CHEMISTRY: ART, SCIENCE, FUN

THEORETICAL
EXAMINATIONSOLUTION and
GRADING SCHEMEJULY 20, 2007
MOSCOW, RUSSIA
Official English version

Problem 1. Proton tunneling1.1.1 The structures of propanedial and two of its isomers
O=CH-CH2-CH=O
1 mark
H
O

O

C
H

C
H

C
H

1 mark

OH
H

C
H

C
C
H

O

1 mark3 marks maximum

1.1.2
Acidic hydrogen atom is in CH2 (in enol forms acidic hydrogen is in OH).
1 markAcidity of СН2 group is caused by the stability of carbanion due to conjugation with two carbonylgroups. The first answer is correct.
2 marks
3 marks maximum
1.2.1 The distance between two minima on the energy curve is 0.06 nm. In a purely aldehyde form
H

O

H

O

such distance between two possible positions of proton is impossible. Tunneling takes place only inenol Z-form:
H

H
O
C
H

O

O

C

C
C
H

O

H

H

C
C
H

H

1 mark for each structure
2 marks maximum

1.3.1 Expressions and plots of probability density
1(а) Y 2 ( x, 0) = éë Y 2L ( x ) + Y 2R ( x) + Y 2L ( x ) - Y 2R ( x ) ùû = Y L2 ( x )2
The probability density is concentrated in the left well:

1 mark

2

2

YL

-0,06

-0,04

-0,02

L

0,00

0,02

0,04

0,06

R

Distance, nm

0.5 marks(b) In the middle of the time intervalæ p ö 1é 2
2
1 markY 2 ç x,
÷ = ë Y L ( x) + Y R ( x) ùû
2
w
2
è
ø
The probability density has a symmetric form, a proton is delocalized between two wells:2

2(YL +YR )/2

-0,06

-0,04

-0,02

L

0,00

0,02

Distance, nm

0,04

0,06

R
0.5 marks

æ pö 1(c) Y 2 ç x, ÷ = éë Y 2L ( x ) + Y 2R ( x) - Y 2L ( x) + Y 2R ( x) ùû = Y 2R ( x)
è wø 2The probability density is concentrated in the right well:

1 mark

2

YR

-0,06

-0,04

-0,02

L

0,00

Distance, nm

0,02

0,04

0,06

R
0.5 marks
4.5 marks maximum

1.3.2 The probability of finding the proton in the left well is 1/2, because probability function issymmetric, and both wells are identical.
2 marks3

1.3.3 The time of transfer from one well to another is t = p / w.
3.14t=
= 4.85 ×10-12 s.
2 marks
11
6.48 ×10
The proton velocity:
0.06 ×10-9
2 marksV == 12 m/s.
4.85 ×10-12
4 marks maximum
1.3.4
The uncertainty of proton position is approximately equal to half of the distance between minima,that is 0.03 nm (0.06 nm will be also accepted).
1 mark
The minimal uncertainty of velocity can be obtained from the uncertainty relation:
1.055 ×10-34h
3 marks
DV =» 1000 m/s.
=
0.001
2mDx
-9
× 0.03 ×10
2×6.02 ×1023
Comparing this uncertainty with the velocity 12 m/s we see that the notion of proton velocity duringtransfer from one well to another is senseless. Therefore, proton tunneling is a purely quantum
phenomenon and cannot be described in classical terms. The second conclusion is correct.
2 marks
6 marks maximum4

Problem 2. Nanochemistry
2.1.1
The Gibbs energy and the equilibrium constant of reaction (1)0
D r G500(1) = DG 0f ,500 (CoO,s) - DG 0f ,500 (H 2O,g) = - 198.4 + 219.1 = 20.7 kJ/mol
K =e

-

0
D r G500
(1)
RT

=e

-

20700
8.314×500= 6.88 ×10-3

0.5 marks
0.5

marks
1 mark maximum

2.1.2
The standard Gibbs energy of the reaction (1) with the spherical cobalt nanoparticles of radius ra is
D Go (1, r ) = G 0(CoO,s) + G 0 (H ,g) - G 0 (H O,g) - G 0 (Co) =
r

500

a

bulk ,500

500

2

500

2

sph

V (Co) ö
2s
æ 0
0
0
0
= G500(CoO,s) + G500
(H 2 ,g) - G500
(H 2 O,gas) - ç G500
(Co,s) + Co-gas
÷=
ra
è
ø
o (1) - 2sCo-gasV (Co) ;= D r G500ra

M Co
10-6 × 59.0
m3=
= 6.6 ×10 -6
;
8.90
mol
r (Co)
for spherical particles with ra = 10–8, 10–9 m one gets, respectively
2s Co-gasV (Co)= 210 and 2100 J/mol.
ra
D Go (1, r ) is equal to 20.5 (a), and 18.6 (b) kJ/mol, respectively.
V (Co) =

r

500

a

The equilibrium constant is calculated from the equation
æ D Go (1, r ) öK (1, ra ) = exp ç - r 500 a ÷ ;
ç
÷
RT
è
ø
-3
-8
K (1, ra ) = 7.22 ´10 ; ra = 10 m
K (1, ra ) = 11.4 ´10-3 ;ra = 10-9 m

2 marks maximum

5

2.2.1The standard Gibbs energy for reaction (1) involving nanoparticles of cobalt iso (1, r ) = D Go (1) - 2s Co-gas V (Co)D r G500
ar 500
ra
D Go (1) is 20.7 kJ/mol. For spherical cobalt particles with r = 1 nm D Go (1, r ) is 18.6 kJ/mol.
r

a

500

r

500

a

Solid cobalt oxide can be formed spontaneously when Gibbs energy of reaction (1) is negative. The
inequality for bulk cobalt is:o (1) + RT ln æ p (H 2 ) ö = D Go (1) - RT ln æ p(H 2O) ö £ 0
D r G (1) = D r G500
ç
÷
ç
÷
r 500
è p(H 2 O) ø
è p (H 2 ) øand for spherical cobalt nanoparticles with ra = 1 nm:
o (1, r ) + RT ln æ p (H 2 ) ö = D G o (1) - 2sCo-gas V (Co) - RT ln æ p (H 2O) ö £ 0
D r G (1, ra ) = D r G500
ç
÷
ç
÷
a
r 500
ra
è p (H 2 O) ø
è p(H 2 ) øo (1) is 20.7 kJ/mol. For spherical cobalt particles with r = 1 nm D Go (1, r ) is 18.6 kJ/mol.
D r G500ar 500
ap(H 2O)are 145.6 (a) and 87.7 (b), respectively.
The minimum ratiosp (H 2 )The hydrogen pressure is
1bar×0.0015 = 1.5×10–3 bar
The minimum pressures of water are
1.5×10–3×145.6 = 0.218 bar (a) and 1.5×10–3×87.7 = 0.132 bar (b), for the bulk cobalt and fornanoparticles, respectively.
H2O%(bulk Co) = 21.8%
We assume that bulk cobalt oxide is formed.

H2O%(nanoparticles with ra = 1*10–9 m) = 13.2%.
4 marks maximum

2.2.2
For the spontaneous oxidationo (1) - 2sCo-gas V (Co) - RT ln æ p(H 2 O) ö £ 0
D r G (1, ra ) = D r G500ç÷ra
è p (H 2 ) øando (1) - 2s Co-gas V (Co) £ RT ln æ p (H 2 O) ö
D r G500ç÷raè p(H 2 ) ø
The left hand side of the last inequality becomes more positive with the increase of ra. At certain
point the inequality will be disturbed and the spontaneous oxidation will not take place. So, toprotect cobalt nanoparticles from the spontaneous oxidation in this case one has to lengthen theradius ra. The answer (a) is correct.
2 marks maximum

2.3.1
The equation for the standard molar Gibbs function of CoO (external layer) reads:
2s
2s
0
Gsph
(CoO,rb ) = Gbulk (CoO) + CoO-gas V (CoO) = Go (CoO,s) + CoO-gas V (CoO)rbrb
1 mark6

2.3.2The equation for the standard molar Gibbs function of Co (internal layer) reads:
æ 2s
ö
2s
0
Gsph
(Co,ra , rb ) = Gbulk (Co) + V (Co) ç CoO-gas + CoO-Co ÷ =
rb
ra
è
ø

æ 2s
ö
2s
= G o (Co,s) + V ( Co ) ç CoO-gas + CoO-Co ÷
ra
è rb
ø
The expression in brackets gives the additional pressure in the internal layer (see the Hint).
5 marks maximum
2.3.3
The standard Gibbs energy for reaction (1) with the double-layered nanoparticles is
D G 0 (1, r , r ) = G 0 (CoO, r ) + G o (H , gas ) - Go (H O,gas) - G 0 (Co, r , r ) =
r

a

b

sph

b

2

2

sph

a

b

= G o (CoO,s) + Go (H 2 , gas) - Go (H 2 O,gas) - Go (Co,s) +æsösV (CoO) - 2V (Co) ç CoO-gas + CoO-Co ÷ =
rb
ra ø
è rb
2s
2s
= D r Go (1) + CoO-gas (V (CoO) - V (Co) ) - CoO-Co V (Co)
rb
ra

+

2sCoO-gas

2 marks maximum
2.3.4.
Under the assumptions made2s
2s
D r Go (1, ra , rb ) = D r G 0 (1, r0 ) = D r G o (1) + CoO-gas (V (CoO) - V (Co) ) - CoO-Co V (Co) =rb
ra2s CoO-gas æ3ö= D r G o (1) +ç V (CoO) - V (Co) ÷r02
èøThe term in brackets in the right-hand side is positive
3
æ
ö
–6
3
ç V (CoO) - V (Co) ÷ = 6.56×10 m
2
è
ø
æ1ö
D r G 0 (1, r0 ) is directly proportional to ç ÷ . The plot (a) is correct.
è r0 ø
3 marks maximum
2.3.5.
The spontaneous forward reaction (1) is possible, when DrG(1,r0) £ 0, andpH 2 O
2s CoO-gas æ3
öD r G 0 (1) +ç V (CoO) - V (Co) ÷ £ RT ln
2
r0
pH 2
è
ø
The term in brackets in the left-hand side is positive. The left hand side of the inequality becomes
more positive with the decrease of r0. At certain point the inequality will be violated and thespontaneous oxidation will not take place.
In order to protect nanoparticles from oxidation in this case one has to shorten the radius r0.
The answer (b) is correct.
2 marks maximum
7

Problem 3. Unstable reaction
3.1.1 The overall reaction equation
B+D®P
The kinetic equation for X
d [X]= k1[B][X]2 - k2 [D][X]dt

1 mark1 mark2 marks maximum

3.1.2 Under the steady-state conditions
d [P]= k2 [D][X] = k1[B][X]2 ,dt
whencek [D][X] = 2
k1[B]d [P] k 2 2 [D]2
3 marks=dt
k1[B]The reaction order is 2 with respect to D, –1 with respect to B; the overall order is 1
0.5 marks for each correct order
4.5 marks maximum
3.2.1 In an open system the initial reaction rate is:d [X]= [B][X] ( k1[X] - k2 )dt1) If [X]0 > k2/k1, then d[X]/dt > 0 at any time, and the concentration of X monotonicallyincreases:[X]t

2 marks
2) If [X]0 < k2/k1, then d[X]/dt < 0 at any time, and the concentration of X monotonicallydecreases:

8

[X]

t

2 marks
4 marks maximum
3.2.2 In a closed system the initial reaction rate is:d [X]= k1[B]0 [X]02 - k2 [D]0 [X]0 = [B]0[X]0 ( k1[X]0 - k2 ) > 0
dt t = 0Hence, at the beginning of the reaction [X] increases but it cannot increase infinitely and finallygoes to its initial value, because the second reaction is irreversible:[X]t

2 marks for maximum
1 mark for the asymptotic value
3 marks maximum
3.3.1 X – C2H6O2, Y – C2H4O, P – C2H6O. Dots denote O2 and H2O.

C 2 H 6 + C2 H 6 O 2 + O 2 ® 2C 2 H 6 O 2
C 2 H 6 O 2 + C2 H 4 O ® 2C 2 H 4 O + H 2O
C 2 H 6 + C2 H 4 O + H 2O ® 2C 2 H 6O
9

0.5 marks for each unknown substance (X, Y, P, three blanks)
3 marks maximum
3.4.1 At the highest possible temperature the rate constants are equal:

æ E ö
æ E ö
A1 exp ç - A,1 ÷ = A2 exp ç - A,2 ÷
è RT ø
è RT ø
E - E A,1
= 354 K
T = A,2A2
R ln
A1
1 mark for the calculation
2 marks for the correct answer
3 marks maximum

10Problem 4. Determination of water by Fischer titration
4.1. Equation:
I2 + SO2 + 2 H2O + 4 Py = 2 PyHI + (PyH)2SO4
1 mark
(0.75 marks for not accounting for the formation of Py salts)
4.2.1. T is equal to:M(Na2C4H4O6.2H2O) = 230.05

2M(H2O) = 36.04m(H2O) = 1.3520 · 36.04 / 230.05 = 0.2118 g = 211.8 mg
1 mark for the formula
T = 211.8 / 12.20 = 17.36 mg/mL
T = 17.36 mg/mL
0.25 marks for the correct
result (within 2 digits)
1.25 marks maximum

4.2.2. T is equal to:Your work:
Volume of iodine spent for 10 mL of pure CH3OH = 2.20·10.00 / 25.00 = 0.88mL
0.5 marks for the correct formula of pure methanol titration)
T = 21.537•0.01•103 / (22.70 – 0.88) = 9.87 mg/mL
More exactly
10.00 mL of the solution contains (1000-21.5)×10.00 / 1000 = 9.785 mL of
methanol
Volume of iodine spent for 9,785 mL of pure CH3OH = 2.20·9,785 / 25.00 =0.86 mLT = 21.537•0.01•103 / (22.70 – 0.86) = 9.86 mg/mL
1 mark for the formula of water titration, only 0.5 marks without subtracting
0.88
T = 9.87 mg/mL

0.25 marks for the correct result
1.75 marks maximum

11

4.2.3. T is equal to:Your work:

Approach 1.
Let 1 mL of CH3OH contain x mg H2O, then 1 mL of A contains ((1.000 –0.006)·x + 5.624) mg H2O.
15.00·T = 22.45·(0.994·x + 5.624) – 1st titration,
10.00·T = 25.00·x + 10.79·(0.994·x + 5.624) – 2nd titration.
Hence, x = 1.13 mg/mL, T = 10.09 mg/mL (10.10 without taking into account0.994 factor)
Approach 2.
Let y mL of B be spent for the titration of water, contained in 1 mL of CH3OH.
22.45 ·5.624
10.79 ·5.624
Then T =(1st titration) =(2nd
15.00 - 22.45·0.994· y
10.00 - 25.00 y - 10.79 ytitration).
Hence, y = 0.1116 and T = 10.10 mg/mL
T = 10.09 mg/mL (10.10 without taking into account 0.994 factor)
2 marks for the correct formulas (with or without taking into account 0.994 factor)
and 0.25 marks for the correct result (10.10 or 10.09)
2.25 marks maximum

4.3. Equation(s):CaO + SO2 = CaSO3
2CaO + 2I2 = CaI2 + Ca(OI)2
6CaO + 6I2 = 5CaI2 + Ca(IO3)2(Instead of CaO, Ca(OH)2 may be written.)
1 mark for ANY correct equation
1 mark maximum

4.4.1 Equation(s):Fe2(SO4)3 + 2HI = 2FeSO4 + I2 + H2SO4

1 markFe2(SO4)3 + H2O + SO2 + CH3OH = 2FeSO4 + CH3OHSO3 + H2SO4
1 mark(or in ionic form)
2 marks maximum
12

4.4.2. Equation:
Fe2(SO4)3·xH2O + (x – 1)I2 + xSO2 + xCH3OH = 2FeSO4 + xCH3OHSO3 +H2SO4 + 2(x – 1)HI
1 mark

4.4.3. The composition of the crystallohydrate is:Your work:
M(Fe2(SO4)3·xH2O) = 399.9 + 18.02x

mH2O (g) =0.6387 ·18.02 x;(399.9 + 18.02 x )1 markx
x -1
1 mark

mH2O (g) = 10.59(mL)´15.46(mg/mL)´0.001(g/mg)´®
0.1637·(399.9 + 18.02x) = 11.51x – 11.51;
x = 8.994
Formula: Fe2(SO4)3.9 H2O

x=90.25 marks (for correct answer)
2.25 marks maximum13

Problem 5. A mysterious mixture (organic hide-and-seek game)
5.1.1 Structure of product D
O

H3C

O

CH3

Ethyl acetate, ethyl ethanoate
Any structural formula or any shorter versions (CH3COOC2H5) including the adopted shortcuts
for organic radicals (Me, Et, Ac), or systematic IUPAC name – 5 marks
5.1.2 Which class of organic compounds does D belong to? Check the appropriate box.
Note! Only one checkmark is allowed. Several checkmarks will lead to 0 mark for this question.
ketonesethersacetals
estersalcohols
aldehydes
glycols
o
o
o
þ
o
o
o
The only correct answer is ester – 5 marks
5.1.3 The expected yield of DYour work:
Statement that the reaction is an equilibrium without any further actions – 1 markAnswer to quantitative question: lower than 85% - 2 marks
Qualitative estimation of yield can be done assuming that the reaction is at equilibrium, and thatthe equilibrium constant is supposed to not vary with temperature and composition of thereaction mixtures.

[AcOEt][H 2O](0.85)2
K=== 4.2[AcOH][EtOH] 0.15 ×1.15
Calculation of yield using this constant in 1:1 mixture gives 67%

yield = 67%
maximum – 10 marks, if the yield is computed within 67±1% error limits
5.2.1 The structures of A, B, and C.
OEt
OEt
OEt

CH3C(OEt)3

OEt

HC≡COEt
ethoxyacetylene, ethynylethyl
ether

COOEt
COOEt

CH2(COOEt)2diethyl malonate

triethyl orthoacetate, 1,1,1triethoxyethaneA
B
C
Each structure represented by structural formula or unambiguous linear notation – 10 marks
Systematic name given in place of structure – 5 marks
5.2.2 Draw in the boxes intermediate compounds formed during the acidic hydrolysis of C, andbasic hydrolysis of B.
a) Malonic acid is formed as intermediate in the hydrolysis of diethyl malonate – 5 marks
14

COOEt H+/H2O

COOH

t

COOEt

COOH

– CO2

CH3COOH + C2H5OH

C

Also accepted will be monoethyl malonate – 2 marks
maximum 5 marksb) Hydrolysis of ethoxyacetylene starts from the addition of hydroxide to the triple bond to give
unstable enolic form of ethylacetate, into which it immediately is transformed
OEt

OH–/H2O

OH

O

OEt

OEt

OH–/H2O

CH3COO– + C2H5OH

B

Indication of any of keto- or enol forms of ethylacetate – 5 marks
Hydrolysis of strong ether bond to give hydroxyacetylene, or any forms coming along this path(ketene, diketene) is impossible and is not allowed – 0 marks
Maximum for a) and b) together – 10 marks

5.3.1 The structure of senecioic acid
From acetone alone the synthesis includes aldol condensation, dehydration, with subsequentiodoform reaction
3 marks
2

O

O

H+
HO

H+

O

-H2 O

I2/OH–OOH

Senecioic acid structure alone – 4 marks, with scheme - maximum 10 marks

5.3.2 The structure of E.
Iodoform, triiodomethane, CHI3 – 5 marks

15

Problem 6. Silicates as the base of the Earth crust
6.1.1
SiO32- + 2CO2 + 2H2O = “H2SiO3”↓ (Silica acid gel) +2HCO3 - or
SiO2(OH)22-+2CO2+H2O=“H2SiO3”↓+2HCO3 - or
SiO32- + CO2 + H2O = “H2SiO3”↓ + CO323 marks maximum
2 marks if silicic acid is written in any form
1 mark if the reaction contains carbonate (or bicarbonate) ions with silicic acid missing
6.1.2а) protonation of ortho-silicate ions leading to the formation of Si-OH groups
SiO44- + H2O = [SiO3(OH)]3- + OH- or
SiO44- + H+ = [SiO3(OH)]3- or[SiO2(OH)2]2-+ H+= [SiO(OH)3]Yesb) formation of hydrated [SiO4(H2O)2]4- anions
SiO44- + 2H2O = [SiO4(H2O)2]4Yesc) polycondensation of ortho-silicate ions leading to the formation of Si-O-Si bonds
2 SiO44- + H2O = [O3Si-O-SiO3]6- +2 OH- or
2 SiO44- + 2H+ = [O3Si-O-SiO3]6- + H2O orYes
2SiO2(OH)22- + Н2O = [O-Si(OH)2-O-Si(OH)2-O]2- + 2 OH9 marks maximum
2 marks for each correct reaction
1 mark for each correct choice of check-boxes

No

No

No

6.2.1 n= 6 (assuming oxidation numbers of silicon (+4) and oxygen (-2), or taking into account its
structure and the charge of orthosilicate ion (-4))
2 marks maximum
1 mark penalty for miscalculation
6.2.2 Si3O9 ≡ 3 [SiO4] – 3 O, i.е. there are 3 oxygen atoms bridging adjacent tetrahedra
2 points maximum
1 point penalty for miscalculation
6.2.3

3 marks maximum
6.2.4
Calculations:
m=4 (assuming oxidation numbers of silicon (+4) and oxygen (-2), or taking into account its
structure and the charge of orthosilicate ion (-4))
Si4O10≡ 4[SiO4] – 6O, i.е. the formula of the tetrahedron is now SiO2.5, which is possible if 1 О
atom belongs to this tetrahedron and the other three are shared between 2 tetrahedra (theircontribution =3/2). This is possible if the tetrahedra are set on a plane and joined together throughall apexes of their bases.
16

10 marks maximum
2 marks for charge determination
3 marks for determination of the number of oxygen bridges
5 marks for the correct structure
1 mark penalty if 6 to 15 tetrahedra shown, the connection being correct
3 marks penalty if less than 6 tetrahedra shown (i.е. it is not clear that polyhedra form layer)
4 marks penalty for connection via apexes, but in 3D network
4 marks penalty for connection via apexes, but in 1D chain
0 mark of 5 for any other structure
6.3.1
рН = 4Cu(H2O)42+ + H2O = Cu(OH)(H2O)3+ + H3O+,[H+]≈ (c KaI)1/2 =1·10-4 M, pH = -log[H+] = 4

5 marks maximum
1 mark penalty for miscalculation
2 marks penalty for wrong expression of [H+] via KaI
2 marks penalty if there are some errors in definition of рН (e.g. using ln instead of lg)
3 marks penalty if the hydrolysis reaction is wrong
6.3.2
СuSO4 + Na2SiO3 + 2H2O = Cu(OH)2↓ + “H2SiO3 ”↓ + Na2SO4
or 2СuSO4 + Na2SiO3 + 2H2O = (CuOH)2SO4↓ + “H2SiO3”↓ + Na2SO4
This (or those) reaction(s) (apart from formation of copper silicate) can be deduced from the factthat the reaction describes mutual (self-amplifying) hydrolysis. It comes from the previous parts ofthe task: рН of LGL is greater than 7 (see questions 6.2), and рН of copper sulfate solution is lessthan 7 (see 6.3.1).
3 marks maximum
2 marks if reaction coefficients are wrong
1 mark if only one of two principal precipitates shown (Cu(OH)2↓ or “H2SiO3”↓)

17

Problem 7. Atherosclerosis and intermediates of cholesterol biosynthesis
7.1.1 Е2-Е4 catalyze one and the same (and only one) reaction type. The only reaction which canbe carried out three times in a row is monophosphorylation (all the rest reaction types are notconsistent with either initial or final products). This is also supported by presence of pyrophosphate
residue in IPP and liberation of inorganic products (including inorganic phosphate) upon
spontaneous decomposition of Х1.
Х is a monocarboxylic acid composed of atoms of three elements: carbon, hydrogen and oxygen. It
can contain neither sulfur which is found in CoA nor phosphorus which is introduced into
intermediates on the pathway from HMG-CoA towards IPP or present in CoA. Thus, Е1 catalyzes
non-hydrolytic removal of CoA from HMG-CoA and is not involved in phosphorylation. Since
water is not a substrate in this reaction, liberation of CoA must be conjugated with another reaction
which affects the carboxylic group esterified in HMG-CoA. The only possible variant is its 4
electron reduction towards hydroxyl group. Е1 can not catalyze dehydration because of opticalactivity of Х (removal of water leads to the loss of sole chiral center). Decarboxylation is excluded,
since Х, being an acid, must contain a carboxylic group. Oxidation of tertiary hydroxyl group in
HMG-CoA according to β-oxidation mechanism is impossible. Further evidence comes from thefact that the carboxylic group initially involved in thioester bond formation is present as the residueof hydroxyl group in IPP. So:
Е1

4,

Е3

6

5

12 marks maximum
E1: 9 marks if 2 variants are given and both are correct.
4 marks if only one variant is given and it is correct
4 marks if two correct and one incorrect variants are given
0 mark if one correct and any number of incorrect variants are given
0 mark if more than three variants are given.
Е3: 3 marks if only correct variant is given. Otherwise 0 mark7.1.2 Based on the reaction types catalyzed by Е1 and configuration of HMG-CoA stereocenter, thestructure of Х is:
HO
HOOC

(R)

OH

Х, mevalonic acid
Note the absolute configuration of the chiral center is changed as a result of HMG-CoA metabolism
into mevalonic acid due to alteration of substituents priority.
HO
HOOC

(S)

S

CoA

E1

HO
HOOC

O

(R)

OH

12 marks maximum
8 marks for correct structural formula
4 marks for correct stereochemistry (only in case both the structure is correct and R-isomer isindicated; otherwise 0 mark).
No penalty for incorrect stereochemistry or absence of stereochemical information

18

7.2.1 Reaction equation for reductive ozonolysis
O
OOPP+ O3 + H3C S CH3
O O
O
O

H
+
O

O

OCH3
H C
O P O + 3 S
O
P
OO
O
O

5 marks maximum
4.5 marks for correct products (1.5 marks each), incorrect structures not penalized
0.5 mark for correct equation coefficients
7.2.2 DAP molecule contains only one carbon atom which can be involved in the formation of С–Сbond during Y biosynthesis. Irrespective of the way this molecule is incorporated in Y, ozonolysisof this fragment will lead to dimethyl ketone (acetone). (See DAP ozonolysis reaction in 7.2.1).
Thus, acetone can be unambiguously attributed to Y1, since it contains 3 carbon atoms (Y2 and Y3
contain 5 and 4 carbon atoms, respectively). Taking into account the ratio between ozonolysis
products, we have:nY(C)= 2*nY1(C)+4* nY2(C)+ nY3(C)= 2*3 + 4*5 + 4 = 30
Y is an acyclic molecule, thus DAP residues can be found only at its ends. Y has only two ends,
since IPP contains only two elongation sites (at least three such sites are needed to get a branchedmolecule). Since reductive ozonolysis of one Y molecule produces two acetone molecules, Y
contains 30 carbon atoms.
To determine the number of hydrogen atoms double bonds in Y should be counted. Formation ofeach double bond reduces by 2 the number of hydrogen atoms in the coupling product as comparedto the sum of atoms of starting substances. The ratio of Y to the sum of its ozonolysis products is
1:7 (2+4+1), which corresponds to 6 double bonds in Y. Then, by using the general formula foralkanes we have:n(H)= 2*nY(C)+2–2*nc=c=30*2 + 2 - 6*2 = 50
Y (squalene) formula – С30Н50.
Work:
Number of carbon atoms 30
nY(C)= 2*nY1(C)+4*nY2(C)+ nY3(C)= 2*3 + 4*5 + 4 = 30
Work:
Number of hydrogen atoms 50
n(H)= 2*nY(C)+2–2*nc=c=30*2 + 2 - 6*2 = 50
Gross formula of Y С30Н50
12 marks maximum
8 marks for correct justification of molecular formula
4 marks for correct molecular formula7.2.3 IPP and DAP are structural isomers containing 5 carbon atoms each. Since all carbon atoms ofthese substances are found in Y, one can calculate the total quantity of IPP and DAP moleculesneeded to synthesize Y:n(IPP&DAP)= nY(C)/5=30/5=6
The number of DAP molecules was determined earlier and is equal to 2. Then, 4 molecules of IPP
are needed.
n(IPP&DAP)= nY(C)/5=30/5=6
Number of DAP molecules 2
Number of IPP molecules 4
7 marks maximum
3.5 marks for calculation of the total number of DAP and IPP molecules
3.5 marks for correct individual numbers of DAP and IPP molecules197.2.4 All possible combinations that do not alter hydrocarbon skeleton are given below(pyrophosphate fragments not shown). Two groups of products differing in carbon atoms involvedin coupling reaction are separated by the dashed line. IPP fragments should be attached to DAP sothat ozonolysis of the product leads to Y2 containing 5 carbon atoms. Only one variant is possible if
stereochemistry is not taken into consideration and two variants with stereochemical details
No

No
*

*
+

*
+
No

No

Yes

(E)

O

O-

P
OO
O
P O
O

or
O-

-

O

(Z)

O

P

O

P O
OO

The upper isomer is geranyl pyrophosphate
8.5 marks maximum
8.5 marks for correct structure
No penalty for stereochemistry, any correct structure is accepted
2.5 marks if ozonolysis of the product leads to acetone, but does not lead to the compound with 5
carbon atoms
2.5 marks if ozonolysis of the product leads to the compound with 5 carbon atoms, but does notlead to acetone
0 mark for any other variant
7.2.5 It is seen from the coupling reaction (Scheme 2) that Y4 contains 15 carbon atoms or 1 DAP
and 2 IPP fragments, the latter being attached to the former consecutively. It is important to notethat Y3 can not be found in two hydrocarbon residues originating for Y4, since Y3 is formed as aresult of ozonolysis in a molar ratio of 1:1 to Y. Thus, geranyl phosphate is the intermediate on theway to Y (all double bonds in trans configuration). Attachment of the next IPP fragment to geranyl
phosphate leads to the product giving 1 molecule of Y1 and 2 molecules of Y2 upon its ozonolysis.
Thus, Y4 structure with stereochemical details:
(E)

(E)

O

O-

P
OO
O
P O
O

Y4, farnesyl pyrophosphate
20Combining two hydrocarbon fragments of Y4 and taking into account that the double bond between
them is being reduced we get the following structure of Y:
(E)

(E)
(E)

(E)

Y, squalene
16 marks maximum
9 marks for farnesyl pyrophosphate (6.5 marks for correct structural formula and 2.5 marks forcorrect stereochemistry)
7 marks for squalene (5 marks for correct structural formula and 2 marks for correctstereochemistry)
2.5 marks penalty for unreduced double bond in squalene

21

Problem 8. ATRP allows new polymers
8.1.1 Expressions for the rates of ATRP elementary stages: activation (vact), deactivation (vdeact),
propagation (vp) and termination (vt)vact = kact·[R-Hal]·[CuHal(Ligand)k]2 marks
vdeact = kdeact·[R·]·[CuHal2(Ligand)k]2 marks·
vp = kp·[R ]·[M]
2 marks
vt = 2kt·[R·]2
2 marks (no penalty for missing 2)
8 marks maximum
8.1.2 Comparison of rates of ATRP elementary stages
Since all the chains grow with equal rate, the process proceeds as living polymerization. Living
radical polymerization is possible only if concentration of active radicals is low to prevent chaintransfer and termination. So:
vdeact >> vact
The portion of active radicals must be small, so the equilibrium is shifted towards dormant species.
vdeact >> vp
Propagation rate should be much slower than that of deactivation to make chains propagate withequal rate.
vdeact >> vt
Termination does not occur since the total number of polymer chains is equal to a certain constantnumber – number of initiator molecules.
vdeact >> vact
3 marks
vdeact >> vp
3 marks
vdeact >> vt
3 marks
9 marks maximum
8.2.1 Calculation of mass (m) of the obtained polymer.
1st variant[M ] = [ M ]0 exp(-k P [ R×]t ) or n( MMA) = n 0 ( MMA) exp(-k P [ R×]t )1 markQuantity of MMA monomer remaining after polymerization during 1400 s is
31.0 × exp( -1616 × 1.76 × 10-7 × 1400) = 20.8 mmol.
2 marks
Quantity of monomer consumed during polymerization: 31-20.8=10.2 mmol
1 markMass of the obtained polymer is m = Dn( MMA) × M (MMA) = (10.2 /1000) ×100.1 = 1.03 g 1 mark2nd variant[M ] = [ M ]0 exp(-k P [ R×]t ) or n( MMA) = n0 ( MMA) exp( -k P [ R×]t )1 markQuantity of MMA monomer consumed during 1400 seconds of polymerization isDn( MMA) = n0 (MMA)(1 - exp(-k p × [ R · ] × t )) = 31.0 × (1 - 1616 ×1.76 ×10 -7 ×1400) = 10.2 mmol
3 marks
Mass of the obtained polymer is m = Dn( MMA) × M (MMA) = (10.2 /1000) ×100.1 = 1.03 g 1 mark3rd variant
æ [M ] ö-7
1 markln ç÷ = - k P [ R×]t = - 1616 ×1.76 ×10 ×1400 = - 0.398
M
[
]
0 ø
è

[M ]
= e -0.398 = 0.672
[M ]0
[M ]
n(MMA)=[M ]0n0 (MMA)n( MMA) = 0.672 × n0 ( MMA) = 20.8 mmol

1 mark1 mark22

Quantity of monomer consumed during polymerization is 31-20.8=10.2 mmol
1 markMass of the obtained polymer is m = Dn( MMA) × M (MMA) = (10.2 /1000) ×100.1 = 1.03 g 1 markm = 1.03 g
5 marks maximum
8.2.2 Calculation of degree of polymerization (DP) of the obtained polymer.
The number of growing chains is equal to the number of TsCl molecules (0.12 mmol) 2 marks
At the first stage, 7.3 mmol of MMA was consumed (0.73/100.1).
The total quantity of monomers at the beginning of the 2nd stage is 23.7 + 23.7 = 47.4 mmol.
2 marks
Since the monomers have the same reactivity, they will be involved in polymerization with thesame rate.
Quantity of monomers consumed during the second stage is
Dn = n0 (1 - exp( - k P [ R×]t )) = 47.4(1 - exp( -1616 × 1.76 × 10 -7 × 1295)) = 14.6 mmol.
4 marks
Totally 7.3+14.6 = 21.9 mmol of monomers was polymerized during two stages.
2 marks
DP = 21.9/0.12 = 182.5
1 markDP = 182-183 (all answers within this range are credited)
11 marks maximum
8.2.3 Structure of the obtained polymer.
The product of polymerization is a block copolymer because it was obtained by sequential
polymerization on living chains.
The first block is built of MMA units solely. The DP is 7.3/0.12 = 60.8 ≈ 61 monomer units.
The second block is obtained by copolymerization of two competing monomers with the same
reactivity. So, it is a statistical copolymer. Fractions of A and B in the 2nd block are equal because
their concentrations in the reaction mixture at the beginning of the 2nd stage were equal. The DP of
the 2nd block is 183-61 = 122 monomer units (121 is also correct if the total DP in 8.2.2 is 182).
Ts-A61-block-(A-stat-B)61-Cl or Ts-A61-block-(A61-stat-B61)-Cl
14 marks maximum
4 marks for block copolymer with blocks A and co-AB
4 marks for an indication of the statistic character of the 2nd block
1 mark for equal fractions of A and B in the 2nd block
2 marks for correct DPs of each block
1 mark for indication of terminal groups
8.3.1 Assignment of NMR signals to substructures given in the Answer Sheet.
** a, b, g 3x1.5 marks
CH2
O

CH2

*
H

c

2 marksd

2 markse

4 marks

H

H

H
H
*

H

H

H

H
*

*
*
H

Cl

23

*

f

4 marksCl
H

H

16.5 marks maximum
8.3.2 Determination of molar fractions of units C and D and molecular weights of P1 and P2.
Intensity of multiplets b and g is 40.2, so intensity per 1 proton is 40.2/4/58=0.173 for both
copolymer spectra
2 marks
Intensity of multiplet с is 13.0, which is equivalent to 13.0/0.173=75 protons. Taking into accountthat each styrene ring has 5 aromatic protons, DP of styrene block is 75/5=15.
2 marks
Molar fraction of styrene units in P1 is 15/(15+58) = 20.5%
1 markIntensity of multiplet d is 10.4, which is equivalent to 10.4/0.173=60 protons. Since each monomer
unit of p-chloromethylstyrene has 4 protons, DP of PCS is 60/4=15.
2 marks
Molar fraction of D is 15/(15+58) = 20.5%
1 markM(P1) = 15.03+58x44.05+72.06+15x104.15+35.45 = 4240
2 marks
M(P2) = 15.03+58 x44.05+72.06+15x152.62+35.45 = 4967
2 marks
M(P1) = 4240
M(P2) = 4967
n(C) = 20.5%
n(D) = 20.5%
12 marks maximum
8.3.3 All possible reactions of activation occurring during the synthesis of P1 and P2.
10 marks maximum
P1:(1.5+2) marks
O
O
H3C

O
CH3

O
58

+

Cu(+)Cl(Ligand)k

O

O
58

H3C

Cl

CH

+

Cu(+2)Cl 2(Ligand)k

CH3

R

R
Cl

CH

+

Cu(+)Cl(Ligand)k+Cu(+2)Cl 2(Ligand)k

Here R is used for the macroinitiator fragment with one or several styrene units attached.

P2:(1.5+2+3) marks

24

O
O
H3C

O
CH3

O
58

+

O

Cu(+)Cl(Ligand)k

O
58

H3C

+

CH

Cl

Cu(+2)Cl2(Ligand)k

CH3

R

R
Cl

CH

+

+

(+)

Cu Cl(Ligand)kCl

Cu(+2)Cl2(Ligand)k

Cl

R

R
Cl

Cl

+

Cu(+)Cl(Ligand)k+Cu(+2)Cl2(Ligand)k

CH2

Cl

Here R is used for the macroinitiator fragment with one or several p-chloromethylstyrene units
attached.

8.3.4 The structure of P1 and one of possible structures of P2
P1 is a block copolymer of PEO and PS. The PS block contains 15 units.
P2 is a block copolymer composed of PEO block and branched styrene block.
The integral intensity of multiplet f is 2.75, so 2.75/0.173=15.9, that is about 16 protons or 8
chloromethyl groups.
d) If there is no branching in molecule P2, it would contain 15 choromethyl groups. Each branching
reduces the number of such groups by 1. Thus P2 has 15-8 = 7 branchings. Every structure with 7
branchings is correct if each monomer unit is linked with not more than 3 other monomer units
Р1

R

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

Cl

25

Р2
Cl
Cl

D
D

Cl
Cl

D

Cl

Cl

D
D

D

Cl

D
Cl

R

D
Cl

D
D

Cl

D

Cl

D
D

D

Cl

Cl

D

Cl

Cl

Cl

13.5 marks maximum
2 marks for P1
7.5 marks for completely correct structure of P2
4 marks for structure of P2 with nonzero but incorrect number of branchings
4 marks penalty if there is a unit linked with more than 3 other monomer units

26