in 1785

Laumontite crystallizes into a monoclinic crystal system of parameters: a = 1.49 nm, b = 1.37 nm, c = 0.76 nm, α = γ = 90°, β = 112°,

Z = 4 Its density is ρ = 2.17 g cm ‒3 After heating in dry air, the mineral loses 15.3% of its mass, and no further mass change is then observed

1 Calculate the stoichiometry y of the water crystallized in laumontite

Hint 1: the volume of a monoclinic unit cell is V = abc × sin β

The mass (m) of the mineral represented as 4 (CaO) x (A) y (B) z ∙ yH2O in a single unit cell can be calculated using the formula m = 4 M/N A, where M denotes the molar mass of the mineral and N A is Avogadro's constant Additionally, this mass can also be expressed as m = ρV, with ρ representing the density and V the volume of the unit cell.

To analyze the mineral composition, 0.500 g of laumontite was combined with 2 mL of concentrated hydrochloric acid and heated to 90 °C After washing the sample with distilled water and drying it at 120 °C, the insoluble residue was transferred to a new crucible weighing 14.375 g This residue was then calcined at 900 °C until a constant weight was achieved, resulting in a final mass of 14.630 g The residue obtained is identified as a pure binary compound, free of chlorine atoms.

2 Determine the nature of A and B and the values of x and z

Some samples of laumontite are orange This coloration is caused by the presence of an impurity, an element E that partly substitutes calcium, yielding the compound of formula:

A 0.500 g sample was dissolved in nitric acid, resulting in the formation of a precipitate that was then filtered The filtrate, upon the addition of a few drops of NH4SCN, developed a bright red color This solution was neutralized with an excess of concentrated ammonia (NH3), leading to the complete formation of another precipitate, which was subsequently filtered, washed, and redissolved in 1 mol L ‒1 sulfuric acid (H2SO4) An excess of zinc powder was added, and after removing the excess metallic zinc by filtration, the solution was transferred to a 100.0 mL volumetric flask and diluted to the mark with distilled water.

A 20.0 mL aliquot was transferred into a titration flask and potentiometrically titrated (using a saturated calomel electrode (SCE) as a reference) by 5.15 mL of a 2.00 mmol L ‒1 solution of Ce(SO4)2 in 1 mol L ‒1 of H2SO4

4 Write the equations of the reactions corresponding to the aliquot preparation and titration

Laumontite from Espira-de-l’Agly deposit, France (© Christian Berbain)

5 Determine the amount of impurity E (mol % compared to Ca)

6 Show that the potential at the equivalence point E e.p can be expressed as:

7 According to the following table, determine which compounds would be the two best indicators in this titration

Oxidized form Reduced form diphenylamine-4-sulfonic acid, sodium salt 0.60 blue colorless

3,3’-dimethoxybenzidine 0.54 red colorless safranin T 0.00 purple colorless

Zeolites are essential materials in heterogeneous catalysis due to their high specific surface area, unique structural framework, and abundance of acid sites Their porous structure creates a molecular sieve effect, enhancing the selectivity of reactions where the kinetic diameter of reactants and products closely matches the zeolite's pore size For instance, laumontite features the largest pores with a diameter of 0.604 nm, while the kinetic diameters of benzene, 1,4-dimethylbenzene, and toluene are 0.585 nm, and that of 1,2-dimethylbenzene is 0.680 nm.

Let us study the following reaction:

8 Draw the two main products F and G

9 This reaction can also be catalyzed by laumontite Determine which product will mainly be formed in the pore system of the mineral

E o (Ox/Red) (V /SCE) = E o (Ox/Red)(V /SHE) ‒ E(SCE)(V)

The mineral of winners

The mineral pyromorphite (from Greek pyro

‒ fire and morpho ‒ form) has the following formula: A 5 (PO 4 ) 3 B It was named after its property to recrystallize after melting Therefore, it is also sometimes called mineral of winners In

France, deposits of this mineral are found in the

Pyromorphite crystallizes into a hexagonal crystal system of parameters: a = b = 0.999 nm, c = 0.733 nm, α = γ = 90°, β = 120°, Z = 2 Its density is ρ = 7.111 g cm ‒3

After complete dissolution of 1.000 g of pyromorphite in concentrated nitric acid, the solution was neutralized with potassium hydroxide up to pH ≈ 5 An addition of 1.224 g of

KI was needed to form 1.700 g of a bright yellow precipitate

1 Determine the formula of pyromorphite

Hint 1: the volume of a hexagonal unit cell is V = abc × sin β

In a unit cell of the mineral 2 A 5(PO4)3 B, the mass (m) can be calculated using the formula m = 2M/N A, where M represents the molar mass of the mineral and N A is Avogadro's constant Additionally, the mass can also be expressed as m = ρV, where ρ denotes the density and V is the volume of the unit cell.

2 Write an equation for a reaction that could occur if the KI was added in excess

In some cases, A is replaced by the impurity C in a significant proportion The atomic mass of

The amount of impurity A is 3.98 times greater than that of C To analyze this, 1.00 g of the mineral was dissolved in HNO3, and a white precipitate formed upon adding Na2SO4, which was then filtered out The remaining filtrate was treated with an aqueous ammonia (NH3) solution, leading to the separation of C(OH)n, which was subsequently dissolved in sulfuric acid (H2SO4) To prepare for the titration of C(+n), the impurity needed to be pre-oxidized to C(+m) by heating the H2SO4 solution with Ag2S2O8 in the presence of Ag+ as a catalyst The resulting solution was then transferred to a 100.0 mL volumetric flask and diluted with distilled water A 10.0 mL aliquot was taken for titration, to which 10.0 mL of a 0.100 mol L ‒1 acidic Fe(NH4)2(SO4)2 solution was added, followed by titration with 15.0 mL of a KMnO4 solution with a concentration of 9.44 × 10 ‒3 mol L ‒1.

3 Identify the impurity C Write an equation for each reaction mentioned in the text

4 Calculate the percentage of C in the studied pyromorphite (w %)

5 Calculate the equilibrium constant of the titration reaction, for one equivalent of permanganate ions, at 298 K

Mn 2+ can be added to the solution to indicate the completeness of the C(+n) pre-oxidation reaction

Pyromorphite from Chaillac Mine, Centre, France (© Didier Descouens)

6 Write the equation of the reaction that indicates the completeness of the C(+n) pre- oxidation reaction Underline the species that allows the detection of the completeness of the reaction

7 Why is Fe(NH4)2(SO4)2 often used in redox titrations instead of FeSO4? Choose the correct answer:

 FeSO4 is not stable and get quickly oxidized by the oxygen in the air

 Fe(NH4)2(SO4)2 is more soluble than FeSO4

 Fe(NH4)2(SO4)2 is a cheaper reagent than FeSO4

Reaction progress kinetics

Kinetic investigations of multistep organic reactions are essential for understanding fundamental mechanisms and enhancing practical applications in organic synthesis Reaction progress kinetic analysis leverages extensive data from continuous reaction monitoring For instance, in a Heck transformation catalyzed by a palladium complex, the substrate reacts with the catalyst to form an intermediate, followed by further reaction with a second substrate to yield the final product while regenerating the catalyst Understanding the rate constants associated with each step is crucial for elucidating the nature of the palladium complexes involved.

1 Express the rate r of the reaction as a function of the rate constant k 2 and the instantaneous concentrations of 2 and 5 ([2] and [5], respectively)

2 Express the total concentration in catalyst [4]tot as a function of [4] and [5]

3 Assuming that intermediate 5 is in a steady-state regime, show that the rate r of the reaction can be written as:

Let us define a parameter called [“excess”], which is equal to the difference in the initial concentrations of the two substrates:

4 Show that the rate can now be written as:

Under specific conditions, the values of [4]tot and [“excess”] remain constant, making [1] the sole variable This establishes a direct correlation between the reaction rate (r) and the instantaneous concentration of reactant [1], which can be conveniently measured through absorbance readings.

Reaction calorimetry is a valuable technique for measuring the heat exchange in a reactor over time while maintaining a constant temperature In this method, we can analyze a specific transformation, such as 1 + 2 → 3, to understand the thermal behavior of the reaction.

5 Express the relationship between the heat flow dq(t) at a given time t evaluated during the period dt, the volume V of the reactor, the reaction enthalpy r H and the rate r

By integrating results from various experimental procedures, researchers can create reaction progress analysis graphs that illustrate the rate (r) as a function of the concentration of 1 The accompanying figure presents experimental data depicting the relationship between the rate of the Heck reaction and the substrate concentration, ArX Two distinct initial conditions, maintaining the same total catalyst concentration and excess values, have been analyzed to provide comprehensive insights into this relationship.

Experiment A: [ArX] 0 = 0.16 mol L ‒1 and [alkene] 0 = 0.24 mol L ‒1 Experiment B: [ArX] 0 = 0.12 mol L ‒1 and [alkene] 0 = 0.20 mol L ‒1

6 For a given concentration [ArX] on the plot, which experiment yielded more product? (Experiment A / Experiment B)

7 For a given concentration [ArX] on the plot, which is the reaction in which the catalyst has completed the more turnovers? (Experiment A / Experiment B)

In the following questions, choose the correct answer (True / False)

8 Product inhibition would reduce more the rate of the reaction in which more product is formed (True / False)

9 Catalyst deactivation would reduce the rate of the reaction where the catalyst has done more turnovers (True / False)

10 Neither catalyst deactivation nor product inhibition is a factor in the Heck reaction shown (True / False)

Experimental results for cobalt-catalyzed epoxide ring-opening reveal that a lower initial concentration of epoxide in Experiment D resulted in a slightly higher reaction rate compared to the higher concentration in Experiment C This observation indicates that factors such as product inhibition or catalyst deactivation may be affecting the reaction dynamics.

Experiment C: [epoxide] 0 = 1.5 mol L ‒1 and [H 2 O] 0 = 2.0 mol L ‒1 Experiment D: [epoxide] 0 = 1.0 mol L ‒1 and [H 2 O] 0 = 1.5 mol L ‒1

Let us assume that a new experiment, Experiment E, is performed with the same initial conditions as in Experiment D but with some product of the reaction added right from the beginning

If the reaction progress kinetic analysis curve matches the one from Experiment D, it indicates that catalyst deactivation is the cause of the observed behavior in the figure.

Nylon 6

Nylon 6 is a synthetic linear polyamide characterized by a repeating unit of six carbon atoms Initially synthesized by P Schlack at IG Farben, this polymer is primarily produced as semi-crystalline fiber yarns Known for its toughness, nylon 6 also exhibits excellent thermal and chemical resistance, making it a versatile material in various applications.

Nylon 6 is synthesized from ε-caprolactam through a catalyzed anionic ring-opening polymerization process The reaction's efficiency can be enhanced by using an acylated lactam known as I A proposed mechanism for this polymerization reaction is illustrated below.

The first acid base reaction will not be considered in the following study We further assume that no reaction other than those listed above occurs

1 Determine the relationship between the following concentrations [I]0 (initial concentration of I), [I], ∑ 𝑛 𝑖=1 [A 𝑖 ] and ∑ 𝑛 𝑖=1 [P 𝑖 ]

2 Apply the steady state approximation to all A n intermediates

3 Derive the rate of disappearance of the monomer MH as a function of the reactant concentrations [I]0, [MNa], [MH], k i and K°

Depending on the rate-limiting step, the partial order concerning the monomer MH can be either 0 or 1 Additionally, it is essential to express the conversion τ, which represents the fraction of the initial monomer concentration that has been consumed.

5 In the two limit cases studied in the previous question, draw the conversion of monomer

The monomer conversion τ vs time curve obtained by Macosco et al is the following:

6 What does the shape of the monomer conversion vs time curve stress out? Choose the correct answer(s)

 An inhibition effect of the monomer

To explain the experimental kinetic data, a competing mechanism was suggested This side reaction decreases the degree of polymerization of nylon:

The chemical structure of E is:

7 Draw possible structures of B ‒ , C and D ‒

The disappearance rate of the monomer MH is directly proportional to the product of its concentration [MH] and the difference between its initial concentration [MH]0 and its current concentration [MH] By plotting the disappearance rate against the monomer concentration [MH], we can identify the concentration at which this rate reaches its maximum value.

Synthesis of block copolymers followed by size-exclusion chromatography

Polymers and polymeric materials possess diverse properties, making them suitable for various applications However, achieving the desired combination of chemical, thermal, and mechanical properties can be challenging One effective approach to this is the creation of block copolymers, which combine different polymers to enhance their performance This article explores a synthetic strategy and characterization methodology for a block copolymer made from polystyrene and polydimethylsiloxane, highlighting its potential benefits and applications.

1 There are three kinds of polymerization initiators: anionic, cationic and free radicals Which of the following is an anionic initiator? Choose the correct answer(s)

The polymerization mechanism for polystyrene synthesis involves the initiation of styrene in the presence of n-BuLi, leading to the formation of reactive anionic species To achieve an ester functional group at the termination stage, a suitable reactive species such as an acyl chloride can be utilized, resulting in the formation of a polystyrene ester and a byproduct of lithium chloride.

In this article, we denote M as the monomer, A as the initiator, and AM_i as the growing polymer with a degree of polymerization i The propagation rate constant, k_p, remains consistent throughout each stage of the chain growth process, while k_a represents the rate constant for the initiation stage We assume that the initiation reaction occurs rapidly and proceeds to completion.

4 Express the disappearance rate of monomer M as a function of the rate constant k p, the growing polymer concentration [AM i ], the monomer concentration [M], and i

We consider that the active species concentration is constant and equal to C

5 Rewrite the disappearance rate of monomer M obtained in question 4 as a function of C,

6 Deduce from this equation the half-life time denoted t ẵ of the polymerization reaction as a function of k p and C

7 We now consider different synthesis conditions for styrene polymerization Fill in with molecules the mechanism presented below

8 Is this polystyrene synthesis regioselective? (Yes/No)

We now examine the synthesis of polydimethylsiloxane (figure below), also known as silicone These polymers are usually used in sealants, adhesives, lubricants, medicine, cooking utensils, and thermal insulation

We will also consider cyclic and short polydimethylsiloxane molecules written D n with n the number of Si atoms For example, D 5 is pictured below

9 We consider a reaction medium with D 4 and hydroxide ions Give the structures of A and

B in the mechanism below, leading to the formation of PDMS

10 During this reaction the synthesis of macrocycles is observed Draw a mechanism or a pattern to explain the formation of such macrocycles

11 Transfer reactions are also observed Draw a mechanism showing what a transfer reaction in such a reaction medium could be

Size-exclusion chromatography and synthesis of a block copolymer

Size-exclusion chromatography (SEC), also known as gel permeation chromatography (GPC), is utilized to study copolymers by introducing a sample into a column filled with microporous packing material that does not react with polymers In this setup, molecules of varying sizes elute from the column at different rates, with smaller molecules being retained longer due to their low hydrodynamic volume A detector, such as a refractive index or infrared absorption detector, measures the concentration of the eluted molecules at the column's end The resulting experimental curve can be converted into a mass fraction curve of polymer chains based on their molar mass, using monodisperse polymer standards for accurate translation.

(4) elution time s ig n a l in te n s it y

PDMS synthesis occurs in THF, initiated by D3 and n-BuLi at a specific concentration of [BuLi]0 The temperature of the reaction medium and variations in [BuLi]0 significantly affect the resulting product Additionally, we define the polydispersity index (I p) as the ratio of the weight average molecular weight (M w) to the number average molecular weight (M n).

12 To better understand the meaning of the polydispersity index, fill in the gaps in the following sentence with the word “low” or “high”:

M n is more responsive to molecules with lower molecular mass, while M w is more influenced by molecules with higher molecular mass Consequently, as the polymer chains become more uniform in length, the value of I p approaches 1.

Three experiments (V Bellas et al., Macromolecules, 2000) in different reaction conditions are performed and the polydispersity index I p is determined in each case

(I) 25 °C until 50% conversion is reached The SEC analysis gives I p =1.06

(II) conditions (I) followed by polymerization at ‒20 °C for 8 days

(III) 25 °C until 100% conversion is reached The SEC analysis gives I p =1.3

The SEC analysis or the three experiments is represented in the figure below

13 A qualitative analysis of the experimental curves obtained with SEC allows to associate which reaction conditions lead to the highest M n Which of the 3 curves is related to the highest M n?

14 Match each curve with reaction conditions (I, II or III with (a), (b) or (c))

We (finally!) synthesize the block copolymer following the procedure depicted below, and monitor the reaction by SEC

15 Suggest a structure for the polymer that is finally obtained

16 The final product is then fractionated (a fraction of the polymer chains are separated from the sample, according to their length) Associate the SEC experimental curves (1, 2, 3 or

4) measured at different stages of the synthesis (figure below) with the corresponding molecules (PS, PS-PDMS precursor, unfractionated product, or fractionated product).

Radical polymerization

Radical polymerization is a preferred method for synthesizing polymers due to its simplicity and compatibility with a variety of functional monomers This process can be conducted under diverse experimental conditions, including in the presence of water It involves three main steps: initiation, propagation, and termination The initiation step occurs through the thermal decomposition of an organic compound via a radical mechanism, generating radical species that initiate the polymerization process.

1 Considering a symmetric unimolecular initiator, give the chemical structures of the initiator and the monomer used for the synthesis of the polymer P1

At a given temperature, the half-life of the initiator t ẵ can be determined experimentally by following the evolution of the concentration of the initiator vs time

2 The table below gives the evolution of A 2 concentration over time at 82 °C in chlorobenzene Determine graphically the value of t ẵ for the initiator A 2 at 82 °C in chlorobenzene

3 Calculate the rate constant for the dissociation of the initiator A 2 , denoted k d, at 82 °C in chlorobenzene

Radical polymerization, while offering numerous benefits, has significant drawbacks, particularly due to irreversible termination reactions such as combination, disproportionation, and transfer reactions These limitations hinder the ability to produce polymers with precise architectures and compositions.

4 Among the possible termination reactions of P1, write its self-combination reaction

Recent advancements in polymerization techniques, particularly Reversible-Deactivation Radical Polymerization (RDRP), aim to minimize the irreversible termination of propagating radical chains One notable method, Nitroxide Mediated Polymerization (NMP), utilizes an alkoxyamine as an initiator When heated, these alkoxyamines dissociate homolytically to generate an alkyl radical that initiates the polymerization process, while simultaneously producing a nitroxyl radical that reversibly caps the polymer chain ends.

5 Give the chemical structure of the alkoxyamine that will be written ALK1 and is used to obtain the polymer P2

In this article, we define key variables related to polymerization: "conv" represents the monomer conversion percentage of the consumed monomer, "m" indicates the mass of styrene, "n" denotes the number of moles of initiator, and "f" is the efficiency factor, which equals 1 for alkoxamine.

6 Knowing that the number average molar mass M n can be expressed as 𝑀 n = 𝑐𝑜𝑛𝑣 × 𝑚

𝑓×𝑛 , give the number of moles and the mass of ALK1 required to obtain 10 g of a polystyrene sample exhibiting M n = 20000 g mol ‒1 at 100% conversion of styrene

RDRP techniques have simplified access to block copolymers, which consist of at least two homopolymer blocks connected by covalent bonds These copolymers merge the distinct properties of their respective homopolymers, such as combining hydrophilic and hydrophobic characteristics A notable example is the poly(hydroxyethyl acrylate)-b-poly(4-vinylpyridine) diblock copolymer (PHEA-b-P4VP, P4), which exhibits surfactant behavior.

Poly(hydroxyethyl acrylate) is hydrophilic, and poly(4-vinylpyridine) is soluble in water for pH < 6 (protonation of pyridine) and insoluble in water for pH > 6

7 In the figure below, assign the state (P4 or P5) of the PHEA-b-P4VP block copolymer in aqueous solution according to the pH values

8 Give the expected number of 1 H NMR signals with their splitting pattern for the C(=O)OCH2CH2OH side chains of the copolymer P4 dissolved in deuterated

P2 tetrahydrofuran In this solvent, the copolymer P4 is perfectly soluble (Note: no coupling will be considered with the terminal OH group)

In the chromatogram derived from Size Exclusion Chromatography (SEC), it is essential to identify the curves corresponding to different polystyrene samples The curve associated with the polystyrene sample produced through conventional free radical polymerization is labeled as P6, while the curve representing the polystyrene synthesized via Nitroxide Mediated Polymerization (NMP) is designated as P7.