Life is a huge lab
THEORETICAL
EXAMINATION
ANSWERS and GRADING SCHEMES
JULY 25, 2015
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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 1 /
= const
V R C
pV +
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 and
standard Gibbs energy = exp
G K RT ∆ − o
Dependence of Gibbs energy of reaction on concentration or pressure prod reag
=
ln
a
G
G
RT
a
∆
∆
o
+
,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
for the system with two chemical reactions 1 and 2 with rates r1 and r2
Syst 1 1 2 2 G G r G r t ∆ = ∆ + ∆ ∆
Problem 1. New and well-forgotten old refrigerants (8 points)
Question 1 2 3 4 Total
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
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 the pressure (p) – internal energy (U) coordinates.
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 = ν∆Hvap = 21.3 kJ; 1p W = νCV(gas)(T2 – T1) = 2.67 kJ. 1p Q = 21.3 kJ W = 2.67 kJ Chlorodifluoromethane Q = ν∆Hvap = 20.0 kJ; 1p W = νCV(gas)·(T2 – T1) = 4.88 kJ. 1p Q = 20.0 kJ
p
U
0
(T1)1
(T1)2
(T2)3
p1 p2liquid liquid + gas gas
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
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 = ∆Hvap / (CV(gas)(T2 – T1)) = 2.80 1p
COP = 2.80
CF3CF=CH2
COP = ∆Hvap / (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.
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 beginning
p
U
0
(T0)1
(T1)2
(T2)3
(T3) p1 p2liquid liquid + gas gas
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.
Calculations:
T0 = 10.4 + 1114 / (4.87 – log p1) = 264 K 2p
T0 = 264 K
∆V = (νRT1 / p1) – (xνRT0 / p1) = 6.7 L 3p
∆V = 6.7 L
Q = ∆H = ∆U+ p1∆V=∆U(liquid fraction)+ ∆U(gas fraction) + p1∆V =
= ∆U(vaporization of liquid fraction at T0) +∆U(heating evaporated liquid fraction up to T1) +
+ ∆U(gas fraction) + p1∆V =
=ν (1–x)(∆Hvap – RTvap + (CV(gas)–CV(liq))(T0–Tvap)) + νCV(gas) (T1–T0) + p1∆V = 19.8 kJ
or
Q = ν (1–x)(∆Hvap + (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
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·(∆Hvap – 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!
Problem 2. Coupling of chemical reactions (7 points)
Question 1 2 3 Total
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:
In the absence of In In the presence of In no reaction! (1) (2) + → + → 1 A Aс A Ас Р
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.
(3 ) ( ) (3 ) ( ) (3 ) ( ) k a a k b b k c c α + β → → + → 2 1 A In R R P R Ас P (3)
α and β are stoichiometric coefficients. Other stoichiometric coefficients and reaction order with
respect 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 n ∆ = ∆ Ас In 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/β 4 points (2 points if only I∞ = 1/β 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 αA + βIn → 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:
(3 )[ ][ ] (3 )[ ] (3 ) (3 )[ ][ ] (3 )[ ] (3 ) (3 )[ ] (3 ) (3 )[ ][ ] (3 )[ ][ ] (3 )[ ] (3 ) k a A In k c Ac r с k с R Ac k c Ac k b k c Ac r a k a A In k a A In k c Ac k b × + = = = +
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
I
= ∆ n A c / ∆ nIn [In]0 [Ac]0 = const6 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
( ) , catalysis ( ) a b α + β → + → 2 1 A In P In A Ас P (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.
(II) «Kinetic coupling»
The standard Gibbs energy of the gas-phase reaction
5 2 5 Br H k HBr H k− → + ← + (5) 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, 5 5 r
r− , 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
66000/8.314/600 6
5 5
1.8 10 /
K =e− = ⋅ − =k k− . 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
6 5 5 2 5 5 5 5 [Br][H ] 1.8 10 [HBr][H] r k k r k k − − − − = = = ⋅ 5 5 r r− = 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: 5 2 5 6 2 Br H HBr H (5) H Br HBr Br (6) k k k − → + ← + + → +
k5, k–5, k6 are rate constants of forward and reverse reaction (5) and forward reaction (6), respectively.
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.
2.2. Calculate ∆G(5) and 5 5 r
r− under such conditions.
Calculations
The steady-state condition is the same for both radicals, e.g. for radical Н
5 2 5 6 2 5 2 5 6 2 [H] [Br][H ] [HBr][H] [H][Br ] 0 [H] [H ] [Br] [HBr] [Br ] d k k k dt k k k − − = − − = = +
The concentrations of all the neutral molecules are the same (they correspond to the pressure of 1 bar), therefore 6 7 5 5 5 5 6 6 5 [H] / 1.8 10 1.6 10 [Br] 1 / 1 10 k k k k k k k − − − − − ⋅ = = = = ⋅ + + + 2 points
The Gibbs energy of reaction (5) under such conditions is:
(
)
3 7 1 2 [H][HBr] ln 66 8.314 10 600 ln 1.6 10 12 kJ mol [Br][H ] G G RT − − − ∆ = ∆ o+ = + ⋅ ⋅ ⋅ ⋅ = − ⋅ 2 pointsThe ratio of rates is:
5 5 2 5 5 6 5 6 5 5 5 5 5 5 5 [Br][H ] [Br] 1 / 1 11 [HBr][H] [H] / r k k k k k k r k k k k k k − − − − − − − + = = = = + = 2 points ∆G(5) = –12 kJ⋅mol–1 5 5 r r− = 11 Total – 6 points
(III) ”Second law of thermodynamics restricts coupling”
According to the Second Law of thermodynamics, two simultaneously occurring chemical reactions should decrease the system’s Gibbs energy GSyst,
Syst 0 G 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:
Syst 7 8 (7) (8) 0 G G r G r t ∆ = ∆ × + ∆ × ≤ ∆ therefore 8 7 1 7 8 (8) 481.2 6.0 10 6.3 10 M min (7) 46.0 G r r G − − − −∆ ≤ = ⋅ ⋅ = ⋅ ⋅ ∆
This is the maximum possible rate of the coupled reaction.
Problem 3. Two binding centers – competition or cooperation? (7 points)
Question 1 2 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:
H + G ←→ HGa [ ] = [ ][ ] a a K HG H G H + G ←→ HGb [ ] = [ ][ ] b b K HG H G 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:
HGa + G ←→ HG2 2 [ ] = [ a][ ] b K β HG HG 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 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: β > 1 1 pt (0.5 pt – for value, not range)
Competition: 0 < β < 1 1 pt (0.5 pt without zero; 0.5 pt – for value, not range)
Independence: β = 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: 2 2 [ ] [ ] [ ] = = = = [ ][ ] [ ][ ] [ ] a a b a b a b b HG HG HG K K K K HG G HG G ⋅ HG β ⋅K β K = βKa 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 β.
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. [ ] 0.1 = = = 2 [ ][ ] 0.1 0.5 a a HG K H G ⋅ 2 [ ] 0.6 = = = 3 [ a][ ] b 0.1 0.5 4 HG HG G K β ⋅ ⋅ [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
β = 3 2 pts
Total 8 pts
If you could not answer this question, for further calculations use reference values Ka = 3.14 and β = 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.
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 β are the same as in question 2.1.
Calculations: 1) [HG2] = [HGa] + [HGb] = 3[HGa] 2 [ ] = = 12 [ a][ ] b HG K HG G β , 3 = 12 [ ]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
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 in question 2.1.
β = 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 Equilibrium constant: = [ ] = 2 [ ][ ] a a HG K 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
β → ∞
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
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
Formation of X2
H2SO4 + BaCl2 = BaSO4↓+ 2HCl
Formation of X2 and X3
Ag2SO4 + 2HCl = 2AgCl↓
BaCl2 + Ag2SO4 = BaSO4↓
Addition of NaOH and formation of
H2SO4 + 2NaOH = Na2SO4
2H3PO4 + 6NaOH + 3Ag2SO
(neutralization of H3PO4 and subsequent reaction with Ag
(50% of points for non-balanced reactions with correct products)
3. In the structural unit of X1 all atoms of structure of X1.
Phosphorus sulfide P4S3 is a molecular cage
(Any reasonable structures with correct valencies of sulfur and phosphorus will also be accepted.
1 pt for the structure with non
4. Predict the products of X1 interaction with: a) excess oxygen;
b) excess of hot concentrated sulfur c) solid KClO3 with grinding Write down the reaction equations.
a) P4S3 + 8O2 = 2P2O5 + 3SO2
+ 2HCl ↓ + H2SO4
↓ + 2AgCl↓
Addition of NaOH and formation of X4
4 + 2H2O
SO4 = 2Ag3PO4↓ + 3Na2SO4 + 6H2O
and subsequent reaction with Ag2SO4 will also be accepted)
balanced reactions with correct products) Total – 8 pts
all atoms of only one element are in equivalent positions. Draw the
is a molecular cage
3 pts
structures with correct valencies of sulfur and phosphorus will also be accepted. for the structure with non-equivalent atoms of both elements)
interaction with:
concentrated sulfuric acid; with grinding.
Write down the reaction equations.
1 pt
1 pt 1 pt
1 pt 2 pts
will also be accepted)
one element are in equivalent positions. Draw the
structures with correct valencies of sulfur and phosphorus will also be accepted. equivalent atoms of both elements)
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
Problem 5. Indispensable glucose (8 points)
Question 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
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 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 a titration flask and Fehling's reagent was added. This reagent was prepared by mixing 50.00 mL of 0.04000 M copper sulfate (solution A) and potassium-sodium tartrate and potassium-sodium hydroxide (solution B). Solution C thus obtained, 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
If C6H12O7 instead of C6H11O7
-Hereinafter if an equation is not balanced, then points/2.
2 points 1 point
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.
2CuSO4 + 4KI = 2CuI + I2 + 2K2SO4
or 2Cu2+ + 4I– = 2CuI + I2 KI +I2 = KI3 or I– + I2 = I3– C6H11O7- + H2SO4 = C6H12O8 + HSO4 -2Na2S2O3 + I2 = 2NaI + Na2S4O6 or 2S2O32– + I2 = 2I– + S4O62– 2 points not marked not marked 1 point
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. 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
6 points
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
IO3– + 5I– + 6H+ = 3I2 +3H2O
2 points
2H2O I2 3NaOH Na 2NaI or 3 OH -I2 2I- 2H 2O Na 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
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
6 points
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
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
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.
a) 18O + 11H → …+ 18F n 0.5 points
b) … + 12D → 18F + α 20Ne 0.5 points
c) 19F + 12D → 20F + … 0.5 points
d) 16O + … → 18F + 11H + n α or 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:
Nucleus 11С 20F 17F 14C
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 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 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 / = 144.7.
Radioactivity is:
A = λN, where N is the number of atoms, λ = ln 2 / t1/2 1 point
( )
( )
18 18 18 18 18 18 18 0 2 2 2 17 17 16 17 16 16 16 0 2 2 2 F ( F) ( O F) (H O) 65 / 60 (H O) (H O) = = 144.7 1.43 ( F) F ( O F) (H O) 109.7 (H O) (H O) A A λ η → χ χ χ ⋅ ⋅ ⋅ ⋅ = η → χ χ χ λAfter 5 minutes the ratio changed due to radioactive decay of fluorine:
18 18 0 18 18 5 300 0 2 17 17 16 17 300 0 2 0 ln 2 ( F) exp 5 ( F) 109.7 ( F) (H O) = = 23.75 = 33.94 = 10 ln 2 ( F) ( F) (H O) ( F) exp 300 65 A A A A A A ⋅ − ⋅ χ ⋅ ⋅ χ ⋅ − ⋅ 1 point 18 2 16 2 (H O) = 2947 (H O) χ χ 1 point
Mass fraction of H218O is:
18 2 2947 20 (H O) = = 0.9997 2947 20 18 ⋅ ω ⋅ + 1 point ω(H218O) =0.9997 = 99.97%. Total – 4 points
2.4. Calculate the yield of labeling D-glucose with 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:
3.5 0 ln 2 = exp 3.5 = 586.9 MBq 109.7 A A ⋅ − ⋅ 1 point η = 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:
(
)
(
)
60 0 1 2 ln 2 ln 2 = exp = 450 exp 600 = 0.32 MBq 109.7 120 A A − λ + λ t ⋅ − + ⋅ 2 points.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. N H 1) (CH3)3COCl (C2H5)2O, 0 C 2) base, rt, then B NaCN H+, rt D 2) base, rt E 1) CH3MgI (1 eq) 2) H3O+, rt X + Y + C N= 25.90% A (C2H5)2O, 0 C 1) (CH3)3COCl (C2H5)2O, -20 C
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: ωC = 72.24%, ωH = 10.91%, ωN = 16.85%.
1. Write down the structural formulae of compounds A-E, X, Y.
A
4 pts
(1pt for any reasonable isomer)
B
4 pts
(Other reasonable structures with molecular formula (C5H9N)n but without 3-fold
axis of symmetry – 2 pts) C 4 pts D 4 pts E 4 pts
(1 pt for any reasonable isomer)
X
4 pts
(other structures will be estimated only if they are consistent with both reaction
schemes, 3 pts if in cell Y)
Y
4 pts
(other structures will be estimated only if they are consistent with both reaction
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.
2. Write down the structural formulae of compounds F and G.
F
2 pts
G
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.
3. Write down the structural formulae of compounds I and J.
I
4 pts
(2 pts for product of substitution of another carbonyl oxygen)
J
4 pts
(other structures will be estimated only if they are consistent with both previous compound I
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 as an anti-helminthic drug. Initially XW (3-(piperidin-2-yl)propanal) was incorrectly 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 (3-(piperidin-2-yl)propanal) 4 pts XS (S)-1-(piperidin-2-yl)propan-2-one 4 pts (2 pts without stereochemistry)
The synthesis of natural pelletierine (XS) based on the transformation of nortropanol A was recently described.
2. Write down the structural formulae of compounds B-G with the stereochemical information. B 4 pts C 4 pts D 4 pts E 4 pts F 4 pts G N Cbz O 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)
4. Write down the structural formulae of compounds I-L with the stereochemical information. I Br HN S CMe3 O 4 pts J 4 pts K 4 pts L 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.)
Problem 8. Oil for Life and Life after Oil (8 points)
Question 1 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) 1 p 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
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
Inequality Formula Equation The first root (q1) The second root (q2) n(H)>n(C)>n(O) СqnHq2nOn 2 12.01 0.5453 12.01 1.008 16.00 qn qn+ q n+ n = 2.00 7.93 n(C)>n(H)>n(O) Сq2nHqnOn 2 2 12.01 0.5453 12.01 1.008 16.00 q n q n+ qn+ n = –1.21 1.32 n(C)>n(O)>n(H) Сq2nHnOqn 2 2 12.01 0.5453 12.01 1.008 16.00 q n q n+ n+ qn = –0.06 1.66
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(4 2) 0.1194 12.01 2 1.008(4 2) 16.00( 1) n n n n + =
⋅ + + + − . Finally, n = 3, and the molecular formula of X is
C6H14O2.
derivation – 2p result – 2p Total 4 pts
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.
Your work
X2 is formed from X composed of three elements (C, H, and O) via a monooxygenase catalyzed
reaction: (C) : (H) : (O) 100 51.56 9.74: 9.74 :51.56 1: 3 :1
12 1.008 16.00
n n n = − − = . 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.
4p
If incorrect, but saturated 0.5p no branch 0.5p alcohol and ether 0.5p
4p If incorrect, but molecular formula 0.5p carboxyl 1p 4p If incorrect, but molecular formula 0.5p aldehyde 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.1 2 3
4 5 6
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):
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.
3. Derive the structures of P–P3. Give the most stable tautomer of P3.
0 6 12 18 24 m as s time, h 0 0 6 12 18 24 m as s time, h 0 0 6 12 18 24 m as s time, h 0 0 6 12 18 24 m as s time, h 0 0 6 12 18 24 m as s time, h 0 0 6 12 18 24 m as s time, h 0
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
(C : H = 91.25 100 91.25: 7 : 8
12.01 1.008
−
= ). Thus, it is toluene. 1p
The molar mass of P3 equivalent containing acidic group(s) is 100 156
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.
P 1p P1 3p (1p for isomer) P2 OH OH 3p (1p for isomer) P3 2p (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).
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 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
ClO3- 1.5p
(wrong central atom 0.5p)
Y2
ClO2- 1.5p
(wrong central atom 0.5p)
Y3 Cl– 1.5p (wrong element 0.5p) Y4 O2 1.5p T1 2p T2 5p If incorrect, but molecular formula 1p aldehyde 1p no identical 0.5p
(5p for hemiacetal, 3p for other
