austrian national chemistry olympiad 1998

8/11/2019 Austrian National Chemistry Olympiad 1998

1/21

24th

Austrian Chemistry Olympiad 1998National Competition - Frstenfeld

Theoretical Part

Problem A 6 Points

Environmental Chemistry

1. We are looking for a particle which plays an important role in the catalyticdegradation of ozone as well as in fotochemical consecutive reactions in thetroposphere. In the latter reactions it causes a backformation of hydroxyl radicals.There are natural and anthropogenic sources for this particle.

a) Name the particle!

b) Name an anthropogenic and natural source each!

c) Write down a balanced equation for the backformation of the hydroxyl radicals

described above!

d) Write down all the single step reactions and also the overall reaction for thecatalytic degradation of ozone where this particle takes part in!

2. Mainly responsible for the degradation of ozone in the stratosphere are the CFCs.Due to the difference stabilities different CFCs have not the same range of action inthe stratosphere. Combine the following compounds with the corresponding curves inthe diagram and give the right formulae and the IUPAC-names of: R11, R12, R13,R14

R11:........................................

R12:........................................

R13:........................................

R14:........................................

3. At the surface of ice cristals of the PSCs heterogenic reactions of chloronitratewhich HCl and H2O molecules take place whereby reservoir gases are formed andfrozen consecutively. In the antarctic spring two particles emerge which catalize theozone degradation.

Write down the balanced equations for the formation of chloronitrate and also for theformation of the two reservoir gases!


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3/21

24th


Theoretical Part

Problem B 7 Points

Noble gases

1. In order to gain noble gases liquified air is fractionated. In a first rectification stepone gets four fractions of which two contain nitrogen and two contain oxygen each asmain part. Assign the non radioactive noble gases to the corresponding fractions:

2. An ice-argon-clathrate has a cubic structure. Its unit cell consists of 46 watermolecules and 8 interstitial sites of which are filled with argon atoms. What is thestoechiometrical formula of the clathrate?

3. In the formation of XePtF6 , discovered by Neil Bartlett in 1962 an intermediateentdeckten dixenon-cation Xe2

+is generated. Sketch the MO-scheme of this cation.(Only take into account the electrons with the main quantum number 5!) Which bond

order occurs in this ion and which magnetic properties does this particle have.

N2

N2

O2

O2

MO-scheme: bond order:

magnetic property:


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Theoretical Part

4. We know of four compounds of xenone which contain fluorine and oxygene or bothof them, where the noble gas has the oxidation number VI. What are the formulae ofthese compounds? Which three dimensional structures do these noble gas

compounds have?

compound

threedimensionalstructure

5. Which noble gas compounds do not exist? Fill in the following table!

compoun d yes/no i f no, reasons

XeF4

NeF2

KrF2

XeCl4

KrBr4

XeOF

XeF9

6. Which redox behaviour of XeF2 is to be expected? Write down a balancedequation for the reaction of this compound with hydrochloric acid!

................................................................................................................................


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Theoretical Part

Problem B- solutions ( 31 b.p. correspond to 7 r.p.; f = 0.22581)

Noble gases

1.

2.

3.

N2 Ne He

N2 Ar

O2 Kr Xe

O2 Ar

3E.23H2O

MO-scheme: bond order:

0,5 1.b.p.

magnetic property:

paramagenetic 1b.p.

1b.p.

1b.p.

1b.p.

1b.p.

1b.p.

x

#

#

b

xb

s#

sb

5s 5s

5p 5p

Xe Xe+

Xe2

+

3 b.p.


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Theoretical Part

4.

compound XeF6 1b.p. XeO3 1b.p. XeO2F2 1b.p. XeOF4 1b.p.

threedimensionalstructure

distortedoctahedral

1b.p.

pyramidal

1b.p.

distortedtetrahedral

1b.p.

squarepyramidal

1b.p.

5.

compoun d yes/no i f no, reasons

XeF4 yes ------------

NeF2 no Ne has too high promotion energy

KrF2 yes ------------

KrBr4 no Br2has too high dissoziation energy

XeOF no Xe-oxidation number 3 not possible

XeF9 no Xe-oxidation number 9 not possible

6.

XeF2strong oxidizing agent: XeF2+ 2 HCl Cl2 + Xe + HF 3b.p.................................................................................................................................

1.bp.

b.p.

1b.p.

b.p.

b.p.

b.p.


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Theoretical Part

Problem C 7 Points

Thermochemistry

1st part

Nitrosylchloride (NOCl) is a very toxic gas, which by dissociation gives nitrogenoxideand chlorine when heated.

The reaction is endothermic: H= 75.3 kJ per mol chlorine (this value remainsconstant up to 600 K)The standard entropy values of the substances involved at 25C are::

substance NOCl NO Cl2

S298(J/mol.K) 264 211 223

Solve the following problems:

a) Write down a balanced equation for the dissociation of nitrosylchloride, wherebyone mol of chlorine should be formed.

b) Calculate the KPof the reaction at 298 K.c) Calculate the KPof the reaction at 475 K.d) Verbally give reasons for the shift of the equilibrium.

2nd part

The equilibrium constant for the formation of hydrogeniodide has the value K = 160.There is a mixture of the three gases with the partial pressures p(H 2) = 1.5 atm, p(I2)= 0.88 atm and p(HI) = 0.065 atm.

Do we find a possible reaction in this mixture at 500 K?If yes, in which direction? Prove your statements by calculations.


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24th


Theoretical Part

Problem C- solutions (15 b.p. correspond to 7 r.p.; f = 0.46667)

Thermochemistry

1st

part

a) 1 b.p.

b) 4 b.p.

c) 3 b.p.

d) 2 b.p. If an endothermic system is heated, the equilibrium will shift to theright side (LeChatelier): KP(475)>>KP(298)

2ndpart:

2 b.p.

3 b.p.

K e G H T S

S G

K K

th

G RT

th P

/ .

. .

J /K J

if p atm atm

298 298 298

298 298

8 8

117 40434

817 10 1 817 10

ln

( )

( )

ln .

.

K T

K T

H

R T T

K ( )

K

P

P

P

P

2

1 1 2

3

1 1

475 500

6 74 10

insertion of numerical values gives

atm

Q

G RTQ

K G

0065

15 0 883 2 10

45

2

3,

. ..

ln

Q K yes, reaction!

kJ the reaction runs in the direction to HI.

2 NOCl 2 NO + Cl2


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24th


Theoretical Part

Problem D 8 Points

Blood as a buffer system

Blood is a very important buffer system. Its buffering effect is 2/3 due to the openbuffer system CO2/HCO3

-. Throughout this problem only this buffer system isregarded.A possibility to get a right value of the proportion CO2/HCO3

- is the equibrilationmethod by Astrup.

Principal of the method:The pH-value of a blood sample is determined. Then the blood sample is exposesdto high and CO2-pressures until equilibrium is reached and the pH-values aremeasured in each case. The pressure of CO2 of the original sample can be

determined graphically:

measured values:

pH-value pCO2(kPa)7.4 x (blood sample)7.2 10.07.3 7.607.5 3.307.6 1.00

Problems:

1. Determine p(CO)2 at pH = 7.4 graphically using the diagram and the measuredvalues.

2. Calculate the concentration of the CO2(=H2CO3) in solution at pH = 7.4; KH= 2.25*10-4mol/L.kPa

3. Calculate the concentration of hydrogencarbonae of the blood at pH = 7.4.pKA1(CO2) = 6.35

4. The deviation of the the required pH of 7.4 must not be to large. pH-values below7.0 are lethal already. Calculate the amount of H3O+-ions (in mol) which may begenerated in 5.0 L of blood so that the pH will not fall below 7.0. Pay attention tothe fact that blood is an open system, that means that CO2-concentration remainsconstant.

5. Calculate the theoretical pH-value which will come up if the above amount of H3O+

is generated in blood which is not an open system i.e. the concentration CO2 insolution increases.

In daily life one speaks about superacidulation by lactic acid. Show by calculationthat under the pH-conditions in blood all the lactic acid exists as lactate. pKA(HLac) =3.86!

P(CO2)kPa

pH


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24th


Theoretical Part

Problem D - solutions (14 b.p. correspond to 8 r.p.; f = 0.57143)

Blood as a puffer system

1. 10.0 p(CO2) = 5,30 kPa

p(CO2) grafics: 1b.p.Pressure: 1 b.p.

6.00

1.00

7.2 7.3 7.4 7.5 7.6 pH

2. c(CO2) = 5.30*2.25*10-4= 1.1925*10-3 c(CO2) = 1.19*10

-3mol/L 2 b.p.

3.

4.

5.

6.

pH pKc(HCO )

c(CO )

xc HCO

S

log

. . log. *

( ) .

3

2

3 37 40 6 35 11925 1000134 mol / L 2 b.p.

7 00 6 3511925 10

0 00533

8 07 10 0 0404

3 3

3 3

3

3

. . log. *

*( ) .

( ) ( ) . * ( ) .

x c HCO

n HCO n H O n H O

mol / L

mol / 5 L 3 b.p.

pH p

6 35533 10

9 263 10

3

3. log. *

. * H = 6.11 3 b.p.

7 40 386. . log( )

( ))

c Lac

c HLac p c(Lac c(HLac) = 3467 H = 6.11

-2 b.p.


11/21

24th


Theoretical Part

Problem E 8 points

Kinetics and Organic Chemistry

A compund with the general formula R-SO2-OR reacts in aqueous solution with OH-

giving the two endproducts B and C. C is volatile.

After solving 1,000 g of A in 50,00 mL of CCl4 (KKR= 30.00 K.kg.mol-1; = 1.594

g/cm3), the melting point of tetrachloromethane is lowered 1,882C.The 1H-NMR-spectrum of substance A is given:

Compund A is brought to reaction with OH- as described above. The initialconcentrations of A and OH- are the same. Observing the change of the OH

concentration by time we receive the following data:

t (min) 0 5.00 10.0 20.0 60.0

c(OH-) (mol/L) 0.0250 0.0155 0.0113 0.00727 0.00301

Solve the following problems:

a) Which family of compounds does A belong to??b) Give a balanced equation for the reaction of A with OH-.c) Calculate the molar mass of A.d) What are A, B and C? Give structural formulae.e) Prove by calculation that we have a second order reaction in b).f) Calculate the velocity constant and the half life of the reaction.

g) What will the value of be if one starts with 1/10 of the initial concentration?h) Calculate the neccessary time till only 1.00% of the initial concentration is present.

(c0= 0.0250 mol/L)?


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24th


Theoretical Part

Problem E - solutions (20 b.p. correspond to 8 r.p.; f = 0,40000)

Kinetics und Organic Chemistry

a) sulfonic acid ester 1 b.p.

b) R-SO2-OR + OH- R-SO3

- + R-OH 1 b.p.

c)

3 b.p.

d)

4 b.p.

1 + 1 b.p.

e)

k remains constant, therefore second order reaction 3 b.p.

f) k = 4.88 L/mol.min 1 b.p.

1

k co. = 8.21 min 2 b.p.

g) = 10 = 82.1 min 1b.p.

h)

1

0 00025

1

0025

1

811. .

k t t min 2 b.p.

T K m m n A

V

n A TK

M mn

M

KR

KR

. * *( )

.

( ) . . .

mol / L g / mol

0 0797 500 10 2003

1 1 1

c c t k

t o

t(min) 5 10 20 60

k(L/mol.min) 4.90 4.85 4.88 4.87

CH3 S

O

O

O CH2 CH3A:

CH3 S

O

O

O CH3 CH2 OHB: C:


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Theoretical Part


14/21

24th


Theoretical Part

Problem F 7 Points

Organic Chemistry - Cyclic systems

There are given:a) the structural formula of pentalene,

b) the structural formula of octalene

and the systematic names ofc) 1-(cyclopenta-2,4-diene-1-ylidene)-cyclohepta-2,4,6-triene,d) tricyclo[3.2.1.01,5]octane,e) 1,6:8,13-propano[14]annulene.

Give the empirical formula of (a) ,

the systematic name of (b)

and the structural formulae (resonating formulae) of c, d and e.

c d e

Give the answer to the following questions:Which of the compounds ae are aromatic?Which of the compounds ae show a significant dipole moment?Which reaction mechanisms take if the compounds ae react with bromine?

arom.: dip.mom. reaction mechanism

J / N J / N SR SN SE AR AN AE(a) (b) (c) (d) (e)


15/21

24th


Theoretical Part

Problem F - solutions (26 b.p. correspond to 7 r.p.; f =0,26923)

Organic chemistry

Cyclic systems

Empirical formula of pentalene: C8H6

Systematic name of octalene: bicyclo[6.6.0]tetradeca-1,3,5,7,9,11,13-heptaene

Structural formulae of c, d and e:

Aromatic are c and e.

Only c gives a significant dipole moment.

Reaction mechanisms:a AEb AE

c SEd SRe SRand SE


16/21

24th


Theoretical Part

Problem G 10 points

Synthesis of ()- Mamanuthachinone

H

O

O

OCH3

HO

The mamanuthachinone was discovered as an secondary metabolite of a sponge.The name is derived from the island Mamanutha, where this organism was found. Asthis substance has an anticancer effect in the case of intestine cancer, the synthesisof this compound is of some interest.The performance of 14 steps the synthesis of Danishefsky (1994) gives racemicmamanuthachinone. The synthesis yields 13% mamanuthachinone taking 1,4-benzochinone as starting material.The main part of the synthesis is a Diels-Alder-reaction between the diene (I) and thedienophile (E).

A) Preparation of the dienophile:

Methanol is added by 1,4-addition to 1,4-benzochinone. The 1,4-addition productrearranges to a substituted hydrochinone (hydrochinone = 1,4-benzenediol). This isfollowed by an oxidation step to compound B, a substituted benzochinone. A second1,4-addition of methanol to B yields compound C, a double substituted hydrochinone.Methylation of dimethylsulfate gives compound D, which is an aromatic compoundbearing 4 identical substituents at the positions 1,2,4 and 5.Performance of a single Friedel-Crafts-acylation of D with E-2-methyl-2-butenoic acidchloride yields the dienophile E.

Answer the following questions:1) Draw the structural formulae of 1,4-benzochinone and the compounds AE.2) Give the name of the compound D.3) The substituents of compound D activates or deactivates the aromatic

compound.4) Give the name of the reaction mechanism of the Friedel-Crafts-acylation.5) Which type of catalyst is taken for Friedel-Crafts-acylations?


17/21


18/21

24th


Theoretical Part

Problem G-solutions (26 b.p. correspond to 10 r.p.; f = 0.38462)

Synthesis of ()- Mamanuthaquinone

I. 1.

.

O

O

OH

OH

OCH3

O

O

OCH3

OH

OH

OCH3

CH3O

O

OCH3

OCH3

OCH3CH3O

OCH3

OCH3

OCH3

CH3O

II. 1.

2.

Wittig-reaction

1.

points : A) 1. 7 b.p. 2. 1 b.p. 3. 1 b.p. 4. 1 b.p. 5. 1 b.p.B) 1. 5 b.p. 2. 3b.p.C) 1. 3 b.p. 2. 1 b.p. 3. 1b.p. 4. 1 b.p. 5. 1 b.p.

O O OOH

OCH2=P(C6H5) O=P(C6H5)33

++

O

CH3O

CH3O OCH3

OCH3

p-benzoquinone A B C D

E

2-methylcyclohexanone F G H I

2. D = tetramethoxybenzene3. activating4. SE5. Lewis-acid

2. e.g.: lithiumaluminiumhydride LiAlH43. Li/NH34. S electron acceptor5. e.g. potassium dichromate K2Cr2O7


19/21

24th


Theoretical Part

Problem H 7 Points

Biochemistry

The amanitines consist of 8 well defined compounds. They are beside thephallotoxines the strongly toxic chemical agents of the green Kn ollenbltterpilz.

.

2.5g of the -amanitine are sufficient to kill a mouse. It is a bicyclic octapeptide withthe empirical formula C39H53N9O15S.Ion exchange chromatography of -amanitine gives:

glycine (double amount of the other amino acids);isoleucine;

,-dihydroxyisoleucine;cysteine (IR shows a sulfoxide group);tryptophane (NMR of the -amanitine gives a hydroxyl group at the

heterocyclic system in position 6 but no hydrogen atom in position 2)aspartic acid;hydroxyproline.

Chymotrypsine A (splits at the carboxylic side of aromatic amino acids) opens one ofthe rings. This is followed by an Edman degradation which gives first glycine thenisoleucine and glycine once more again. The consecutive hydrolysate cannot bedetermined because of its size.Staphylococcus-protease (splits at the carboxylic side of asp und glu) opens a ring

too. Following Edman degradation gives hyroxyproline, ,-dihydroxyisoleucine and abig hydrolysate (containing the remaining amino acids).

Determine the structure of the octapeptide and give the structure by amino acid codeand by structural formula as well (carboxylic group as COOH).A table of amino acids is included.


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24th


Theoretical Part

Problem H- solution (12 b.p. correspond to 7 r.p.; f = 0.58333)

Biochemisty

The endopeptidases give the peptides: Trp-Gly-Ileu-Gly and Asp-Hypro-Ileu(OH)2.Cystein-sulfoxide is missing. The bridge is built by Trp (as H misses in pos. 2) andCys-sulfoxide:

Ileu(OH)2TrpGly

Hypro SO Ileu

AspCysGly

resp.

NH

O

NHOO

HONH

O

NH

NHO

OHNH

NH

OO

NH

HOOC

ONH O

OH

SO


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24th


Theoretical Part

austrian national chemistry olympiad 1998

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