25th Austrian Chemistry Olympiad 1999 1

National Competition – Theoretical part

Problem 1 4 points

Complex Chemistry
Several different locations can be imagined for the ligands of a complex of the type MeX 3Y2
(Me = centre atom; X, Y = ligands).

Draw all possible diastereomeres for the above described complex including contingent
optically active isomeres.
Additionally state, which of the possible complexes outlined will be the most stable one
(assuming that X is a small and Y a large ligand).

As an assistant device (so to say as a master copy) you can choose among several different
coordination polyhedrons.
25th Austrian Chemistry Olympiad 1999 2
National Competition – Theoretical part

Problem 2 10 points
Environmental and Inorganic Chemistry
A) Determination of oxygen in a sample of water from the river “Schwechat”

The determination of oxygen in water samples is carried out in the iodometric analysis
described below (method by WINKLER) :
1st step: “fixation of oxygen”. The oxygen in solution oxidizes Mn2+ to Mn(IV) in alkaline
medium giving MnO(OH)2
nd
2 step: By addition of acid the above mentioned manganese compound reacts with a
surplus of Mn2+ to give Mn3+-ions (comproportion reaction).
3rd step: The Mn3+-ions developed oxidise added iodide to give iodine and are themselves
reduced to Mn2+ .
4th step: The amount of iodine generated in step 3 is titrated with a solution of
thiosulfate.

Task 1: Write balanced ionic equations for the four reactions described above!

1st step:

2nd step:

3rd step:

4th step:

The analysis of water samples from the river “Schwechat” showed the following data:

1. Standardization of the sodium thiosulfate solution: the standardisation was ensued with
KIO3 in acidic medium, whereby iodate was reduced to iodide. 25.00 mL of KIO 3 -
solution ((KIO 3) = 174.8 mg/L) consumed 12.45 mL of thiosulfate.
2. Immediately after sampling, the determination of oxygen was carried out using
WINKLERs method described above. 11.80 mL of the standardised sodium thiosulfate
solution were consumed for 103.50mL of water sample at 20.0 °C. The saturation
concentration of O2 in water amounts to 9.08 mg/L at 20.0°C.
3. A second sample (V = 102.20 mL, T = 20.0 °C) was incubated after sampling for 5 days
in a climatic cabinet at 20.0°C. The analysis after this incubation gave a consumption of
6.75 mL of thiosulfate.

task 2:
a) Write balanced ionic equation for the standardisation of the thiosulfate solution!

b) Calculate the concentration of the standardised thiosulfate solution!

c(S2O32-)= mol/L
25th Austrian Chemistry Olympiad 1999 3
National Competition – Theoretical part
c) Calculate the oxygen content of the water which was analysed immediately after sampling
using the unit mg/L!

(O 2)= mg/L

d) Calculate the oxygen saturation index for this water sample!

OSI= %

e) What is the content of oxygen in the incubate water sample?

(O 2)= mg/L

f) Which of the characteristic parameters of water analysis was thus determined? What is
the value of it?

B) The peculiar chemistry of a well known element

An oxygen containing anion of an allotropic element is characteristic for water pollution; the
element has a lower electronegativity than oxygen. With halogenes it forms only molecular
compounds. Additionally to two monomolecular oxides also high molecular ones are known.
Element X is also of great importance in biochemistry. Its p-atomic orbitals are half occupied
each.

Task 1: What element is talked about? Write down its electron configuration!

Element: electron configuration:

Element X also forms numerous covalent compounds with hydrogen with the general formula
XaHb. A series of such compounds is analogous to the homologous series of the alkanes.

Task 2: Draw the structural (constitutional) formulae of the first four homologous X-H-
compounds!

1 2 3 4

One of these four substances consists of 3 diastereomeres (similar to tartaric acid).

Task 3: Give the number of this compound!

compound nr.:

Element X forms oxoacids with the general composition H 3 XO n , where n=2,3, and 4.
25th Austrian Chemistry Olympiad 1999 4
National Competition – Theoretical part

Task 4: Draw the structural (constitutional) formulae of these three acids. Mark the acidic
Mark the acidic H-atoms (asterisks or arrows) and give the oxidation numbers of element X in
each of these compounds.

A heterocyclic compound of element X with plane structure, which was already synthesised
by J. Liebig and F. Wöhler in 1834, is gained starting with NH 4Cl and the pentachloro
compound of X. As by-product a gas evolves, which is readily soluble in water and reacts as
strong acid.

Task 5:
a) Write a balanced equation for the above described reaction!

b) Draw the structural (constitutional) formula of the compound (NXCl 2)3 !

The just described inorganic substance shows a peculiar behaviour when heated: it boils
at 256°C when heated rapidly. If one heats slowly, the substance will start to melt at
250°C. Cooling the liquid rapidly, one gains a substance similar to rubber.

Task 6: Explain this peculiar behaviour!

25th Austrian Chemistry Olympiad 1999 5
National Competition – Theoretical part

Problem 3 5 points
Cerimetric Analysis
In cerimetric analysis the oxidising agent are Ce4+-ions, which are easily reduced to Ce3+-ions.
The value of the standard redox potential of these ions depends on the anions present. As an
original titer for standardisation of Ce4+-solutions As2O3 is used, which gives with sodium
hydroxide and subsequent acidification arsenite (AsO 3 3-), which the is oxidised by Ce4+ to
form arsenate (AsO43-). As a catalyst one uses small amounts of OxO 4. Ferroin acts as redox
indicator.

Write a balanced ionic equation for the described redox titration of arsenite with Ce 4+ and
calculate the equivalent potential when working at pH = 1!

E1 °(AsO43-/AsO33-) = 0.56 V
E2 °(Ce4+/Ce3+; HClO4) = 1.70 V
25th Austrian Chemistry Olympiad 1999 6
National Competition – Theoretical part

Problem 4 7 points

Physical Chemistry

A. Vapour pressure of a mixture

The vapour pressure of pure ethylethanoate amounts to 9706 Pa at 20°C, the one of
ethylpropanoate to 3693 Pa at the same temperature.
Due to the similarity of these two compounds the mixture shows an ideal behaviour.

Calculate the vapour pressure of a mixture 25 g of ethylethanoate and 50 g of ethyl-

propanoate at 20°C.

B. A kinetic problem

Bromomethane may react with OH- according to a SN -mechanism.

a) Write a balanced equation for this substitution.

The following table shows the initial rates of the reaction together with the corresponding
initial concentrations of CH3Br and KOH. All experiments were carried out at 25°C.

c(CH3 Br) c(KOH) vo (mol/L*s)

1st experiment 0.10 mol/L 0.10 mol/L 2.80*10-6
2nd experiment 0.10 mol/L 0.17 mol/L 4.76*10-6
3rd experiment 0.033 mol/l 0.20 mol/L 1.85*10-6

b) Determine the reaction order relative to the individual starting substances and also the
overall reaction order using a calculation.

c) Calculate the velocity constant of the reaction.

d) How long will it last to reach a concentration of KOH of 0,05 mol/L in the first
experiment?

e) Give the more accurate name of the mechanism which may possibly apply to the reaction.
25th Austrian Chemistry Olympiad 1999 7
National Competition – Theoretical part

Problem 5 7 points

Thermochemistry

The following table shows a series of reactions and reaction enthalpies at standard conditions:

# Reaction H298 (kJ)

1 2 NH3 + 3 N2O 4 N 2 + 3 H 2O - 1011
2 N2O + 3 H2 N2H4 + H2O - 317
3 2 NH3 + 0,5 O2 N 2H4 + H2O -143
4 H2 + 0,5 O2 H2O -286

Additionally the entropies of the listed substances are given at standard conditions:

S298 (N2H4) = 240 J/mol.K S298 (N2) = 191 J/mol.K

S298 (H2O) = 66.6 J/mol.K S298 (O2) = 205 J/mol.K

a) Calculate the formation enthalpies H298 of the compounds hydrazine, dinitrogenoxide

and ammonia.

b) Write a balanced equation for the combustion of hydrazine to give water and nitrogen.

c) Calculate the reaction heat of the above combustion at constant pressure and at 298 K.
Also calculate G298 and Kth.

d) What will be the reaction heat of reaction #3 at constant volume, if one starts with 2 mol
ammonia and 0,5 mol oxygen?
25th Austrian Chemistry Olympiad 1999 8
National Competition – Theoretical part

Problem 6 6 points

Ionic Equilibrium

0.85 g of AgNO3 are solved in water to give 500 mL of solution. A second solution of 500
mL of NH3(aq) with c0 (NH3) = 2.0 mol/L is at your hand.

a) Calculate the concentration of Ag +.

Now the two solutions are mixed. In this case the complex [ Ag(NH 3)2]+ is generated
primarily, the dissociation constant of it amounts to KD = 5.9*10-8 mol2 /L2 .

b) Write a balanced equation for the formation of the complex.

c) Calculate the equilibrium concentration of free silver.

The formation constant of [ Ag(NH3 )]+ is K1 = 1.8*103 L/mol.

d) Calculate the equilibrium concentration of [Ag(NH3)]+ using the data from c).

e) What is the value for K2 of the reaction: [Ag(NH3 )]+ + NH3 [Ag(NH3)2]+ ?

Hint: Use in your calculations appropriate simplifications.

25th Austrian Chemistry Olympiad 1999 9
National Competition – Theoretical part

Problem 7 11 points

Organic Chemistry
A. Spectra of an antifungal agent:

A fungicide and oxidizing substance A is described by an 13C-NMR-spectrum (containing a

solvent peak) and by the mass spectrum.
The infrared spectrum shows a distinct peak at 1685 cm-1 and another peak at 1570 cm-1.
The 1H-NMR-spectrum gives no signal.

The production of A starts from C 6H4O2. The cycle

addition / rearrangement forming the aromatic compound / oxidation
has to be done four times.

a) Give the structural formula of A.

b) Give the equations for the first three-step-cycle of the preparation.
c) Which is the reaction mechanism of the very first step.
d) Give an explanation why A is an oxidizing substance.
25th Austrian Chemistry Olympiad 1999 10
National Competition – Theoretical part

B. Organic synthesis:

One of the substances obtained by cracking is the symmetric compound A. A is a gas which
consists of 85,7% C and 14,3 % H (m/m). The geometric isomerism of A can be neglected.
 If HBr is added to A the compound B will be formed.
 B reacts with KCN forming C; B reacts with KOH forming D.
 Hydrolysis of C yields E.
 D can be oxidized by H2O2 resulting H.
 D reacts with concentrated sulfuric acid under various conditions yielding four different
products: F, G, I and a substance already mentioned.
 H is a solvent for paints and is used for the dewaxing of lubricants. H reacts with
phenylhydrazine to K.

1. Write down the constitution formulae of all compounds.

2. Which properties have B, C, D, E and F in common?
3. Which reaction mechanism leads from A to B? Which one from B to C?
4. Draw the configuration formula of the R-enantiomer of one of the compounds.
A

C D

E H I

R-enantiomer:

K
25th Austrian Chemistry Olympiad – 1999
National Competition – Theoretical part

Problem 8 9 points

Biochemistry
A. Drug against lung emphysem:

Lung can be damaged (especially by smokers) by the enzyme electrophilic elastase. 1 -

antitrypsin is an enzyme which is able to bind the elastase. 1 -antitrypsin has the amino acid
methionine in position 358. The methionine is oxidized by ozone or by the radicals of the
smoke of cigarettes. The oxidized form of the 1 -antitrypsin is no longer able to block the
elastase.
If the methionine in position 358 is substituted by gentechnological methods by valine, a
modified 1 -antitrypsin is obtained which is a drug against lung emphysems:
At first the original 1 -antitrypsin gene is introduced into a vector. The important base
sequence at the codogenic strand in the new recombinant DNA is 5´-TTTAAGCATGGC-3´ .
On the other hand an oligonucleotide is synthesized which is complementary to the sequence
mentioned. But this synthetic oligonucleotide should express the valine and not the
methionine. Therefore one of the bases is not complementary!
The oligonucleotide is attached to the singular strand of the recombinant DNA. The build-up
of the complete double-stranded DNA is done according to the singular strand template. The
“base error” leads after cloning and expressing in E.coli to a wild type and the desirable
mutant.

 Why is the oxidized form of the 1-antitrypsin not effective?

 Which agent is necessary for the inserting of the gene into the plasmid?
 Give the sequence of the synthetic oligonucleotide.
 Why are wild type and mutant not obtained in a ratio of 1:1 ?
 Which method is used to the select the mutant?
 Write down the sequence of the amino acids for the given sector of the 1-antitrypsin
mutant.
 Draw the formula for the sequence of the amino acids.
25th Austrian Chemistry Olympiad – 1999
National Competition – Theoretical part

Cut of the genetic code :

1. position 2. position 3. position
(5´-end) U C A G (3´-end)
Leu Pro His Arg U
Leu Pro His Arg C
C Leu Pro Gln Arg A
Leu Pro Gln Arg G
Ile Thr Asn Ser U
Ile Thr Asn Ser C
A Ile Thr Lys Arg A
Met Thr Lys Arg G
Val Ala Asp U
Val Ala Asp C
G Val Ala Glu A
Val Ala Glu G

The important amino acids in human proteins:

25th Austrian Chemistry Olympiad – 1999
National Competition – Theoretical part
B. Discovering of the genetic code:

One step in the discovering of the genetic code was done by mixing a DNA free culture of a
synthetic polyribonucleotide, the amino acids and ATP. According to the polyribonucleotide
template the culture makes peptides. The amino acid composition of the peptides is examined.
If for instance a poly(C)ribonucleotide is used the culture will build up a peptides which
contain proline only.
In one of the experiments a polyribonucleotide was used which contained U and G in a molar
ratio of 3:1.
The investigation of the resulting peptides brought following molar ratio of the amino
acids:
100 Phe : 37 Val : 36 Leu : 35 Cys : 14 Trp : 12 Gly
The other amino acids could not be detected in the peptides.

 One of the amino acid codons was exactly determined by this experiment. Which one?
 What could be said about the other 19 amino acids? Which information could not be
obtained?