CHEMISTRY: ART, SCIENCE, FUN

THEORETICAL
EXAMINATIONPROBLEMSJULY 20, 2007
MOSCOW, RUSSIA
24863 characters in Problems and Answer Sheets

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Constants and useful formulas

Gas constantAvogadro constantPlanck constants
Speed of light
Uncertainty relation

R = 8.314 J⋅K–1⋅mol–1
NA = 6.022·1023 mol–1
h = 6.626·10–34 J⋅s
h = 1.055·10–34 J⋅sc = 3.00·108 m⋅s–1∆x ⋅ ∆p ≥h2

Gibbs energy of a condensed phase at
G = pV + const
pressure p
Excess pressure caused by surface tension ∆Pin = 2σ / r
Relation between equilibrium constant andRT ln K = −∆ r Go
Gibbs energy∆G = ∆H − T ∆S
Gibbs energy at constant temperature∆G = ∆G° + RT·ln Q
product of c(products)
Isotherm of a chemical reaction
with Q =product of c(reactants)
 E 
k = A exp  − A 
Arrhenius equation
 RT 
Osmotic pressure of a solution
p =c RT
Beer- Lambert law

A = logP0= ε·lcP

V(cylinder) = πr2h
S(sphere) = 4πr2
V(sphere) =

4 3
πr
3

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Problem 1. Proton tunnelingProton tunneling through energy barriers is an important effect, which can be observed in many
complex species containing hydrogen bonds (DNA, proteins, etc.). Propanedial (malonaldehyde)
is one of the simplest molecules for which intramolecular proton transfer can occur.
1.1.1 Draw the condensed formula of propanedial and the structures of two of its isomers, which
can exist in equilibrium with propanedial.
1.1.2 In a water solution propanedial is a weak acid, its strength being comparable with that ofacetic acid. Specify the acidic hydrogen atom. Explain its acidity (choose one version in theAnswer Sheet).

Energy, arb. units

On the plot below an energy profile of the intramolecular proton transfer is given (thedependence of energy on the distance of proton motion (in nm)). Energy curve has a symmetricdouble-well form.

-0,06

-0,04

L

-0,02

0,00

Distance, nm

0,02

R

0,04

0,06

1.2.1 Draw the structures corresponding to two minima on this curve.

A proton is delocalized between two atoms and oscillates between two minima L and R with anangular frequency ω = 6.48⋅1011 s–1. Probability density for a proton depends on time as follows:1Ψ 2 ( x, t ) =  Ψ 2L ( x) + Ψ 2R ( x) + ( Ψ 2L ( x) − Ψ 2R ( x) ) cos ( ωt )  ,2
wavefunctions Ψ L ( x) and Ψ R ( x) describe a proton localized in the left and right wells,
respectively:

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Ψ

2

2

2

ΨL

-0,06

-0,04

-0,02

ΨR

0,00

0,02

0,04

0,06

Distance, nm

1.3.1 Write down the expressions for the probability density at three moments: (a) t = 0, (b) t =π/(2ω), (c) t = π/ω. Sketch the graphs of these three functions.
1.3.2 Without calculations, determine the probability of finding the proton in the left well at t =π/(2ω)
1.3.3 How much time is required for a proton to move from one well to another? What is the
proton mean speed during the transfer?
1.3.4 From the energy curve, estimate the uncertainty of the position of proton forming hydrogenbonds. Estimate the minimal uncertainty of the proton speed. Compare this value with thatobtained in 1.3.3 and draw a conclusion about the proton tunneling (choose one of the versions
in the Answer Sheet).

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Problem 2. Nanochemistry
Metals of the iron subgroup are effective catalysts of hydrogenation of СО (Fischer-Тropsch
reaction)CO+ 3H 2

Fe, CoCH 4+ H 2O

Catalyst (e.g. cobalt) is often used in the form of solid nanoparticles that have a sphericalstructure (fig.1). The reduction in size of the catalyst increases catalytic activity significantly.
The unwanted side-reaction however involves the oxidation of the catalyst:
Co(s)+ H2O (gas)CoO(s)+

H2 (gas)(1)

Solid cobalt oxide (bulk) is formed in the reaction vessel. This causes an irreversible loss of thecatalyst’s mass. Solid cobalt oxide can also be deposited on the surface of Co(s). In this case the
new spherical layer is formed around the surface of the catalyst (see figure 2) and the catalytic
activity drops.

Let us see how formation of nanoparticles affects the equilibrium of reaction (1).
2σG 0 (r ) = G 0 (bulk) + V
r0
2.1.1 Calculate the standard Gibbs energy ∆ r G (1) and the equilibrium constant for thereaction (1) at T = 500 K.
2.1.2 Calculate the equilibrium constant for reaction (1) when the cobalt catalyst is dispersedin the form of spherical particles (Fig.1) of radius(a) 10–8 m,⋅(b) 10–9 m.
The surface tension at the Co-gas interface is 0.16 J/m2. CoO forms a bulk phase.

The mixture of gases involved in the Fischer-Tropsch (CO, CН4, Н2, Н2O) reaction was put intoa reaction vessel containing the cobalt catalyst. The total pressure is р = 1 bar, temperature is T =500 K. The mole fraction of hydrogen (%) in the mixture is 0.15%.

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2.2.1 At what minimum mole fraction of water (%) in the gas mixture the unwanted
spontaneous oxidation of the catalyst becomes possible so that solid bulk CoO may appear in thesystem? Assume that cobalt catalyst is in the form of(a) a bulk phase(b) spherical nanoparticles with ra = 1 nm (Fig. 1).
2.2.2 What would you suggest to protect Co nanoparticles from the spontaneous oxidation withthe formation of bulk CoO at a constant ratio p (H 2 O) / p (H 2 ) and a constant temperature:(a) to increase ra;(b) to decrease ra;(c) change of ra has no effect.

Assume now that solid cobalt oxide forms a spherical layer around a nanoparticle of cobalt. In
this case the nanoparticle contains both a reactant (Co) and a product (CoO) (fig. 2).
In the following problems denote surface tensions as σCoO-gas, σCoO-Co, radii as ra, rb, molar
volumes as V(Co); V(CoO).
2.3.1 Write down the expression for the standard molar Gibbs function of CoO.
2.3.2 Write down the expression for the standard molar Gibbs function of Co.

Hint. If two spherical interfaces surround a nanoparticle, the excess pressure at its centre is givenby the expressionσσPin − Pex = ∆P = ∆P1 + ∆P2 = 2 1 + 2 2
r1
r2
ri, σi are radius and surface tension at the spherical interface i, respectively.
2.3.3 Express the standard Gibbs energy of the reaction (1) ∆ r G 0 (1, ra , rb ) in terms of σCoO-gas,

σCoO-Co, ra, rb, V(Co); V(CoO) and ∆ r G 0 (1) .
2.3.4 When spontaneous oxidation of Co begins the radii of two layers in the nanoparticle (Fig.
2) are almost equal, ra = rb = r0, and ∆ r G 0 (1, ra , rb ) = ∆ r G 0 (1, r0 ) . Assume that σCoO-gas = 2σCoO-Co .

Which plot in the Answer Sheet describes correctly the dependence of ∆ r G 0 (1, r0 ) on r0 ?
2.3.5 What would you choose to protect Co nanoparticles from the spontaneous formation of
the external layer of CoO at a constant ratio p (H 2 O) / p (H 2 ) and a constant temperature:a) increase r0b) decrease r0c) change of r0 has no effect.
.

Reference data:
Substance
Co (s)CoO (s)H2O (gas)ρ, g/cm3
8.90
5.68

o , kJ/mol∆ f G500–198.4–219.1
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Problem 3. Unstable chemical reactions
Many chemical reactions display unstable kinetic behavior. At different conditions(concentrations and temperature) such reactions can proceed in various modes: stable, oscillatory
or chaotic. Most of these reactions include autocatalytic elementary steps.
Consider a simple reaction mechanism involving autocatalytic step:k1
B + 2X → 3X
k2
X + D →P
(В and D are reagents, X is an intermediate and P is a product).

3.1.1 Write down the overall reaction equation for this two-step mechanism. Write the rateequation for X.
3.1.2 Deduce a rate equation using steady-state approximation. Find the orders:(i) a partial reaction order with respect to B;(ii) a partial reaction order with respect to D;(iii) the overall order of a reaction.

Let the reaction occur in an open system where reagents B and D are being continuously added
to the mixture so that their concentrations are maintained constant and equal: [B] = [D] = const.
3.2.1 Without solving the kinetic equation draw the kinetic curve [X](t) for the cases: 1) [X]0 >k2/k1; 2) [X]0 < k2/k1.
3.2.2 Without solving the kinetic equation draw the kinetic curve [X](t) for the case when thereaction proceeds in a closed vessel with the initial concentrations: [B]0 = [D]0, [X]0 > k2/k1.

Much more complex kinetic behavior is possible for the reactions with several intermediates.
Consider a simplified reaction mechanism for cold burning of ethane in oxygen:
k1
C2 H 6 + X + ... → 2X
k2
X + Y →2Y + ...
k3
C2 H 6 + Y + ... → 2PUnder specific conditions this reaction displays oscillatory behavior:Intermediates are peroxide C2H6O2 and aldehyde C2H4O, P is a stable product.
3.3.1 Identify X, Y, and P. Fill the blanks in the reaction mechanism.

Behavior of unstable reactions is often controlled by temperature which affects the rate constants.
In the above oxidation mechanism oscillations of concentrations are possible only if k1 ≥ k2.
Parameters of the Arrhenius equations were determined experimentally:
Step
EA, kJ⋅mol–1
A, cm3⋅mol–1⋅s–1
90
1
1.0⋅1011
12100
2
3.0⋅10
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3.4.1 What is the highest temperature at which oscillatory regime is possible? Show yourcalculations.

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Problem 4. Determination of water by Fischer titraton
Determination of water by the classical Fischer method involves titration of a sample solution (orsuspension) in methanol by a methanolic iodine solution, containing also an excess of SO2 andpyridine (C5H5N, Py) – Fischer reagent. The following reactions occur during the titration:
SO2 + CH3OH + H2O + I2 = 2HI + CH3OSO3H
Py + HI = PyH+IPy + CH3OSO3H = PyH+CH3OSO3Iodine content is usually expressed in mg of water reacting with 1 mL of the titrant solution(hereunder T, mg/mL), which equals the mass of water (mg) reacting with 1.00 mL of the iodine
solution. T is determined experimentally by titration of a sample with a known water content.
The sample may be, for example, a hydrated compound or a standard solution of water inmethanol. In the latter case it should be taken into account that methanol itself can containcertain amount of water.
In all calculations please use the atomic masses accurate to 2 decimal points.
4.1. Sometimes titration of water is performed in pyridine medium without methanol. How would
the reaction of I2 with SO2 and H2O occur in this case? Write down balanced reaction equation.

Calculate the T values of iodine solution in each of the following cases:
4.2.1. 12.20 mL of Fischer reagent solution were used for titration of 1.352 g of sodium tartrate
dihydrate Na2C4H4O6.2H2O.
4.2.2. A known amount of water (21.537 g) was placed into a 1.000 L volumetric flask which
was filled by methanol up to the mark. For titration of 10.00 mL of the obtained solution, 22.70
mL of Fischer reagent solution were needed, whereas 2.20 mL of iodine were used for titrationof 25.00 mL of methanol.
4.2.3. 5.624 g of water were diluted by methanol up to a total volume of 1.000 L (solution A);
22.45 mL of this solution were used for titration of 15.00 mL of a Fischer reagent (solution B).
Then 25.00 mL of methanol (of the same batch as used for the preparation of solution A) and10.00 mL of solution B were mixed, and the mixture was titrated by the solution A. 10.79 mL ofthe latter solution were spent.
4.3. An inexperienced analyst tried to determine water content in a sample of CaO using Fischer
reagent. Write down the equation(s) of reaction(s) demonstrating possible sources of errors.

For the titration of 0.6387 g of a hydrated compound Fe2(SO4)3·xH2O, 10.59 mL of iodine
solution (T = 15.46 mg/mL) were consumed.
4.4.1. What other reaction(s), beside those given in the problem, can occur during the titration?
Write down the equations of two such processes.
4.4.2. Write down an equation of the overall reaction of Fe2(SO4)3·xH2O with the Fischer
reagent.
4.4.3. Calculate the composition of the hydrate Fe2(SO4)3·xH2O (x = integer).

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Problem 5. A mysterious mixture (organic hide-and-seek game)An equimolar mixture X of three colorless organic liquids A, B, C was treated by water with adrop of hydrochloric acid at heating to give, after separation from water, a 1:2 (molar ratio)
mixture of acetic acid and ethanol without any other components. To the mixture after hydrolysis
a catalytic amount (one-two drops) of concentrated sulfuric acid was added, and after long reflux(boiling with reflux condenser) a compound D, a volatile liquid with pleasant smell, was formed
in 85% yield. Compound D is not identical to any of A, B, C.
5.1.1 Draw the structure of compound D?
5.1.2 Which class of organic compounds does D belong to? Choose the proper variant fromthose given in the Answer Sheet.
5.1.3 Even if the reflux is continued twice as long, the yield of D would not exceed 85%.
Calculate the expected yield of D if 1:1 (molar ratio) mixture of ethanol and acetic acid is taken.
Assume that: a) volumes do not change during the reactions; b) all concomitant factors, such as
solvent effects, non-additivity of volumes, variation of temperature, etc. are negligible. If you
cannot make a quantitative estimate, please indicate whether the yield will be: a) the same(85%); b) higher than 85%; c) lower than 85%.
1

H NMR spectra of compounds A, B, C look very similar and each shows singlet, triplet andquartet with the ratio of integral intensities equal to 1:3:2.
The same mixture X was subjected to alkaline hydrolysis. A remained unchanged, and wasseparated. The remaining solution after acidification and short boiling gave 2:3 (molar ratio)
mixture of acetic acid and ethanol with evolution of gas.
The mixture X (3.92 g) was dissolved in diethyl ether and underwent hydrogenation in thepresence of Pd on charcoal catalyst. 0.448 L (standard conditions) of hydrogen were absorbed,
but after the reaction A and C were isolated unchanged (3.22 g of mixture were recovered) while
neither B, nor any other organic compounds except diethyl ether could be identified afterhydrogenation.
5.2.1 Determine and draw the structures of A, B, and C.
5.2.2 Which intermediate compounds are formed during the acidic hydrolysis of C, and basic
hydrolysis of B.

The reaction of either B or C with acetone (in the presence of a base) with subsequentacidification by dilute HCl at gentle heating gives the same product, senecioic acid (SA), a
compound widely occurring in Nature. Alternatively, senecioic acid can be obtained fromacetone by treating it with concentrated HCl and subsequent oxidation of the intermediate
product by iodine in alkaline solution. The latter reaction gives, besides sodium salt of senecioic
acid, a heavy yellow precipitate E (see the scheme 2).

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B or CO

1. Me2CO/base
2. HCl, t
1. HCl cat.
2. I2, NaOHSA
C5H8O2(1)

SA (sodium salt) + E(2)

5.3.1 Determine the structure of senecioic acid and draw the reaction scheme leading tosenecioic acid from acetone.
5.3.2 Give structure of E.

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Problem 6. Silicates as the base of the Earth crust
Silica and compounds derived from it, silicates, constitute ca. 90 % of the Earth crust substances.
Silica gives rise to a beautiful material – glass. Nobody knows exactly how glass was discovered.
There is a well-favored story related to Phoenician sailors who fused occasionally sea sand andsoda ash. It is likely that they discovered the secret of “liquid glass” (LGL) – sodium metasilicate(Na2SiO3) soluble in water.
6.1.1 The solution of LGL was used earlier as office glue. Write down the net ionic equationaccounting for the ability of LGL to set in air.
Hydrolysis of LGL in water allows obtaining a colloidal solution of silicic acid.
6.1.2. Complete the Table in the Answer Sheet. Write down the net ionic equations matching theprocesses enumerated in the Table. For each process check the “Yes” box if it leads to changesof pH. Otherwise check the “No” box.
The structure of species occurring in aqueous solutions of silicates is rather complex. However,
it is possible to distinguish the main building block of all species – orthosilicate tetrahedron(SiO44-, 1):

n-

(1)

For [Si3O9] ion found in aqueous solution of silicates:
6.2.1 Determine the charge (n).
6.2.2 Determine the number of oxygen atoms bridging adjacent tetrahedra.
6.2.3 Depict its structure joining together several tetrahedra (1). Take into account that any
adjacent tetrahedron shares one vertex.
Charged monolayers with the composition [Si4O10]m- are found in kaolinite (clay).
6.2.4 Using the same strategy as in 6.2.1-6.2.3, depict a fragment of the layered structure joining
16 tetrahedra (1). Note that 10 tetrahedra have shared vertices with 2 neighbors each, and therest 6 have shared vertices with 3 neighbors each.
Being placed into the LGL solution, salts of transition metals give rise to fancy “trees” tinted
relevant to the color of the salt of the corresponding transition metal. Crystals of CuSO4·5H2O
produce “trees” of blue color, whereas those of NiSO4·7H2O form green “trees”.
6.3.1 Determine the pH of 0.1 M aqueous solution of copper sulfate at 25°С, assuming that its
hydrolysis occurs in small degree only. Use the value of the first acidity constant of[Cu(H2O)4]2+ KaI=1·10-7 M.
6.3.2 Write down equation of a reaction between aqueous solutions of CuSO4 and sodium
metasilicate (LGL). Take into account the pH values of aqueous solutions of the salts.

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Problem 7. Atherosclerosis and intermediates of cholesterol
biosynthesis
Cholesterol is a lipid wide-spread in living nature. Disruption of its metabolism leads toatherosclerosis and related potentially fatal diseases.
Substances Х and Y are two key intermediates of cholesterol biosynthesis in animals.
Х is an optically active monocarbonic acid composed of atoms of only three elements. It is
formed in organisms from (S)-3-hydroxy-3-methylpentanedioyl-coenzyme A (HMG-CоА). This
reaction is catalyzed by enzyme Е1 (which catalyses two types of reactions) and does not
involve water as a substrate. Х is further metabolized into Х1 through a three-stage process
requiring enzymes E2, E3, E4, which catalyze reactions of one and the same (and only one) type.
Finally, Х1 spontaneously (non-enzymatically) decomposes to give isopentenyl pyrophosphate(3-methylbut-3-enyl diphosphate, IPP) and inorganic products:
HO

S
O

OH O

CoA

E1

X

E2, E3, E4

X1*

Scheme 1

HMG-CoA

O
O
OP
P
O
* O O
OIPP

7.1.1 In the Answer Sheet, choose the reaction type(s) for Е1 and Е3.
7.1.2 Draw the structure of X with stereochemical details and indicate absolute configuration (Ror S) of the stereocenter.
Y is an unsaturated acyclic hydrocarbon. Its reductive ozonolysis leads to a mixture of only threeorganic substances Y1, Y2 and Y3 in a molar ratio of 2:4:1. Y is formed as a result of a numberof successive coupling reactions of two isomeric substances: IPP and dimethyl allyl
pyrophosphate (3-methylbut-2-enyl diphosphate, DAP) with subsequent reduction of a doublebond in the final coupling product Y5. Carbon atoms IPP and DAP involved in the formation of
C-C bonds during biosynthesis of Y are marked with asterisks.
O
O
OP
P
* O
O
O- ODAP7.2.1 Write down the overall reaction equation for reductive ozonolysis of DAP, if dimethyl
sulfide is used as the reducing agent.

The product of the final coupling reaction (hydrocarbon Y5) is formed when two hydrocarbon
residues (R) of intermediate Y4 are combined:OO
2 R O P O P OOO-

Y4

Y5

+2H
R R

Y
2PPi

Scheme 2

At each coupling stage but that shown in Scheme 2, pyrophosphate is released in a molar ratio of
1:1 to the coupling product.
7.2.2 Determine molecular formula of Y, if it is known that Y2 and Y3 contain 5 and 4 carbon
atoms, respectively.

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7.2.3 Calculate the number of IPP and DAP molecules needed to give Y5, if it is known that all
carbon atoms of isomeric pyrophosphates are incorporated into Y.
7.2.4 Draw the product of coupling reaction of one IPP molecule with one DAP molecule (C-Cbond can be formed only by carbon atoms marked with asterisks), if it is known that subsequent
reductive ozonolysis of the product of the coupling reaction gives Y1, Y2 and one more product,
the latter containing phosphorus.

The only double bond reduced in Y5 during its metabolism into Y was formed in the reaction
described in Scheme 2. All double bonds in Y and Y4 exist in trans configuration.
7.2.5 Draw structures of Y and Y4 with stereochemical details.

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Problem 8. ATRP allows new polymers
ATRP (Atom Transfer Radical Polymerization) is one of the most promising novel approaches
towards polymer synthesis. This modification of radical polymerization is based on a redox
reaction of organic halides with complexes of transition metals, Cu (I) in particular. The process
can be described by the following scheme (M – monomer, Hal – halogen):
R-Hal+Cu(+)Hal(Ligand)k

.

kact

R +Cu(2+)Hal2(Ligand)kkdeactkreinic(+)

R-M-Hal+Cu Hal(Ligand)kk'deact

kP +M.

R-M +Cu(2+)Hal2(Ligand)k...
kpkreinic
R-Mn-Hal+Cu(+)Hal(Ligand)k. .

R-M y +R-M x

k'deactkt(n-1)M.

R-Mn +Cu(2+)Hal2(Ligand)kR-M (y+x)R

The reaction rate constants are:kact - all activation reactions, kdeact – all reversible deactivation reactions, kp - chain propagation,and kt - irreversible termination.
8.1.1 Write down expressions for the rates of ATRP elementary stages: activation (vact),
deactivation (vdeact), propagation (vp) and termination (vt). Write down generalized equation
assuming just one reacting species R’X.

Consider that the total number of polymeric chains is equal to that of initiator molecules.
Assume that at each moment throughout polymerization all chains are of the same length.
8.1.2 Compare the rate of deactivation to the rates of ATRP elementary stages.

Dependence of monomer concentration ([M]) on reaction time (t) for ATRP is: [M ]  = − k p ⋅ [ R ⋅ ] ⋅ t ,ln
 [ M ]0 
[M]0 - initial monomer concentration, kp – rate constant of propagation, [R·] – concentration ofactive radicals.

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To prepare a polymer sample by using ATRP, catalytic amounts of CuCl, organic ligand (L) and
31.0 mmol of monomer (methylmethacrylate, or MMA) were mixed. The reaction was initiatedby adding 0.12 mmol of tosyl chloride (TsCl). Polymerization was conducted for 1400 s. kp is1616 L·mol-1s-1, and the steady state concentration of radicals is 1.76·10-7mol·L-1.
H2 C

CH3

CH3

O
O
CH3

H3 C

O

H2 C

O
OSi(CH 3 ) 3

SO2Cl

MMA

TsCl

HEMA-TMS

8.2.1 Calculate mass (m) of the polymer obtained.

In another experiment the time of MMA polymerization was changed, all the rest reaction
conditions being the same. The mass of the obtained polymer was 0.73 g. Then 2(trimethylsilyloxy)ethyl methacrylate, HEMA-TMS (23.7 mmol) was added to the mixture andpolymerization was continued for another 1295 s. MMA and HEMA-TMS reactivities are thesame under reaction conditions.
8.2.2 Calculate degree of polymerization (DP) of the obtained polymer.
8.2.3 Depict the structure of the obtained polymer (including end groups), showing MMA and
HEMA-TMS units as A and B, respectively. If necessary, use the symbols in the copolymer
structure representation: block (block), stat (statistical), alt (alternating), grad (gradient), graft(grafted). For example, (A65-graft-C100)-stat-B34 means that chains of polymer C are grafted onunits A in the statistic copolymer of A and B.
ATRP was applied to synthesize two block copolymers, P1 and P2. One block in both blockcopolymers was the same and was synthesized from mono-(2-chloropropionyl)-polyethyleneoxide used as a macroinitiator:
O
O
H 3C

CH3
58O

Cl

The other block in P1 consisted of styrene (C), and in P2 of p-chloromethylstyrene (D) units.
1

H NMR spectra of the macroinitiator, P1 and P2 are given below. Integral intensities ofcharacteristic signals can be found in the table.
8.3.1 Assign 1H NMR signals to substructures given in the Answer Sheet.
8.3.2 Determine molar fractions of units C and D and molecular weights of P1 and P2.
8.3.3 Write down all possible reactions of activation occurring during the synthesis of P1 and P2.
You may use R symbol to depict any unchanged part of the macromolecule, but you should
specify what substructure you use it for.
8.3.4 Draw the structure of P1 and one of possible structures of P2 representing poly(ethyleneoxide) chain by a wavy line and showing units of co-monomers as C and D, respectively.

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