19th – 29th July 2018
Bratislava, SLOVAKIA
Prague, CZECH REPUBLIC
www.50icho.eu

THEORETICAL PROBLEMSCountry:
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Student code:
Language:50th IChO 2018
International Chemistry Olympiad
SLOVAKIA & CZECH REPUBLICBACK TO WHERE IT ALL BEGAN

XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

Table of Contents
Instructions ......................................................................................................................................2
Physical constants and equations ...................................................................................................3
Problem 1. DNA ..............................................................................................................................5
Problem 2. Repatriation of remains in the middle ages .................................................................. 14
Problem 3. Emerging electro-mobility ............................................................................................ 22
Problem 4. Column chromatography of radioactive copper ........................................................... 30
Problem 5. Bohemian garnet ......................................................................................................... 34
Problem 6. Let’s go mushrooming ................................................................................................. 41
Problem 7. Cidofovir ...................................................................................................................... 47
Problem 8. Caryophyllene ............................................................................................................. 54

THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

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Instructions•

This theoretical exam booklet contains 62 pages.

•

You may begin writing as soon as the Start command is given.

•

You have 5 hours to complete the exam.

•

All results and answers must be clearly written in pen in their respective designed areas on theexam papers. Answers written outside the answer boxes will not be graded.

•

You were provided with 3 sheets of scratch paper. If you need more, use the backside of theexam sheets. Remember that nothing outside the designed areas will be graded.

•

The periodic table and visible light spectrum are not part of this booklet; they are provided
separately.

•

Use only the pen and calculator provided.

•

The official English version of the exam booklet is available upon request and serves
for clarification only.

•

If you need to leave the exam room (to use the toilet or have a snack), wave the blue IChO card.
The exam supervisor will come to accompany you.

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The supervisor will announce a 30-minute warning before the Stop command.

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You must stop your work immediately when the Stop command is announced. Failure to stop
writing by ½ minute or longer will lead to nullification of your theoretical exam.

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After the Stop command has been given, place your exam booklet back in your exam envelope
and wait at your seat. The exam supervisor will come to collect the envelope.

THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

Physical constants and equations
Avogadro's constant:

NA = 6.022 × 1023 mol−1

Universal gas constant:

R = 8.314 J K−1 mol−1

Speed of light:c = 2.998 × 108 m s−1

Planck's constant:h = 6.626 × 10−34 J s

Faraday constant:

F = 9.6485 × 104 C mol−1Standard pressure:p = 1 bar = 105 Pa

Normal (atmospheric) pressure:patm = 1.01325 × 105 Pa

Zero of the Celsius scale:273.15 K

Mass of electron:me = 9.109 × 10−31 kg

Unified atomic mass unit:u = 1.6605 × 10−27 kgÅngström:

1 Å = 10−10 m

Electronvolt:

1 eV = 1.602 × 10−19 J

Watt:

1 W = 1 J s−1

Ideal gas equation:

pV = nRTThe first law of thermodynamics:

ΔU = q + W

Power input for electrical device:

P =UI
where U is voltage and I electric current

Enthalpy:

H = U + pV

Gibbs free energy:

G = H – TS
ΔGo = – RT lnK = – zFEocell
ΔG = ΔGo + RT lnQ

Reaction quotient Q
for a reaction a A + b B ⇌ c C + d D:

Q=

Entropy change:ΔS =Heat change
for temperature-independent cm :Δq = ncm ΔT[C]c [D]d[A]a [B]bqrev
Twhere qrev is heat for the reversible process

where cm is molar heat capacity

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

Van ’t Hoff equation:d lnK Δr Hm
=
dT
RT2

⇒

Henderson–Hasselbalch equation:

pH = pKa + log

[A– ]
[HA]

Nernst–Peterson equation:

E = Eo –Energy of a photon:

E=

Relation between E in eV and in J:

E⁄eV =

Lambert–Beer law:A = log

Wavenumber:

ν̃ =

ν
1
k
√
=
c 2πc μReduced mass µ for a molecule AX:

μ=

mA mX
mA + mX

Energy of harmonic oscillator:

1
En = hν (n + )2Arrhenius equation:

k = A e RTK2
Δr Hm 11ln ( ) = –(– )K1
R T2 T1

RT
lnQ
zF

hc
λ
E⁄J
qe ⁄C

I0
= εlc
I

E
− a

Rate laws in integrated form:
Zero order:[A] = [A]0 – kt

First order:ln[A] = ln[A]0 – kt

Second order:

1
1
=
+ kt
[A] [A]0

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

Theoretical
Problem 1

Question

1.1

1.2

1.3

1.4

1.5

1.6

Total

Points

5

5

4

12

12

24

62

7% of the total

Score

Problem 1. DNA
Palindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the 5′→3′ direction matches
the 5′→3′ reading on the complementary strand. Hence, a palindromic dsDNA consists of two
identical strands that are complementary to each other. An example is the so-called
Drew–Dickerson dodecanucleotide (1):5′-CGCGAATTCGCG-3′| | | | || | | | | | |3′-GCGCTTAAGCGC-5′(1)
1.1

How many different palindromic double-stranded DNA dodecanucleotides (i.e., dsDNA species
with twelve base pairs) exist?
The choice of the first, second, third, fourth, fifth and sixth nucleotide in one strand isarbitrary; the seventh through twelfth nucleotides in the same strand are determined by thecondition of self-complementarity. There are always 4 options (C, G, A, T) for the first throughsix nucleotide. Hence, the total number of palindromic DNA hexanucleotides is
4 × 4 × 4 × 4 × 4 × 4 = 46 = 4 096.
5 points in total
Any numerical result above scores full marks.
No explanation of the calculation is needed to score full marks as long as the result is correct.
Example of a partially correct solution:
4 points if the student writes 4 × 4 × 4 × 4 × 4 × 4 explicitly, but he/she gives an incorrect final result(e.g., due to an improper use of a calculator).
Examples of incorrect solutions that do not score any marks:
The student counts uracil as the fifth possibility, yielding 15 625 octanucleotides.
The student counts four possibilities for each of the twelve nucleotides, i.e. they ignore the conditionof self-complementarity, yielding 16 777 216 dodecanucleotides.
0 points for any other result.

1.2

How many different palindromic double-stranded DNA undecanucleotides (i.e., dsDNA species
with eleven base pairs) exist?
0. There is no palindromic dsDNA with an odd number of base pairs.
5 points
No explanation is needed.
0 points for any other result.

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

The melting temperature of dsDNA, Tm is defined as the temperature at which 50% of the originalamount of DNA double strands are dissociated into separate strands.
1.3

Consider the Drew–Dickerson dodecanucleotide (1). Assume that a G–C nucleobase paircontributes to the DNA duplex stability more than an A–T pair does. What is the probability
that its Tm increases when a single randomly selected base pair is replaced by a G–C pair?
Probability
The thermodynamic stability, and thus the melting temperature increases whenever an A–T
pair is replaced by a G–C pair. The probability of randomly drawing one of the 4 A–T pairs
from the Drew–Dickerson dodecanucleotide containing 12 base pairs is
4/12 = 1/3 ~ 0.333 ~ 0.33 or 33.3% ~ 33%
4 points in total
Any of the numerical results above scores full marks.
No additional explanation is needed to score full marks.
Example of a partially correct solution:
2 points if the student declares 4/12 but they present an incorrect final result (e.g. 4/12 = 0.25).
Examples of incorrect solutions that do not score any marks:
The student counts only two A–T pairs rather than four, yielding the probability of 2/12.
The student incorrectly counts the total number of nucleotide pairs.

Let us analyze the thermodynamics of formation of double-helical DNA from single strands, and its
dependence on the length of the DNA and on the temperature. The equilibrium constant ofassociation of single strands to form dsDNA differs for palindromic and non-palindromic dsDNA.
A solution of dsDNA with the initial concentration of cinit = 1.00 × 10−6 mol dm−3 was heated to Tm andequilibrium was reached.
1.4

Calculate the equilibrium constant of association of single strands at Tm for both nonpalindromic and palindromic DNA.
Non-palindromic dsDNA
Calculation:
The association reaction of a non-palindromic dsDNA reads
ssDNA1 + ssDNA2 ⇌ dsDNA
and the equilibrium constant of association takes the form
[dsDNA]c0
Knp =[ssDNA1] [ssDNA2]×c0c0
where c0 is the standard concentration of 1.00 mol dm−3 and the lower index np stands for“non-palindromic”.
At Tm, one half of the initial dsDNA concentration has melted to ssDNA, so[dsDNA] = [ssDNA1] = [ssDNA2] = 1/2 cinit

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

which yields
1/2 cinitc0
1.00 {mol dm−3 }c0(T)
Knp m ===
= 2.00 × 106
1/2 cinit 1/2 cinit 1/2 cinit 1/2 × 1.00 × 10−6 {mol dm−3 }
× c
c0
0
K=
Palindromic dsDNA
Calculation:
The association reaction of a palindromic dsDNA reads
2 ssDNA ⇌ dsDNA
and the equilibrium constant of association is[dsDNA]c0
Kp =[ssDNA] 2( c)0
where c0 is the standard concentration, and the lower index p stands for “palindromic”.
At Tm, one half of the initial dsDNA concentration has melted into two ssDNAs with identicalsequences, so[dsDNA] = 1/2 cinit[ssDNA] = 2 × [dsDNA] = cinit
which yields
1/2 cinit
1/2 c0 1/2 × 1 × 1.00 mol dm−3c0
Kp (Tm ) ==== 5.0 × 105
−6
−3
cinit
cinit 2
1.00
×
10
mol
dm
(c )
0
K=
2 points for the association reactions (1 point each)10 points for the calculation of the association constant with cinit (5 points each)Some of the steps may be condensed or performed implicitly, and the association reactions may not
be presented explicitly. As long as the derivation is correct, full marks are scored for all of theelementary steps.
K is considered correct if given as unitless number or with the unit of mol−1 dm3. Thus, failure to
introduce unitless concentrations explicitly does not lead to any deduction provided the result isnumerically correct.
Examples of grading partially correct solutions:Dissociation is considered rather than association, and the calculation is otherwise correct
deduction of 3 points
Wrong numeric value of K(Tm) in spite of the correct symbolic result, due to, e.g., an improper use ofa calculator
deduction of 2 points each
A correct expression for K is written initially, but there is an error in the derivation
deduction of 2 pts per each K affected, max. 2 points
K is correct numerically, but the unit is wrong (e.g., mol dm−3)
deduction of 2 points, enforced max. once
The score for any sub-task may not be negative.

THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

The mean contributions to the Gibbs energy of association of two single strands to form dsDNA were
estimate over a certain range of experimental conditions, and they amount to −6.07 kJ mol−1 per one
G–C pair, and −1.30 kJ mol−1 per one A–T pair present in a dsDNA.
1.5

How many base pairs are there in the shortest dsDNA oligonucleotide that has Tm above
330 K? At this Tm, consider the following values of the equilibrium constant of association of
single strands to form a dsDNA: Knp = 1.00 × 106 for a non-palindromic dsDNA, Kp = 1.00 × 105
for a palindromic dsDNA. Is the shortest oligonucleotide palindromic or non-palindromic?
Calculation of the number of base pairs:
K is related to the association Gibbs energy as ∆G°np = – RT ln Knp , so the Gibbs energy at
Tm is∆G°np (Tm ) = – RT ln Knp (Tm ) == – 8.314 J K–1 mol–1 × 330 K × ln(1.00 × 106 ) = – 37.9 kJ mol–1
The shortest oligonucleotide will be obtained if C–G-only dsDNA is considered, because
C–G pairs are more stable than A–T pairs.
Then, the smallest number of base pairs is obtained by dividing the Gibbs energy by thecontribution of one C–G pair, which for a non-palindromic dsDNA is:
nnp =– 37.9 kJ mol–1– 6.07 kJ mol–1= 6.2

The needed length of a non-palindromic dsDNA: 7
For a palindromic dsDNA, the Gibbs energy at Tm is∆G°p (Tm ) = – RT ln Kp (Tm ) == – 8.314 J K–1 mol–1 × 330 K × ln (1.00 × 105) = – 31.6 kJ mol–1,
and the number of base pairs is
np =– 31.6 kJ mol–1– 6.07 kJ mol–1= 5.2

The needed length of a palindromic dsDNA: 6
The shortest oligonucleotide is☐ palindromic (P)☐ non-palindromic (NP).
A smaller number of base pairs was obtained under the assumption of a palindromic dsDNA.
Therefore, we conclude that the shortest dsDNA that has Tm above 330 K has a palindromic
sequence of 6 C–G pairs.
5 points for the calculations of Gibbs energies
2 points for the consideration of C–G-only DNA (either stated explicitly, of just considering−6.07 kJ mol−1 for the contribution of a single base pair to the Gibbs energy of association; notawarded if the value for A–T DNA is considered)
3 points for the calculation of the number of base pairs
THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

8XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

2 points for the decision on a palindromic sequence
Full marks will be given if palindromic dsDNA is considered correctly. The non-palindromic case is
not required explicitly.
If only non-palindromic case is considered, 50% of the marks may be scored for the calculation (max.
5 points in total)−2 points for the wrong numeric value of np in spite of the correct symbolic result and logics, e.g. dueto an improper use of a calculator, or due to a gross rounding error

Finally, let us leave the simplified idea of base pairs contributing individually to the associationof DNA strands. The Gibbs energy of this process may be considered explicitly dependent
on temperature. The dependence of the inverse Tm of the Drew–Dickerson dodecanucleotide (1) on
the logarithm of the initial duplex concentration cinit is shown below. (Note: a standard concentration
c0 = 1 mol dm−3 is introduced.)

THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

9XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

cinit / 10−6 mol dm−3

0.25

0.50

1.00

2.0

4.0

8.0

Tm / K

319.0

320.4

321.8

323.3

324.7

326.2

1.6

Calculate the standard enthalpy ΔH° and the standard entropy ΔS° of the association of DNA
single strands to form the palindromic double-stranded Drew–Dickerson dodecanucleotide (1).
Assume that ΔH° and ΔS° do not vary with temperature.
Calculation:
Start from the definition of ΔG° and its relation to the equilibrium constant:°°∆G = ∆H° – T∆S°∆G = – RT ln K
Combination of the expressions leads to°∆H° – T∆S = – RT ln K
Division by T × ΔH° yields
°

1 ∆S
R
–
=
–
ln K
T ∆H°
∆H°
so that, specifically for Tm
°

1
∆S
R
=
–
ln Km
°
Tm
∆H
∆H°
where Km is the equilibrium constant at the melting temperature.
For the palindromic Drew–Dickerson dodecanucleotide, the concentrations of dsDNA andssDNA at the melting temperature may be expressed in terms of the initial duplex
concentration cinit[dsDNA] = 0.5 cinit[ssDNA] = cinitand the equilibrium constant at the melting temperature follows as
Km =c
0.5 cinit
02c( cinit )0=

1c
2 cinit0This may be cast into the equation above for the inverse melting temperature, yielding
a linearized concentration dependence (note the “plus” sign), which corresponds to the plot
in this task directly:
1
∆S°
R
cinit
=
+
ln (2
)
°
°
Tm
c0
∆H
∆H
The slope R/ΔH° and the intercept ΔS°/ΔH° may be obtained
THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

10XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018••

either graphically from the plot – method 1,
or by solving a system of 2 linear equations using any 2 data points – method 2.

Method 1: The dependence in the plot is strongly linear, and it is very easy to draw a straightline passing the centre of each of the circles representing data points. Then, values of thecclinear function may be read, e.g., at ln (2 cinit ) = –15, ca. 0.003145 K−1, and at ln (2 cinit ) = –11,
00

ca. 0.003065 K−1. From these numerical values, the slope follows as
slope =

0.003065 K–1 – 0.003145 K–1= – 0.000020 K–1–11 – (–15)

and the intercept has to be re-calculated to abscissa equal to zero, e.g., asintercept = 0.003065 K–1 + (0 – (–11)) × (–0.000020 K–1 ) = 0.002845 K–1
Method 2: For instance, considering the first and the last point, we obtain:
the first point:11== 3.135 × 10–3 K–1
Tm
319.0 Kln (2 init ) = ln (2 × 0.25 × 10–6 ) = –14.5cc0

the last point:

1
1== 3.066 × 10–3 K–1
Tm
326.2 Kln (2 init ) = ln (2 × 8.0 × 10–6 ) = –11.0cc0

A system of two linear equations follows as–33.135 × 10

–33.066 × 10

–1

K

–1

K

°

=

∆S

°

∆H

– 14.5 ×

R
∆H°

°

=

∆S

°

∆H

– 11.0 ×R∆H°

which yields the solution∆S°= 2.849 × 10–3 K–1
°
∆H

and

R
°

∆H

= – 2.0 × 10–5 K–1

Common to Methods 1 and 2: The desired thermodynamic quantities follow from there:∆H° =∆S° = R ×

R
8.314
=
J mol–1 = – 416 kJ mol–1
slope – 0.000020

intercept
0.002845= 8.314 ×
J K–1 mol–1 = –1.18 kJ K–1 mol–1
slope–0.000020

Alternative route:
It is also possible to consider the linearized dependence implicitly.
After the combination of equalities for ΔG°, Km may be expressed in terms of cinit:∆H° – Tm ∆S° = RTm ln (2

cinit
)
c0

Then, a set of 2 such equations for 2 different concentrations∆H° – Tm1 ∆S° = RTm1 ln (2THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

cinit,1)c0

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

∆H° – Tm2 ∆S° = RTm2 ln (2cinit,2)c0

may be solved for 2 unknowns, ΔH° and ΔS°.
For instance, multiplication of the equations with Tm2 and Tm1, respectively, followed by thesubtraction of equations leads to∆H° = R

Tm1Tm2cinit,1ln
Tm2 – Tm1 cinit,2and a plain subtraction of equations leads to°∆S = Rccinit,2
Tm1 ln (2 init,1c ) – Tm2 ln (2 c )00Tm2 – Tm1

Or, ΔS° may be obtained by casting ΔH° into one of the equations being solved:
∆S° =

∆H°
cinit,1+ R ln 2
Tm1
c0

The desired quantities follow as ΔH° = −420 kJ mol−1 and ΔS° = −1.2 kJ mol−1 K−1.

24 points in total
15 points for the symbolic derivations
3 points for the combination of the expressions for ΔG°
3 points for the rearrangement to yield the linear relationship between T m and Km
3 points for the concentrations of the dsDNA and ssDNA at T m
3 points for the expression of Km in terms of cinit for a palindromic sequence
3 points for the linear relationship between Tm and ln 2cinit
9 points for the numerical calculations
Note: the least-squares linear fit to the data series leads to the following results including thestandard deviations: slope = (−19.97 ± 0.08) × 10−6, intercept = (2.845 ± 0.001) × 10−3.
Method 1:
Perform a least-squares linear fit on the pocket calculator, or
read-off of the slope & intercept for the linear dependence (ref.: least-sq. linear fit)
6 points within a margin of 5σ (i.e., −20.39 < 106 slope < −19.55 & 2.840 < 103 intercept < 2.850)
3 points within a margin of 10σ (i.e., −20.81 < 106 slope < −19.13 & 2.835 < 103 intercept < 2.855)
partial marks will be scored in case the slope & intercept are given with different
deviations or if one is missing
3 points for the calculation of ΔH° & ΔS° from the slope & intercept
Method 2:
3 points for the setup of the system of two equations
up to 6 points for the calculation of ΔH° and ΔS°, considering the accuracy requirements given
above
Alternative route – symbolic derivations up to 18 points, numerical calculations up to 6 points
Symbolic derivations:
3 points for the combination of the expressions for ΔG° (a)3 points for the concentrations of the dsDNA and ssDNA at the melting temperature (b)3 points for the expression of Km in terms of cinit for a palindromic sequence (c)THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

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3 points for the setup of the system of two equations (d)3 points for the expression for ΔH° (e)3 points for the expression for ΔS° (f)9 points for a correct application of Van ‘t Hoff equation, yielding an expression for ΔH° (a+d+e)
Numerical calculations: maximum 6 points considering the accuracy requirements given above
General remarks
Some of the steps in the derivation and in the calculation may be taken implicitly or simultaneously.
As long as they are correct, full marks will be scored for each of the partial steps.
Failure to write unitless concentrations explicitly and consistently does not lead to any deduction
as long as the concentrations are in mol dm−3 and numerically correct.

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Theoretical
Problem 2

Question

2.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

Total

Points

1

4

4

2

6

10

17

14

58

8% of the total

Score

Problem 2. Repatriation of remains in the middle ages
At ambient temperatures, racemization is a slow reaction. As such, it can be used for dating
biological objects and, moreover, for studying their thermal history. Let us take L-isoleucine (L-Ile)((2S,3S)-2-amino-3-methylpentanoic acid) as an example. It isomerizes on the -carbon and forms(2R,3S)-2-amino-3-methylpentanoic acid, also known as D-allo-isoleucine. As the configuration
changes on only one of the two stereogenic centres, this process is called epimerization rather than
racemization.
2.1

Choose all true statements.
☐ D-allo-isoleucine and L-isoleucine have the same values of specific optical rotation but
they have different melting points.
☐

D-allo-isoleucine has an identical absolute value of specific optical rotation as L-isoleucine

but the sign is opposite. The melting point is the same for both isomers.
☐

D-allo-isoleucine and L-isoleucine have different values of specific optical rotation but they

have the same melting points.
☒

D-allo-isoleucine and L-isoleucine have different values of specific optical rotation anddifferent melting points.
☐

D-allo-isoleucine is not optically active.

1 point for the correct answer−1 point for each incorrect answer
0 points minimum score

2.2

Assign the absolute configurations for each stereoisomer of isoleucine.

C

2S,3R (L-allo-isoleucine)

D

2R,3S (D-allo-isoleucine)A

2S,3S (L-isoleucine)B

2R,3R (D-isoleucine)

1 point for each correct answer
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2.3

The equilibrium constant Kep for L-isoleucine epimerization has the value of 1.38 (at 374 K).
If we set molar Gibbs free energy of L-isoleucine G°m = 0 kJ mol−1, determine the Gibbs freeenergies for all structures A–D from question 2.2 at 374 K.
A

kJ mol−1

BkJ mol−1

C

kJ mol−1

D

kJ mol−1

A G°m = 0 kJ mol−1°

B Gm = 0 kJ mol−1(L-isoleucine)(B is a mirror image of A with identical physical properties)

C G°m = − RT ln Kep = −1.00 × 103 J mol−1°

D Gm = − 1.00 × 103 J mol−1 (D is a mirror image of C with identical physical properties)
4 points in total, 1 point for each numerical value A, B, C, D
No explanation is needed to score full marks as long as the results are correct.
0 points for incorrect sign.

2.4

If we take into account stereoisomerism at all stereocentres, what is the maximum possible
number of the stereoisomers of the tripeptide Ile-Ile-Ile?
The number of stereoisomers is 2N, where N is the number of stereogenic centres.
For Ile-Ile-Ile, N = 6, so there are 26 = 64 stereoisomers.
2 points for the numerical result (either 26 or 64). No explanation is needed.
1 point if the student correctly declares 26 but the final result is numerically incorrect.
Example of an incorrect solution that do not score any marks:
The student incorrectly counts the number of stereocenters.

At the start of the epimerization, we can neglect the reverse reaction. The epimerization then follows
the first-order kinetics:

The value of the rate constant at 374 K is k1(374 K) = 9.02 × 10−5 h−1 and at 421 K it isk1(421 K) = 1.18 × 10−2 h−1.
In the following calculation, shorten the concentration of L-isoleucine to [L] and of D-allo-isoleucine
to [D].
We can define a quantity de (diastereomeric excess):
de =
2.5

[L] – [D]× 100(%).
[L] + [D]Let us boil L-isoleucine for 1 943 hours at 374 K. What is the value of de (with three significant
figures) for L-isoleucine a) before boiling and b) after boiling?

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Calculation:
Before boiling, the de is:de =[L ]0 − [D ]0× 100(%)[L ]0 + [D ]0Because [D ]0= 0, the de of native L-isoleucine before boiling is 100%.
de =% (with three significant figures)b) After boiling
Calculation:
First, we have to calculate the concentration ratio[L]from the first-order rate equation
[L]0

[L]
= e–k1 t = 0.839[L]0so [D] = 0.161 [L]00.839 – 0.161

and de = 0.839 + 0.161 × 100% = 67.8% (with three significant figures)de =% (with three significant figures)

Number of significant figures is not evaluated
2 points for the de before boiling
No additional explanation is needed to score full marks.
After boiling:
1 point for the equation for the concentration ratio from the first-order rate equation
1 point for the relation between the concentration of D-allo-isoleucine and L-isoleucine
2 points for the numerical result
If de is given as 0.678, full marks are scored.
Some of the steps may be performed implicitly. As long as the result is correct, full marks are scoredfor all elementary steps.
No points are scored if the calculations are not provided

2.6

How long does it take to convert 10% of L-isoleucine to D-allo-isoleucine at 298 K?
Calculation:
We need to evaluate the rate constant at 298 K. It can be evaluated from the Arrhenius
equation:Ea
ln k = ln A –RT
We get ln A and Ea from the values of k at two temperatures (374 K and 421 K)
ln k (374 K) = ln 9.02 × 10–5 [h–1 ] = – 9.313 = ln A –Ea
R × 374

ln k (421 K) = ln 1.18 × 10–2 [h–1 ] = – 4.444 = ln A –Ea
R × 421

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It follows thatln A = – 9.313 +Ea
Ea= – 4.444 +R × 374
R × 421from which Ea = 136 kJ mol−1.
ln A is from the equation at 374 K
ln (A[h–1 ]) = – 9.313 +Ea= 34.4
R × 374

For k at 298 K we then have:Ea
136 × 103
= 34.4 –= – 20.5
R × 298
R × 298ln( k (298 K)[h–1 ]) = ln A –k(298 K) = 1.25 × 10–9 h–1
It follows from the rate equation thatln[L]
= – kt[L]0

ln 0.90 = – 1.25 × 10–9 t
t = 8.42 × 107 h = 9 610 years
t=

years

2 points for the formulation of the set of two equations
2 points for the correct formulae for ln A and EA (1 point each)
2 points for the correct numerical values of ln A and EA (1 point each)
2 points for the calculation of the rate constant at 298 K (1 point for the formula, 1 point for thenumerical value)
2 points for the calculation of time
Full marks will be given if the time is calculated correctly using a wrong value of k based on the correct
formula.
Some of the steps may be performed implicitly. As long as the result is correct, full marks are scoredfor all elementary steps. Full marks will be given for numerical calculations of time and rate constants
within margins k(298 K) = 1.25 × 10−9 h−1 ± 0.1 × 10−9 h−1, t = 9 610 years ± 850 years.
No points are scored if the calculations are not provided.

In fact, the reverse reaction cannot be neglected. The correct kinetic scheme is expressed as

Let us define the deviation of concentration from its equilibrium value [L]eq
x = [L] – [L]eq
It is possible to derive that x evolves with time according to the following equation:x = x(0) × e – (k1 + k2 )t ,where x(0) is the deviation from equilibrium at t = 0 h.
2.7

Let us boil 1.00 mol dm−3 L-isoleucine solution for 1 943 hours at 374 K. The rate constant for
the forward reaction is k1(374 K) = 9.02 × 10−5 h−1, Kep for L-isoleucine epimerization has the

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value of 1.38 (at 374 K). In the following calculation, shorten the concentration of L-isoleucine
to [L] and of D-allo-isoleucine to [D]. Evaluate (with three significant figures) a) [L]eq,b) diastereomeric excess (de) after boiling.
a) Calculation:
From the definition and mass balance we know thatx = [L] – [L]eq = [D]eq – [D]We can derive the formulae for [L]eq (and also for [D]eq) in terms of [L]0
[L]0 = [L]eq + [D]eq[D]eq = [L]0 – [L]eq
We also know thatKep =k1 [D]eq=k2 [L]eq

from which[D]eq =

k1
[L]eq
k2

The substitution for [D]eq in [D]eq = [L]0 – [L]eq yields[L]eq =k2
[L]0
k1 + k2

The rate constant k2 is calculated from the epimerization constant Kep as
k1
9.02 × 10–5 h–1
k2 === 6.54 × 10–5 h–1
Kep
1.38
After the substitution of the numerical values, we get (with three significant digits)[L]eq =

6.54 × 10–5 h–1
9.02 × 10–5 h–1 + 6.54 × 10–5 h–1

× 1 mol dm–3 = 0.420 mol dm–3

mol dm−3 (with three significant figures)[L]eq =b) Calculation:

de is defined asde =[L] – [D]× 100(%)[L] + [D]We can express de in terms of the equilibrium values asde =[L]eq – [D]eq + 2x
[L]eq + x – [D]eq + x
× 100(%) =× 100(%)[L]eq + x + [D]eq – x[L]0

We already have the formula for [L]eq from task a). [D]eq in terms of [L]0 is then[D]eq =k1
k1
[L]0
[L]eq =
k2
k1 + k2

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Finally, we need to evaluate x in terms of [L]0. We know that x(0) = [D]eq, so
x = [D]eq × e – t(k1 + k2 ) =k1[L]0 × e – t(k1 + k2 )k1 + k2

We can now express de as
de =
de =

–t(k1 + k2)
k2
k1
k1[L]0 –[L]0 + 2[L]0 ek1 + k2
k1 + k2
k1 + k2[L]0

× 100(%) =–t(k + k
k2 – k1 + 2k1 e 1 2)
× 100(%)
k1 + k2–5 + 6.54 × 10–5 )6.54 × 10–5 – 9.02 × 10–5 + 2 × 9.02 × 10–5 e–1 943 (9.02 × 10
9.02 × 10–5 + 6.54 × 10–5

× 100(%) = 69.8%(with three significant digits)de =% (with three significant figures)a)
1 point for the mass balance equation
2 points for the equation for the equilibrium constant in terms of the rate constants and concentrations
2 points for the derivation of the formula for [L]eq in terms of [L]0
1 point for the numerical value of k2
1 point for the numerical value of [L]eq
No points are scored if the calculations are not provided.
b)
2 points for the correct equation for de in terms of the equilibrium values
2 points for the derivation of the formulae for [D]eq in terms of [L]0
2 points for the derivation of the formula for x in terms of [L]0
2 points for the correct equation for de in terms of [L]0
2 points for the correct numerical value of de based on correct formula
Some steps in the derivation may be taken implicitly or simultaneously. As long as they are correct,
full marks will be scored for each of the partial steps. If the numerical calculation of k 2 in point a) is
incorrect but k2 is correctly used in task b), full marks are scored in task b).
No points are scored if the calculations are not provided.

Amino acids with a single chiral centre undergo racemization, e.g. L-arginine racemizes:

The time evolution of concentrations is governed by
[D]
[L]
ln
= 2k1 t + C
[D]
1–
[L]
1+

Here [D] and [L] are concentrations of D- and L-arginine at time t, k1 is the rate constant, and the term
C is set according to the initial concentrations.
Holy Roman Emperor Lothar III passed away during his journey to Sicily in 1137. To facilitate therepatriation of the remains, his body was, immediately after his death, boiled in water (373 K) fora certain time. Let us try to estimate the boiling time with the help of chemical kinetics. We know that

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the rate constant k1 of arginine racemization within the protein at 373 K and pH = 7 has the value of
5.10 × 10−3 h−1.
In order to analyse the isomeric composition of arginine in Lothar’s bones, we need to start withtransferring arginine into solution. Lothar’s bones were hydrolyzed in a highly acidic environment for[D]

4 hours at 383 K. The ratio of the optical isomers was [L] = 0.090. Lothar’s wife Richenza was notboiled after her death. Her bones were hydrolyzed using the same procedure and in this case the[D]ratio was [L] = 0.059. (Note that the racemization also takes place during the hydrolysis, with the rate
constant k'1 , different from k1).
2.8

How long was the Holy Roman Emperor Lothar III boiled in water in 1137?
Note: The racemization of arginine is an extremely slow process at temperatures typicallyencountered in graves. As both bodies are only some 880 years old, we can neglect the natural
racemization during this time.
Calculation:
Let us assume the following schemes for Lothar and Richenza
Lothar:'k1

k1

0 → 1 → 2'

Richenza:

k1

0 → 3
[D]

where 0 is the state before boiling with [L] ratio X0, 1 corresponds to the Lothar’s state after[D]boiling in water in 1137 with the [L] ratio X1, 2 is the Lothar’s state after the acidic hydrolysis[D][D]ratio X2, and 3 is the Richenza’s state after the hydrolysis with theratio X3. k1[L][L]is the rate constant for racemization at 373 K, k'1 is the rate constant for acid-catalyzed

with theracemization at 383 K.
We can write the following equations:ln

1 + X3= 2 k'1 thydrolysis (eq. 1)
1 – X3ln
ln

1 + X1= 2 k1 tboiling (eq. 2)
1 – X1

1 + X2
1 + X1= 2 k'1 thydrolysis + ln
(eq. 3)
1 – X2
1 – X1

Combining the equations leads toln

1 + X2
1 + X3= ln+ 2 k1tboiling
1 – X2
1 – X3

From which the time of boiling tboiling isln(1 + X2 ) (1 – X3 )= 2 k1 tboiling(1 – X2 ) (1 + X3 )Using the values X2 = 0.090 and X3 = 0.059 we get

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tboiling =1–3

2 × 5.10 × 10

ln

(1 + 0.090) (1 – 0.059)= 6.11 h(1 – 0.090) (1 + 0.059)

tboiling =h8 points for rate equations describing the time evolution of the ratio between the optical isomers for
Lothar and Richenza (2 points for each eq. 1 and 2; 4 points for eq. 3; 1 point for eq. 3 with an incorrect
evaluation of C)
4 points for the combination of the equations to get the time of boiling
2 points for the evaluation of the boiling time
Some steps may be taken implicitly. If the correct boiling time is provided with correct assumptions in
calculations, full marks will be scored.
Example of a partially correct solution:
1 point if the boiling time is derived correctly but the result is incorrect
No points are scored if the calculations are not provided.

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Question

3.1

3.2

3.3

3.4

3.5

3.6

Points

2

6

7

3

7

8

Question

3.7

3.8

3.9

3.10

3.11

Total

Points

6

10

5

2

6

62

Theoretical
Problem 3

Score

8% of the total

Score

Problem 3. Emerging electro-mobility
Contemporary means of transportation rely on burning fossil fuels, although the efficiency of real
combustion engines is inherently limited and typically ranges between 20 and 40%.
3.1

Mark the factors that can make the efficiency of a heat engine higher:☐ Increasing the friction in the mechanical parts of the engine☒ Increasing the burning temperature of the fuel in the engine☐ Narrowing the working temperature interval of the engine☒ Increasing the working pressure of the gas

2 points in total; 1 points for each correct answer (the second and fourth options in the list), −1 points for eachincorrect answer, total points scored in this question may not be negative

Fuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency
can be improved by using hydrogen-based fuel cells.
3.2

The standard enthalpy of formation of liquid water is ΔfH°(H2O,l) = −285.84 kJ mol−1, and thestandard combustion enthalpy of isooctane is ΔcH°(C8H18,l) = −5 065.08 kJ mol−1 (both at
323.15 K). Calculate the values of specific (per unit of mass) combustion enthalpy at 232.15 Kof pure liquid isooctane and pure gaseous hydrogen.
Δc Hs °(C8 H18 ) =Δc Hs °(C8 H18 ) =Δc H°(C8 H18 ) – 5 065.08 × 103 J mol–1== – 44.34 MJ kg–1
–3
–1
M(C8 H18 )114.23 × 10 kg mol

Δc Hs °(H2 ) =Δc Hs °(H2 ) =Δc H°(H2 ) – 285.84 × 103 J mol–1== – 141.50 MJ kg–1
–3
–1
M(H2 )
2.02 × 10 kg mol

6 points in total. Some of the steps may be condensed or performed implicitly. As long as this isperformed correctly, full marks are scored for all of the elementary steps. Zero points will be scoredfor any numerical (sub)result given with wrong or missing units.
1 point for each numerically correct molar mass (including unit)
2 points for the correct way of calculation of the specific combustion enthalpy
1 point for each numerically correct specific combustion enthalpy (including unit)−1 point for the wrong sign(s)THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

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3.3

Calculate the standard electromotive force (EMF) of a fuel cell using gaseous oxygen andhydrogen, both ideal gases at 100 kPa and 323.15 K, to produce liquid water. Use thefollowing entropy data for 323.15 K: S°(H2O,l) = 70 J K−1 mol−1, S°(H2,g) = 131 J K−1 mol−1,
S°(O2,g) = 205 J K−1 mol−1.
Calculations:
The oxidation of molecular hydrogen requires the transfer of z = 2 electrons according to theequation: H2 + ½ O2 → H2O
Δr S° = S°(H2 O,l) – (S°(H2 ,g) +1
S°(O2 ,g)) = – 164 J K–1 mol–1
2

Δr G° = Δr H° – TΔr S° == –285.84 × 103 J mol–1 + 323.15 K × 164 J K–1 mol–1 = – 233 kJ mol–1
EMF = –
EMF =

Δr G°
– 233 × 103 J mol–1=–= 1.21 V|z|F
2 × 96 485 C mol–1

V

7 points in total for the correct result calculated in any correct way. Some of the steps may becondensed or performed implicitly. As long as this is performed correctly, full marks are scored for allof the elementary steps. Zero points will be scored for any numerical (sub)result given with wrong
or missing units.
1 point for the correct value of z resulting from the correctly written chemical equation
1 point for the correct formula for the reaction entropy according to the written chemical equation
1 point for the numerically correct value of the reaction entropy
1 point for the correct formula for the reaction Gibbs energy
1 point for the numerically correct value of the reaction Gibbs energy
1 point for the correct formula for the EMF
1 point for the numerically correct value of the EMF−1 point for incorrect sign

3.4

Determine the ideal thermodynamic efficiency (η) of a fuel cell producing liquid water at353.15 K. At this temperature, the enthalpy of formation of water is ΔfH°(H2O,l) = −281.64 kJ
mol−1 and the corresponding reaction Gibbs energy change is ΔrG° = −225.85 kJ mol−1.
Δr G° – 225.85 kJ mol–1
η==× 100% = 80.19%
Δf H° – 281.64 kJ mol–1
η=%

3 points in total
2 points for the correct way of calculation
1 point for the numerically correct result

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3.5

A polymer membrane electrolyzer facility operates at the voltage of 2.00 V and is powered by
a 10.0 MW wind turbine plant which was running at full power from 10 pm to 6 am. The
electrolysis yielded 1 090 kg of pure hydrogen. Calculate the electrolysis yield defined as themass of produced hydrogen divided by its theoretical produced mass.
Calculations:Electric energy produced:
W = P × τ = 10 × 106 W × 8 × 3 600 s = 2.88 × 1011 J
is sufficient to transfer the charge:
W
2.88 × 1011 J
Q=== 1.44 × 1011 C
U
2V
between the electrodes. Water formation is a two-electron process. Thus, the theoreticalmass of hydrogen produced is:
QM
1.44 × 1011 C
mtheory ==× 2.02 × 10–3 kg mol–1 = 1 507 kg–1|z| F
2 × 96 485 C mol
The efficiency of the electrolysis can be evaluated as the ratio:ηelectrolysis =ηelectrolysis =

mreal
1 090
× 100% =× 100% = 72.3%.
mtheory
1 507%

7 points in total for the numerically correct final result calculated in any correct way. Some of the steps
may be condensed or performed implicitly. As long as this is performed correctly, full marks will bescored for all elementary steps. If only the final equation for mtheory is derived correctly, but its
numerical value is wrong, 4 points will be scored for mtheory. If a wrong numerical subresult is used insubsequent steps which are performed correctly but yield a wrong numerical value, 4 points will bescored for mtheory. 0 points will be scored for any numerical (sub)result given with wrong or missing
units.
1 point for the correct formula for W
1 point for the numerically correct value of W
1 point for the correct way of calculation of Q
1 point for the numerically correct value of Q
1 point for any correct way of calculation of the theoretical amount of hydrogen1 point for the numerically correct theoretical amount of hydrogen1 point for the numerically correct electrolysis efficiency

3.6

Calculate the mass of hydrogen required to drive the distance between Prague and Bratislava(330 km) at the average speed of 100 km h−1 with a car fitted with a 310 kW electric engine
running on average at a 15% rate of its maximum power. Assume that the efficiency of thehydrogen cell producing electrical energy is 75%, the efficiency of the electric engine is 95%,
and the Gibbs energy change for combustion of hydrogen fuel is ΔrG = −226 kJ mol−1.

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Calcualtions:
Driving time:τ=s
330 km== 11 880 s
v 100 km h–1 / 3 600 s

determines the ideal amount of energy required to cover the given distance:
Wideal = P × f × τ = 310 × 103 W × 0.15 × 11 880 s = 5.52 × 108 J.
Assuming the overall non-unity efficiency, the real energy required is:
Wreal =

Wideal5.52 × 108 J== 7.75 × 108 J.
ηelectrolysis × ηengine
0.75 × 0.95

This can be combined with the standard Gibbs energy of liquid water formation to yield theamount of hydrogen:Wreal
7.75 × 108 Jm=
×M =× 2.02 × 10–3 kg mol–1 = 6.93 kg.
5–1|Δr G°|
2.26 × 10 J molm=kg

8 points in total for the numerically correct final result calculated in any correct way. Some of the steps
may be condensed or performed implicitly. As long as this is performed correctly, full marks will bescored for all of the elementary steps. If only the final relation for m is derived correctly, but its
numerical value is wrong, 6 points will be scored in total . If a wrong numerical subresult is used insubsequent steps which are performed correctly but yield a wrong numerical value, 6 points will bescored for mtheory. 0 points will be scored for any numerical (sub)result given with wrong or missing
units.
1 point for the correct formula for 𝜏
1 point for the correct value of 𝜏1 point for the correct formula for W ideal
1 point for the numerically correct value of W ideal
1 point for the correct formula for W real
1 point for the numerically correct value of W real
1 point for the correct formula for m
1 point for the numerically correct value of m

The low efficiency of hydrogen production and the safety issues connected with its storage impede
spreading the hydrogen-based transportation technology. Hydrazine (N2H4) fuel cells might bea suitable alternative.
The following standard reduction potentials for aqueous hydrazine systems are available:

N2(g) + 5 H+(aq) + 4 e− → N2H5+(aq)N2H5+(aq) + 3 H+(aq) + 2 e− → 2 NH4+(aq)N2(g) + 4 H2O(l) + 4 e− → N2H4(aq) + 4 OH− (aq)
N2H4(aq) + 2 H2O(l) + 2 e− → 2 NH3(aq) + 2 OH− (aq)
2 H2O(l) + 2 e− → H2(g) + 2 OH− (aq)THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

E° = −0.23 V
E° = +1.28 V
E° = −1.16 V
E° = +0.10 V
E° = −0.83 V.
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3.7

Fill in the following Latimer diagrams with the forms of hydrazine and ammonia prevailing at thegiven conditions and write the redox potential value for each arrow representing theelectrochemical half-reaction. Record all the necessary calculations.
a)Acidic environment (pH = 0)N2b)Basic environment (pH = 14)
N2

Acidic environment:

N2−0.23 V

NH4+

N2H5+
+1.28 V
+0.27 V

Basic environment:

N2−1.16 V

N2H4

+0.10 V

NH3

−0.74 V

Calculations:
Redox potentials for N2 to NH4+ and N2 to NH3, respectively, are calculated as follows:
4 E°(N2 /N2 H+5 ) + 2 E°(N2 H+5 /NH+4 ) 4 × (– 0.23) + 2 × 1.28+
E°(N2 /NH4 ) ==V = 0.27 V
6E°(N2 /NH3 ) =6

4 E°(N2 /N2 H4 ) + 2 E°(N2 H4 /NH3 ) 4 × (– 1.16) + 2 × 0.10=
V = – 0.74 V
66

6 points in total
1 point for each diagram completely correctly filled with chemical species, no partial points given
if any mistake occurs
1 point for the numerically correct value of each of the standard redox potentials E°(N2 /NH+4 ) andE°(N2 /NH3 ), no partial points given if any mistake occurs
1 point for each diagram correctly filled with the remaining standard redox potentials, no partial
points given if any mistake occurs

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Due to the toxicity, odour and its environmental impact, it is extremely unfavourable to produce
ammonia in fuel cells.
3.8

Write down the net reaction for the decomposition of hydrazine under basic conditions to(i) ammonia and nitrogen and (ii) nitrogen and hydrogen and calculate the corresponding
equilibrium constants at T = 298.15 K.
Equations for hydrazine decomposition:
N2H4(aq) → N2(g) + 2 H2(g)3 N2H4(aq) → 4 NH3(aq) + N2(g)Calculations:|z|FE°)
RT
Hydrazine decomposition to NH3 and N2 in a basic environment:
Δr G° = – RT ln K = – |z|FE° → K = exp (E° = E°(N2 H4 ⁄NH3 ) + (– E°(N2 ⁄N2 H4 )) = (0.10 + 1.16) V = 1.26 V|z|FE°
4 × 96 485 C mol–1 × 1.26 V
K = exp () = 1.6 × 1085
) = exp (–1–1
RT
8.314 J K mol × 298.15 K
K=
Hydrazine decomposition to H2 and N2 in a basic environment:
E° = E°(H2 O⁄H2 ) + (– E°(N2 ⁄N2 H4 )) = (– 0.83 + 1.16) V = 0.33 V
K = exp (|z|FE°
4 × 96 485 C mol–1 × 0.33 V) = 2.1 × 1022
) = exp (–1–1
RT
8.314 J K mol × 298.15 K

K=
10 points in total
2 points for each correctly balanced reaction (4 points in subtotal)
4 points for the correct way of calculation of the equilibrium constant (2 points will be scored if thecorrect formula for the related standard reaction Gibbs energy is given)1 point for each numerically correct value of equilibrium constant (2 points in subtotal), no points will
be scored if a wrong value of z is used in the calculation

Rechargeable lithium-based batteries are an alternative to fuel cells. Lithium-ion batteriescommonly use graphite for one of the electrodes, in which lithium clusters intercalate in betwe enthe graphite sheets. The other electrode is made of lithium cobalt oxide, which can reversibly
absorb lithium ions moving from one electrode to the other during the charge and discharge
processes. The half-reactions relevant for the system can be formally written as:

3.9(C)n + Li+ + e– → Li(C)nE° = – 3.05 V,

CoO2 + Li+ + e– → LiCoO2

E° = +0.19 V.

Using the formalism given above, write down the overall chemical reaction occurring in thebattery during the discharge process. Give the oxidation states of the cobalt atom.

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Because E°lower > E°upper, the upper reaction occurs spontaneously in the opposite direction.
Therefore, the discharge of the battery occurs when lithium leaves the graphite structure
and its ions intercalate in the cobalt oxide:
Li(C)n + CoIV O2 → LiCoIII O2 + (C)n5 points in total
1 point for each oxidation state of cobalt atom
3 points for the overall reaction
No points will be scored for the reaction written in the opposite direction.

3.10 Tick the boxes to get the correct statements which are valid for the discharge of the lithium-based
battery described in 3.9:

Li(C)n electrode is cathode✓ anodebecause lithium ions are reduced here.
✓ because lithium atoms are oxidized here.

LiCoO2 electrode is✓ cathodeanode✓ because cobalt ions are reduced here.
because cobalt ions are oxidized here.

2 points in total
1 point for each correct pair of boxes marked

3.11 Assume that a C6 unit, a CoO2 unit and Li atom form the active battery mass required to transfer
one electron between the electrodes. Using the corresponding standard EMF, calculate thetheoretical specific reversible charge capacity (in mAh g−1) and the energy density (in kWhkg−1) of such a model lithium ion battery related to the whole active battery mass.
Calculations:
The transfer of 1 mol of electrons requires at least the molar mass of the active ingredients
Mtotal = 169.94 g mol−1, meaning that the specific charge capacity is:F
96 485 C mol–1
cq,s === 567.76 C g–1 ≈ 567.76 A s g–1 ≈ 157.71 mAh g–1
M
169.94 g mol–1
Charge capacity (cq,s ) =mAh g−1

Calculations:
Assuming the standard EMF (voltage) of the battery is:
U = E°(Lix1 CoO2 /Lix2 CoO2 ) – E°(Li+ /Li0 ) = 0.19 + 3.05 V = 3.24 V,
the energy density of the battery can be calculated as:ρel = U × cq,s = 3.24 V × 567.76 C g–1 = 1 839.6 W s g–1 ≈ 0.511 kWh kg–1
Energy density (ρel )=THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

kWh kg−128

XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

6 points in total for the numerically correct final result. Some of the steps may be condensed or
performed implicitly. As long as this is performed correctly, full marks will be scored for all elementary
steps.
1 point will be scored if only the correct numerical value for the Gibbs energy associated with theelectron transfer is given. Zero points will be scored for any numerical (sub)result given with wrong or
missing units.
1 point for the numerically correct value of the active molar mass
1 point for the correct formula for the specific charge capacity
1 point for the numerically correct value of the specific charge capacity
1 point for the numerically correct value of EMF
1 point for the correct formula for the energy density
1 point for the numerically correct value of the energy density

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Theoretical
Problem 4

Question

4.1

4.2

4.3

4.4

4.5

4.6

4.7

4.8

Total

Points

2

5

1

2

7

2

3

2

24

6% of the total

Score

Problem 4. Column chromatography of radioactive copper
64

Cu for positron emission tomography is prepared by the bombardment of a zinc target with
deuterium nuclei (further referred to as the activated target).
4.1

Write down the balanced equation for the 64Zn nucleus bombardment with deuterium nuclei,
giving 64Cu. Specify the corresponding atomic and mass numbers of all species. Disregard thecharges.
64
1
Zn
…
30

1
+…

2
1

H

1
→ …

64

Cu

29

1

2
+ …

1

p

2 points (1 point if any mass number is missing or the reaction is written with a wrong
stoichiometry; 0 points if both or other mistakes are made)The activated target is dissolved in concentrated hydrochloric acid (HCl (aq)) to give a mixture
containing Cu2+ and Zn2+ ions and their respective chlorido complexes.

4.2

Calculate the mole fraction of negatively charged copper species with respect to the amountof copper prepared by zinc target activation. Assume [Cl−] = 4 mol dm−3. For the overall
complexation constants, β, see Table 1.
Before you start the calculation, write down the charges in the upper right boxes:

Cu

2+[CuCl]+[CuCl2]0[CuCl3]−[CuCl4]2−

1 point for correct charges for all species; 0 points for 1 or more incorrect charge(s)Table 1. Overall complexation constants β of Cu species (charges were omitted in the formula).
βi =[CuCli ][Cu] × [Cl]ii in [CuCli]

βi

1

2

3

4

2.36

1.49

0.690

0.055

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Calculation:
The mole fraction is the sum of the distribution coefficients of [CuCl3]− and [CuCl4]2−:([[CuCl3]−] + [[CuCl4]2−]) / c(Cu2+) = (3 [Cl−]3 + 4 [Cl−]4) / (1 + 1 [Cl−] + 2 [Cl−]2+ 3 [Cl−]3 +4 [Cl−]4) = (0.69 × 43 + 0.055 × 44) / (1 + 2.36 × 4 + 1.49 × 42 + 0.69 × 43 + 0.055 × 44) == 58.24 / 92.52 = 0.63

Mole fraction: 0.63
(answer with 2 digits after decimal point)
4 points for completely correct answer; −1 point for the wrong mole fraction obtained by the correct
procedure

The mixture containing Cu2+ and Zn2+ ions and their respective chlorido complexes was separated
with an anion exchange resin. Dry resin in OH− form was dispersed in water and the suspension wastransferred into a column. To occupy all sites with Cl− ions (i.e. to obtain resin in a Cl− form), the resin
was washed with hydrochloric acid and then with deionized water to wash out all the unbound Cl−
ions.
4.3

Everything was initially at laboratory temperature before washing with hydrochloric acid. Does
the column temperature change during the washing with hydrochloric acid?

☐ No.
☐ Yes, the temperature decreases.
☒ Yes, the temperature increases.
1 point for the correct answer

The mixture containing Cu2+ and Zn2+ ions and their respective chlorido complexes was transferred
onto the resin-filled column. Hydrochloric acid solution was used as an eluent.
Using the simple experimental formula below, you can calculate quantities that determine average
elution properties of both copper species and zinc species on the column.
The retention volume VR (the mobile phase volume at which 50% of the compound has been eluted
from the column) can be calculated as follows:
VR = Dg × mresin,dry,OH form + V0
4.4

Using the average mass distribution coefficients Dg (Dg(Cu species) = 17.4 cm3 g−1,
Dg(Zn species) = 78.5 cm3 g−1), calculate the retention volumes VR in cm3 of both copper
species and zinc species. The mass of dry resin in OH− form mresin,dry,OH form = 3.72 g and thevoid volume of a column V0 = 4.93 cm3.

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Calculation:

VR(Cu species) = 69.7 cm3 (answer with 1 digit after the decimal)
VR(Zn species) = 297 cm3 (answer with 0 digit after the decimal)
2 points in total (1 point for each VR)

If you cannot find the answer, use VR(Cu species) = 49.9 cm3 and VR(Zn species) = 324 cm3 forfurther calculations.
Using the simple experimental formula, separation of two sets of species, A and B, can beconsidered complete if
V0.001(A) – V0.999(B) > 10Vc
where V0.001 is the mobile phase volume at which 0.1% of A has been eluted from the column, andV0.999 is the mobile phase volume at which 99.9% of B has been eluted from the column.
V0.001 (A) = VR (A) × (1 – 6.91√dp /Lc )V0.001 (B) = VR (B) × (1 – 6.91√dp /Lc)
V0.999 (B) = 2VR (B) – V0.001 (B)
4.5

Based on a calculation, decide whether copper species were separated completely from zincspecies. The volume of the column filled with the swollen resin Vc = 10.21 cm3, the resin particle
diameter dp = 0.125 mm, and the height of the wet resin in a swollen state in the column
Lc = 13.0 cm.
According to the retention volumes (VR), V0.001(A) corresponds to V0.001(Zn species) andV0.999(B) corresponds to V0.999(Cu species)
V0.001(A) = 297 cm3 × (1 – 6.91 × √0.125 mm/130 mm) = 233 cm32 points (1 point if Lc and dp are used with different units;)

V0.999(B) = 2 × 69.7 cm3 − 54.8 cm3 = 84.6 cm3
1 point (even with the wrong VR and V0.001 for Cu species)

where V0.001(Cu species) = 69.7 cm3 × (1 – 6.91 × √0.125 mm/130 mm) = 54.8 cm3
2 points (1 point if Lc and dp are used with different units)

It is possible to separate copper species from zinc species.

☒ True☐ False2 points for the correct decisions based on V0.001 and V0.999 calculations

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4.6

Calculate the theoretical value of the total ion exchange mass capacity of the dry resin used inthis problem, Qm,theor, in mmol g−1. Consider tetralkylammonium groups were the only ones
responsible for ion exchange of the resin. No other nitrogen containing groups were present.
The mass fraction of nitrogen in the dry resin was 4.83%.
Qm,theor = w(N) / M(N) = 0.0483 / (14.01 g mol−1) = 3.45 mmol g−1
2 points (1 point if the value is in the wrong order of magnitude)(answer with 2 digits after decimal point)
If you cannot find the answer, use Qm,theor = 4.83 mmol g−1 for further calculations.
In reality, not all tetraalkylammonium groups are involved in the ion exchange. To determine the total
ion exchange volume capacity, Qv, the column filled with 3.72 g dry resin converted to the Cl− form
was washed with the excess of sodium sulfate solution. The effluent was collected in a 500 cm3
volumetric flask, which was then filled with water to the mark. An aliquot of 100 cm3 waspotentiometrically titrated with 0.1027 mol dm−3 silver nitrate. The silver nitrate solution volume at
the equivalence point was 22.20 cm3. The volume of the column filled with the swollen resin, Vc, was10.21 cm3.
4.7

Calculate the Qv of the swollen resin in mmol of active tetraalkylammonium groups per cm3 ofthe swollen resin.
Qv = V(AgNO3) × c(AgNO3) × Vflask / (Valiquot × Vc) == 0.0222 dm3 × 0.1027 mol dm−3 × 0.500 dm3 / (0.100 dm3 × 0.01021 dm3) = 1.12 mmol cm−3
3 points−1 point if dilution is forgotten−1 point for the wrong order of magnitude(1 point for titration calculation)(answer with 2 digits after decimal point)
If you cannot find the answer, use Qv = 1.00 mmol cm−3 for further calculations.
4.8

Calculate the mole fraction (x) of the tetraalkylammonium groups actively involved in the ion
exchange.
x = Qv × Vc / (Qm,teor × mresin) = 1.12 mmol cm−3 × 10.21 cm3 / (3.45 mmol g−1 × 3.72 g) =0.891
2 points(answer with 3 digits after decimal point)

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Question

5.1

5.2

5.3

5.4

5.5

5.6

5.7

5.8

5.9

Theoretical
Problem 5

Points

3

3

1

5

3

2

4

1

2

8% of thetotal

Question

5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17

Total

Points

5

50

Score

7

3

2

6

1

1

1

Score

Problem 5. Bohemian garnet
Bohemian garnet (pyrope) is a famous Czech blood coloured semi-precious stone. The chemicalcomposition of natural garnets is expressed by the general stoichiometric formula of A3B2(SiO4)3,
where AII is a divalent cation and BIII is a trivalent cation. Garnets have a cubic unit cell that contains
8 formula units. The structure comprises 3 types of polyhedra: the AII cation occupies a dodecahedral
position (it is surrounded with eight O atoms), the BIII cation occupies an octahedral position (it issurrounded with six O atoms) and SiIV is surrounded with four O atoms arranged into a tetrahedron.
The most common garnet mineral is almandine with the formula of Fe3Al2(SiO4)3. Its unit cellparameter is a = 11.50 Å.
5.1

Calculate the theoretical density of almandine.
M(almandine) = 497.75 g mol−1

 = m / V = (8 × M) / (NA × a3) = (8 × 497.75) / [6.022 × 1023 × (11.50 × 10−8)3] = 4.35 g cm−3
3 points in total
correct molar mass 1 point, correct equation 1 point, correct numerical calculation starting fromcorrect equation 1 point

The Bohemian garnet has the composition of Mg3Al2(SiO4)3. Pure compound is colourless and thecolour of natural garnets comes from chromophores – transition metal cations that substitute thehost material cations. The red colour of the Bohemian garnet comes from trace amounts of CrIII ionsin the octahedral sites and FeII ions in the dodecahedral sites.
5.2

Draw the splitting diagram for the [CrIIIO6]oct d-orbitals and fill it with electrons.
Splitting diagram – configuration d3:

3 points in total
correct orbital splitting 1 point, correct number of electrons 1 point, correct configuration according

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to Hund’s rule 1 point, −1 point if only a line not indicating the electron spin is used (deducted only
once per Problem); orbital marking is not evaluated

5.3

Identify the 1st transition row element(s) whose trivalent cation(s) MIII placed in an octahedral
position is/are diamagnetic in the low-spin arrangement and paramagnetic in the high-spinarrangement.
Co
1 point
correct answer 1 point, incorrect answer −1 point, minimum 0 points (the score may not be
negative)

5.4

The figure below shows d-orbitals splitting in the dodecahedral crystal field. Fill in the electrons
for the [FeIIO8]dod chromophore for both existing arrangements.
a) high-spin arrangementb) low-spin arrangement

5 points in total
correct number of electrons 1 point, correct high-spin configuration 2 points, correct low-spin
configuration 2 points (any other configuration 0 points: high spin and low spin configurations differ
in total spin. Any other configurations do not fulfil this condition.)
If incorrect number of electrons is considered, then the correctness of such a configuration is taken
into account.

5.5

Derive the inequalities (e.g. P < E1 + E2 + E3) for the pairing energy (P) magnitude in relation
to energies E1, E2 and E3 for both arrangements.
a) high-spin arrangement:

P > E1 − E3b) low-spin arrangement:

P < E1 − E3

3 points in total
correct BOTH inequality signs 1 point (half-points are not allowed), correct right-side terms(according to the answer in 5.4) 1 + 1 points

5.6

Assuming that P > E3, identify the 1st transition row element(s) whose divalent cation MII placedin dodecahedral position is diamagnetic in the low-spin arrangement and paramagnetic in thehigh-spin arrangement.

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Cr, Ni2 points
each correct metal 1 point, each incorrect metal −1 point, minimum 0 points (the score may not be
negative)

A

400

450

500

550

600

650

700

λ (nm)

C

300

350

400

450

500

550

600

650

700

B

300

350

400

500

450

600

550

650

Absorbance

350

Absorbance

300

5.7

AbsorbanceAbsorbance

The figures below show simplified absorption spectra of four coloured minerals – blood-coloured
Bohemian garnet, green uvarovite, blue sapphire and yellow-orange citrine.

λ (nm)300

700

λ (nm)

D

350

400

450

500

550

600

650

700

Match the spectra with the minerals.
Bohemian garnet:

B

Sapphire:D

Uvarovite:C

Citrine:A

4 points in total, each correct assignment 1 point

5.8

If illuminated with monochromatic blue-green light, how will the Bohemian garnet look?
 Red

 Blue Yellow-orange

 Black Yellow Blue-green

 Violet White1 point

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Andradite is another garnet mineral; its chemical composition is Ca3Fe2(SiO4)3. A double cation
substitution – TiIV for FeIII in the octahedral position and FeIII for SiIV in the tetrahedral position – gives
rise to blackschorlomite. Its chemical composition can be expressed as
tet
Ca3[Fe,Ti]oct
2 ([Si,Fe] O4)3.
5.9

Calculate the percentage of SiIV ions in a sample of schorlomite that must be substituted withFeIII, if we know that 5% of FeIII ions in octahedral position are substituted with TiIV.
Both substitutions change the charge by 1 unit per ion. Taking total number of ions in formula
into consideration we get:5%  2 = p  3
p = 3.33 %2 points
correct starting consideration in the form of an equation 1 point, correct numerical calculation 1
point

The colour of the mineral is caused by two chromophores: [FeIIIO6]oct and [FeIIIO4]tet. The central ionsof both chromophores have equal number of unpaired electrons.
5.10 Draw the d-orbitals splitting diagrams for both chromophores and fill in the electrons.
[FeIIIO6]oct:[FeIIIO4]tet:

5 points in total
correct octahedral splitting 1 point, correct tetrahedral splitting 1 point, correct number of electrons
1 point, correct high-spin configurations 1 + 1 points; orbital marking is not evaluated.

4

Tetrahedral field causes a smaller splitting than the octahedral field (∆tet = 9 ∆oct). Surprisingly for
the FeIII ion, the energy of the first d–d transition (although very weak) for the octahedralchromophore is smaller (11 000 cm−1) than for the tetrahedral one (22 000 cm−1).
5.11 Calculate the size of pairing energy (P) and the sizes of Δoct and Δtet splitting. Assume that thepairing energy is equal in both chromophores.
Because Δoct < P (high-spin configuration in tetrahedral as well as in octahedral field), thewavenumber corresponds to electron transition from the upper to the lower level in bothcases and thus:
11 000 cm−1 = P − Δoct

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22 000 cm−1 = P − Δtet = P − 9 Δoct
By solving the system of these equations we get:P = 30 800 cm−1
Δoct = 19 800 cm−1
Δtet = 8 800 cm−1

Alternative solution comes from the assumption (independent of the result of 5.10) of the
low-spin arrangement in octahedral field and high-spin arrangement in tetrahedral field:
11 000 cm –1 = Δoct – P
4

22 000 cm−1 = P − Δtet = P − 9 Δoct
By solving the system of these equations we get:P = 48 400 cm−1
Δoct = 59 400 cm−1
Δtet = 26 400 cm−1

7 points in total
formulation of equations 2 + 2 points, correct energy values (i.e. correct numerical calculation
starting from correct assumption) 1 + 1 + 1 points

Synthetic garnet YAG (Yttrium Aluminium Garnet), used in optoelectronics, has the compositionof Y3Al5O12. Its structure is derived from the general garnet structure A3B2(SiO4)3 by placing the ions
YIII and AlIII to the A, B and Si positions.
5.12 Based on your knowledge of the relative ion radii, determine which cation occupies which
position.
A: YIII

B: AllII

Si: AlIII

3 points in total, each correct assignment 1 point

5.13 For the use in LED technology, YAG is doped with CeIII. Determine the values of x and y in theformula of YAG in which 5% of yttrium atoms are substituted with cerium.
YxCeyAl5O12

x = 2.85

y = 0.15

2 points in total, each value 1 point

If you don’t get result, use x = 2.25 and y = 0.75.

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 20185.14 The CeIII-doped YAG is prepared by annealing the mixture of Y2O3, Al2O3 and CeO2 in H2
atmosphere. Use the formula from 5.13, write down a balanced equation for this reaction withthe smallest whole-number stoichiometric coefficients.
57 Y2O3 + 6 CeO2 + 100 Al2O3 + 3 H2  40 Y2.85Ce0.15Al5O12 + 3 H2O(or 2 Y57Ce3Al100O240 + 3 H2O)
for x = 2.25 and y = 0.75:
9 Y2O3 + 6 CeO2 + 20 Al2O3 + 3 H2  8 Y2.25Ce0.75Al5O12 + 3 H2O(or 2 Y9Ce3Al20O48 + 3 H2O)

complete set of correct coefficients 6 points−1 point if correct stoichiometric coefficients are only on one side of the equation−1 point if stoichiometric coefficients are correct and whole-number but not the smallest−2 points if stoichiometric coefficients are correct but not whole-numbers
if hydrogen+water are omitted from the equation or when redox change was incorrect, then penalty−3 point is applied
if incorrect formula in 5.13 is given, the answer is judged with respect to the incorrect formula

Doping the YAG structure with rare-earths ions enables the production of lasers with emission
wavelengths ranging from the UV to the mid-IR region. In the scheme below, simplified f–f energy
transitions of selected rare-earths ions are shown.

5.15 Which cation has a transition which corresponds to blue light emission.
 Er3+

 Sm3+

 Tm3+

 Yb3+

 Nd3+

 Tb3+

 Pr3+

1 point

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XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 20185.16 Calculate the emission wavelength of this light.
1

λ = 2 127 700 m = 4.70 × 10−7 m = 470 nm
1 point. The result of 5.16 is evaluated with respect to the answer in 5.15

5.17 According to a legend, Noah used a stick with a garnet stone for illumination during his voyage.
Assuming only the photoluminescence effect, determine the colour of the laser light emitted
from his stick if the stone were the blood-coloured Bohemian garnet.
 Red

 Blue Yellow-orange

 Black Yellow Blue-green

 Violet White1 point

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Theoretical
Problem 6

Question

6.1

6.2

6.3

6.4

6.5

6.6

6.7

6.8

Total

Points

18

4

8

3

4

12

16

3

68

7% of the total

Score

Problem 6. Let’s go mushrooming
Mushrooming belongs to Czech and Slovak traditional pastimes. While some of our mushroom
species are edible, some are inedible or even poisonous.
Inky cap (Coprinopsis atramentaria) is considered edible and delicious. It contains a naturalcompound called coprine (E), which can be easily synthesized from ethyl 3-chloropropanoate (1).

6.1

Draw the formulae of compounds A–E including stereochemistry when necessary. Hint: The firstreaction affording compound A proceeds via an organometallic compound which then cyclizes.
ABC

4 points

2 points for TMS deprotection

4 points for ethoxy group
substitution by amino groupDE

4 points for amide formation
2 points for the corresponding ester formation−2 points for wrong or undefined stereo-chemistry(2 rules may apply at once)

4 points for Boc and tBu deprotection (zwitterionicform is also accepted with no penalty)−2 points for wrong or undefined stereo-chemistry

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No points will be given for structure A if it is acyclic.
For wrong isomeric structures of compounds A–E, 0–2 points will be given depending on the rationality of theanswer.
No points will be given for structures A–C or E if they do not match the given molecular formulae, except forminor mistakes (see below).
−1 point for each atom with abnormal valence or for incorrect number of hydrogens on heteroatoms.

In the human body, coprine undergoes hydrolysis to L-glutamic acid (3) and compounds C and 4,
which are responsible for the coprine adverse side-effects. They inhibit the enzyme acetaldehyde
dehydrogenase, which is involved in the metabolism of alcohol. When the enzyme isinhibited, acetaldehyde formed by alcohol dehydrogenase accumulates in the body, causing strong
symptoms of hangover (so called antabuse effect). The active site of the enzyme contains a cysteine
SH group, which is blocked either by compound C or 4.

Enzyme = acetaldehyde dehydrogenase
6.2

Using the pictogram for acetaldehyde dehydrogenase above, draw the structure F of the
enzyme inhibited by compound 4.
F

4 points; 0 points for structures likeNo penalty if two molecules of the enzyme are blocked by one molecule of 4.

The antabuse effect got its name after antabuse (5), the most known drug used inalcohol-addiction treatment. This drug can be synthesized according to the following scheme.

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6.3

Draw the formulae of compounds G and H. Hint: Compound H contains five carbon atoms.
GH

4 points
No penalty for other meaningful reagents, such as
halodithioformic acid, its ester or salt.
0 points for xanthates.

6.4

4 points
3 points for the above-pictured salt if the cation iswrong or missing
2 points for the corresponding acidMark all possible reagents which could be used for I from the following list.
☒ m-chloroperbenzoic acid (mCPBA)☒ diluted H2O2

☐ Zn/CH3COOH☐ NaBH4☒ I2☐ hot concentrated H2SO4☐ K2CO3, H2O☐ AlCl3

1 point for each correct answer (3 points in total).
−1 point for each incorrect answer. The total score in task 6.4 may not be negative.

The way antabuse inhibits acetaldehyde dehydrogenase is similar to the effect of compounds
C and 4.

Enzyme = acetaldehyde dehydrogenase
6.5

Using the pictogram for acetaldehyde dehydrogenase above, draw the structure J of the
enzyme inhibited by antabuse (5). Hint: Three sulfur atoms should be in the structure.
J

4 points (the only acceptable answer)

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False morel (Gyromitra esculenta) is another interesting mushroom. Although it was considered
edible in the past (esculentus means edible in Latin), there is clear evidence that this mushroom ispoisonous due to the content of gyromitrin (M). This natural compound can be prepared fromN-methylhydrazine (6):6.6Draw the formulae of compounds K–M.
K

4 points
4 points for formic anhydride
2 points for formyl chlorideLM

4 points for the correct structure−1 point if acetylated−1 point for multiple acylation−2 points if acylated on theother nitrogen atom

4 points for the correct structure(cis or trans isomer)−1 point for aminal (exception
down below)−3 points if the reaction withacetaldehyde was carried out
on the amidic nitrogen atom

−1 point for each atom with abnormal valence or for each missing hydrogen atom.
If L is acylated with wrong regioselectivity, then in M aminals, enamines, and iminium salts are accepted withno penalty.

In human body, gyromitrin (M) hydrolyzes and provides N-methylhydrazine (6), which is strongly
hepatotoxic. Gyromitrin (M) hydrolysis occurs as soon as it enters the acidic environment in humanstomach where both its amide and imine groups are hydrolyzed.
Let us focus on the hydrolysis of the amide moiety within the gyromitrin molecule. The vibrationalwavenumber of the stretching mode of the relevant C−N bond amounts to 1 293.0 cm−1 and thepotential energy surface does not significantly alter its shape with isotope substitution effect.

6.7

Calculate the highest possible hypothetical kinetic isotope effect at the temperature of human
body, 37 °C, for the given hydrolysis reaction assuming that both relevant nitrogen and carbon
atoms were simultaneously substituted, 14N with the 15N isotope and 12C with the 13C isotope.
Consider that only the zero point vibrational energy affects the rate constants. Assume that themolar masses of all isotopes are integers. In all steps consider five significant digits.

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C−N bond reduced mass:
14.000 g mol–1 × 12.000 g mol–1N
μ12 == 1.0730 × 10–26 kg
–1
23
–1
C
(14.000 + 12.000) g mol × 6.0221 × 10 mol
14

14

N
C

Alternatively, (μ 12 )’ = 6.4615 g mol–1
15.000 g mol–1 × 13.000 g mol–1N
μ 13 == 1.1565 × 10–26 kg
–1
23
–1
C
(15.000 + 13.000) g mol × 6.0221 × 10 mol
15

15

N
C

Alternatively, (μ13 )' = 6.9643 g mol–1
2

14

N

14

N
C

C−N bond force constant: k = (2π c ν̃ 12C ) × μ12 = 636.48 kg s–2
C−N substituted bond wavenumber:15

Nν̃ 13C =1k= 1 245.5 cm–1
15
√
N
2πc
μ13
C

Hypothetical kinetic isotope effect:
14

N

k 12C

14
15
hc
N
N
̃
̃
=
exp
(
(ν
−
ν
12
13 )) = 1.1166
15
C
C
N
2kB T

k 13C

16 points in total
2 points for the correct formula for the reduced mass, 1 point for each numerical value of the reduced
mass (4 points in total), full marks will be scored if the reduced mass is given in units such as kg, g,
amu, kg mol−1 or g mol−1
2 points for the correct formula for the C–N force constant, 2 points for its numerical value
2 points for the correct formula for the wavenumber, 2 points for its numerical value
2 points for the correct formula for the KIE, 2 points for its numerical value
Alternative to force constant and wavenumber one can derive formula for KIE as follows
14

15

N

μ12

14

N

N
C

ν̃ 13C = ν̃ 12C √ 15 = 1 245.5 cm–1
N
μ13
C

14

N

E 14N =hcν̃ 12C
2

ν̃ 12

C

= 1.2851 × 10–20 J

15

E 15N =
ν̃ 13

C

N
C

hcν̃ 13
2

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E 14N − E 15N

14

N

k 12C
15

N

= exp (

k 13C

ν̃ 12

C

kB T

ν̃ 13

C

) = 1.1166

2 points for the correct formula for the wavenumber, 2 points for its numerical value
2 points for the correct formula for the ZPVE, 1 point for each numerical value
2 points for the correct formula for the KIE, 2 points for its numerical value
Some of the steps in the derivation may be taken implicitly or simultaneously. As long as they arecorrect, full marks will be scored for each of the partial steps. If the correct KIE (a dimensionlessquantity) is provided, full marks will be scored for the whole 6.7 problem. If the calculation of the KIE
is not completed or its wrong value is given, 0 points will be scored for any of the numerical results ofthe partial steps if these are given with wrong units or no units at all.

6.8

After making these isotopic changes, the rates of hydrolysis are not significantly different.
Which of the following is the most likely the rate determining step?
☒ Nucleophilic attack of water on a protonated amidic moiety☐ C−N bond cleavage☐ Protonation of the gyromitrin molecule

3 points for the correct answer.
−3 points for each incorrect answer. The total score in task 6.8 may not be negative.

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Theoretical
Problem 7

Question

7.1

7.2

7.3

7.4

7.5

7.6

Total

Points

10

6

15

9

11

6

57

7% of the total

Score

Problem 7. Cidofovir
Cidofovir (1), originally designed and prepared by the group of Professor Holy in former
Czechoslovakia, is a nucleotide analogue with antiviral activity. It is used to treat viral infections,mostly in patients with AIDS.

The key intermediate in the synthesis of cidofovir is optically pure diol 2, which can be prepared fromL-mannitol (3).
The total for each structure/sub-problem may not be negative.

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7.1

Draw the structures of compounds A–D, including stereochemistry. One molecule of A
produces two molecules of B.
A

C12H22O6B

3 points for the correct structure
2 points if the molecular formula is correct but
different OH groups are protected
2 points if the correct OH groups are protected
but the stereochemistry is wrong or missing
1 point if different adjacent OH groups areprotected and the stereochemistry is wrong
0 points if the molecular formula is not correct

3 points for the correct structure
3 points if any wrongly assigned compound A iscleaved between two adjacent free OH groups
2 points if stereochemistry is wrong or missing
1 point if compound A is cleaved properly but
the oxidation state of C1 is wrong
1 point if A is not determined and B contains
one CHO group
1 point if B contains one CHO group, but theprotecting groups are cleaved-1 point if stereochemistry is wrong or missingCD

2 points for aldehyde reduction to alcohol(ONa scores full marks)
1 point if B is not determined and C contains a
primary OH group-1 point if stereochemistry is wrong or missing

2 points for alcohol protection (“Bn” also
accepted)
If C is not correct:
2 points if all primary OH groups are benzylprotected
1 point if D contains a benzyloxy- group (and not
all primary OH groups are benzyl-protected)-1 point if stereochemistry is wrong or missing

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7.2

Draw the structural formulae of all alternative stereoisomers of compound 3 which could be
used in the same reaction sequence to afford only the same product 2.

3 points for each correct compound (maximum 6 points)−2 point for each incorrect compound−2 point for a compound which is drawn twice (i.e. 1 point for a pair of compounds which are identical)

Diol 2 is further modified to provide compound I. The synthesis of phosphonate 4 used to convert
compound F to G will be discussed later.

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7.3

Draw the structures of compounds E–I, including stereochemistry. Use the abbreviation MMT
for the (4-methoxyphenyl)diphenylmethyl group.
The same scheme as on the previous page, foreasier orientationE

C30H30O4

3 points for the correct structure
1 point if the secondary OH group is protected
and the configuration on C2 is correct
0 points if the secondary OH group is protected
and the configuration on C2 is incorrect
0 points if the molecular formula is not correct-1 point if stereochemistry is wrong or missingF

3 points for the correct structure)
2 points if written without the Na+ counter ion
1 point if the alkoxide on C2 is formed, but theconfiguration is inverted
1 point if the alkoxide on C2 is formed and anyof the protecting groups is cleaved-1 point if stereochemistry is wrong or missingG

3 points for the correct structure
1 point if the substitution was performed on
phosphorus and the configuration on C2 iscorrect-1 point if stereochemistry is wrong or missingHI

3 points for MMT cleavage
1 point if either benzyl or both the protecting
groups are cleaved and the phosphonate isretained with the correct configuration-1 point if stereochemistry is wrong or missing

3 points for the mesylate formation (“Ms” also
accepted)
1 point if the molecular formula is correct, but I
contains a mesylate group
0 points if molecular formula is not correct-1 point if stereochemistry is wrong or missing

THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

C16H27O8PS

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Phosphonate 4 can be prepared according to the following scheme:

7.4

Draw the structures of compounds J–L.
J

3 points for any phosphite or dialkyl
phosphonate salt Transesterification in thesubsequent step would provide the required
diethyl phosphonate.
1 point if all R = aryl in P(OR)3K

3 points for the correct structureL

3 points for the correct structure

The reaction of I (from question 7.3) with cytosine (5) leads to a 3:1 mixture of isomeric compounds
M and N. The formation of these two products may be understood by realizing that cytosine (5) can
also exist as an aromatic tautomer P. The reaction of M with cyclohexa-1,4-diene and palladiumhydroxide on carbon leads to compound O. The phosphonic ester moiety in compound O reacts with
bromotrimethylsilane to provide cidofovir (1).

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7.5

Draw the structures of the two isomers M, N, and of compound O, including stereochemistry
and the structure of the aromatic tautomer P of cytosine (5). Transformation of M to O is theremoval of a protecting group.
M (75%)N (25%)

3 points for the correct structure-1 point if the stereochemistry is wrong or missing

3 points for the correct structure
2 points if cytosine is linked via the amino group-1 point if the stereochemistry is wrong or missing
3 points in total if M and N are interchanged

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7.6

O

P

3 points for the benzyl cleavage-1 point if the stereochemistry is wrong or missing

2 points for the correct aromatic structure

Draw the structures of the two simple organic side products Q and R formed during theconversion of M to O.
Q

from cyclohexadiene

3 points for the correct structure

THEORETICAL PROBLEMS, OFFICIAL ENGLISH VERSION

R

from the protecting group

3 points for the correct structure53

XXX-XINTERNATIONAL CHEMISTRY OLYMPIAD / SLOVAKIA & CZECH REPUBLIC, 2018

Theoretical
Problem 8

Question

8.1

8.2

8.3

8.4

8.5

8.6

8.7

8.8

Total

Points

14

14

2

16

6

8

9

6

75

9% of the total

Score

Problem 8. Caryophyllene
-Caryophyllene (3) is a naturally occurring sesquiterpene present in clove tree and in sometraditional Czech and Slovak plants, such as the hop plant or small-leaved linden.
The synthesis of -caryophyllene starts from a single enantiomer of dienone A. The reaction of A
with silyl ketene acetal 1, followed by immediate reduction and aqueous work-up affords ketone 2.
This intermediate then undergoes reaction with tosyl chloride, providing B. Basic cyclization of thiscompound affords C. Finally, the reaction of C with ylide D provides -caryophyllene.

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8.1

Draw the structures of compounds A–D, including the appropriate stereochemistry. Hint: Intransformation A → 2, the silyl ketene acetal acts as a nucleophile.
A C10H14O

B

4 points
2 points if configuration on C2-C3 double bondis (E)
1 point for the β,γ-unsaturated isomer of A

4 points
2 points if stereo configuration is incorrect
or unclear

CD

4 points
2 points if stereo configuration is incorrect
or unclear

2 points

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Due to the large ring size, both compounds 2 and 3 are stable, even though they contain a doublebond in the trans configuration. trans-Cyclooctene (4) is the smallest ring that can accommodatea trans double bond. It can be prepared according to the following scheme:

8.2

Draw the structures of reagent E and intermediates F and G, including the appropriate
stereochemistry. For F and G, tick the box indicating the stereochemical outcome.
EF

4 points for the structure (O-Li+ and OH are
also accepted)
3 points when Li cation is missing from OLi
2 points if stereochemistry is incorrect
or unclear

2 points
Other common epoxidation agents are also
accepted.
1 point if a two-step epoxidation is given
0 points for H2O2 (with no acid indicated)

2 points for ticking the right box (these points
are additive to the points for the structure)

☐
☐
☒
☐

achiralsingle enantiomer
racemic mixture
mixture of diastereoisomersG

4 points for the structure
4 points for the oxidized product with acetylated -OH
3 points if charges and/or multiple bonds
on phosphine oxide are incorrect or missing
2 points if stereochemistry is incorrect or unclear
1 point for oxidizing the alcohol to a ketone
instead of or in addition to the phosphine
0 points for only acetylating -OH
2 points for ticking the right box (these points are
additive to the points for the structure)

☐
☐
☒
☐

achiralsingle enantiomer
racemic mixture
mixture of diastereoisomers

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8.3

Draw the structure of the enantiomer of cycloalkene 4.

2 points (other correct drawings are also accepted)

The two double bonds in -caryophyllene display different reactivity: the double bond in the ring(endocyclic) is more reactive than the other one (exocyclic) due to the ring strain.

8.4

Draw the structures of compounds Ha + Hb, I and Ja + Jb, including the appropriate
stereochemistry. Hint: Ha + Hb and Ja + Jb are pairs of diastereomers.

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Ha + Hb

6 points if both diastereomers are given
4 points if both diastereomers are given but stereochemistry of one of the two newly formed
stereocenters is missing, unclear or incorrect
4 points if only one diastereomer is given (the second is missing) but stereochemistry is correct
3 points if one diastereomer is drawn correctly and the second structure is incorrect
2 points if both diastereomers are given but stereochemistry is incorrect or missing
2 points if both diastereomers are given but incorrect double bond or both double bonds are synepoxidized
1 point if only one diastereomer is given and stereochemistry is incorrect or missing
1 point if only one diastereomer is given and incorrect double bond or both double bonds are synepoxidized
the above partial credit points are not additiveI

4 points
2 points if primary or secondary ozonide is drawn
1 point if the correct ozonolysis product of the exocyclic bond or of both bonds is given

Ja + Jb

6 points if both diastereomers are given
4 points if both diastereomers are given but stereochemistry of one of the two newly formed
stereocenters is missing, unclear or incorrect
4 points if only one diastereomer is given (the second is missing) but stereochemistry is correct
3 points if one diastereomer is drawn correctly and the second structure is incorrect
2 points if both diastereomers are given but stereochemistry is incorrect or missing
2 points if both diastereomers are given but regioselectivity is incorrect
2 points if both diastereomers are given but correct hydration of the wrong double bond or of both
double bonds is given
1 point if only one diastereomer is given and stereochemistry is incorrect or missing
1 point if only one diastereomer is given and regioselectivity is incorrect
1 point if only one diastereomer is given and incorrect double bond or both double bonds arecorrectly syn-hydrated
the above partial credit points are not additive

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Interestingly, the reactivity of the double bonds is reversed when isocaryophyllene (5) is used insteadof -caryophyllene (3).

8.5Draw the structures of compounds Ka and Kb. Hint: Ka + Kb are a pair of diastereomers.
Ka + Kb

6 points if both diastereomers are given
4 points if only one diastereomer is given (the second is missing) but stereochemistry is correct
3 points if one diastereomer is drawn correctly and the second structure is incorrect
2 points if both diastereomers are given but stereochemistry is incorrect or missing
2 points if both diastereomers are given but regioselectivity is incorrect
2 points if both diastereomers are given but correct hydration of the wrong double bond or of both
double bonds is given
1 point if only one diastereomer is given and stereochemistry is incorrect or missing
1 point if only one diastereomer is given and regioselectivity is incorrect
1 point if only one diastereomer is given and incorrect double bond or both double bonds arecorrectly syn-hydrated
the above partial credit points are not additive

Isotope-labelled compounds are invaluable tools for reaction mechanism investigation, structure
determination, and mass or NMR spectroscopy studies. Let us have a look at the synthesis of somelabelled analogues of -caryophyllene.

8.6

Draw the structures of compounds L and M, including the appropriate stereochemistry.

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L

M C14H20D2O

4 points

4 points

2 points for the structure below

-Caryophyllene (3) undergoes acid-catalyzed cyclization, which leads to a complex mixtureof products. Among them, the pair of diastereomers Na + Nb and the pair of diastereomers 7a + 7b
are the most abundant. The reaction starts with protonation of the more reactive internal double bondaffording cation O. This cyclizes without the cleavage of a carbon-carbon single bond to yield
diastereomeric tricyclic cations Pa and Pb, which undergo hydration to give the target alcohols Na
and Nb. Alternatively, the cations Pa and Pb rearrange with the cleavage of a carbon-carbon single
bond to cations Qa and Qb, which deprotonate to compounds 7a and 7b.

8.7

Draw the structures, including the appropriate stereochemistry, of the three intermediates O,
Pa, Qa leading to the diastereomer 7a, including the appropriate stereochemistry.

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O

3 points for the correct structure
2 points if stereochemistry is missing or
incomplete
1 point if the wrong double bond is protonatedwith correct regioselectivity
0 points if either double bond is protonated withincorrect regioselectivityPa

3 points for the correct structure
2 points for the carbocation with incorrect
regioselective ring closure
1 point if stereochemistry is missing or
incomplete
2 points for structure Pb

Qa

3 points for the correct structure
2 points for the carbocation with incorrect
regioselective rearrangement
1 point if stereochemistry is missing or
incomplete
2 points for structure Qb
2 points for incorrect cation Qa'

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8.8 Draw the structures of diastereomers Na + Nb.
Na + Nb C15H26O

6 points if both diastereomers are given
4 points if only one diastereomer is given
4 points if both diastereomers are given which are tricyclic compounds resulting froma regioisomeric attack on the correct carbocation (see example below)
2 points if both diastereomers are given which are tricyclic compounds resulting from an attackon incorrect carbocation (see example below)
2 points if only one diastereomer is given which is a tricyclic compound resulting froma regioisomeric attack on the correct carbocation (see example below)
1 point if only one diastereomer is given which is a tricyclic compound resulting from an attack
on an incorrect carbocation (see example below)•

incorrect regioisomeric attack on the correct carbocation intermediate•

attack on the incorrect carbocation intermediate

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