Life is a huge lab

THEORETICAL
EXAMINATION
ANSWERS and GRADING SCHEMES

JULY 25, 2015

BAKU, AZERBAIJAN

1
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-
- Formulas necessary for solution of some problems can be found on the next page.

2
 Physical Constants, Units, Formulas and Equations

Universal gas constant R = 8.3145 J∙K–1∙mol–1

Standard pressure p = 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
W = nCV (T2 – T1)
process
Dependence of internal energy on temperature U(T2) = U(T1) + CV (T2 – T1)
Relation between molar isobaric and isochoric
Cp = C V + R
heat capacities for an ideal gas
Gibbs energy G = H – TS
 G  
Relation between equilibrium constant and K = exp   
standard Gibbs energy  RT 
 
aprod
G = G   RT ln
Dependence of Gibbs energy of reaction on areag
,
concentration or pressure
a = c / (1 mol/L) for the substances in
solution, a = p / (1 bar) for gases
Change of Gibbs energy per unit volume in time GSyst
for the system with two chemical reactions 1 and  G1r1  G2 r2
t
2 with rates r1 and r2

3
Problem 1. New and well-forgotten old refrigerants
(8 points)

1 2 3 4 Total
Question
1.1 1.2 1.3 2.1 2.2 2.3 4.1 4.2 4.3 4.4
Marks 4 2 2 1 1 1 3 10 2 6 1 33

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

Thermodynamic properties of various refrigerants

ΔHvap /
CV(gas) /
Refrigerant “Generation” kJ·mol–1
J·K–1·mol–1
(at 280 K)
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 the
pressure (p) – internal energy (U) coordinates.

4
 p

p2
3 2 (T2)
liquid liquid + gas
gas
p1
0 (T1) 1 (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 constant
pressure p1 and temperature T1 (boiling temperature) until it completely evaporates. At this step, the
refrigeration 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 the
compressed 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
Note: here and below in this problem, only correct VALUES are marked except 4.1, 4.3
Ammonia
Q = νΔHvap = 21.3 kJ; 1p
W = νCV(gas)(T2 – T1) = 2.67 kJ. 1p

Q = 21.3 kJ
W = 2.67 kJ

Chlorodifluoromethane
Q = νΔHvap = 20.0 kJ; 1p
W = νCV(gas)·(T2 – T1) = 4.88 kJ. 1p

Q = 20.0 kJ

5
 W = 4.88 kJ

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

U H S G V

2p for the correct answer

Minus 1p for every incorrect option, total – no less than 0.

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 a
cooled 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 = Q/W = 7.98 1p

COP = 7.98

Chlorodifluoromethane
COP = Q/W = 4.10 1p

COP = 4.10

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

c
1p

6
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)
a) ozone molecule easily adds to C–F bond
b) C–F bond is easily broken by radiation, which leads to the formation of free radicals
c) ozone molecule easily adds to C–Cl bond
d) C–Cl bond is easily broken by radiation, which leads to the formation of free radicals

d
1p

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 2,3,3,3-
tetrafluoropropene 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 decompose
b) it easily reacts with ozone
c) it is better soluble in water

a
1p

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

Calculations

CF3CH2F
COP = ΔHvap / (CV(gas)(T2 – T1)) = 2.80 1p

COP = 2.80

Yes No 0.5p

7
 CF3CF=CH2
COP = ΔHvap / (CV(gas)(T2 – T1)) = 1.59 1p

COP = 1.59

Yes No 0.5p

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 have
a 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 heat
exchange 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.

3 (T3) 2 (T2)
p2
liquid liquid + gas
gas
p1
0 (T0) 1 (T1)

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 ΔHvap = 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 beginning
of the process (point 0), the molar fraction of gaseous ammonia is x = 0.13.

8
4.1. Calculate the initial temperature of refrigerant T0, its volume change ΔV and the amount of heat
Q absorbed by refrigeration unit during this step. Take into account that the dependence of ΔHvap
from the temperature cannot be neglected.

Calculations:
T0 = 10.4 + 1114 / (4.87 – log p1) = 264 K 2p
T0 = 264 K

ΔV = (νRT1 / p1) – (xνRT0 / p1) = 6.7 L 3p

V = 6.7 L

Q = ΔH = ΔU+ p1ΔV=ΔU(liquid fraction)+ ΔU(gas fraction) + p1ΔV =

= ΔU(vaporization of liquid fraction at T0) +ΔU(heating evaporated liquid fraction up to T1) +
+ ΔU(gas fraction) + p1ΔV =
=ν (1–x)(ΔHvap – RTvap + (CV(gas)–CV(liq))(T0–Tvap)) + νCV(gas) (T1–T0) + p1ΔV = 19.8 kJ
or
Q = ν (1–x)(ΔHvap + (CV(gas)+R–CV(liq))(T0–Tvap)) + ν(CV(gas)+R)(T1–T0) = 19.8 kJ
5p
(3p for a correct equation for calculation and 2p for correct value)

Q = 19.8 kJ

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 to
find Q in 4.1, use Q = 20.15 kJ.

Calculations:
W = νCV(gas) (T2– T1) = 3.15 kJ 1p

W = 3.15 kJ

COP = Q/W = 6.3 1p

COP = 6.3

9
At the next step corresponding to the line 2-3 in diagram, the compressed refrigerant is cooled in a
condenser at constant pressure. Then it returns to the initial state through adiabatic expansion with
zero work (line 3-0).

4.3. Determine the temperature T3 at point 3 to which the refrigerant is cooled in a condenser.

Calculations:

The internal energies of the refrigerant are equal in points 0 and 3. Thus,
x·(ΔHvap – RTvap + (CV(gas) – CV(liq))(T0– Tvap)) + CV(liq) (T0– T3) = 0,
T3 = 298 К.
6p
(3p for a correct equation and 3p for correct value)
T3 = 298 K

In the production of refrigeration units it is necessary to consider climatic factors. If a condenser is

cooled 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) Increase
b) Remain the same
c) Decrease

c
1p

Comment: It will decrease because the length of 0-1 line decreases or because x (see 4.3) increases
and less liquid is in the equilibrium mixture at T0, so less heat Q is necessary to evaporate it!

10
Problem 2. Coupling of chemical reactions
(7 points)

1 2 3 Total
Question
1.1 1.2 1.3 2.1 2.2
Marks 4 6 4 3 6 2 25

I.Prigogine (left) N. Shilov W. Ostwald

When in the system one reaction allows another one to proceed they say that these two reactions are
coupled. Ilya Prigogine, Nobel prize winner in chemistry (1977) in his books widely used the
concept 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 the
following 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:

A  Aс 
In the absence of In
 no reaction! (1)
A  Ас 
 Р1
In the presence of In
(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.

11
 k (3a )
(a ) A   In  R
k (3b)
(b) R   P2
k (3c )
(c ) R  Ас   P1 (3)

 and  are stoichiometric coefficients. Other stoichiometric coefficients and reaction order with
respect to all reactants in all three reactions are unity.

n Ас
I
In the Shilov’s experiments the ratio of the consumed amounts of Аc and In, nIn increased up
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

The value of I should increase with the increase of [Ac]0 at [In]0 = const, because the larger fraction
of the intermediate product R will enter the reaction (3c). The maximum value of I will be achieved
if all R reacts in (3c), therefore I = 1/β.

I = 1/ 4 points (2 points if only I = 1/ is given!)

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
Shilov’s mechanism includes the initial reaction

A + In  R (3a)

and two competitive reactions

R + Ac  P1 (3c)
R  P2 (3b)

The rates of conversion of In and Ас are determined by the rates of the reactions (3а) and (3c),
respectively:

12
 k (3a )[ A][ In ]
k (3c )[ Ac ] 
r (3с ) k (3с)[ R ][ Ac ] k (3c)[ Ac ]  k (3b) k (3c)[ Ac]
  
r (3a ) k (3a )[ A][ In ] k (3a )[ A][ In ] k (3c)[ Ac]  k (3b)

in steady-state approximation for [R]. We see that the ratio of two rates does not depend on the
initial concentration [In]0 and I will also not depend on it. This gives the straight line parallel to
the [In]0 axis on the graph.

Graph

0
[Ac] = const
I = n Ac /n In

[In] 0

6 points (2 points for the graph + 4 points for the steady-state equations)

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 ) A  In 
 P2
In, catalysis
(b) 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
In this case I will permanently increase with the increase of [Ac]0   at [In]0 = const. The rate
of the reaction (4b) may be so high that conversion of In in reaction (4а) will be negligible.
Hence I  ∞ if [Ac]0   at [In]0 = const.

I =  (infinity) 4 points (2 points if only I =  (infinity) is given).

13
(II) «Kinetic coupling»

The standard Gibbs energy of the gas-phase reaction

k5

Br  H 2   HBr  H

k 5 (5)
is positive, G(5) = 66 kJmol at Т = 600 К.
–1

r5
2.1. What is the ratio of the rates of forward and reverse reactions, r5 , at this temperature, standard
pressures of H2 and HBr and equal pressures of H and Br?

Calculations
The standard Gibbs energy of reaction (5) at 600К is 66 kJ/mol. The equilibrium constant is
K  e 66000/8.314/600  1.8  106  k5 / k 5 . 1 point
Reaction is considered at standard pressures of all the reactants and products. The ratio of the rates
of forward and reverse reactions is
r5 k [Br][H 2 ] k
 5  5  1.8  106
r5 k5 [HBr][H] k 5

r5
r5 = 1.810–6 2 points
Total – 3 points
If you could not answer this question, for further calculations use 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

Br  H 2   HBr  H
 (5)
k 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.

14
Let pressures of neutral molecules keep standard values p(H2) = p(Br2) = p(HBr) = 1 bar, and
pressures of radicals p(H), p(Br) reach steady-state values. Rate constant k6 is 10 times larger than
k–5.

r5
2.2. Calculate G(5) and r5 under such conditions.

Calculations
The steady-state condition is the same for both radicals, e.g. for radical Н

d [H]
 k5 [Br][H 2 ]  k 5 [HBr][H]  k6 [H][Br2 ]  0
dt
[H] k5 [H 2 ]

[Br] k 5[HBr]  k6 [Br2 ]

The concentrations of all the neutral molecules are the same (they correspond to the pressure of 1
bar), therefore

[H] k5 k /k 1.8  10 6
  5 5   1.6  107
[Br] k5  k6 1  k6 / k 5 1  10 2 points

The Gibbs energy of reaction (5) under such conditions is:

[H][HBr]
G  G   RT ln  66  8.314  10 3  600  ln  1.6  10 7   12 kJ  mol 1
[Br][H 2 ]
2 points
The ratio of rates is:

r5 k [Br][H 2 ] k [Br] k5 1  k6 / k 5 k
 5  5   1  6  11
r5 k5 [HBr][H] k5 [H] k 5 k5 / k5 k5 2 points

G(5) = –12 kJmol–1

r5
r5 = 11
Total – 6 points

15
(III) ”Second law of thermodynamics restricts coupling”

According to the Second Law of thermodynamics, two simultaneously occurring chemical reactions
GSyst
0
should decrease the system’s Gibbs energy GSyst, 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 (7)
G(7) = 46.0 kJmol–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,
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 is
coupled to reaction (8)?

Calculations
According to the Second law the following condition has to be met:
GSyst
 G (7)  r7  G (8)  r8  0
t
therefore

G (8) 481.2

r7  r8   6.0  108  6.3  107 M  min 1
G (7) 46.0

This is the maximum possible rate of the coupled reaction.

r7(max) = 6.310–7 Mmin–1 2 points

16
Problem 3. Two binding centers – competition or cooperation?
(7 points)
1 2
Question Total
1.1 1.2 2.1 2.2 2.3 2.4
Marks 3 2 8 3 6 6 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 ]
Ka =
 [ H ][G ]
H+G 
 HGa
[ HGb ]
Kb =
 [ H ][G ]
H+G 
 HGb 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”  which reflects the influence of one binding
center on another and is defined as follows:
[ HG2 ]
=  Kb
 [ HGa ][G ]
HGa + G   HG2
where HG2 is the completely bound complex.

1.1. Determine the range of values (or one value, if necessary) of  which correspond to three
possible 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:  > 1 1 pt (0.5 pt – for value, not range)

Competition: 0 <  < 1 1 pt (0.5 pt without zero; 0.5 pt – for value, not range)
Independence:  = 1 1 pt
Total 3 pts


1.2. Find the equilibrium constant for the process: HGb + G 
 HG2 in terms of binding
constant(s) and interaction factor.

Calculations:
[ HG2 ] [ HG2 ] [ HGa ] K
K= =  = K b  a = K a
[ HGb ][G ] [ HGa ][G ] [ HGb ] Kb

17
 K = Ka 2 pts

2.1. The solution was prepared with the initial concentrations [H]0 = 1 M and [G]0 = 2 M. After the
reactions 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 .

Calculations:
From Kb = 2Ka it follows: [HGb] = 2[HGa] 1 pt
Material balance with respect to H: [H] + [HGa] + [HGb] + [HG2] = [H]0 = 1 M, or
0.1 + 3[HGa] + [HG2] = 1 M 0.5 pt
Material balance with respect to G: [G] + [HGa] + [HGb] + 2[HG2] = [G]0 = 2 M, or
0.5 + 3[HGa] + 2[HG2] = 2 M. 0.5 pt

Solving the system of two equations, we find: [HGa] = 0.1 M, [HG2] = 0.6 M, hence [HGb] = 0.2
M.
[ HGa ] 0.1
Ka = = =2
[ H ][G ] 0.1  0.5
[ HG2 ] 0.6
= = =3
[ HGa ][G ]Kb 0.1  0.5  4

[HGa] = 0.1 M [HGb] = 0.2 M [HG2] = 0.6 M

(1 pt for [HGa], 2 pts for [HG2], and [HGb] is not marked if [HGb] = 2[HGa] was given 1 pt,
otherwise 1 pt)

Ka = 2 1 pt

=3 2 pts
Total 8 pts
If you could not answer this question, for further calculations use reference values Ka = 3.14 and 
= 2.72.

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

18
 Go
H
H Ga
H Gb

H G2
1 pt – for the highest G(H),
1 pt – for the lowest G(HG2)
1 pt – for G(HGa) > G(HGb)
Total 3 pts

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 the
total number of single-bound molecules HG. Find the initial amount of G (in mol). The constants
Ka and Kb and the factor  are the same as in question 2.1.

Calculations:

1) [HG2] = [HGa] + [HGb] = 3[HGa]

[ HG2 ] 3
= K b = 12 = 12
[ HGa ][G ] , [G ] , [G] = 0.25 M
2) Material balance with respect to H: [H] + 3[HGa] + [HG2] = 1 M
[H] + 6[HGa] = 1 M
[H] + 12[H][G] = 1 M
[H] = 0.25 M.
3) [HGa] = Ka[H] [G] = 0.125 M.
[HG2] = 3[HGa] = 0.375 M.
4) Material balance with respect to G: [G]0 = [G] + 3[HGa] + 2[HG2] = 1.375 M

n0(G) = 1.375 mol

Correct determination of [G], [H], [HGa], [HG2] – 1 pt for each concentration
n0(G) – 2 pts

19
 Total 6 pts

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

=0
Calculations:
In this case, no HG2 is formed.
Material balance with respect to H: [H] + [HGa] + [HGb] = 1 M, or
[H] + 3[HGa] = 1 M
Material balance with respect to G: [G] + [HGa] + [HGb] = 2 M, or
[G] + 3[HGa] = 2 M
[ HGa ]
Ka = =2
Equilibrium constant: [ H ][G ]
Solving the system of three equations, we get:
[H] = 0.129 M [G] = 1.129 M [HGa] = 0.290 M [HGb] = 0.580 M
[HG2] = 0
1 pt for each concentration except HGb, maximum – 4 pts

Calculations (or arguments):
In this case, formation of HG2 is practically irreversible, so only HG2 is present in the solution.
[H] = 0 [G] = 0 [HGa] = 0 [HGb] = 0
[HG2] = 1 M
2 pts
(any calculation which gives similar result – full mark)
Total 6 pts

20
Problem 4. From one yellow powder to another: A simple inorganic riddle
(6 points)
Question 1 2 3 4 Total
Marks 8 8 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 the
solution 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 by
filtration. To the new filtrate the solution of sodium hydroxide was being added drop-wise until the
solution 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 the
first portion of X2.

1. Determine the chemical formulae of X1 – X4.

Calculations:
The precipitate X2 formed by addition of barium chloride in acidic medium is barium sulfate
BaSO4. 1 pt
The precipitate X3 formed by addition of silver sulfate is silver chloride AgCl 1 pt

The yellow precipitate X4 formed by addition of alkali can be mercury oxide HgO or silver
phosphate Ag3PO4. The ratio of molar masses X4 : X2 is 0.931 for HgO : BaSO4 which is not
valid and 1.798 for Ag3PO4 : BaSO4 which gives 2.4 being multiplied by 4/3. So, the molar ratio
is 4Ag3PO4 : 3BaSO4 which corresponds to P : S = 4:3, i.e. to formula of X1 P4S3.

X1 = P4S3 X2 = BaSO4 X3 = AgCl X4 = Ag3PO4

2 pts for Ag3PO4
4 pts for P4S3 (0 pts without calculations)
Total – 8 pts

2. Determine the chemical formula of the gas and provide equations for all reactions in ionic or non-
ionic form.

Calculation
The gas evolved has a molar mass 1.586 × 29 = 46 g/mol, that is NO2. 1 pt
Chemical formula of the gas ________

Dissolution of X1
P4S3 + 38HNO3 = 4H3PO4 + 3H2SO4 + 38NO2+ 10H2O 2 pt
21
Formation of X2
H2SO4 + BaCl2 = BaSO4+ 2HCl 1 pt
Formation of X2 and X3
Ag2SO4 + 2HCl = 2AgCl + H2SO4 1 pt
BaCl2 + Ag2SO4 = BaSO4 + 2AgCl 1 pt
Addition of NaOH and formation of X4
H2SO4 + 2NaOH = Na2SO4 + 2H2O 1 pt
2H3PO4 + 6NaOH + 3Ag2SO4 = 2Ag3PO4 + 3Na2SO4 + 6H2O 2 pts
(neutralization of H3PO4 and subsequent reaction with Ag2SO4 will also be accepted)

(50% of points for non-balanced reactions with correct products)

Total – 8 pts

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

Phosphorus sulfide P4S3 is a molecular cage

3 pts
(Any reasonable structures with correct valencies of sulfur and phosphorus will also be accepted.
1 pt for the structure with non-equivalent atoms of both elements)

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) P4S3 + 8O2 = 2P2O5 + 3SO2 1 pt

22
b) P4S3 + 16H2SO4 = 4H3PO4 + 19SO2 + 10H2O 2 pts
(oxidation of sulfide to S instead of SO2 is full mark)

c) 3P4S3+ 16KClO3 = 16KCl + 6P2O5 + 9SO2 2 pts

(50% of points for non-balanced reactions with correct products)
Total – 5 pts

23
Problem 5. Indispensable glucose
(8 points)

1 2 Total
Question
1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 2.5
Marks 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 O

includes the use of Fehling's reagent. A 10.00-mL aliquot of fruit juice (assuming
HO
the initial sample contained only glucose and fructose) was transferred into a
OH
titration flask and Fehling's reagent was added. This reagent was prepared by HO

mixing 50.00 mL of 0.04000 M copper sulfate (solution A) and potassium- OH

sodium tartrate and sodium hydroxide (solution B). Solution C thus obtained,
OH
was then heated and red precipitate was formed.
Glucose

1.1. Write the balanced ionic equation of chemical reaction occurring upon heating of the
solution C. Use Cu2+ for initial copper solution.
C6H12O6 + 2 Cu2+ + 5OH-= C6H11O7- + Cu2O+ 3H2O 2 points
If C6H12O7 instead of C6H11O7- 1 point
Hereinafter if an equation is not balanced, then points/2.

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
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 in
the flask.
24
 2CuSO4 + 4KI = 2CuI + I2 + 2K2SO4
or 2Cu2+ + 4I– = 2CuI + I2 2 points
KI +I2 = KI3
or I– + I2 = I3– not marked
C6H11O7- + H2SO4 = C6H12O8 + HSO4- not marked
2Na2S2O3 + I2 = 2NaI + Na2S4O6 1 point
or 2S2O32– + I2 = 2I– + S4O62–

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.
Total amount of copper(II) is 50.00 mL * 0.04000 M = 2.0000 mmol.
Obviously, there is an excess of iodine and the remaining iodine was titrated with
sodium thiosulphate: 11.87 mL * 0.05078 M = 0.6028 mmol.
2.0000 – 0.6028 mmol = 1.3972 mmol of copper(II) was required to oxidize the sugars. 6 points
ν(sugars) = ν(Cu2+)/2 = 0.6986 mmol in 10.00 mL
C(sugars) = 0.6986 mmol/10.00 mL = 0.06986 M
mass content = 180.16 g/mol * 0.06986 M = 12.6 g/L

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. After
the mixture turned brown, an excess of sodium hydroxide solution was added. The flask was then
covered with a watch glass and put into a dark place. The obtained solution was acidified and
titrated with 0.01089 M solution of sodium thiosulphate. The average titrant volume used for
titration 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.

KIO3 + 5KI + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O

2 points
IO3– + 5I– + 6H+ = 3I2 +3H2O
Only glucose was oxidized with iodine 2 points

25
 O O

–
O Na

HO HO

OH
+ I2 + 3NaOH OH
+ 2NaI + 2H2O
HO HO

OH OH

OH OH

or

O O

–
O Na

HO HO

OH
+ I2 + 3 OH- OH + 2I- + 2H2O
HO HO

OH OH

OH OH
+ -
H + OH = H2O not marked
2Na2S2O3 + I2 = 2NaI + Na2S4O6 not marked

1.5. Calculate the mass content of each sugar (in g/L) in the juice.
Total amount ν(I2) = 3ν(IO3–) =3*0.01502 M * 10 mL = 0.4506 mmol 1 pt
ν(S2O32-)=23.43 mL*0.01089 M = 0.2552 mmol 1
pt
ν(S2O32-)/2=ν(I2) = 0.1276 mmol 1 pt
6 points
0.4506 mmol – 0.1276 mmol = 0.3230 mmol of iodine was used to oxidize glucose
C(glucose) = 0.3230 mmol/10.00 mL = 0.03230 M 1 pt
mass content of glucose = 180.16 g/mol *0.03230 M = 5.82 g/L 1 pt
mass content of fructose = 12.6 – 5.82 = 6.78 g/L 1 pt

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?
0.2 L*5.82 g/L = 1.16 g of digestible carbohydrates, it is 0.1 BEU 1 point
Or 0.2 L*12.6 g/L = 2.52 g, it is 0.2 BEU 1 point

2. Diagnosis of diseases

26
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 by
nucleophilic 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 of
fluorine-18. This radionuclide is a β+ emitter – nucleus emits a positron (anti-electron). Positron
interacts 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.

1
a) 18
O+ 1 H  …+ 18F n 0.5 points
2
b) …+ 1 D  18F +  20
Ne 0.5 points
2
c) 19
F+ 1 D  20F + … 1
1 H 0.5 points
1
d) 16
O + …  18F + 1 H +n 4
 or 2He 0.5 points

2.2. The decay mode of unstable light nuclei depends on the ratio between the number of neutrons
and 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:

11 20 17 14
Nucleus С F F C
Decay mode β+ β- β+ β–
0.5 points 0.5 points 0.5 points 0.5 points

When nuclear reaction (a) is used for fluorine-18 preparation, the target material is presented as
water enriched with H218O. The presence of usual water H 216O 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 of
radioactivities of 18F and 17F is 105. Assuming that irradiation time is short, the radioactivity of each
isotope is proportional to the nuclear reaction yield and the mole fraction of a component in 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 η18 O −18 F / η16 O −17 F = 144.7.

27
 Radioactivity is:
A = N, where N is the number of atoms,  = ln 2 / t1/2 1 point
The initial ratio of radioactivities:

A0 (18 F)   18 F  ( 18 O  18 F) (H 218O) 65 / 60 (H 218O) (H 218O)

=   =  144.7   1.43
A0 (17 F)   17 F  ( 16 O  17 F) (H 216O) 109.7 (H 216O) (H 216O)

After 5 minutes the ratio changed due to radioactive decay of fluorine:

 ln 2 
A0 (18 F)  exp    5
 109.7  = 23.75  A0 ( F) = 33.94  (H 2 O) = 105
18 18 18
A300 ( F)
=
A300 (17 F)  ln 2  A0 (17 F) (H 216O)
A0 (17 F)  exp    300 
 65  1 point
(H 218O)
= 2947
(H 216O) 1 point
Mass fraction of H218O is:
2947  20
(H 218O) = = 0.9997
2947  20  18 1 point

(H218O) = 0.9997 = 99.97%.

Total – 4 points

2.4. Calculate the yield of labeling D-glucose with fluorine-18, if initial radioactivity of a fluorine-
18 sample was 600.0 MBq and radioactivity of the obtained 2-deoxy-2-( 18F)fluoro-D-glucose is
528.2 МBq. Synthesis time is 3.5 minutes.
During the synthesis, the radioactivity will decrease:
 ln 2 
A3.5 = A0  exp    3.5  = 586.9 MBq
 109.7  1 point
 = 528.2 / 586.9 = 0.900 = 90.0% 1 point

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.
Radioactivity is excreted by radioactive decay and through the excretory organs (e.g. kidneys).
The excretion process may be considered as two competitive first-order reactions. Activity after
one hour is:
  ln 2 ln 2  
A60 = A0 exp    1   2  t  = 450  exp       600  = 0.32 MBq
  109.7 120  

28
 2
points.

29
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 one of
the novels said on a similar occasion: "If you tell me
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 equilibrium
with its tautomer Y in a 2:1 ratio. Unfortunately, both
forms 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 the
reaction 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.

B + C

1) (CH3)3COCl 1) (CH3)3COCl 1) CH3MgI (1 eq)

(C2H5)2O, 0 C NaCN (C2H5)2O, 0 C (C2H5)2O, -20 C
A D E X + Y
N
2) base, H+, rt 2) base, rt 2) H3O+, rt
H rt, then  N = 25.90%

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 same
elemental composition: C = 72.24%, H = 10.91%, N = 16.85%.

30
1. Write down the structural formulae of compounds A-E, X, Y.

A B C

H H
N N N N N
H H
4 pts N N
H H
(1pt for any reasonable
isomer) 4 pts 4 pts
(Other reasonable structures
with molecular formula
(C5H9N)n but without 3-fold
axis of symmetry – 2 pts)

D E X

N CN CH3
H N N
N
H O
4 pts 4 pts
4 pts
(1 pt for any reasonable
(other structures will be
isomer)
estimated only if they are
consistent with both reaction
schemes, 3 pts if in cell Y)
Y

O
N
CH3

4 pts
(other structures will be
estimated only if they are
consistent with both reaction
schemes, 3 pts if in cell X)

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 workup of
this precipitate led to the mixture of compound E and its tautomer G.

31
2. Write down the structural formulae of compounds F and G.

F G

Li
Li
N N N
N N H N
2 pts
2 pts

Another 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.

Scheme 2.

OC2H5 Cp2Ti(CH 3)2 1) H3O+ O2

N I J X + Y
O toluene 2) pH 7
O O 85 C C H NO
14 25 3
C(CH3)3 Cp = cyclopentadienyl
H

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

I J

OC2H5 CH3
N N
CH2 H O
O O
C(CH3)3 4 pts

4 pts (other structures will be estimated only if they

(2 pts for product of substitution of another are consistent with both previous compound I

carbonyl oxygen) and products X and Y)

32
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 in
various 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
N
as an anti-helminthic drug. Initially XW (3-(piperidin-2-yl)propanal) was incorrectly H
piperidine
proposed for pelletierine. But now it is accepted that natural pelletierine is (S)-1-
(piperidin-2-yl)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
O
O
N N
H H
4 pts 4 pts
(2 pts without stereochemistry)

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

1) (C2H5)3SiH
H 1) (COCl)2
H N CbzCl (CH3)2SO m-CPBA BF3 O(C2H5)2
N OH B C D
K2CO3 2) (C2H5)3N Na2HPO4 2) H3O+
C15H17NO3 C15H17NO4
OH
A

1) SOCl2 1) CH3MgBr H2, Pd/C

E F G pelletierine
2) CH3NHOCH3 2) H2O
C15H19NO4 C17H24N2O4

O O
Cl
O Cl OOH
CbzCl = m-CPBA =

33
2. Write down the structural formulae of compounds B-G with the stereochemical information.

B C D
Cbz Cbz
Cbz
N N
N
OH O
O O
4 pts 4 pts 4 pts
E F G
O O O

HO N N N N
Cbz O Cbz Cbz
4 pts 4 pts 4 pts
(For the structural formulae without stereochemistry (or with bad stereochemistry): 3 pts for each.
Comments: a) for compound B, product of hydroxyl group acylation in A is estimated by 2 pts; b)
for wrong isomeric structures of compounds C–F the mark will be in the range of 0-2 pts depending
on the credibility of answer; c) for the wrong structures of compound G the mark will be in the
range of 0-4 pts depending on the credibility of answer (4 pts will be given if under the specified
conditions compound G can be obtained from F and can be transformed into (S)-1-(piperidin-2-
yl)propan-2-one.)

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: 4 .

2 pts (0.5 pts for 8)

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

34
 1) Ti(OC2H5)4
O 2) Br
Br O , indium KHMDS
S + I J
(H3C)3C NH2 3) H2O THF, 0 C
H HCl
C12H24BrNOS
CH3OH

H H Cl
1) HCl O2 Boc2O N
( R)-pelletierine L K
2) NaOH Pd(OAc)2 NaOH
Cu(OAc)2
O O
OO
KHMDS = [(CH3)3Si]2NK; THF = Ac = ; Boc2O = O O O
O

4. Write down the structural formulae of compounds I-L with the stereochemical information.

I J
Br CMe3 CMe3
S S
HN O N O

4 pts 4 pts
K L
O O
C(CH3)3
N O N
Boc

4 pts 4 pts
(For the structural formulae without stereochemistry (or with bad stereochemistry): 3 pts for each.
Wrong structures will be estimated depending on the credibility of answer.)

35
Problem 8. Oil for Life and Life after Oil
(8 points)
1
Question 2 3 4 Total
1a 1b 1c 1d 1e
Marks 1 4 4 3 12 5 13 13 55

Azerbaijan is known for its vast oil and gas fields. The first drilling
for 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 is
regarded as a starting point of contemporary oil industry, the
leading sector of today’s world economy. Currently, on-land and
shelf 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 most
promising clean up approaches. Organic substance X (11.94% of H by mass) is a typical component
of 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 geometric
progression (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 and
oxygen atoms in X1.
n(C) > n(O)
n(C) < n(O) n(C) = n(O) Data insufficient

  
1p

1b. Derive the empirical formula of X1 (hereafter always show your work where required). Be sure
you prove the answer by calculations.
Your work
With regard to 1a, three variants (n(H)>n(C)>n(O), n(C)>n(H)>n(O), and n(C)>n(O)>n(H)) are
possible for X1. For each inequality, one can write down the corresponding formula using elements
of a geometric progression (q is the progression common ratio), equations for calculation of mass
fractions of carbon and its roots

36
 The first The second
Inequality Formula Equation
root (q1) root (q2)
n(H)>n(C)>n(O 12.01qn
СqnHq2nOn  0.5453 2.00 7.93
) 12.01qn  1.008q 2 n  16.00n
n(C)>n(H)>n(O 12.01q 2 n
Сq2nHqnOn  0.5453 –1.21 1.32
) 12.01q 2 n  1.008qn  16.00n
n(C)>n(O)>n(H 12.01q 2 n
Сq2nHnOqn  0.5453 –0.06 1.66
) 12.01q 2 n  1.008n  16.00qn
There is only one positive integer root, thus the empirical formula is C2H4O.
problem formulation – 1p
derivation – 1p
result – 2p
Total 4 pts
(alternative approaches are possible)
Empirical formula of X1: C2H4O

The biotransformation of X into X1 occurs in two enzymatically catalyzed steps according to the
hereunder reaction balanced equations (NAD+ and NADH are the oxidized and reduced forms of
nicotinamide adenine dinucleotide, respectively):
Х + NAD+ → X0 + NADH + H+ (1)
+ +
X0 + NAD + H2O → X1 + NADH + H (2)

1c. Derive the molecular formula of X.

Your work
Since (1) and (2) are the reaction equations, one can write down the formula of X as: C2nH4nOn + 2H
 1O = C2nH4n+2On-1. With an account for the known mass fraction of hydrogen:
1.008(4n  2)
 0.1194
12.01  2n  1.008(4n  2)  16.00( n  1) . Finally, n = 3, and the molecular formula of X is
C6H14O2.
derivation – 2p
result – 2p
Total 4 pts
Molecular formula of X: C6H14O2

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% of
hydrogen by mass) and X3.

1d. Derive the molecular formula of X2 and draw its structure.

37
 Your work
X2 is formed from X composed of three elements (C, H, and O) via a monooxygenase catalyzed
100  51.56  9.74 9.74 51.56
n(C) : n(H) : n(O)  : :  1: 3 :1
reaction: 12 1.008 16.00 . 1p
Since the number of hydrogen atom is necessarily even, the molecular formula of X2 is C2H6O2.
Other variants with a higher even number of hydrogen are not valid. Ethylene glycol
HOCH2CH2OH is the only stable substance with the molecular formula deciphered above.

Molecular formula of X2: C2H6O2 1p Structure of X2: HO–CH2–CH2–OH 1p

X contains only primary and secondary carbon atoms. X0 and X3 contain common functional
group.

1e. Draw the structural formulae of X, X1, and X3.

O O O
OH
O
OH X3 4p
X 4p
X1 4p
If incorrect, but
If incorrect, but
saturated 0.5p If incorrect, but
molecular formula 0.5p
no branch 0.5p molecular formula 0.5p
aldehyde 1p
alcohol and ether 0.5p carboxyl 1p

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 a
volunteer participated in this experiment. Write down the number of the correct graph.

38
 1 2 3

mass
mass

mass
0 0.01 6 1212.011818.0124
0 3 6 9 12 15 18 21 24
0

18

24
12
time, h
time, h time, h

4 5 6
mass

mass

mass
0 6 12 18 9
3 .5
2 0 6 12 18 24 0 3 6 9 12 15 18 21 24
time, h time, h time, h

Number of graph: 1 (5 pts), if 5 (2.5 pts)

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):
O2,2[H] NAD+ NADH,H+ O2
OH OH CO2H
CO2
CO2H
D OH OH D
here and hereinafter
D - aromatic dioxygenase
Under the same conditions, a monocyclic aromatic hydrocarbon P (91.25% of carbon by mass)
undergoes the following transformation (first three steps are balanced):
O2,2[H] NAD+ NADH,H+ O2

P1 P2 P3 CO2
P
D D

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

39
 Your work
Dioxygenase incorporates two oxygen atoms in vicinal positions of the substrate, which can be
followed by chemical bonds reorganization. The empirical formula of the hydrocarbon Р is C7H8
91.25 100  91.25
:  7:8
(C : H = 12.01 1.008 ). Thus, it is toluene. 1p
100
 156
The molar mass of P3 equivalent containing acidic group(s) is 6.41  0.100 g/mol. 1p
Two dioxygenase steps suggest the composition of C7H8O4. 1p
P3 must be a monocarboxylic acid if it still contains seven carbon atoms. Fragments containing a
СH3CO– group (or a СH3CH(OH)– group further transforming into СH3CO– one) (1p) are involved
into the iodoform reaction. This suggests splitting of the benzene moiety during the second
oxygenase step at the carbon connected to the methyl group.
OH
no way to re-aromatization
OH

O2,2[H] NAD+ NADH,H+ O2

O
OH OH
CO2
OH
D OH OH D
CO2H
P P1 P2 P3

OH OH
no way to acetylcarbonic acid
OH OH

P P1 P2 P3
O
OH OH
OH
OH OH CO2H
1p 3p 3p
2p
(1p for isomer) (1p for isomer)
(1 p for tautomer)

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 are
balanced).
Y4
Y1 Y2 Y1 Y2
CO2
T1 T2
OH D2
OH
D2 is aromatic dioxygenase different from D
The intermediate anion Y2 is enzymatically decomposed according to the balanced reaction
equation:

40
 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 the
ammonia solution of Ag2O, whereas Y3 does not.

4. Deduce and give formulas of Y1-Y4. Draw the structures of T1-T2. Give the most stable
tautomer of T2.
Y1 Y2 Y3
ClO3- ClO2- Cl–
1.5p 1.5p 1.5p
(wrong central atom 0.5p) (wrong central atom 0.5p) (wrong element 0.5p)
Y4 T1 T2
OH
O2
1.5p HO2C
OH
O
2p 5p
If incorrect, but
molecular formula 1p
aldehyde 1p
no identical 0.5p
(5p for hemiacetal, 3p for other
tautomers)

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