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2002年第34届ICHO预备试题答案(英文)1-29 2



34th International Chemistry Olympiad | Preparatory Problems

Worked Solutions to the Problems
Important general remark: For the sake of clarity, most answers are more elaborat

e than strictly necessary. In many cases the explaining text can be replaced by arrows, e.g. ? . Problem 1
1-1

Production of Ammonia

n[H2,"] = 3 x 1000:2 = 1500 mol s-1 n[N2,#] = 1000:2 = 500 mol s-1 n[CH4,$] = 1000:2 = 500 mol s-1 n[H2O,$] = 1000:2 = 500 mol s-1 n[CO,%] = 1000:2 = 500 mol s-1 n[O2,&] = ? x 1000:2 = 125 mol s-1 n[CO,'] = n[CO,%] - 2 n[O2,&] = 250 mol s-1 n[N2,(] + n[H2,(] = 2 n[NH3,(] = 2 n[NH3,!] n[N2,(] = 500 mol s-1 n[H2,(] =1500 mol s-1 ?Gr = 2G(NH3) - G(NH2) - 3 G(H2) ?Gr = [2 x 24.4 + 8.3 + 3 x 8.3] x 103 = 82 x 103 J mol-1 ?Gr = - RTln Kr , Kr = 4.4×10-6

1-2

1-3

1-4 1-5

pN2 = (1/4) (1 – x) ptot pH2 = (3/4) (1 – x) ptot Kr = x2 ? 44 ? ? 33 ? ?? p0 ? ?? ? ?? p ? ?? tot ?
2

1-6

(1 ? x )4

1-7

? 33 ?? p tot ? ? ?? ? = 0.0418 → x = (1 ? x )2 0.0418 = K r 4 4 ?? p ? ? (1 ? x ) ? 4 ?? 0 ? ? 0.204 x 2 + 1.408 x ? 0.204 = 0 → x = 0.148 x2

2

Problem 2
2-1

Myoglobin for Oxygen Storage
100 100 + K p

Take at the X-axis 100, this corresponds with 0.5 at the Y-axis, thus

0.5 =
2-2

50 +0.5 Kp = 100

Kp = 100 Pa
3 -27 3

Volume of Mb: VMb = 0.5 x 4.5 nm x 3.5 nm x 2.5 nm = 19.6875 nm = 19.6875 x 10 -27 23 -1 Molecular weight of Mb = VMb x NA = 19.6875 x 10 x 6.02 x 10 = 16.6 kg mol

m

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31

2-3 2-4

1 kg Mb is 1/16.6 mol; 20% is 0.2/16.6 = 0,012 mol kg 400 kJ mol
-1

-1

→ 400,000 x 0.012 J kg-1 of muscle tissue. W = J s-1 ? per second per kg of 400000 × 0.012 = 9,600 s = 2 hours and 40 minutes. muscle tissue 0.5 J. Time = 0.5 → (3n + 9) CO2 + (3n + 7) H2O

2.5

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

Problem 3
3-1

Lactose Chemistry

H

C

O

H

C

O

H C OH HO C H HO C H H C OH CH2OH D-galactose

H C OH HO C H H C OH H C OH CH2OH D-glucose

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

Answer box a. 3-3
Sorbitol CH2OH H C OH HO C H H C OH H C OH CH2OH

Answer box b.

Answer box c

Galactitol CH2OH H C OH HO C H HO C H H C OH CH2OH Optically active: yes / no

Optically active: yes / no

32

34th International Chemistry Olympiad | Preparatory Problems

3-4

CH2OH HO OH OH Lactitol O O

OH OH CH2OH OH CH2OH

CH2OH HO OH OH O O OH O CH2OH OH CH2OH HO

CH2OH O OH OH O

OH CH2OH OH O
OH

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
= 100 = 115 E-factor =115/100 = 1.15 E-factor = 0/100 = 0

The molecular weight of methyl methacrylate The molecular weight of NH4HSO4

Classical route: Atom utilization = 100/(100+115) =0.47 or 47% Catalytic route: Atom utilization = 115/115 = 1 or 100% 5-2

Classical chlorohydrin route: Atom utilization = 44/173 = 0.25 or 25% Modern petrochemical route: Atom utilization = 44/44 = 1 or 100% Classical route: E-factor = 133.4/39.6 = 3.37 product : 44 -10 % = 39.6 by-products : 111+ (10% of 62) + (18 - 10%) = 111 + 6.2 + 16.2 = 133.4 Modern route: E-factor = 18.6/37.4 = 0.49 product : 44 -15% = 37.4 by-products : 2 CO2 + 2 H2O per mole of C2H4 (15%) ? 2 x 15% of 44 + 2 x 15% of 18 = 18.6

Problem 6
6-1
2+

Selective Solubility
2-

The relevant equations are: Ba Sr (aq) + SO4 (aq) (aq) +
2SO4

BaSO4 (s) SrSO4 (s)

2+

(aq)

Precipitation of BaSO4 will start, when

[SO42-] =

Ksp(BaSO4) [Ba2+]

=

1 x 10-10 = 10-8 M 10-2

(1)

Precipitation of SrSO4 will start, when

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33

[SO42-] =

Ksp(SrSO4) [Sr2+]

=

3 x 10-7 = 3 x 10-5 M 10-2

(2)

If there are no kinetic complications (for example when the formation of BaSO4 would be very slow) first 2+ BaSO4 will be formed. This results in a decrease of the concentration of Ba ions. If the concentration 2+ 2SO4 satisfies equation (2), the concentration of Ba can be calculated from the formula: Ksp(BaSO4) = 1 x 10 = [Ba ][3 x 10 ] 1 10-10 [Ba2+] = = x 10-5 M -5 3 x 10 3 At the starting point the concentration of Ba
1/3 x 10-5 10-2 x 100% = 0.033%
2+ -10 2 -5

was 10 M. This means that the loss amounts to

-2

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)

Total:

AgCl (s) + 2 NH3 (aq)
[Ag(NH3)2+][Cl-] [NH3]2
-1

Ag(NH3)2+ (aq) + Cl- (aq)
= 1.7 x 10-10 x 1.5 x 10+7 = 2.6 x 10-3

Koverall = Ksp Kf =

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
+

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 can be written as: x .x K overall = 2.6 x 10-3 = 2 (1.0 - 2x ) and 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

x2 (1.0 - 2x )2

or 0.051 =

x (1.0 - 2x )

Problem 7
7-1

UV-spectrometry as an Analytical Tool
10

2% decrease in light intensity implies: I/I0 = 98/100 and A = - log 98/100 = 0.01 -6 -1 This absorption corresponds with 0.01 = 10500 x cmin x l ? cmin = 0.95 x 10 mol L . 2% light throughput means that: I/I0 = 2/100 and A =
10

7-2

log 100/2 = 2 - 0.3010 = 1.6990

For ferroin this absorption at 512 nm corresponds with: -4 -1 1.6990 = 10500 x cmax x l ? c max = 1.618 x 10 mol L . 7-3 Minimum in curve at xM = 0.33: cM = 0.33 cM + 0.33 cL or 0.67 cM = 0.33 cL The composition of the complex is ML2

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34th International Chemistry Olympiad | Preparatory Problems

7-4

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. xM = 0, cL = 1, AL = εL cL l xM = 1, cM = 1, AM = εM cM l l has the same value in both formulas, just as cL and cM, so that εL / εM = AL / A M = 0.5 / 1, thus εM = 2 εL For xM = 0 holds 0.5 = -log I/I0, For xM = 1 holds 1 = -log I/I0, I/I0 = 0.32 thus 32% has been transmitted I/I0 = 0.1 thus 10% has been transmitted

7-5

7-6

Problem 8
8-1

Reaction Kinetics

Due to the Arrhenius equation: log k = log A - Ea /2.3RT we can substitute the values of k and T: log k1 = log A - Ea /2.3RT1 Subtraction gives: and log k2 = log A - Ea /2.3RT2

Ea ? 1 1 ? ? + ? ? 2.3R ? ? T1 T2 ? T1T2 k 300 × 400 4.9 ? 10 ?4 Ea = 2.3R ( ) log 2 = 2,3 x 8.314 ( ) log (T2 ? T1 ) 400 ? 300 k1 2.6 ? 10 ?8
log k1 - log k2 = Ea = 98.225 kJ mol 8-2
-1

The slow step is rate determining; this is the second reaction. d[NO2] = k2 [NO3] [NO] The expression for s is: dt [NO3] k The equilibrium gives: K = 1 = [NO][O ] 2 k -1 Rewritten this is: [NO3] = K [NO] [O2] 2 Substitution gives for s: s = k2 K [NO] [O2]

8-3

b. The mechanism is correct.

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-

The energy loss in the first step is 464 kJ mol -1 The energy gain in the second step is –(I.E. of Na + E.A. of Cl) = -136 kJ mol -1 Overall loss = 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
2+ -1

The (ionic) bond energy of Ca Cl = -429 x 2 / 0.91 = -943 kJ mol (The measured value for CaCl is –429, but the charge of Ca is now +2 and the bond length has decreased by a factor of 0.91).
35

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The energy loss in the first step = -(bond energy of CaCl2) = 2 x 942 minus the Cl-Cl repulsion. -1 The Cl-Cl repulsion is (429 / 2)x(1 / 0.91) = 236 kJ mol , so the energy loss in the first step = -1 +1650 kJ mol . -1 The energy gain in the second step is –(2 x E.A. of Cl + total I.E. of Ca) = -1020 kJ mol . -1 Dissociation energy into atoms = 630 kJ mol .

Problem 10
10-1

The Nature of Phosphorus

Spatial structures Et Me O H Me O HO P : O H Me Et

Compound A Et H H O Me Me O Et H

Compound B Et H O Me

: P OH
O Me

: P OH
O H Me H H

: P OH
O Me Me

: P OH
O H Et Only (S,S) [derived from optically pure (S)-butan-2-ol]

Et Et meso-I meso-II P is pseudo-asymmetric center diastereomeric structures different 31P signals Fischer projections Et Me O H Me O HO P : O H Et meso-I Me Et meso-II H

Et Et (S,S) (R,R) P is not an asymmetric center in (S,S) and (R,R); enantiomers have same 31P signals (this is the signal in the middle)

Compound A Et H H O Et Me Me O Et H

Compound B Et H O Me

: P OH
O Me

: P OH
O Me Et (S,S) H H

: P OH
O Me Et (R,R) Me

: P OH
O H Et 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)

10-3

H CH3CH2C OH (b) CH3 (a) OH CH CH2 CH3 a CH3b

36

34th International Chemistry Olympiad | Preparatory Problems

10-4

An Ph

.. P

An .. P Ph

Ph An

.. P

An ..P Ph

meso not chiral not suitable as catalyst

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

It should be noted that this is extra information, which was not part of the question. The essence is that the nonchiral meso isomer is recognized 10-5 Option 1 and option 3; P is the asymmetric center. Phosphorus compounds are pyramidal and they are configurationally very stable (no inversion).

10-6
PPh2 H3C Ph H N N PPh2 H Ph CH3 PPh2 Ph H3C H N N PPh2 H Ph CH3

meso compound not suitable as chiral catalyst

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

10-7 10-8

Option 2 . 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 Ph O MeO Ph N H H O OMe

CF3

11-2 11-3

Option 2

(a) ratio 1:3

(b) ratio 1:1:3:3 Long range coupling of the CH3O protons with the CH is very small (close to zero). The essence of the question is to indicate the peak ratio’s.

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37

11-4

ratio 1:3

Problem 12
12-1
4

Polylactic Acid
OH CH3 OH O HO * O CH3 + H2O O O * O O CH3 * O O O O * CH3 O
2

H3C *

OH

+
CH3 * O

3 H2O

CH3 2 CH3 HO *

12-2
2

O

O

*

OH

H3C * O

12-3

The aim here is working with reactive groups. [Ester ][Water ] = p 2U2 = 4 ? p K= =2 2 2 [- OH][- COOH] (1 ? p ) U 1? p

?

p = 2/3 ?

P=

1 =3 1? p

12-4

First the remaining amount of water at a chain length of 100 units is calculated: ? [Ester ][Water ] = p 2UW K =4= ? 2 2 [- OH][- COOH] (1 ? p ) U ? ? ? ? W = 0.0040404 mol ? 1 ? p = 0.99 ? U = 10 and P = 100 = 1? p ? ? Water formed : pU = 0.99x10 = 9.9 mol Water removed : 9.9 ? 0.0040404 = 9.896 mol = 178 g of H2 O

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

A Chemical Puzzle

Nitrogen atom -NH2 or -NHR -CH2CH3 -NO2

38

34th International Chemistry Olympiad | Preparatory Problems

13-5

No AgNO3 reaction
Cl NO2 or N

no aliphatic chlorine substituent
Cl H N CH2CH3

H

CH2CH3

NO2

Problem 14
14-1 14-2

Delft Blue and Vitamin B 12
2+ 2 2 6 2 6 7

Electron configuration of Co : 1s 2s 2p 3s 3p 3d

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 ? -log 0.1=20*c*0.1 ? c = 1/(20*0.1) = 0.5 M
+

14-4 14-5

r(Co ) < r(Co ) < r(Co ). 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 ) / (58.93 * 1.67 * 10 ) = 3 * 10 Co ions. 1 2 3 4 5
-6 -27 19 6 7 8

3+

2+

14-6 14-7

# 7 8

Problem 15
15-1

Synthesis of a local anaesthetic
O O OH O2N CH3CH2CH2O A O O2N OCH2CH2N(C2H5)2 H2N CH3CH2CH2 O D E 39 B O OCH2CH2N(C2H5)2 OH O2N CH3CH2CH2O C O Cl

O2N HO

CH3CH2CH2O

Groningen | The Netherlands | 5 - 14 July 2002

15-2
NO2 O OH and O 2N OH OH O OH

15-3

Decomposition of the tert-C4H9Cl.

Problem 16
16-1
Ph H2N O FA H N

Structure of peptides
Ph O OH H2N O AF H N O OH Ph H2N O FF H N

O OH Ph H2N O AA H N

O OH

16-2

This is a nucleophilic aromatic substitution
NO2 O2N F + 2H2NR O2N NO2 NHR + + RNH3 F

16-3
H N O O N H H N O O OH Ph

H2N O

H N

O N H GALF

H N O

O H2N OH Ph

GLAF

Only the positions of G and F are determined, the other two are in the middle, but no information is provided if this is AL or LA.

Problem 17
17-1
S

Ribonuclease
S + HSCH2CH2OH

SH

S-SCH2CH2OH

A First product + HSCH2CH2OH SH SH

+ HOCH2CH2S-SCH2CH2OH C

B Final product

17-2 17-3

Electrostatic forces, hydrogen bonds and van der Waals forces. There are 8 Cys residues. The probability that any residue is coupled to its correct partner is 1:7. Next there remain 6 residues to consider. The chance that any of those is coupled to its correct partner is 1:5, etc. Therefore the fraction of active molecules is: 1/7 x 1/5 x 1/3 x 1/1 = 1/105.

40

34th International Chemistry Olympiad | Preparatory Problems

Problem 18
18-1

Enzyme Kinetics

KA = [E] [A]/[EA] KB = [E] [B]/[EB] K'A = [EB] [A]/[EAB] K'B = [EA] [B]/[EAB]

18-2

v=

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

18-3 18-4 18-5

If [A] If [A]

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

18-6

18-7 18-8

Maltose functions as a competitive inhibitor

Problem 19
19-1 a.

Dendrimers: Tree-like Macromolecules
NH3 +
H2/Pd

H2 C

CHCN

N(CH2CH2C N)3

b. c. d.

N(CH2CH2CH2NH2)3

N[CH2CH2CH2N(CH2CH2CN)2]3 H2 C C C NH2 H2 H2 R = Same chains at these positions C C C NH2 H2 H2 H2 41

H2C N R

C C N H2 H2 R

Groningen | The Netherlands | 5 - 14 July 2002

19-2

After the first cycle there are 3 amine groups (see answer). Then, the number of amine groups is doubled after each cycle (see answer 19-1c and d). Thus, after 5 full cycles, the total number of amine end-groups is 48. a. After 5 full cycles 3 + (3 x 2) + (6 x 2) + (12 x 2) + (24 x 2) = 93 moles of H2 have been used. b. Idem for acrylonitrile (93 moles). 3 c. Radius is diameter/2 = 50/2 = 25 ?. Volume: 4/3 π r .

19-3

Problem 20
20-1

Carvone

Number of C-atoms: nC = (Mr x %C)/12 = (150 x 0.8)/12 = 10 Number of H-atoms: nH = (Mr x %H)/1 = (150 x 0.0933)/1 = 14 Number of O-atoms: nO = (Mr x %O)/16 = (150 x 0.1067)/16 = 1 A fully saturated hydrocarbon with 10 carbon atoms has the formula C10H22 Carvone has the formula C10H14O, for calculation of the unsaturation, the O is not relevant. Subtraction gives a shortage of 8H for carvone, this is equivalent to 4 unsaturated sites (either double bonds or rings). C=O group -OH (-CO2H is not a correct answer! Carvone only has one oxygen atom). There is no relevant -1 strong absorption above 3000 cm . That means no -OH group present. Carvone is a 6-membered ring, this leaves three more unsaturated sites. The IR show the presence of a C=O group, this leaves 2 more unsaturated sites, these must be C=C bonds. The strong UV-absorption suggests a conjugated system, most likely C=C-C=O. 1 The singlets at 1.63 and 1.68 ppm in the H-NMR are two–CH3 groups with no vicinal coupling. The multiplet from 1.9-2.2 ppm consists of a –CH and a –CH2- group (from the question) The multiplet from 2.2-2.5 ppm is most likely a –CH2- group with many neighboring H-atoms. The doublets at 4.75 and 4.93 ppm are indicative of two =CH- groups, it might even be =CH2 given the small and identical coupling constants (see the enlarged area). The triplet at 6.73 ppm is indicative of a =CH- group which is situated next to a –CH2- group. Combining all this information with the given 1,2,4 substitution pattern gives the structure below as the most likely structure of carvone. CH3 HC H 2C H H 3C C C C C O

20-2

20-3 20-4

20-5

CH2 CH2

Problem 21
21-1 21-2 21-3

Experiment

At the cathode oxygen is reduced to hydroxide, i.e., half-reaction (1) At the anode hydrogen is oxidized to water, i.e. reaction (2). Anode: 2 H2 (g) → 4 e + 4 H + Cathode: 4 H + 4 e + O2 (g) → 2 H2O (g) Fuel cell reaction: 2 H2 (g) + O2 (g) → 2 H2O (g)
+

21-4

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 The total number of electrons transferred in the reaction = 4

42

34th International Chemistry Olympiad | Preparatory Problems

?Go = - n F E = -4 x 96487 C x (1.23 V – 0 V) = -474,716 J 21-5 CH4 (g) + 2 (O , electrolyte) → 2 H2O (g) + CO2 + 4 e 2O2 (g) + 4 e → 2 (O , electrolyte) CH4 (g) + 2 O2 (g) → 2 H2O (g) + CO2 (g)
2-

21-6

Anode: Cathode: Fuel-cell reaction is:

2 H2 (g) + 2 CO3 (l) → 2 H2O (g) + 2 CO2 (g) + 4 e 2O2 (g) + 2 CO2 (g) + 4 e → 2 CO3 (l) (cathode) 2 H2 (g) + O2 (g) → 2 H2O (g)
2-

-

Problem 22
22-1

Experiment
3 3

Volume SDS micelle = 4/3 π (16.6 + 4.6) = 39911.33 ? 3 3 Volume of the core = 4/3 π (16.6) = 19160.77 ? 3 Volume of the Stern layer = volume SDS micelle – volume core = 20750.56 ? [M] [S]n[B]n [M] RT RT = - n (ln[M] - nln[S] - nln[B]) ln n [S]n[B]n

22-2

The equilibrium constant KM = Substitution in ?GM: ?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 22-3 Average number of amphiphiles per micelle = relative micelle mass / relative amphiphile mass 3 For SDS (Mr = 288): n =18 x 10 / 288 = 62.5 3 For TDAB (Mr = 308): n =15 x 10 / 308 = 48.7

Problem 23
23-1 23-2

Experiment → BP + 6 HBr

BBr3 + PBr3 + 3 H2

23-3

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43

23-4

A FCC-structure of the B-atoms and that gives: Angular points: 8 x 1/8 =1 Planes: 6 x 1/2 =3 Total = 4 In each cell 4 phosphorus atoms are present which are tetrahedrally surrounded by boron.

23-5

23-6 23-7

Atom masses of boron and phosphorus are 11 and 31, respectively. 4 × 42 m (n Mbp ) = = 2554 kg/m 3 R= = 3 V Na 6.022 ? 10 26 4.78 ? 10 ?10 1/2 Distance B-P is 1/2 x 3 x 1/3 x a = 2.069 ?

(

(

))

Lattice energy of BP:

=-

1390 × 3 × 3 × 1.638Z + Z - Ae 2 6 = 8489kJ/mol 2.069 7

23-8 23-9

The order of the reaction is 2 r = k [BBr3][PBr3]
2 -1 -1 mol s -1 = 2272 L mol s mol 2L-2 -1 -8 -6 -6 mol s 2 -1 -1 = 9679 L mol s k880 = r880 / [BBr3][PBr3] =19.60 10 / 2.25 10 x 9 10 2 -2 mol L -1 ΔH = -R ln (k2/k1) x (1/T2-1/T1) T1= 880 + 273 = 1073 K and T2 = 880 + 273 = 1153 K -1 -1 ΔH = -R ln (9679 / 2272) x (1/1153 -1/1073) = 186 kJ mol

k800 = r800 / [BBr3][PBr3] = 4.60 10 / 2.25 10 x 9 10

-8

-6

-6

23-10

Problem 24

Experiment
o

- the yield will be ca. 75%, mp =104-105 C 24-1 24-2 tert-butyl cation: (CH3)3C
+

Methoxy group is strongly activating in electrophilic aromatic substitution reactions and will direct the tert-butyl to ortho-para positions.
t

Bu OMe
t

MeO
t

OMe
t

MeO

Bu

Bu

Bu

much less likely due to steric hindrance

Problem 25

Experiment

- 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

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34th International Chemistry Olympiad | Preparatory Problems

Problem 26

Experiment
o

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

C6H5CH

NC6H5

Problem 27

Experiment
o

- 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

+N H
H

H N H H + O C O

H Pd

H

Problem 29
29-1 29-2 29-3 29-4 Yes

Experiment

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

45



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