47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

Life is a huge lab

THEORETICAL
EXAMINATIONJULY 25, 2015
BAKU, AZERBAIJAN

1

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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General Directions-

You have 5 hours to fulfill the tasks. Failure to stop after the STOP command may resultin zero points for the current task.

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Write down answers and calculations within the designated boxes. Give your work whererequired.

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Use only the pen and calculator provided.

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If you need draft paper use the back side of the paper. It will not be marked.

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There are 40 pages in the booklet including the answer boxes, Cover Sheet and Periodic
Table.

-The official English version is available on demand for clarification only.

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Need to go to the restroom – raise your hand. You will be guided there.

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After the STOP signal put your booklet in the envelope (don’t seal), leave at your table.
Do not leave the room without permission.

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You have additional 15 minutes to read the whole set.

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Formulas necessary for solution of some problems can be found on the next page.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Physical Constants, Units, Formulas and Equations

Universal gas constantR = 8.3145 J·K–1·mol–1

Standard pressurep° = 1 bar = 105 Pa = 750 mmHg

Atmospheric pressure

1 atm = 1.013×105 Pa = 760 mmHg

Zero of the Celsius scale

273.15 K

Reversible adiabatic process for an ideal gas

pV 1+ R / CV = const

Work made on an ideal gas in an adiabatic
process

W = nCV (T2 – T1)

Dependence of internal energy on temperature

U(T2) = U(T1) + CV (T2 – T1)

Relation between molar isobaric and isochoric
heat capacities for an ideal gas

Cp = CV + R

Gibbs energy

G = H – TS

Relation between equilibrium constant andstandard Gibbs energy ∆Go K = exp  − RT a∆G = ∆G o + RT ln prod ,areag

Dependence of Gibbs energy of reaction on
concentration or pressurea = c / (1 mol/L) for the substances insolution, a = p / (1 bar) for gases

Change of Gibbs energy per unit volume in time
for the system with two chemical reactions 1 and
2 with rates r1 and r2

3∆GSyst∆t= ∆G1r1 + ∆G2 r2

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Problem 1. New and well-forgotten old refrigerants(8 points)
1

Question

2

1.1

1.2

1.3

2.1

2.2

2.3

4

2

2

1

1

1

Marks

3
3

4
4.1

4.2

4.3

4.4

10

2

6

1

Total
33

The problem of choosing a refrigerant for refrigeration and air conditioning systems attracted theattention of scientists and technologists throughout the last century. It has been suggested thatduring this time refrigerants progressed through four generations. Ammonia, which was ascribed to
the first generation, had been used in most of theoldest refrigeration units. It was later replaced bychlorofluorocarbons (CFCs) – derivatives ofmethane and ethane with the hydrogen atoms
replaced by fluorine and chlorine.
In Baku, at "Bakkonditsioner" factory, productionof the first Soviet serial household air conditioners
BK-1500 had been launched. A second-generation
refrigerant chlorodifluoromethane CHF2Cl wasused in them. In this problem, we compare various
refrigerants in terms of thermodynamics.
First air conditioner of Baku factory in asouvenir shop in the Old City (“Icheri Sheher”)
Thermodynamic properties of various refrigerants

Refrigerant“Generation”∆Hvap /
kJ·mol–1(at 280 K)

CV(gas) /J·K–1·mol–1

NH3

1

21.3

26.7

CHF2Cl

2

20.0

48.8

CF3CH2F

3

22.1

79

CF3CF=CH2

4

19.1

120

Consider a model refrigeration cycle consisting of 4 steps schematically shown below in thepressure (p) – internal energy (U) coordinates.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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p
p2

3
liquidp1

2 (T2)
liquid + gas
gas

1 (T1)

0 (T1)U
Diagram 1(dashed line indicates the phase boundaries)
During the first step of the cycle (line 0-1 in diagram 1), a liquid refrigerant is boiling at constantpressure p1 and temperature T1 (boiling temperature) until it completely evaporates. At this step, therefrigeration unit absorbs heat from surrounding objects. At the second step, the refrigerant
undergoes reversible adiabatic compression and heats up to temperature T2 (line 1-2). After that thecompressed refrigerant is cooled in a condenser at constant pressure p2 (line 2-3) and then returns to
the initial state (line 3-0).
Let the cycle involve 1 mole of refrigerant, which is initially (point 0) completely liquid, T1 = 280
К, T2 = 380 К, assume that the vapor of any refrigerant behaves like an ideal gas. The
thermodynamic characteristics of refrigerants are listed in the table above.
1.1. For each of refrigerants, ammonia and chlorodifluoromethane, calculate the amount of heat Q
absorbed by refrigeration unit during heat exchange (line 0-1) and the work W required to compress
its vapor adiabatically (line 1-2).
Calculations
Ammonia

Q=

W=

Chlorodifluoromethane5

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
Q=

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W=

1.2. Which quantity(ies) remain(s) constant during the adiabatic compression step? Indicate by thecircle(s).
U

H

S

G

V

To compare the energy efficiency of refrigeration cycles with different parameters and refrigerants,
the coefficient of performance (COP) is used, which is defined as a ratio of heat removed from acooled system to the work of compressor: COP = Q/W.
1.3. Calculate the values of COP in a considered cycle for ammonia and chlorodifluoromethane.
Calculations
Ammonia

COP =Chlorodifluoromethane

COP =2.1. Why was ammonia replaced by CFCs in household refrigeration units? (Choose only one
option)a) to increase the energy efficiency of refrigeration cyclesb) because the density of ammonia is less than that of air under the same conditionsc) for user safety reasons

A search for replacement of CFCs as refrigerants started when it was shown that their use can cause
irreparable damage to the protective ozone layer of the atmosphere. The third, ozone-friendly
generation of refrigerants came on the scene. Its typical representatives are fluoroalkanes.
2.2. What is the cause of the damage made by CFCs to the ozone layer? (Choose only one option)
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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

ABC-1a) ozone molecule easily adds to C–F bondb) C–F bond is easily broken by radiation, which leads to the formation of free radicalsc) ozone molecule easily adds to C–Cl bondd) C–Cl bond is easily broken by radiation, which leads to the formation of free radicals

However, under the 1997 Kyoto Protocol, fluoroalkanes also had to be replaced because they
accumulate in the atmosphere and rapidly absorb infrared radiation, causing a rise in temperature of
the atmosphere (the greenhouse effect). The refrigerants of the fourth generation such as 2,3,3,3tetrafluoropropene CF3CF=CH2 have been suggested and are coming into use.
2.3. Why does this compound enhance the greenhouse effect less than fluoroalkanes? (Choose only
one option)a) it is more reactive and easier to decomposeb) it easily reacts with ozonec) it is better soluble in water

3. Calculate the values of the COP in the refrigeration cycle considered above for two refrigerantsof the third and fourth generations – CF3CH2F and CF3CF=CH2. Did the energy efficiency improvein comparison with CHF2Cl? Choose “Yes” or “No”.
Calculations
CF3CH2F
COP =
Yes

No

CF3CF=CH2

COP =

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
Yes

ABC-1

No

Unlike household appliances, industrial refrigeration systems are often still using ammonia. It does
not contribute to the greenhouse effect nor does it destroy the ozone layer. Industrial units can havea huge size and a large cost. Prior to their construction, they should be carefully modeled taking into
account many different factors. In real systems, some part of the refrigerant at the start of the heatexchange with the environment is in the vapor phase (point 0 in the diagram below), and at the end(point 1) it is always overheated above the boiling point.

p
3 (T3)

p2
liquidp1

2 (T2)

liquid +gasgas

1 (T1)0 (T0)U
Diagram 2(dashed line indicates the phase boundaries)
Consider a cycle with 1 mole of ammonia. Its thermodynamic properties are the following: enthalpy
of vaporization ∆Hvap = 23.35 kJ·mol–1 at Tvap = 239.8 К (boiling temperature at 1 bar pressure).
Heat capacity of the liquid phase CV(liq) = 77 J·K–1·mol–1, of the gas phase CV(gas) = 26.7 J·K–1·mol–1.
Assume that the heat capacities are temperature-independent and the vapor behaves like an ideal
gas. The temperature dependence of the saturated vapor pressure of ammonia can be described by
the empirical equation:log (p/bar) = 4.87 – 1114 / (T/K – 10.4).
During the first step of the cycle (line 0-1 in diagram 2), the equilibrium mixture of liquid
refrigerant and its vapor receives heat from the environment at constant pressure p1 = 3.0 bar. The
refrigerant completely evaporates and overheats up to the temperature T1 = 275 K. In the beginningof the process (point 0), the molar fraction of gaseous ammonia is x = 0.13.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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4.1. Calculate the initial temperature of refrigerant T0, its volume change ∆V and the amount of heat
Q absorbed by refrigeration unit during this step. Take into account that the dependence of ∆Hvap
from the temperature cannot be neglected.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Calculations:

T0 =

∆V =

Q=

Then the refrigerant is reversibly and adiabatically compressed. It heats up to the temperature T2 =393 К (line 1-2).
4.2. Find the work W required for compression and the COP of the system. If you were not able tofind Q in 4.1, use Q = 20.15 kJ.
Calculations:

W=

COP =

At the next step corresponding to the line 2-3 in diagram, the compressed refrigerant is cooled in acondenser at constant pressure. Then it returns to the initial state through adiabatic expansion withzero work (line 3-0).
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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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4.3. Determine the temperature T3 at point 3 to which the refrigerant is cooled in a condenser.
Calculations:T3 =In the production of refrigeration units it is necessary to consider climatic factors. If a condenser iscooled by atmospheric air, the temperature T3 increases as the air temperature increases.
4.4. How will the COP change if T3 increases while T0, T1, T2 remain the same?
a) Increaseb) Remain the samec) Decrease11

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Problem 2. Coupling of chemical reactions(7 points)
Question
Marks

I.Prigogine (left)

1

2

1.1

1.2

1.3

2.1

2.2

4

6

4

3

6

N. Shilov

3

Total

2

25

W. Ostwald

When in the system one reaction allows another one to proceed they say that these two reactions arecoupled. Ilya Prigogine, Nobel prize winner in chemistry (1977) in his books widely used theconcept of “coupled reactions”. Coupling of reactions is an essential feature of living systems,
including human body.
How one reaction makes another one to occur? In this problem we are going to discuss several
possible mechanisms of coupling.
(I) “Chemical coupling”“On Chemical coupling” was the title of the dissertation defended by Russian chemist N.Shilov in
1905. N. Shilov was the graduate student of famous professor W. Ostwald. Dr. Shilov described thefollowing set of reactions.
The substance А does not react with Ac. In the presence of the third reagent (called inductor), In,
however, the reaction of А with Ac takes place:In the absence of In
A + Aс 
→ no reaction!

(1)

In the presence of In
A + Ас 
→ Р1

(2)

In is not a catalyst! Its concentration decreases in the course of the reactions.
According to the scheme proposed by Shilov, Ас reacts not with A itself, but with the intermediate
product R of the reaction of А with In. There is another, competing reaction of R that forms P2.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

(b)k (3a )
αA + β In →Rk (3b)
R 
→ P2(c )k (3c )R + Ас 
→ P1

(a )

ABC-1

(3)

α and β are stoichiometric coefficients. Other stoichiometric coefficients and reaction order withrespect to all reactants in all three reactions are unity.
In the Shilov’s experiments the ratio of the consumed amounts of Аc and In, I =∆n Ас
increased up∆n In

to the constant value with the increasing initial concentration [Ac]0 at [In]0 = const.
1.1. What was this limiting constant value of I at [Ac]0 → ∞, [In]0 = const?
Brief explanation

I∞ =1.2. Derive an expression for I using the steady-state approximation if necessary. Plot the graph of
I vs [In]0 at [Ac]0 = const. Assume that In was completely consumed and Аc was in excess..
Calculations

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Graph

I =∆nAc/∆nIn[Ac]0 = const[In]0

What if Shilov’s mechanism is not valid and In is a conventional catalyst of the reaction (2)?
Simultaneously In reacts with А and its concentration decreases. The reaction scheme in this case is(a )αA + βIn 
→ P2

(b)

In, catalysis
A + Ас → P1(4)

1.3. What is the limiting value of I for the reaction scheme (4) at [Ac]0 → ∞, [In]0 = const?
Brief explanation

I∞ =(II) «Kinetic coupling»
The standard Gibbs energy of the gas-phase reactionk5
→ HBr + H
Br + H 2 ←

k −5
14

(5)

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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is positive, ∆G°(5) = 66 kJ⋅mol–1 at Т = 600 К.
2.1. What is the ratio of the rates of forward and reverse reactions,r5, at this temperature, standard
r−5

pressures of H2 and HBr and equal pressures of H and Br?
Calculations

r5
=
r−5

If you could not answer this question, for further calculations use the reference value r5/r–5 =3.14⋅10–7.
Reaction (5) proceeds in the forward direction due to the reaction (6) which simultaneously occurs
in the system:k5→ HBr + H
Br + H 2 ←

k −5

(5)

k6
H + Br2 →
HBr + Br(6)

k5, k–5, k6 are rate constants of forward and reverse reaction (5) and forward reaction (6),respectively.
This is the kinetic coupling of two reactions.
Let pressures of neutral molecules keep standard values p(H2) = p(Br2) = p(HBr) = 1 bar, andpressures of radicals p(H), p(Br) reach steady-state values. Rate constant k6 is 10 times larger than
k–5.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
2.2. Calculate ∆G(5) and

ABC-1

r5
under such conditions.
r−5

Calculations∆G(5) =r5=r−5(III) ”Second law of thermodynamics restricts coupling”
According to the Second Law of thermodynamics, two simultaneously occurring chemical reactions∆GSystshould decrease the system’s Gibbs energy Gsyst,< 0.
∆t
One of these reactions may have positive Gibbs energy and still proceed in the forward direction
due to the coupling with the second reaction. This second reaction must have negative Gibbs energy
and the requirements of the Second law must be fulfilled! Consider the example.
The synthesis of urea under specific conditions
2NH3 + CO2 → (NH2)2CO + H2O∆G(7) = 46.0 kJ⋅mol

(7)

–1

is supposed to be coupled with the complete oxidation of glucose (under the same conditions)
1/6 C6H12O6 + O2 → CO2 + H2O(8)∆G(8) = –481.2 kJ⋅mol–1,

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r(8) = 6.0⋅10–8 M⋅min–1.
Both reactions are presented schematically. No other reactions are considered.
3. What is the maximum rate of the reaction (7) permitted by the Second Law if this reaction iscoupled to reaction (8)?
Calculations

r7(max) =17

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
Problem 3. Two binding centers – competition or cooperation?
(7 points)
1
Question
1.1 1.2 2.1
Marks

3

2

8

2
2.2

2.3

2.4

3

6

6

ABC-1

Total
28

Many chemical reactions in living organisms include the formation of “host-guest” complexes
where the host molecule reversibly binds one or several guest molecules. Consider a host molecule
H with two binding centers – say, a and b which have different affinities for the guest molecules G:

→ HGa
H+G ←


→ HGb
H+G ←


[ HGa ]
[ H ][G ][ HGb ]Kb =[ H ][G ]Ka =Kb ≠ Ka.

where HGa and HGb denote a complex where guest is bound to a center and b center, respectively.
Ka and Kb are the binding constants for the centers a and b, brackets denote molar concentrations.
Attachment of one G molecule to H can change the binding ability of the second centre.
This change is described by the “interaction factor” β which reflects the influence of one binding
center on another and is defined as follows:[ HG2 ]
→ HG2= βK b
HGa + G ←
[ HGa ][G ]where HG2 is the completely bound complex.
1.1. Determine the range of values (or one value, if necessary) of β which correspond to threepossible ways of interaction between binding centers: a) cooperation (binding by one center
facilitates subsequent binding); b) competition (first binding complicates the second); c)
independence (no interaction).
Cooperation:
Competition:
Independence:18

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
→ HG2 in terms of binding
1.2. Find the equilibrium constant for the process: HGb + G ←constant(s) and interaction factor.
Calculations:K=

2.1. The solution was prepared with the initial concentrations [H]0 = 1 M and [G]0 = 2 M. After thereactions were completed, the concentration of H decreased by 10 times and that of G by 4 times.
For these host and guest, Kb = 2Ka. Determine the concentrations of all other species in the solution
and find the binding constant Ka and the factor β.
Calculations:[HGa] =

[HGb] =

[HG2] =Ka =β=
If you could not answer this question, for further calculations use reference values Ka = 3.14 and β= 2.72.

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2.2. Find the correct order of standard molar Gibbs energies of formation of host H and all hostguest complexes from H and G. In the scheme below, write the corresponding chemical formula
near every line.

Go

2.3. Some amount of G was added to 1 mole of H and the mixture was dissolved in water to obtain
1 liter of the solution. The number of the totally bound molecules HG2 in the solution is equal to thetotal number of single-bound molecules HG. Find the initial amount of G (in mol). The constants
Ka and Kb and the factor β are the same as in question 2.1.
Calculations:n0(G) =

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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2.4. What would be the equilibrium composition of the solution if: a) β = 0; b) β is very large (β →∞). The constants Ka and Kb as well as the initial concentrations of H and G are the same as inquestion 2.1.

β=0Calculations:[H] =[HG2] =[G] =[HGa] =[G] =[HGa] =

[HGb] =

β→∞Calculations (or arguments):[H] =[HG2] =

21

[HGb] =

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
Problem 4. From one yellow powder to another: A simple inorganic riddle(6 points)
Question
1
2
Marks

8

8

ABC-1

3

4

Total

3

5

24

The yellow binary compound X1 was completely dissolved in concentrated nitric acid by heating,the gas evolved is 1.586 times denser than air. Upon adding an excess of barium chloride to thesolution formed a white solid X2 precipitates. It was filtered. The filtrate reacts with an excess of
silver sulfate solution forming a precipitate of two solids X2 and X3, also separated from solution byfiltration. To the new filtrate the solution of sodium hydroxide was being added drop-wise until thesolution became nearly neutral (about pH 7). At this time a yellow powder X4 (77.31 wt.% of Ag)
crystallized from the solution. The mass of X4 is nearly 2.4 times larger than that the mass of thefirst portion of X2.
1. Determine the chemical formulae of X1 – X4.
Calculations:

X1 =

X2 =

X3 =

X4 =

2. Determine the chemical formula of the gas and provide equations for all reactions in ionic or nonionic form.
Calculation
Chemical formula of the gas ________
Dissolution of X1
Formation of X2

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Formation of X2 and X3

Addition of NaOH and formation of X4

3. In the structural unit of X1 all atoms of only one element are in equivalent positions. Draw thestructure of X1.

4. Predict the products of X1 interaction with:a) excess oxygen;b) excess of hot concentrated sulfuric acid;c) solid KClO3 with grinding.
Write down the reaction equations.

a)
b)
c)

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Problem 5. Indispensable glucose(8 points)

Question
Marks

1

2

Total

1.1

1.2

1.3

1.4

1.5

1.6

2.1

2.2

2.3

2.4

2.5

2

3

6

4

6

1

2

2

4

2

2

34

Carbohydrates are the most important providers of energy for living cells. Monosaccharide glucose
is a source of energy for the living cell, but for persons who suffer from diabetes glucose may be
dangerous. High level of glucose may lead to cardiovascular diseases and even death. That is why
people avoid consuming too much carbohydrates and glucose particularly.

1. Determination of reducing sugars in fruit juice

One of the technique for determination of reducing sugars in different samples
includes the use of Fehling's reagent. A 10.00-mL aliquot of fruit juice (assuming
the initial sample contained only glucose and fructose) was transferred into atitration flask and Fehling's reagent was added. This reagent was prepared bymixing 50.00 mL of 0.04000 M copper sulfate (solution A) and potassiumsodium tartrate and sodium hydroxide (solution B). Solution C thus obtained,
was then heated and red precipitate was formed.
Glucose1.1. Write the balanced ionic equation of chemical reaction occurring upon heating of thesolution C. Use Cu2+ for initial copper solution.

After that 10 mL of 10% solution of potassium iodide and 1 M sulfuric acid were added to the flask.
The mixture was covered with watch glass and was then placed in a dark place. An excess of iodine

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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was then titrated with 0.05078 М sodium thiosulphate solution. 11.87 mL of the titrant was required
to reach the endpoint.

1.2. Write the balanced equation(s) in molecular or ionic form for all the reactions taking place inthe flask.

1.3. Consider all fructose was transformed into glucose under the experimental conditions; calculate
the total mass content of sugars (in g/L) in a fruit juice. Mw = 180.16 g/mol.

mass content =A new 10.00-mL aliquot of the same juice was treated with a 10.00-mL portion of acidified
potassium iodate(V) solution (0.01502 М) and 10 mL of 10 % solution of potassium iodide. Afterthe mixture turned brown, an excess of sodium hydroxide solution was added. The flask was thencovered with a watch glass and put into a dark place. The obtained solution was acidified andtitrated with 0.01089 M solution of sodium thiosulphate. The average titrant volume used fortitration was 23.43 mL. Note that fructose is not converted into glucose under these conditions.

1.4. Write all the balanced equations for the described reactions in molecular or ionic form.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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1.5. Calculate the mass content of each sugar (in g/L) in the juice.

mass content of glucose =mass content of fructose =1.6. One bread exchange unit (1 BEU) corresponds to the content of 12 g of digestible
carbohydrates in product. How many BEU are in one glass (200 mL) of juice?

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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2. Diagnosis of diseases
The derivative of glucose, 2-deoxy-2-(18F)fluoro-D-glucose (FDG), is the most common
radiopharmaceuticals for diagnosis of cancer using positron emission tomography. The first step of
FDG preparation is to produce a radionuclide fluoro-18 by nuclear reaction in a cyclotron. The next
step is the radiochemical synthesis. Fluorine-18 is introduced into D-glucose molecule bynucleophilic substitution. 2-deoxy-2-(18F)fluoro-D-glucose once injected into the patient actively
accumulates in cells of malignant tumors; this process is accompanied by decomposition offluorine-18. This radionuclide is a β+ emitter – nucleus emits a positron (anti-electron). Positroninteracts with an electron and after that annihilation occurs, which can be detected. This allows
determining precisely the tumor sizes and type.

2.1. Complete the nuclear reactions leading to various fluorine isotopes.
a)18→ …+ 18Fb)… + 12 D → 18F + αc)

19

F + 12 D → 20F + …

d)16

O + … → 18F +O++n2.2. The decay mode of unstable light nuclei depends on the ratio between the number of neutronsand protons in them. If this ratio is greater than that for a stable isotope then the nucleus decays in a
β–-mode, if it is smaller – in a β+-mode.
Determine the type of decay for the nuclei in the table:

Nucleus

11

С

20

17

F

F

14

C

Decay mode

When nuclear reaction (a) is used for fluorine-18 preparation, the target material is presented aswater enriched with H218O. The presence of usual water H216O leads to a side nuclear reaction with
16

O, leading to the formation of isotope 17F.

2.3. It is known that within five minutes after completion of irradiation of the target the ratio ofradioactivities of 18F and 17F is 105. Assuming that irradiation time is short and the radioactivity ofeach isotope is proportional to the nuclear reaction yield and the mole fraction of a component in27

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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the irradiated target, calculate the mass fraction of H218O in the target. t1/2(18F) = 109.7 minutes,
t1/2(17F) = 65 seconds. The ratio between nuclear reactions yields is/= 144.7.

2.4. Calculate the yield of labeling D-glucose with fluorine-18, if initial radioactivity of a fluorine18 sample was 600.0 MBq and radioactivity of the obtained 2-deoxy-2-(18F)fluoro-D-glucose is528.2 МBq. Synthesis time is 3.5 minutes.

η=
2.5. Biological half-life (through the excretory organs) of 2-deoxy-2-(18F)fluoro-D-glucose is 120.0
minutes. How much radioactivity (in MBq) will remain in the patient ten hours after injection of
FDG with the initial radioactivity of 450.0 MBq.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Problem 6. Bread is the stuff of life
(8 points)
Question

1

2

3

Total

Marks

28

4

8

40

When you pass by the bakery, you are stopped
by the smell of freshly baked bread. The hero of oneof the novels said on a similar occasion: "If you tellme that this is not perfect, you are my enemy forever."The principle bread flavour component was identified
in 1969 as compound X which occurs in equilibriumwith its tautomer Y in a 2:1 ratio. Unfortunately, bothforms are labile, and after some hours bread has no the same nice smell.
This tautomeric mixture of X and Y was synthesized in 1993 from piperidine by thereaction sequence given in Scheme 1. It is noteworthy that the initial ratio of X and Y was 1:4; on
standing this ratio gradually changed to an equilibrium one.
Scheme 1.

Compound B which is characterized by 3-fold axis of symmetry (i.e., rotation by 120° results
in a molecule indistinguishable from the original) occurs in equilibrium with its diastereomer C.
The interconversion of these two forms proceeds via intermediate A which is also intermediate in B
and C formation as well as their transformation to D. Compounds A, B, and C have the sameelemental composition: ωC = 72.24%, ωH = 10.91%, ωN = 16.85%.
1. Write down the structural formulae of compounds A-E, X, Y.
A

B

29

C

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
D

E

ABC-1

X

Y

Treatment of compound E with CH3Li⋅LiBr complex in (C2H5)2O at 0 °C failed to produce
the target products X and Y. Instead, a yellow precipitate F was initially formed. Aqueous workupof this precipitate led to the mixture of compound E and its tautomer G.

2. Write down the structural formulae of compounds F and G.

FGAnother approach to compound X is based on the use of pipecolinic acid derivative H. It was
shown that X can be synthesized by reaction sequence presented in Scheme 2.
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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

Scheme 2.

3. Write down the structural formulae of compounds I and J.

I

J

31

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Problem 7. Not by bread alone(8 points)
Question

1

2

3

4

Total

Marks

8

24

2

16

50

Pomegranate is called in Azerbaijan, which is famous for its
vegetables, as the “king of all fruits”. Pomegranate is honored invarious religions as a “fruit of Paradise”, symbol of righteousness,
wealth, hope for eternal life.
In 1878 alkaloid pelletierine was isolated from the bark of pomegranate tree
(Punica granatum L., Lythraceae). This alkaloid is traditionally used as an antihelminthic drug. Initially XW (3-(piperidin-2-yl)propanal) was incorrectly proposed forpelletierine. But now it is accepted that natural pelletierine is (S)-1-(piperidin-2yl)propan-2-one (XS).
1. Write down the structural formulae of XW and XS (the latter – with the stereochemical
information).
XW

XS(3-(piperidin-2-yl)propanal)(S)-1-(piperidin-2-yl)propan-2-one

The synthesis of natural pelletierine (XS) based on the transformation of nortropanol A wasrecently described.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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2. Write down the structural formulae of compounds B-G with the stereochemical information.

B

C

D

E

F

G

3. Nortropanol A was used in this reaction as a single stereoisomer. How many stereoisomers can
exist for compound A (including A)? Ignore nitrogen chirality.

The number of possible stereoisomers of A

Enantiomer of XS was synthesized using chiral tert-butanesulfinamide (H):O

4. Write down the structural formulae of compounds I-L with the stereochemical information.
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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

I

J

K

L

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Problem 8. Oil for Life and Life after Oil(8 points)
Question
Marks

1
1a

1b

1c

1d

1e

1

4

4

3

12

2

3

4

Total

5

13

13

55

Azerbaijan is known for its vast oil and gas fields. The first drillingfor oil was done in Bibi-Heybat in 1846, 13 years before
establishment of the first commercial oil well in Pennsylvania(USA). This remarkable date in the history of Azerbaijan isregarded as a starting point of contemporary oil industry, theleading sector of today’s world economy. Currently, on-land andshelf sea oil production is being developed in Azerbaijan. Though serious precautions are taken,
there is always a risk of hydrocarbon pollution of the environment during production,transportation, and processing of oil. In this task we will consider diverse technologies of oil spills
clean up and specific features of metabolic pathways involved.
Application of complex solvents (dispersants) leading to capture of marine oil spills is among mostpromising clean up approaches. Organic substance X (11.94% of H by mass) is a typical componentof such dispersants. Safety of X to human is fiercely debated. X1 (54.53% of carbon by mass)composed of three elements and excreted with urine is the major metabolite of X in humans. The
numbers of atoms of different elements in X1 are three consecutive terms of a geometricprogression (n, nq, nq2), whereas the sum of these numbers does not exceed 25.
1a.

Decide on the relationship (tick the correct variant) between the numbers of carbon andoxygen atoms in X1.
n(C) > n(O)n(C) < n(O)n(C) = n(O)Data insufficient









1b. Derive the empirical formula of X1 (hereafter always show your work where required). Be sure
you prove the answer by calculations.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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Your work

Empirical formula of X1:
The biotransformation of X into X1 occurs in two enzymatically catalyzed steps according to thehereunder reaction balanced equations (NAD+ and NADH are the oxidized and reduced forms of
nicotinamide adenine dinucleotide, respectively):Х + NAD+ → X0 + NADH + H+
X0 + NAD+ + H2O → X1 + NADH + H+
1c.

Derive the molecular formula of X.

Your work

Molecular formula of X:

36

(1)
(2)

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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A minor metabolic transformation of X is catalyzed by cytochrome P450-dependent
monooxygenase. This reaction leads to two compounds X2 (51.56% of oxygen and 9.74% ofhydrogen by mass) and X3.
1d.

Derive the molecular formula of X2 and draw its structure.

Your work

Molecular formula of X2:

Structure of X2:

X contains only primary and secondary carbon atoms. X0 and X3 contain common functionalgroup.
1e.

Draw the structural formulae of X, X1, and X3.

X

X1

37

X3

47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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In a medical study, personnel permanently exposed to X-based solvents without proper protection
was found to have a stationary concentration of X in blood.
2.
X1 is excreted with urine. Choose the graph of X1 daily mass content in the body of avolunteer participated in this experiment. Write down the number of the correct graph.

Number of graph:

The use of different bacteria is also considered as a promising way for the removal of hydrocarbon(even aromatic) contaminants from sea water and soil. Under aerobic conditions, benzene
undergoes biodegradation as follows (first three steps are balanced):

Under the same conditions, a monocyclic aromatic hydrocarbon P (91.25% of carbon by mass)undergoes the following transformation (first three steps are balanced):

P3 gives a positive iodoform test. A 100 mg sample of P3 requires 6.41 mL of 0.100 M KOH
solution for complete neutralization.
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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.
3.

ABC-1

Derive the structures of P–P3. Give the most stable tautomer of P3.

Your work

P

P1

P2

P3

Microorganisms Alicycliphilus are capable of biodegradation of aromatic hydrocarbons even in soil.
The process requires a suitable electron acceptor such as inorganic anion Y1 (first three steps arebalanced).

The intermediate anion Y2 is enzymatically decomposed according to the balanced reaction
equation:Y2(aq)  Y3(aq) + Y4(g),wherein each of Y3 and Y4 is composed of atoms of only one element. T2 does not contain two
identical oxygen-containing functional groups. T2 gives a precipitate when treated with theammonia solution of Ag2O, whereas Y3 does not.

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

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4.
Deduce and give formulas of Y1-Y4. Draw the structures of T1-T2. Give the most stable
tautomer of T2.
Y1

Y2

Y3

Y4

T1

T2

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47th International Chemistry Olympiad. Baku, Azerbaijan, July 20-29, 2015.

41

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