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2002年第34届ICHO预备试题答案(英文)1-29


34th International Chemistry Olympiad | Preparatory Problems

Answers to the Problems

Important notice: In accordance with the decision of the International Jury of the C

hemistry Olympiad only partial answers will be published on the website. Worked solutions to the problems are included in the booklet that has been sent to all Olympiad organizations of participating countries. The worked solutions will also be published on the web in due time. For any questions you can contact us by e-mail (Icho34@chem.rug.nl or Ich034@sci.kun.nl). Problem 1
1-1

Production of Ammonia

n[H2,"] = 1500 mol s-1 n[N2,#] = 500 mol s-1 n[CH4,$] = 500 mol s-1 n[H2O,$] = 500 mol s-1 n[CO,%] = 500 mol s-1 n[O2,&] = 125 mol s-1 n[CO,'] = 250 mol s-1 n[N2,(] = 500 mol s-1 ?Gr = 82 x 103 J mol-1 Kr = 4.4×10-6 J mol-1 pN2 = (1/4) (1 – x) ptot pH2 = (3/4) (1 – x) ptot Kr = x2 ? 44 ? ? 33 ? ?? p0 ? ?? ? ?? p ? ?? tot ?
2

1-2 1-3 1-4 1-5

n[H2,(] =1500 mol s-1

1-6

(1 ? x )4

1-7

(1 ? x )4

x2

= 0.0418

x = 0.148

Problem 2
2-1 2-2

Myoglobin for Oxygen Storage

Kp = 100 Pa Volume of Mb: 19.6875 x 10 m -1 Molecular weight of Mb = 16.6 kg mol 0.012 mol kg
-1 -27 3

2-3 2-4

2 hours and 40 minutes.
31

Groningen | The Netherlands | 5 - 14 July 2002

2.5

(CH2)3n+6 (CO2)3 H2 + (4,5n + 9,5) O2



Problem 3
3-1

Lactose Chemistry

Answers in textbook

Formula

Formula

D-galactose

D-glucose

3-2
CH2OH O OH OH O OH O CH2OH OH OH HO OH CH2OH O OH O OH O CH2OH OH OH HO OH CH2OH O OH O OH O CH2OH OH OH

HO

Answer box a. 3-3
Sorbitol

Answer box b.

Answer box c

Galactitol

Answers in textbook

Formula

Formula

Optically active: yes / no

Optically active: yes / no

32

34th International Chemistry Olympiad | Preparatory Problems

3-4

Find answer in textbook

formula

formula

formula

Lactitol

Lactulose furanose form

Lactulose pyranose form

Lactulose is a mixture of the furanose and pyranose form in the ration 4 : 6. When a student gives either the furanose or the pyranose form, he/she will receive full marks.

Problem 4
4-1 4-2 4-3

Atom Mobility (Dynamics) in Organic Compounds

a >> c, d > b
y>x No

Problem 5
5-1

Towards Green Chemistry: The E-factor
E-factor = 1.15 E-factor = 0

Classical route: Atom utilization =0.47 or 47% Catalytic route: Atom utilization = 1 or 100%

5-2

Classical chlorohydrin route: Atom utilization = 0.25 or 25% Modern petrochemical route: Atom utilization = 1 or 100% Classical route: E-factor = 3.37 Modern route: E-factor = 0.49

Problem 6
6-1 Ba Sr
2+

Selective Solubility
(aq) + SO4 (aq) (aq) +
22SO4 -5 2-

The relevant equations are: BaSO4 (s) SrSO4 (s)
2+

(aq)

[SO4 ] = 3 x 10 M x 10-5 M 3 2+ -2 At the starting point the concentration of Ba was 10 M. This means that the loss amounts to 0.033% The separation meets the criterion. 6-2 The following equilibrium reactions have to be considered: AgCl (s) Ag+ (aq) + Cl- (aq) + Ag (aq) + 2 NH3 (aq) Ag(NH3)2+ (aq) [Ba2+] = 1

Total:

AgCl (s) + 2 NH3 (aq)

Ag(NH3)2+ (aq) + Cl- (aq)

Koverall = Ksp Kf = 2.6 x 10-3
Groningen | The Netherlands | 5 - 14 July 2002 33

If x is the molar solubility of AgCl (mol L ) then the changes in concentration of AgCl as the result of the formation of the complex ion are
AgCl (s) + 2 NH3 (aq) Starting point: Change: Equilibrium: 1.0 M -2x M (1.0 - 2x) M
+

-1

Ag(NH3)2+ (aq) + Cl- (aq) 0.0 M +x M +x M 0.0 M +x M +x M

Kf is quite large, so most of the Ag ions exist in the complexed form. + In absence of NH3 at equilibrium holds [Ag ] = [Cl ] + Complex formation leads to: [Ag(NH3)2 ] = [Cl ]
Koverall = x.x (1.0 - 2x)2

x = 0.046 M This result means that 4.6 x 10 M of AgCl dissolves in 1 L of 1.0 M NH3. Thus the formation of + the complex ion Ag(NH3)2 enhances the solubility of AgCl, because in pure water the molar -5 solubility amounts to only 1.3 x 10 M.
-2

Problem 7
7-1 7-2 7-3 7-4

UV-spectrometry as an Analytical Tool
-4 -1

cmax = 1.618 x 10 mol L . cmin = 0.95 x 10 mol L
-6 -1

The composition of the complex is ML2 For xM = 0: cM / cM + cL = 0, cM = 0 and cL = 1 For xM = 1: cM / cM + cL = 1, cM = 1 and cL= 0 M and L both absorb and have an absorption of AM = 1.0 and AL = 0.5, respectively. εM = 2 εL For xM = 0 32% has been transmitted For xM = 1 0% has been transmitted

7-5 7-6

Problem 8
8-1

Reaction Kinetics

Arrhenius equation: log k = log A - Ea /2.3RT we can substitute the values of k and T: Ea = 98.225 kJ mol
-1

8-2

The expression for s is: s = k2 K [NO] [O2]
2

d[NO2] = k2 [NO3] [NO] dt

8-3

b. The mechanism is correct.

34

34th International Chemistry Olympiad | Preparatory Problems

Problem 9
9-1 NaCl

Bonding and Bond Energies
Na+ + ClNa + Cl
-1

Born-Haber cycle for the dissociation of NaCl into Na + Cl:

Na+ + Cl-

Dissociation energy = 328 kJ mol 9-2

Born-Haber cycle for the dissociation of CaCl2 into Ca + 2 Cl: Ca2+ + 2 ClCaCl2 Ca2+ + 2 Cl Ca + 2 Cl
-1

Dissociation energy into atoms = 630 kJ mol .

Problem 10
10-1

The Nature of Phosphorus
Compound A Compound B

Spatial structures Et Me O P OH O Me H meso-II Et meso-I H formula

formula

formula

formula

(S,S)

(R,R)

Only (S,S) [derived from optically pure (S)-butan-2-ol]

Fisher projections Et Me O P OH O Me H meso-II Et meso-I H formula

Compound A

Compound B

formula

formula

formula

(S,S)

(R,R)

Only (S,S)

(Hint: You may wish to compare the 2 meso structures of 2,3,4-trihydroxypentane)

10-2

C: (CH3O)2P-OH D: [(CH3)2CHO]-P-OH E: (Ph-(CH3)CH2O)2P-OH

one signal one signal three signals as in 10-1 ratio 1:2:1 (meso-I : RR + SS : meso-II)

Groningen | The Netherlands | 5 - 14 July 2002

35

10-3

H CH3CH2C OH (b) CH3 (a) OH 10-4 CH CH2 CH3 a CH3b

An Ph

.. P

An .. P Ph

formula

meso not chiral not suitable as catalyst

(S,S) will give D(+) DOPA F is (R,R) and gives L(-) DOPA

10-5

Option 1 and option 3; P is the asymmetric center. Phosphorus compounds are pyramidal and they are configurationally very stable (no inversion).
PPh2 N H N PPh2 H

10-6

formula
meso compound not suitable as chiral catalyst

Ph H3C

Ph CH3

(S,S) will give D(+) DOPA G is (R,R) and gives L(-) DOPA

10-7 10-8

Option … and option …. One signal, substituents have the same chirality (R). No splitting.

Problem 11
11-1 From P

Optical purity
From Q O Ph O MeO Ph N H OMe

formula

CF3

11-2

Option …

36

34th International Chemistry Olympiad | Preparatory Problems

11-3

Spectrum

(a) ratio 1:3 Spectrum ratio 1:3

(b) ratio 1:1:3:3

Small splitting due to coupling with CH

11-4

Problem 12
12-1
4

Polylactic Acid
OH CH3 OH O HO * O CH3 + H2O O O * O O CH3 * O OH

H3C *

+

3 H2O

CH3 2

12-2
2

O

O

*

H3C * O

12-3 12-4

P=

1 =3 1? p

First the remaining amount of water at a chain length of 100 units is calculated: [Ester ][Water ] K =4= [- OH][- COOH] U = 10 and P = 100 Water formed : pU Water removed = 178 g of H 2 O

Problem 13
13-1 13-2 13-3 13-4 … … … …

A Chemical Puzzle

Groningen | The Netherlands | 5 - 14 July 2002

37

13-5

No AgNO3 reaction
Cl NO2 or N

no aliphatic chlorine substituent

formula

H

CH2CH3

Problem 14
14-1 14-2

Delft Blue and Vitamin B 12
2+

Electron configuration of Co :

dyz 14-3

dxy

dxz

dz

2

dx

2 2 -y

If 90% of the light is absorbed, the transmission T is 0.1 (10% transmitted). Fill out: A=-log I/I0 A= ε * c * l c = 0.5 M

14-4 All three oxidation states have unpaired d-electrons (d , d and d ) in the high spin configuration and thus for all three oxidation states an EPR spectrum can be measured. + Co yes / no 2+ Co yes / no 3+ Co yes / no 3 * 10
19 6 7 8

14-5

14-6 14-7

Co ions.

1 2 3 4 5

# 7 8

Problem 15
15-1

Synthesis of a local anaesthetic
formula
B O O2N OCH2CH2N(C2H5)2

formula
A

formula
C

formula

CH3CH2CH2O D E

38

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Problem 18
18-1 KA = KB = K'A = K'B =

Enzyme Kinetics

18-2

v=
If [A] If [A]

Vmax 1 + KA/[A]
0 ∞ v = Vmax [A]/KA. This corresponds with first order kinetics. v = Vmax. This corresponds with zero order kinetics.

18-3 18-4 18-5

A high affinity corresponds with a small KA. v = ? Vmax when [A] = KA.

18-6 figure

18-7 18-8

Maltose functions as a competitive inhibitor

Problem 19
19-1 a. b. c. d.

Dendrimers: Tree-like Macromolecules
NH3 +
H2/Pd

H 2C

CHCN

structure

structure

H2C R 40 N

C C N H2 H2 R

H2 C C C NH2 H2 H2 C C C NH2 H2 H2 H2 R = Same chains at these positions

34th International Chemistry Olympiad | Preparatory Problems

19-2

After the first cycle there are 3 amine groups. After 5 full cycles, the total number of amine endgroups is 48. a. After 5 full cycles 93 moles of H2 have been used. b. Idem for acrylonitrile (93 moles). 3 c. Radius is 25 ?. Volume: 4/3 π r .

19-3

Problem 20
20-1

Carvone

Number of C-atoms: nC = 10 Number of H-atoms: nH = 14 Number of O-atoms: nO = 1 Carvone has the formula C10H14O and 4 unsaturated sites. C=O group -OH (-CO2H is not a correct answer! Carvone only has one oxygen atom) Look up structure in a textbook or encyclopedia, then interpret the spectrum

20-2 20-3 20-4 20-5

Problem 21
21-1 21-2 21-3 At the cathode, half-reaction (1) At the anode, reaction (2). Anode: Cathode: Fuel cell reaction: 2 H2 (g) + O2 (g) → 2 H2O (g) The standard electrode potential of the reaction at the anode = 0 V The standard electrode potential of the reaction at the cathode = + 1.23 V ?Go = - n F E = -474,716 J CH4 (g) + 2 (O , electrolyte) → 2 H2O (g) + CO2 + 4 e 2O2 (g) + 4 e → 2 (O , electrolyte) …………………………………………………………………….. Anode: Cathode: Fuel-cell reaction is:
2-

21-4

21-5

21-6

2 H2 (g) + O2 (g) → 2 H2O (g)

Problem 22
22-1 Volume SDS micelle = 39911.33 ? 3 Volume of the core = 19160.77 ? 3 Volume of the Stern layer = 20750.56 ?
3

22-2

The equilibrium constant KM = .... Substitution in ?GM: ....
At the CMC there are no micelles: [M] =0 and [S] ≈ [B] thus: ?GM = 2 RT ln[S] -1 For SDS: ?GM = -23.86 kJ mol -1 For TDAB: ?GM = -21.01 kJ mol

Groningen | The Netherlands | 5 - 14 July 2002

41

22-3

Average number of amphiphiles per micelle For SDS (Mr = 288) = 62.5 For TDAB (Mr = 308) = 48.7

Problem 23
23-1 23-2 BBr3 + PBr3 + 3 H2 figure BBr3 planar and trigonal 23-3

→ BP + 6 HBr
figure PBr3 trigonal pyramidal

23-4

A FCC-structure of the B-atoms and that gives: Angular points: =1 Planes: =3 Total = 4 In each cell 4 phosphorus atoms are present which are tetrahedrally surrounded by boron. Atom masses of boron and phosphorus are 11 and 31, respectively. -3 R = 2554 kg m Distance B-P is 2.069 ? Lattice energy of BP: -1 = 8489 kJ mol The order of the reaction is 2 r = k [BBr3][PBr3] k800= 2272 L mol s k800= 9679 L mol s
2 -1 2 -1 -1

23-5

23-6 23-7

23-8

]23-9

-1 -1

23-10

ΔH = -R ln (k2/k1) x (1/T2-1/T1) -1 ΔH = 186 kJ mol

Problem 24
- the yield will be ca. 75%, mp =104-105 C 24-1 24-2 tert-butyl cation Methoxy group is strongly activating in electrophilic aromatic substitution reactions and will direct the tert-butyl to ortho-para positions.
MeO
t

o

OMe

formula
Bu
t

Bu much less likely due to steric hindrance

42

34th International Chemistry Olympiad | Preparatory Problems

Problem 25
- 2.5 mmole of diacid, ca 5 mL of 1.0 M NaOH is needed in procedure 1; ca 2.5 mL in procedure 2. - color changes: colorless to violet in procedure 1, red to yellow in procedure 2. 25-1 a: pKa phenolphthalein b: pKa methylorange pKa > 6.1 pKa > 1.8

25-2

Explanation according to option a

Problem26
- the yield will be ca 360 mg, m.p. = 125 C 26-1
C6H5C O H + H2NC6H5
o

Problem 27
- The yield will be ca. 64%, m.p. = 103.5-104.5 C 27-1 27-2 27-3 From the experiment From the experiment Catalytic cycle -O H
H O H H Pd H
o

+N H
H

H H N H

+

H O C O

Pd

H

Problem 29
29-1 29-2 29-3 29-4 Yes optically enriched optically pure When the enzyme is highly selective: no When the enzyme is not highly selective: yes. In this case the preferred enantiomer will be hydrolyzed very fast and the other enantiomer will be converted more slowly.

Groningen | The Netherlands | 5 - 14 July 2002

43

Sponsors of the 34th International Chemistry Olympiad

34th International Chemistry Olympiad | Nijenborgh 4 | 9747 AG Groningen | The Netherlands telephone +31 50 363 46 15 | fax +31 50 363 45 00 | e-mail: icho34@chem.rug.nl | www.chem.rug.nl/icho34

ISBN 90 806903 1 7


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