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2014年第46届IChO国际化学奥林匹克预备题答案

时间:2015-04-27


SOLUTIONS TO Preparatory Problems
46th International Chemistry Olympiad (IChO - 2014)
Editorial Board Nguyen Tien Thao, Editor in Chief Nguyen Minh Hai Nguyen Van Noi Truong Thanh T

u Hanoi University of Science, Vietnam National University, Hanoi Tel: 0084 435406151; Fax: 0084 435406151 Email: icho2014prep@hus.edu.vn

July 17th, 2014

46th International Chemistry Olympiad Hanoi, Vietnam – 2014

Preparatory Problem Solutions

PART I. THEORETICAL PROBLEMS
Problem 1. Polar and non-polar molecules
1. The net dipole moment ? is calculated as follows:
B ?1 α ?

A

?2 C

? 2 = ?12 + ? 22 + 2 ?1? 2 cos α

(1)

2. 2.1 The geometry of CO2 : other
2

O

C

O

Because two bond moments of ?CO have opposite directions and cancel each out, the net dipole moment for CO2 is zero. Therefore:

?CO = 0
H H

2.2 The geometry of H2S:

α==104 92 5 α

o0

O S

From the general equation (1),

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2

Preparatory Problem Solutions
α
2

2 2 2 2 2 2 ?H S = ? HS + ? HS + 2 ? HS ? HS cosα = 2 ? HS (1 + cosα ) = 4 ? HS cos

?H S = 2 ? HScos
2

α
2

Therefore,

?H S
2

2.61×10?30 92 = 2× × cos = 1.09 D ?30 3.33 ×10 2

3. 3.1
H
120
0

H
sp2 sp2

C

O H

H

3.2 Since χ C > χ H , ? C? H has the direction showed in the above plot, and
? H ?C ? H = 2? C? H cos 120 = 2 × 0.4 × 0.5 = 0.4 D 2

? C = O also has the direction toward the O atom. Therefore, the net dipole moment ?

of the molecule is
?HCH = ?HCH + ?C=O = 0.4 + 2.3 = 2.7 D

4. We can plot the geometry of the three molecules involved in this problem in the following scheme:
H O 1050 H O CH3 1100 CH3 O CH3

α α?
H

H2O

CH3-O-CH3

CH3-O-H

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Preparatory Problem Solutions

- The dipole moment ? is a vector which can be calculated by adding individual bond moments ?1 and ?2
2 ? 2 = ?12 + ? 2 + 2 ?1? 2 cos α

(1)

α is the angle formed by the individual bond moments. - The dipole moment ? in the water molecule with the bond angle α formed by the two bond moments of O-H can be calculated as follows. From equation (1) we have
2 2 ? = ( 2 ?OH + 2?OH cos α ) 2 =? ? 2?OH (1 + cos α ) ? ? 1/2 1/2

α? ? 2 cos 2 ? = ? 4 ?OH 2? ?

1/2

= 2?OH cos

α
2

Given α = 105o, the bond moment ?OH in water can be calculated:
1.84 = 2 ?OH cos 105 → ?OH = 1.51 D 2

Similarly, we can calculate the bond moment for O-CH3 in dimethylether:
1.29 = 2 ? OCH3 cos 110 → ?OCH3 = 1.12 D 2

- In methanol, the individual bond moments are given as ?1 = ?OH and ?2 = ?OCH as in water and dimethylether. The bond angle α is formed by the two
3

individual bond moments. From equation (1), cosα is:
cos α =

(?

2

?? ??
2 1

2 2

2 ×1.51×1.12 α = 101 57
o

(1.69 =

2

? 1.512 ? 1.122 )

2?1? 2

)


CH3

O

α=? / H

Therefore, the bond angle C – O – H in methanol is of 101o57 46th IChO Preparatory Problem Solutions, Hanoi, Vietnam, July 2014 95

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Preparatory Problem Solutions

Problem 2. Calculations of lattice energy of ionic compounds
1. 1.1 Lithium reacts with water: 2 Li(s) + 2 H2O(l) 2 Li(s) + Cl2(g) 2 Li+(aq) + 2 OH-(aq) + H2(g) 2 LiCl(s) 2 Li+(aq) + SO42-(aq) + H2(g) 2 LiHSO4 + SO2(g) + 2 H2O(l) 1.2 Lithium reacts with chlorine: 1.3 Lithium reacts with sulfuric acid: With dilute sulfuric acid: 2 Li(s) + H2SO4(aq) With concentrated acid: 2 Li(s) + 3 H2SO4(c) 2. 2.1 To calculate Uo in accordance with Born-Haber cycle, the following cycle is constructed:
Ag(r) Li (s) +
?SH Li(g) Ag(k)

1 Cl2 (g) (k) 2

?H fH ? f

LiCl (s) AgCl

1 1 ?DH ?HD 2 2

U0 I E
Li (g) (k) Li Ag (k)
+ +

+

Cl_ (g)

2.2 Based on this cycle and Hess’s law, we have:
?f Η = ?S Η + 1 ?D Η + I + Ε + U 0 2

or

1 ? ? U0 = ? f H ? ? ?S H + ?D H + I + E ? 2 ? ?

After converting all the numerical data to the same unit, we have: U0 = - 402.3 – 159 – 121 – (5.40 – 3.84)×1.6×10-19×10-3×6.022×1023 U0 = - 832.56 kJ/mol. 96

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 3. U0 = - 287.2
Z+ Z?

Preparatory Problem Solutions

∑ν

r+ + r ?

? 0.345 ? ? ?1 ? r + r ? ? + ? ? ?

For LiCl crystal, we have:

U o = ?287.2

1 × 1× 2 ? 0.345 ? ?1 ? ? = ?201.43 kcal/mol 0.62 + 1.83 ? 0.62 + 1.83 ?

To conveniently compare the results, we convert the obtained result to SI units:
U o = – 201.43 × 4.184 = – 842.78 kJ/mol

4. According to the Born-Haber cycle and Kapustinskii empirical formula for lithium chloride crystal structure, both methods are close to the experimental value. 5. 5.1 The geometry diagram for octahedral holes is shown below. X

RClrLi+ where, R and r are the radii of Cl- and Li+ ions, respectively. Based on the diagram, we have: cos 45o = 0.707 =
2R R = 2R + 2r R + r

R → R+r

R = 0.707(R + r) → r = 0.414 R

- The body edge length of the unit cell LiCl = 2R + 2r = 5.14 ? 2R + 2(0.414 R) = 5.14 ? → R = 1.82 ? (radius of Cl-) 46th IChO Preparatory Problem Solutions, Hanoi, Vietnam, July 2014 97

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Preparatory Problem Solutions

2(1.82 ?) + 2r = 5.14 ? → r = 0.75 ? (radius of Li+) 5.2 Based on the experimental and theoretical data for the radii of Li+ and Cl- ions, it is realized that: ? Both calculated radii of lithium and chloride ions are close to the experimental values.

? Only the calculated radius of lithium ion is close to the experimental value. ×

? Only the calculated radius of chloride ion is close to the experimental value.

Problem 3. A frog in a well
1. 1.1 The general expression is given by:
2 2 h2 ?? N ? ? N? ? h2 ?ELUMO?HOMO = N+1) ?? +1? ?? ? ? = 2( 8mL2 ? ?? 2 ? ? 2 ? ? ? 8mL

E5
?E

LUMO HOMO

E4 E3 E2 E1

h2 ?ELUMO?HOMO = ( N+1) 8mL2

(1)

1.2 From Planck’s quantum theory:
?E =
hc =

hc

λ

(2) λ can be given by:

λ

h2 8mc L2 N + 1 → λ = × ( ) 8mL2 h ( N + 1)

( 3)

2. 2.1 For BD: L = (2×2 +1)0.140 nm = 5×0.140×10-9 m = 7×10-10 m = 7.0 ? For HT: L = (2×3 +1)0.140 nm = 7×0.140×10-9 m = 9.8×10-10 m = 9.8 ? Chemistry: The flavor of life 98

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Preparatory Problem Solutions

For OT: L = (2×4 +1)0.140 nm = 9×0.140×10-9 m = 12.6×10-10 m = 12.6 ? 2.2 From the general equation (3), the wavelength λ for each of the dyes are given:
λ=
8mc L2 8(9.11× 10?31 )(3 × 108 ) L2 L2 12 × = = 3.30 × 10 h 6.626 × 10?34 ( N + 1) ( N + 1) ( N + 1) L2 (7 × 10?10 ) 2 = 3.30 × 1012 = 3.234 × 10?7 m = 323.4 nm ( N + 1) ( 4 + 1)
L2 (9.8 × 10?10 ) 2 = 3.30 × 1012 = 4.528 × 10?7 m = 452.7 nm ( N + 1) ( 6 + 1) L2 (12.6 × 10?10 ) 2 = 3.30 × 1012 = 5.82 × 10?7 m = 582.0 nm ( N + 1) (8 + 1)

BD: λ = 3.30 ×1012 HT: λ = 3.30 ×1012 OT: λ = 3.30 ×1012

3. The box length can be calculated based on the geometry of the C – C – C chain as follows: C lC-C C
d 2

60o C d
d which 2

The box length is a combination of a number of the length of is given by
d = lC-C× sin60 = (0.140×10-9)×sin 60 = 1.21×10-10 m. 2

Therefore, the box length for the three dye molecules can be calculated as follows: For BD, the box length is consisted of 5 lengths of
d : 2

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Preparatory Problem Solutions

BD:

L = 1.21×10?10 m × 5 = 6.05×10?10 m = 6.05 ?
d : 2

Similarly, the box length for HT has 7 lengths of HT:

L = 1.21×10?10 m × 7 = 8.47×10?10 m = 8.47 ?
d : 2

The box length for OT has 9 lengths of OT: 4. 4.1 From equation (3), λ =

L = 1.21×10?10 m × 9 = 10.89×10?10 m = 10.89 ?

8mc L2 , and therefore: × h ( N + 1)

L=

λ × h × ( N + 1)
8mc

For BD:
L=

λ × h × ( N + 1)
8mc

=

(328.5 × 10 ?9 )(6.626 × 10 ?34 )5 = 7.06 × 10 ?10 = 7.06 ? ?31 8 8(9.11× 10 )(3 × 10 )

For HT:
L=

λ × h × ( N + 1)
8mc

=

(350.95 × 10 ?9 )(6.626 × 10 ?34 )7 = 8.63 × 10 ?10 = 8.63 ? 8(9.11 × 10 ?31 )(3 × 108 )

For OT:
L=

λ × h × ( N + 1)
8mc

=

(586.1× 10 ?9 )(6.626 × 10 ?34 )9 = 12.64 × 10 ?10 = 12.64 ? 8(9.11 × 10 ?31 )(3 × 108 )

4.2 The following table shows the values of the box length for the Chemistry: The flavor of life 100

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Preparatory Problem Solutions

investigated dyes calculated with different methods. L Substance L=(2k+1)0.140 nm (1) calculated based on the bent chain (2) BD HT OT 7.0 9.8 12.6 Method (1) is the best fit Method (2) is the best fit Method (3) is the best fit All methods (1), (2) , (3) are best fit × 6.05 8.47 10.89 L calculated from λexp. (3) 7.06 8.63 12.64 7.66 8.64 Experimental L

Problem 4. Electron in a 2 or 3 – Dimensional Box 1. 1.1 ?E = n 2
h2 h2 h2 2 ? 1 = (n 2 ? 12 ) 8mL2 8mL2 8mL2

1.2 According to Planck’s equation:
?E = hc (6.626 ×10-34 J s)(2.9979 ×108 m/s) = = 1.446 ×10-20 J λ 1.374 ×10-5 m

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 So
?E = 1.446 × 10-20 J =

Preparatory Problem Solutions

(6.626 ×10-34 J s) 2 8(9.109 ×10-31 kg)(10.0 ×10?9 m) 2

(n

2

?1

)

1.446 × 10-20 J = 6.025 × 10-22 n 2 ? 1

(

)

→ n 2 -1= 24.00 → n 2 = 25.00 → n = 5.00

n=5 2. 2.1 The quantum numbers are: Ground state (E11) First excited state (E21) Second excited state (E12) → → → nx = 1, ny = 1 nx = 2, ny = 1 nx = 1, ny = 2

Since the energy levels, Exy, are inversely proportional to L2, then the nx = 2, ny = 1 energy level will be lower than the nx = 1, ny = 2 energy level since Lx > Ly. The first three energy levels, Exy, in order of increasing energy are: E11 < E21 < E12 2.2 Calculate the wavelength of light necessary to promote an electron from the first excited state to the second excited state. E21 → E12 is transition. E xy
E 12 = E 21
h2 = 8m ? n2 ? n2 y x + ? ? L2x L2y ? ? ? ?

h2 ? 12 22 + ? 8 m ? (8.00 × 10 ?9 m) 2 (5.00 × 10 ?9 m) 2 h2 ? 22 12 = + ? 8 m ? (8.00 × 10 ?9 m) 2 (5.00 × 10 ?9 m) 2

? 1.76 × 1017 h 2 ?= 8m ? ? 1.03 × 1017 h 2 ?= 8m ?

1.76 × 1017 h 2 1.03 × 1017 h 2 7.3 × 1016 h 2 ? = 8m 8m 8m 16 ?2 ? 34 2 (7.3 × 10 m )(6.626 × 10 Js) ?E = = 4.4 × 10 ?21 J 8(9.11 × 10 -31 kg) ? E = E 12 ? E 21 = λ= hc (6.626 × 10 -34 Js) (2.998 × 10 8 m/s) = = 4.5 × 10 -5 m ?E 4.4 × 10 -21 J

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 3. 3.1 E =
2 2 (n12 + n2 + n3 )h2 n 2 h2 = = 6.173 ×10?21 J 2 2 8mL 8mL 2 8mL n2 = 2 E h

Preparatory Problem Solutions

If L3 = 8.00 m3, then L2 = 4.00 m2
h2 (6.626 ×10?34 )2 = = 2.582 ×10?43 J 2 0.032 8mL ? ? 8? 4 23 ? 6.022 × 10 ? ? 6.173 ×10?21 = 2.39 × 1022 ; n = 1.55 × 1011 n2 = ?43 2.582 ×10

3.2 ?E = En +1 ? En = E1.55×10

11

+1

? E1.55×1011

h2 h2 11 ?E = (2n + 1) = [2(1.55 × 10 ) + 1] = 8.00 ×10?31 J 2 2 8mL 8mL

4. The energy levels are
En1 ,n2 ,n3 =
2 2 (n12 + n2 + n3 )h2 2 2 = E1 (n12 + n2 + n3 ) 2 8mL

where E1 combines all constants besides quantum numbers. The minimum value for all quantum numbers is 1, so the lowest energy is E1,1,1 = 3E1 The question asks about an energy 21/3 times this amount, namely 21E1. This energy level can be obtained by any combination of allowed quantum numbers such that
2 2 (n12 + n2 + n3 ) = 21 = 4 + 2 + 1

2

2

2

The degeneracy, then is 6, corresponding to (n1, n2, n3) = (1, 2, 4), (1, 4, 2), (2, 1, 4), (2, 4, 1), (4, 1, 2), or (4, 2, 1).

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Preparatory Problem Solutions

Problem 5. Tug of war
1/ 2 PSO2 PO 2

1. ?rnx G = ?rnx G + RTln

o

PSO3
1/ 2 PSO2 PO 2

At equilibrium: ?rnx G = ?rnx Go + RTln ?rGo = - RTln
1/ 2 PSO2 PO 2

PSO3

=0

PSO3

= - RTlnKp1

T /K = T/oC + 273; T/ K lnKp1 800 -3.263 825 -3.007 900 - 1.899 953 -1.173 100 -0.591

2. Plot lnKp against 1/T:
0 -0.5 -1 -1.5 lnKp1 -2 -2.5 -3 -3.5 -4 9.0E-04 9.5E-04 1.0E-03 1.1E-03 1.1E-03 1.2E-03 1.2E-03 1.3E-03 1.3E-03 1/T (K) y = -10851x + 10.216 R = 0.9967
2

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Preparatory Problem Solutions

Assuming that ?rHo is temperature independent, the slope of this plot is -?rHo/R, so that ?rnxHo = 90.2 kJ/mol. 3. 2SO3(g) 2SO2 (g) + O2 (g) (2)

A best-fit equation is lnKp1 = - 10851(1/T) + 10.216 with R-squared value of 0.9967. We can use this equation to estimate the Kp1 at (651.33 + 273) = 924.33 K because ?rnxHo is temperature independent. lnKp1 = -10851(1/924.33) + 10.216 → lnKp1 = -1.523313881 → Kp1 = 0.218 For reaction (2), the equilibrium constant is expressed as:
K p2 =
2 PSO PO2 2 2 PSO 3

= ( K p1 ) 2 = (0.218) 2 = 0.047524

4.

Reaction (3): Decomposition: Equilibrium: Reaction (4): Initial P: Change Equilibrium a/2 = 0.028 atm
2 PSO PO2 2

2 FeSO4 (s) 2 SO3(g) P -a P-a

Fe2O3 (s) + SO3 (g) + SO2 (g) P-a 2SO2 (g) + P +a P+a O2 (g) 0 +a/2 a/2 P+a

At equilibrium: partial pressure of oxygen = 21.28/760 = 0.028 atm → a = 0.056 atm Equilibrium constant for (4):
K p4 = P
2 SO3

= ( K p1 ) 2 = (0.218) 2 = 0.047524 ( P + a) 2 (a / 2) ( P + 0.056)2 0.028 = = 0.047524 ( P ? a)2 ( P ? 0.056)2

K p4 =

2 PSO P 2 O2 2 PSO 3

=

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014
( P + 0.056) 2 0.028 = 0.047524 ( P ? 0.056) 2

Preparatory Problem Solutions

→ →

( P + 0.056) 2 = 1.6973 ( P ? 0.056) 2 ( P + 0.056) = 1.303 ( P ? 0.056)

→ P + 0.056 = 1.303P - 0.073 → P = 0.425 atm 2 FeSO4 (s) Fe2O3 (s) + SO3 (g) + SO2 (g)

→ 0.303P = 0.12896

Equilibrium constant for (3)

Kp3 = PSO3PSO2 = (P-a)(P+a) = (0.425 - 0.056)(0.425 + 0.056) = 0.177 5. Calculate the percent of FeSO4 decomposed? Mole number of SO3 = SO2 comes from the decomposition of FeSO4: PV = nRT, n = PV/PT = (0.425)1 /(0.082×924.33) = 5.6×10-3 moles Molar number of FeSO4 decomposed = 2nSO3 = 0.0112 mol Mass of FeSO4 decomposed = 0.0112 ×151.91 = 1.70 grams Percent of FeSO4 decomposed = 1.70/15.19 = 11.21 %.

Problem 6. Radiochemistry
1.
204

Pb x

206

Pb

207

Pb

208

Pb

2. Assume that the mineral initially contained n1,0 moles of
206

238

U, n2,0 moles of

Pb, and n3 moles of 204Pb; and at present, it contains n1 moles of 238U, n2 moles 106

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 of
206

Preparatory Problem Solutions
238

Pb, and n3 moles of

204

Pb (this isotope is not generated by the decay of

U

and 235U). The age of the zircon mineral is usually very large, and we can consider that the century equilibrium for the decay process has been reached (i.e. loss of 1 mole of
238

U will lead to formation of 1 mole of (1)

206

Pb). By conservation of mass,

we have the following equation: n1 + n2 = n1,0 + n2,0 Dividing (1) by n3: n1/n3 + n2/n3 = n1,0/n3 + n2,0/n3 → n2/n3 = n1,0/n3 - n1/n3 + n2,0/n3 age of the mineral, → n2/n3 = n1eλt/n3 - n1/n3 + n2,0/n3 = (n1/n3)( eλt - 1) + n2,0/n3 → e-λt -1 = → e λt = 1 +
1
n2 / n3 ? n2,0 / n3 n1 / n3 n1 / n3

(2)

In addition, we have n1,0 = n1eλt, where λ is the decay constant of 238U, and t is the (3)

n2 / n3 ? n2,0 / n3

→ t = ln(1 +
λ

n2 / n3 ? n2,0 / n3 n1 / n3

)

(4)

According to the data given:
14.30 n2 = 206 = 51.12; n3 0.277 204
99.275 n1 = 238 = 307.19 0.277 n3 204 4.47 ×109 51.12 ? 17.05 t= ln(1 + ) = 6.78 ×108 years 0.693 307.19

n2,0

24.10 = 206 = 17.05 1.4 n3 204

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 m0( U) = 0.721 ×
238 235

Preparatory Problem Solutions

0.693 ×6.78×108 8 e 7.038×10
0.693 ×6.78×108 8 e 4.47×10

= 1.406 g = 110.28 g

m0( U) = 99.275 ×

m0(235U)/ m0(238U) = 1.406/110.28 = 0.0127 3. After 99% of Fe3+ precipitated, the concentration of the remaining Fe3+ in the solution is: [Fe3+] = 2 × 0.05×10?2 = 1×10?3 M The concentration of hydroxide ions necessary to maintain a Fe3+ concentration of 10?3 M in the solution is:
1 ? TFe ( OH )3 ? 3 ? 3.8 ×10?38 ? 3 ?12 3 =? [OH ] = ? ? = (38) ×10 M 3+ ? ?3 ? ? Fe ? ? 10
?

1

1

Thus, the pH value of the solution can be calculated as follows: pH = - log{10-14/[(38)1/3×10?12]} = 2 + (1/3)log38 = 2.53 At this pH, the reaction quotient of the dissociation of UO2(OH)2 in 0.01 M of solution is: [UO22+][OH?]2 = 0.01×[(38)1/3×10?12]2 = 1.13×10?25 < 10?22 Since the ionic product is much smaller than the solubility product of UO2(OH)2, we can conclude that uranium cannot precipitate under these conditions. 4. Volume ratio of the two phases: Vaq/Vorg = 1000 : 500 = 2 Let x represent the equilibrium concentration of UO2(NO3)2 in the aqueous phase. Let Co represent the initial concentration of UO2(NO3)2 in the organic phase.

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Preparatory Problem Solutions

The equilibrium concentration of UO2(NO3)2 in the organic phase is calculated as follows: Corg = (Vaq/Vorg)( C0 - x) D=
Corg x = 2(Co ? x) = 10 x

(6)

x : Co = 1 : 6 = 16.67%. 5. 500 mL of organic solvent may be divided into n equal portions for extraction. Volume ratio of the two phases: Vaq/Vorg = 1000 : (500/n) = 2n - After the first extraction: D=
Corg x1 = 2n(Co ? x1 ) = 10 x1

(7) (8)

→ x1 =

2nCo D + 2n

- For the second extraction, the initial concentration of the aqueous phase is x1, while the equilibrium concentration is x2. Using equation (8), we replace x2 with x1, and x1 with Co to obtain the following expression:
2nx1 2n ? x2 = =? ? ? Co D + 2 n ? D + 2n ?
2

(9)

- After n extractions, the concentration of UO2(NO3)2 remaining in the aqueous phase is:
? xn = ? ? ? Co ? D + 2n ?
2n
n

(10)

% UO2(NO3)2 remaining in the aqueous phase after n extractions is:
xn ? 2n ? 100% = ? ? 100% C0 ? D + 2n ?
n

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 n=
? 2n ? ? ? 100% = ? D + 2n ?
n

Preparatory Problem Solutions
4 3.9 5 3.1 6 2.63

1 16.67
n

2 8.16

3 5.27

x 2n ? n = 5 → n 100% = ? ? ? < 4% C0 ? D + 2n ?

Thus, the optimal approach is to divide 500 mL of solvent into 5 portions and extract 5 times. Other schemes are acceptable, if all calculations and justifications are reasonable.

Problem 7. Applied thermodynamics 1. 1.1 Based on the above data: ?GoT = ?HoT - T?SoT Reaction (1): ?GoT (1) = (- 112298.8 + 5.94T) – T(54.0 + 6.21lnT) ?GoT (1) = - 112298.8 – 48.06T – 6.21TlnT ?GoT (1) decreases with an increase in temperature. Reaction (2): ?GoT (2) = (- 393740.1 + 0.77T) – T(1.54 - 0.77lnT) ?GoT (2) = - 393740.1 – 0.77T + 0.77TlnT 1.2 ?GoT (2) increases with an increase in temperature. 2. 2.1 C(graphite) + ? O2(g) → CO(g) C(graphite) + O2(g) → CO2(g) (2) – (1) → CO(g) + ? O2 → CO2(g) We have, ?GoT (3) = ?GoT (2) - ?GoT (1) (1) (2) (3)

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 2.2 Substitute the values in:

Preparatory Problem Solutions

?GoT (3) = (- 393740.1 – 0.77T + 0.77TlnT) – (- 112298.8 – 48.06T – 6.21TlnT) ?GoT (3) = - 281441.3 + 47.29T – 6.98TlnT At 1673 K: ?GoT (3) = -115650 J/mol 2.3 Since ?Go = -RTlnKp, the equilibrium constant Kp for reaction (3) can be calculated as follows:
ln K p,1673 (3) = ?
0 ?G1673 (3) 115650 = = 8.313457 RT 8.314 ×1673

→ Kp,1673 (3) = 4083 3. 3.1 CO(g) + ? O2(g) → CO2(g) NiO(s) + CO(g) → Ni(s) + CO2(g) (4) – (3) 3.2 At 1673 K, we have: For reaction (4): K p (4) = For reaction (3): K p (3) =
p CO2 p CO pCO2 p p
1/ 2 CO O2

(3) (4) (5)

NiO(s) → Ni(s) + ? O2(g)

=

99 1
2 = 4083 or p1/ O2 =

pCO2 p CO 4083

=

K p (4) 4083

=

99 4083

2 For reaction (5): K p (5) = p1/ O =
2

p CO2 p CO K p (3)

=

K p (4) K p (3)

or

K p (5) = p1/2 O2 =

K p (4) K p (3)

=

99 = 0.024247 = 2.4247 × 10?2 4083
2 2

?2 Hence, pO = ? ? K p (5) ? ? = ( 2.4247 × 10 )
2

?4 and pO2 = 5.88 ×10 bar = 58.8 Pa

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Problem 8. Complex compound
1. How many atoms of EDTA are capable of binding with the metal ions upon complexation? 1.1 2 1.2 4
X

6

8

2.
2.1 Let [H+] be h and at pH = 10.26:
β′ = β α Mg 2+ α Y 4-



1

1 + *βh ?1 h + K a4

K a4

= 108.69

1

10?10.26

1 + 1.58 × 10?13 × 1010.26 10?10.26 + 10?10.26

β′ = β α Mg 2+ α Y 4- = 108.69 × 1 × 0.5 = 2.45 × 108

2.2 At pH = 10.26:

(Mg 2+ )' + (Y4- )' → MgY20.05 0.05

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Preparatory Problem Solutions

MgY 2- ? (Mg 2+ )' + (Y4- )'
0.05 - x x
x

(β')?1 = (2.45 × 108 )?1

CMg 2+ = C'Mg 2+ = x = 1.43 × 10-5 (M)
CMg 2+ ? OH ? ? = 1.43 × 10?5 10?3.74 ? ?
2

(

)

2

= 10?12.32 < K s(Mg ( OH ) = 10?10.95
2

Hence no Mg(OH)2 precipitate appears X Precipitation No precipitation

3.
3.1
[CN - ] = CCN Ka 1.400 × 1000 10-9.35 = = 1.00 (M) h + Ka 65 × 20.00 10-10.50 +10-9.35

α Hg 2+ =

1 1 + β Hg(CN)2- [CN ]
4

- 4

=

1 1 + 10
38.97

×1

4

= 1.00 × 10?38.97

α Y 4- ≈

K a4 h + K a4

=

10?10.26 10?10.50 + 10?10.26

= 0.635

β 'HgY 2- = β HgY 2- α Hg 2+ α Y 4- = 1021.80 (1.00 × 10?38.97 )0.635 = 4.29 × 10?18 β 'HgY 2- is very small, Hg2+ is cannot be titrated in the experiment 2.

3.2 Chemical equations: Experiment 1:

Hg2+ + Y4Mn2+ + Y4-

→ HgY2→ MnY2-

Mg2+ + Y4-(excess) → MgY2( CMn + CHg )×20.00 = 25.00 × 0.040 – 12.00×0.025
2+

2+

(1)

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Experiment 2: Hg2+ + 4CNMn2+ + Y4-

Preparatory Problem Solutions
→ Hg(CN)42→ MnY2-

Mg2+ + Y4-(excess) → MgY2CMn2+ × 20.00 = 25.00×0.040 – 20.00×0.025

(2)

According to (1) and (2): CMn = 0.025 M; CHg = 0.010 M
2+

2+

4.
4.1 As Ka3/Ka4 > 1×104 and Ka4 < 10?9 only one endpoint can be determined for the titration of H2Y2?: Titration reaction: H2Y2? + OH? → HY3? + H2O 4.2 pHEP = pH(HY3?) = (pKa3 + pKa4)/ 2 = 8.21 4.3 pHEP = pH(phenol red), hence the most suitable indicator is phenol red X Bromothymol blue 4.4 If the final pH is 7.60 the percentage of H2Y2? that is titrated: [HY3- ] [H 2 Y 2- ] + [HY3- ] = [H + ] + K a3 K a3 100 = 10-6.16 10-6.16 + 10-7.60 100 = 96.5% Phenol red Phenolphtalein

The volume of NaOH solution needed to reach pH of 7.60 is: VNaOH = V1 = (0.25 ×10 × 0.965)/0.2 = 12.06 (mL) VEP = V2 = (0.25 × 10)/0.20 = 12.50 (mL)

q=

12.06 ? 12.50 × 100% ≈ ?3.5% 12.50 (As 96.5% of H2Y2- is titrated, 3.5% of the analyte has not been titrated, or the

error q = - 3.5%.)

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Preparatory Problem Solutions

Problem 9. Lead compounds
1.
X (I) (II) (III) (IV)

2.
X (I) (II) (III) Other

3.
3.1 Condition for precipitation of:
10-7.66 PbSO4 : CPb2+ (1) ≥ = 1.09 × 10-6 (M) 0.02

PbC2O4 : CPb2+ (2) ≥ PbI2 : CPb2+ (3) ≥

10-10.05 5.0 × 10
?3

= 1.78 × 10-8 (M)

10-7.86 (9.7 × 10 )
(0.001)
2

?3 2

= 1.47 × 10-4 (M)
= 2.45 × 10-7 (M)

Pb(IO3)2: CPb2+ (4) ≥ PbCl2: CPb2+ (5) ≥

10-12.61
2

10-4.8 (0.05)

= 6.34 × 10-3 (M)

CPb 2+ (2) < CPb2+ (4) < CPb 2+ (1) < CPb 2+ (3) < CPb 2+ (5) → The order of precipitation:

PbC2O4, Pb(IO3)2, PbSO4, PbI2 and PbCl2. 46th IChO Preparatory Problem Solutions, Hanoi, Vietnam, July 2014 115

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Preparatory Problem Solutions

3.2 When PbI2 begins to precipitate (assume I- has not reacted)

[SO24 ]

=

K s(PbSO4 ) CPb2+ (3)

=

10-7.66 1.47 × 10
-4

= 1.49 × 10-4 (M)

= K s(PbSO4 ) = 1.48 × 10-4 (M) = SPbSO4 (S is the solubility of PbSO4 in saturated solution). Hence PbC2O4, Pb(IO3)2 and PbSO4 have precipitated completely. → 21.60 × CPb(NO3 )2 = 20.00 × ( CC O2- + 2× CIO- + CSO2- ) 2 4 3 4 = 20.00(5.0 × 10?3 + 2 × 0.0010 + 0.020) → CPb(NO3 )2 = 0.025 (M)

4.

PbCrO4 C CH3COOH Pb2+ + CH3COO? Pb2+ + 2CH3COO? Pb2+ + H2O
CrO 24 + H
+

Pb2+ + CrO 24 S S CH3COO? + H+ Pb(CH3COO)+ Pb(CH3COO)2 PbOH+ + H+
HCrO -4
2-

Ksp Ka = 10?4.76
β1 = 102.68 β2 = 104.08
*

β = 10?7.8

Ka-1 = 106.5 K-1 = 1014.64
(1)

+ Cr2O7 + H2O 2 CrO 24 + 2H Let h be [H+], a conservation of mass requires that:
24

22-1 ?1 2 2- 2 S = CCrO = [CrO 24 ] + [HCrO 4 ] + 2.[Cr2 O 7 ] = [CrO 4 ](1 + K a . h) + 2 . K . h . [CrO 4 ]

S = CPb = [Pb 2+ ] + [PbOH + ] + [Pb(CH 3COO)+ ] + [Pb(CH3COO) 2 ]
2+

= [Pb 2+ ](1 + ? β h ?1 + β1[CH 3COO ? ] + β 2 [CH 3COO ? ]2 )

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→ [Pb 2+ ] =
? ?1

Preparatory Problem Solutions
(2)

S 1 + β h + β1[CH3COO ? ] + β 2 [CH 3COO ? ]2
3

Because S = 2.9×10?5 M << CCH COOH = 1 M → pH of the solution is largely dependent on the dissociation of CH3COOH: CH3COOH [] 1–h
-

H+ + CH3COO?
h
+

Ka = 10?4.76

h
?2.38

→ [CH3COO ] = [H ] = h = 10

(M)

Substitute [CH3COO?] = [H+] = h = 10?2.38 (M) and S = 2.9×10?5 into (1) and (2), we have: ?9 2+ ?6 [ CrO 2(M) 4 ] = 2.194 × 10 (M) and [Pb ] = 9.051 × 10
?14 → Ksp = [Pb2+][ CrO 2. 4 ] = 1.99 × 10

5. Cathode:

PbO2 + 4 H

+

+ 2e HSO4-

Pb

2+

+ 2 H2O

2(1.455) 10 0.0592

SO42? + H+ PbSO4 PbSO4 + 2 H2O K1 Pb
2+

10-2 107.66 (*)
?2( ?0.126) 10 0.0592

Pb2+ + SO42Cathode reaction: PbO2 + HSO4- + 3H+ + 2e Anode: Pb HSO4Pb2+ + SO42Anode reaction: Pb + HSO4-

+ 2e

SO42- + H+ PbSO4 PbSO4 + H+ + 2e

10-2 107.66 K2 (**)

Overall reaction as the battery discharges: PbO2 + Pb + 2 HSO4- + 2 H+ Cell diagram: 2 PbSO4 + 2 H2O (***) (a) Pb│PbSO4, H+, HSO4-│PbO2 (Pb) (c)

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2E 0 PbO2 /PbSO4

Preparatory Problem Solutions

6.1 According to (*): 10

0.0592

= K1 =

2(1.455) 10 0.0592 10?2

107.66

→ E0 PbO 2 /PbSO 4 = 1.62 (V)
According to (**):
?2E 0 PbSO 4 /Pb

10

0.0592

= K2 =

?2( ?0.126) 10 0.0592 10-2

107.66 → E 0 PbSO 4 /Pb = - 0.29 (V)

6.2 According to (***):
0 V = E(c) – E(a) = E 0 PbO 2 /PbSO 4 - E PbSO 4 /Pb +

0.0592 2

log[HSO 4 ] [H ]

- 2

+ 2

+ In which [HSO4 ], [H ] are calculated as follows:

HSO 4

-

H+ 1.8 + x

+

SO4

2-

Ka = 10?2

[]

1.8 – x

x

-3 + [ SO24 ] = x = 9.89×10 (M) → [H ] = 1.81 (M); [ HSO4 ] = 1.79 (M)

V = 1.62 + 0.29 +

0.0592 2

log(1.79) (1.81) = 1.94 (V)

2

2

Problem 10. Applied Electrochemistry
1.

1.1 0 +7 +2 +4
?? → K2SO4 + MnSO4 + CO2 + H2O C6H12O6 + KMnO4 + H2SO4 ←? ?

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Preparatory Problem Solutions

5 × 24 ×

?? → 6 C + 24e– 6 C ←? ?
Mn + 5e
+7

0

+4



?? → Mn ←? ?

+2

?? → 12K2SO4 + 24MnSO4 + 30CO2 + 66H2O 5 C6H12O6 + 24 KMnO4 + 36 H2SO4 ←? ?

1.2 5× 2×
?? → 2Fe3+ + 2e 2Fe2+ ←? ? ?? → Mn2+ + 4H2O MnO4- + 8H+ + 5e ←? ? ?? → Mn2+ + 10Fe3+ + 8H2O 2 MnO4- + 10Fe2+ + 16H+ ←? ?

Overall reaction:
?? → 2MnSO4+ 5Fe2(SO4)3+ 8H2O 2 KMnO4+ 10FeSO4+ 8H2SO4 ←? ?

1.3
?? → 2 Fe3+ + 2e– - At anode: 2 Fe2+ ←? ? ?? → Mn2+ + 4H2O At cathode: MnO4- + 8 H+ + 5e– ←? ?

The cell diagram:
? Pt Fe3+ , Fe 2+ MnO 4 , Mn 2+ , H + Pt

1.4 Electromotive force E of the cell can be calculated as follows:
3+ ? Mn 2 + ? 0.059 ?? ? Fe ? ? [ H 2O] E=E ? log ? ? 2+ 5 + 8 5 ? ? MnO 4 ? ?? ? Fe ? ? ? ?H ? ? 5 4 0

2.
2? 2.1 In order to determine the reduction potential of the pair MnO 4 MnO we need 2

to use the below diagram: According to Hess’ Law:

M nO 4 -

? G2
M nO 4 2-

M nO 2
? G3

? G1

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?G02 = ?G01 + ?G03 ?G03 = ?G02 - ?G01
0 0 = ? nFE 0 → E 3 = 2.27 V = E 0 We have ?G 3 MnO
2? 4

Preparatory Problem Solutions

MnO 2

2.2 Similarly, we have:
?G04 = ?G05 + ?G06

?G 4 MnO 2 ??? → Mn 3+

?G5

?G6

We have ?G0 = - nFE0 .
? E MnO
2

Mn

3+

= 0.95 V = E 0 4

Mn2+

3.

3.1 According to the standard reduction potential diagram, we have: MnO42- + 2e + 4H+ 2MnO4- + 2e 3MnO42- + 4H+ MnO2 + 2H2O 2MnO422MnO4- + MnO2 + 2H2O (3) (1)
?G03 (E03 = 2.27 V) ?G01 (E01 = 0.56 V) ?rG0

3.2 In order to know if the reaction is spontaneous, ?G must be considered.
The reaction that is considered can be obtained by subtracting (1) from (3):

?rG0 = ?G03 - ?G01. We have ?G0 = - nFE0 where ?E0reaction = 1.71 V, or ?G3 < 0

and the reaction is spontaneous. 3.3 The equilibrium constant can also be calculated:
log K 3 =
n?E 0 2 × 1.71 → log K 3 = → K 3 = 9.25 × 10 57 0.059 0.059

The large value of K confirms the reaction to be spontaneous.

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Preparatory Problem Solutions

Problem 11. Phosphoric acid
1. H+ is used instead of H3O+ for clarity. The activities of the ions are ignored. [H+]

is abbreviated as h in all calculations and acid constants for H3PO4 are written as
K1, K2 and K3.

As K1 >> K2 >> K3, only first dissociation step is considered.
H3PO4 H+ + H2PO4-

As
? [ H + ][ H 2 PO4 ] h2 (10?1.14 ) 2 ?2.14 K1 = = = 10 = [ H 3 PO4 ] Co ? h Co ? 10?1.46

Solving for Co gives Co = 0.200 M The concentrations of the forms:

h3Co [H3PO4 ] = 3 2 (hK1K2 + K1K2 K3 is ignored) h + h K1 + hK1K2 + K1K2 K3 h3Co hCo 10?1.46 × 0.2 = 3 2 = = = 0.1653 M h + h K1 h + K1 10?1.46 +10?2.14
Similarly, we have:
? [H 2 PO 4 ]=

h 2 K1Co K1Co 10?2.14 × 0.2 = = = 0.0346 M h3 + h 2 K1 h + K1 10?1.46 + 10?2.14

hK1 K 2Co K1 K 2Co 10?2.14 ×10?7.20 × 0.2 [HPO ] = 3 = 2 = = 6.29 × 10?8 M 2 ?1.46 2 ?1.46 ?2.14 h + h K1 h + hK1 (10 ) + 10 × 10
2? 4

K1 K 2 K 3Co 10?2.14 ×10?7.20 ×10?12.38 × 0.2 [PO ] = 3 = = 7.56 × 10?19 M 2 ?1.46 3 ?1.46 2 ?2.14 h + h K1 (10 ) + (10 ) × 10
3? 4

2. We have:

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n H3PO4 = 0.2 × 0.050 = 0.010 mol

Preparatory Problem Solutions

n NH3 = 0.4 × 0.050 = 0.020 mol

Hence the following reaction occurs: H3PO4 + 2 NH3 → (NH4)2HPO4 And [(NH 4 ) 2 HPO 4 ] =
0.010 = 0.1 M 0.100

In solution B: (NH4)2HPO4 → 2 NH4+ + HPO420.2 M We have the following equilibria: NH4+ NH3 + H+ H2PO4H3PO4 H+ + PO43(1) HPO42- + H+ H2PO4- + H+ HPO420.1 M

A conservation of protons requires: [H+] + 2[H3PO4] + [H2PO4-] = [OH-] + [PO43-] + [NH3] In which [NH3] + [NH4+] = 0.2 M [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-] = 0.1 M We also have:
[NH3 ] = K NH+ × 0.2
4

h + K NH +
4

[H 3PO 4 ] =
? 4

h3 × 0.1 h3 + h 2 K1 + hK1 K 2 + K1 K 2 K 3

h 2 K1 × 0.1 [H 2 PO ] = 3 h + h 2 K1 + hK1 K 2 + K1 K 2 K 3

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? [HPO 2 4 ]=

Preparatory Problem Solutions

hK1 K 2 × 0.1 h + h K1 + hK1 K 2 + K1 K 2 K 3
3 2

? [PO3 4 ]=

K1 K 2 K 3 × 0.1 h3 + h 2 K1 + hK1 K 2 + K1 K 2 K 3

As pH of the solution is of about 7 – 9 so we can ignore the [H+], [OH-], [H3PO4] and [PO43-] in the equation (1): [H2PO4-] = [NH3]
K NH+ × 0.2 h 2 K1 × 0.1 4 = 3 2 h + h K1 + hK1 K 2 + K1 K 2 K 3 h + K NH+
4

h × 0.1 h × 0.1 10?9.24 × 0.2 3 = = (h + K1 K 2 K 3 is ignored) h + K 2 h + 10?7.20 h + 10?9.24
Solving for h gives h = 8.81 × 10-9 M and pH = 8.06.
3. Mixing of B and Mg(NO3)2 solution leads to precipitation reaction:

NH4+(aq) + Mg2+(aq) + PO43-(aq) → NH4MgPO4(s) [Mg2+] = 0.2/2 = 0.1 M As B is a buffer solution when it is diluted to twice the original volume, pH is virtually unchanged and is 8.06.
[NH + 4]= hC 10?8.06 × 0.1 = ?8.06 = 0.094 M h + K NH+ 10 + 10?9.24
4

? [PO3 4 ]=

K1K 2 K 3 × Co K 2 K 3Co 10?7.20 ×10?12.38 × 0.05 = = = 2.06 ×10?8 M 2 2 ?8.06 2 ?8.06 ?7.20 h K1 + hK1 K 2 h + hK 2 (10 ) + 10 ×10

The ionic product: [NH4+][Mg2+][PO43-] = 0.1 × 0.094 × 2.06 × 10-6 = 1.93 × 10-8 > 2.5 × 10-13 Therefore, the precipitation occurs.
4. We have: Ca3(PO4)2

3Ca2+ + 2PO43-

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Preparatory Problem Solutions

Assume that the hydrolysis of PO43- can be ignored, the solubility So of Ca3(PO4)2 can be calculated as follows:
Ksp = [Ca2+]3[PO43-]2 = (3So)3(2So)2 = 108So5 = 2.22 × 10?25

Solving for So to gives So = 4.6 × 10?6 M However, the hydrolysis of PO43- cannot be ignored due to its rather strong basicity (pKb = 14 – pKa = 14 – 12.38 = 1.62) PO43- + H2O = 14 – 2.14 = 11.86). According to (1): [HPO42-] + [PO43-] = 2S (2) As [PO43-] is very small (the calculation above), it can be ignored in (2). It can alternatively be calculated as follows: Let x be the concentrations of HPO42- and OH-, [HPO42-] = [OH-] = x We have:
x2 = 10?1.62 = 0.024 ?6 2 × 4.6 ×10 ? x

HPO42- + OH- (1)

We can ignore the hydrolysis of HPO42- (pKb = 14 – 7.20 = 6.80) and H2PO4- (pKb

Solving for x gives x = 9.19 × 10-6 → [PO43-] = 0.01 × 10?6 M Therefore we can assume that [HPO42-] = [OH-] = 2S and [PO43-] is determined based on K3:
K 3 = 10?12.38 = [H + ][PO310?14 [PO34 ] 4 ] = × 22[HPO 4 ] [OH ] [HPO4 ]

? [PO34 ]=

10?12.38 × 2S? S = 167S2 10?14

The solubility S of Ca3(PO4)2: Ksp = 2.25 × 10-25 = (3S)3(2 × 167S2)2 = 3012012S7
? S = 3.6 × 10-5 M.

We can see that solubility of Ca3(PO4)2 increases about 10 times due to the hydrolysis of PO43-.
Note: Students may use logarithmic concentration diagram to get the relationship

[HPO42-] = [OH-] = 2S.
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Preparatory Problem Solutions

Problem 12. Kinetic Chemistry
1. To determine t1/2, the time taken from the initial concentration of N2O5 (3.80×10-3 mol.dm-3) to fall to one-half of its value: t1/2 ≈ 180 s corresponding to [N2O5]t1/2 = 1.90×10-3 mol/dm3 2.

2.1
3.5 3 2.5 2 1.5 1 0.5 0 0 100 200 300 400 500 600 700 800 900 Time/ s

Figure 2. A re-plot of the data in Figure 1 as function of ln {[N2O5]0/[N2O5]t} versus time The plot of ln {[N2O5]0/[N2O5]t} versus time is linear for a first order reaction. 2.2 r = k [N2O5]
[ N 2O5 ]0 = kt [ N 2O5 ]t

The form of integrated rate equation:
ln

ln{[N2O5]0/[N2O5]t}

or

[N2O5]t = [N2O5]0e-kt

3. The 1st order reaction:

k = ln2/t1/2 = ln2/180 s = 3.85 × 10-3 s-1

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Preparatory Problem Solutions

Ea is independent of temperature:
ln k336.6 K Ea ? 1 Ea 1 ? 3.85 ×10?3 ? = ? ? ln = 3.145 ×10?3 ? 2.97 ×10?3 ? ? 4 ? ? ? ? k318 K R ? 318 336.6 ? 5.02 ×10 8.314 J / mol.K

Ea = 97.46 kJ Pre-exponential factor (A): k = A. e-Ea/RT at 336.3 K, A = k. eEa/RT = 3.85 × 10-3 e97460/8.3145×336.3 = 5.28 ×1012 s-1.
5. The intermediate concentrations can be treated by the steady-state

approximation:
rNO = d [ NO ] = k 2 [ NO2 ][ NO3 ] ? k 3 [ NO ][ NO3 ] = 0 → dt [ NO] = k2 [ NO2 ] k3

(Eq.1)

Substituting this equation into the below equation:
rNO3 = d [ NO3 ] = k1[ N 2O5 ] ? k ?1[ NO2 ][ NO3 ] ? k 2 [ NO2 ][ NO3 ] ? k3[ NO][ NO3 ] = 0 dt k2 [ NO2 ] [ NO3 ] = 0 k3

(Eq. 2)

→ k1[ N 2O5 ] ? k?1[ NO2 ][ NO3 ] ? k2 [ NO2 ][ NO3 ] ? k3

→ k1[ N 2O5 ] ? k?1[ NO2 ][ NO3 ] ? 2k2 [ NO2 ][ NO3 ] = 0 →
k1[ N 2O5 ] = [ NO2 ][ NO3 ] k?1 + 2k2

(Eq.3)

The reaction rate:
r2 = rN 2O5 = ? d [ N 2O5 ] = k1[ N 2O5 ] ? k?1[ NO2 ][ NO3 ] dt k1[ N 2O5 ] k?1 + 2k2

= k1[ N 2O5 ] ? k?1
=

2k1k2 [ N 2O5 ] k?1 + 2k 2

= k[ N 2O5 ]

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Preparatory Problem Solutions

Problem 13. Kinetics of the decomposition of hydrogen peroxide
1. Chemical reaction: 2 H2O2

2 H2O + O2

The reaction rate is proportional to the volume of oxygen gas released in a unit of time. In experiments #1, #2, and #3 when the volume of H2O2 solution doubles while keeping the same volume of KI solution, the reaction rate also doubles. Therefore, the rate is directly proportional to the concentration of H2O2. Hence, the reaction is the first-order with the respect to H2O2. Similarly, from experiments #2, #4, and #5 the rate is directly proportional to the concentration of I-. Hence, the reaction is the first-order with the respect to I?.
2. Chemical reaction: 2 H2O2

2 H2O + O2

The rate law: v = kCH O CI
2 2

?

3. In the experiment #4, the solution of H2O2 is diluted three times; therefore, the

concentration of H2O2 was reduced three times. C0 = 10 g H2O2/ 1 L = 10/34 = 0.294 M. Because the reaction proceeds slowly, the reaction rate (or the rate of releasing oxygen gas) is considered to be unchanged after of short period of time (4 min). The volume of oxygen released after 4 min is equal to 4.25 × 4 = 17 (mL O2). Hence, nO =
2

PV (1)(17 × 10?3 ) = = 0.695 × 10? 3 (mol) RT (0.082)(298)
2 2

At the beginning, n H O = (0.294)(0.15) = 44.1×10-3 (mol) After 4 min, n H O = 44.1×10-3 – 2(0.695×10-3) = 42.71×10-3 (mol)
2 2

Therefore, after 4 min C H O =
2 2

0.04271 = 0.285 M. 0.15

4. The overall reaction: 2 H2O2

2 H2O + O2

(*)

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v=? 1 d [ H 2O2 ] 2 dt

Preparatory Problem Solutions

Consider three different cases: a) If step (1) is slow and determines the overall rate, the rate of the overall reaction (*) will be the same as the rate of step (1):
v=? 1 d [ H 2O2 ] = k1[ H 2O2 ][ I ? ] 2 dt

which corresponds to the overall rate law as determined in section 2. b) If step (2) is slow, hence
v=? 1 d [ H 2O2 ] = k2 [ H 2O2 ][ IO ? ] 2 dt

(a)

Assume that the steady-state approximation is applied for IO-, we have
d [ IO ? ] k = k1[ H 2O2 ][ I ? ] ? k2 [ IO ? ][ H 2O2 ] = 0 → [ IO ? ] = 1 [ I ? ] dt k2

(b)

Replace [IO-] from (b) in (a), we have:
v=? 1 d [ H 2O2 ] = k1[ H 2O2 ][ I ? ] 2 dt

which is also appropriate to the overall rate law. c) If the two steps have similar rates:
v=? 1 d [ H 2O2 ] 1 = (k1[ H 2O2 ][ I ? ] + k2 [ H 2O2 ][ IO ? ]) 2 dt 2

Let us assume that the concentration of IO- is in steady-state condition. Similar to the case b), we have:
v=? 1 d [ H 2O2 ] = k1[ H 2O2 ][ I ? ] 2 dt

which corresponds to the overall rate law. Among the three cases, case a) is the most appropriate to the overall rate law because no assumption is made. Besides, in the case b) the assumption of the steady-state IO- is not valid since the step (2) is considered slow.

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Problem 14. Magnetism of transition metal complexes
1. Two compounds are octahedral complexes of Mn2+ (d5).

1.1 K4[Mn(CN)6].3H2O is low spin, 1 unpaired electron. K4[Mn(SCN)6] is high spin, 5 unpaired electrons. 1.2 CN- is strong field ligand, electronic configuration is (t2g)5(eg)0 SCN- is weak filed ligand, electronic configuration is (t2g)3(eg)2
2. Ni2+ (d8) in octahedral field has electronic configuration of (t2g)6(eg)2 with two

unpaired electrons. The spin only χeff is 2.83 MB.
3.
?eff = 2.83 × ?1 ?
? ?

4(?315) ? ? (BM) 8500 ?

Thus, ?eff is 3.25 MB
4.

4.1 d8 in square planar field is diamagnetic. 4.2 C is neutral, DBM is monoanionic form. Mc = 504 (g/mol). A should be hydrate form of C, MA = MB / 0.932 = 540.8 (g/mol), corresponding to two molecules of H2O per [Ni(DBM)2]. Thus, the formula is [Ni(DBM)2].2H2O 4.3 Water should coordinate to Ni center due to the change of color and magnetic property. ?eff value of A is close to that of [Ni(H2O)6]Cl2. So, an octahedral complex is expected for A. 4.4 There are three isomers, the trans isomer and two optical cis isomers. 4.5 B should be an octahedral complex, due to the color and magnetic moment are similar to those of A. Octahedral geometry can be formed by oligomerization / polymerization of B on heating, the DBM may play as bridging ligand.

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Problem 15. Structure and synthesis of Al-Keggin ion
1.

1.1 Al2Cl6 has sp3 hybridization 1.2 Al ? Al =
2.
2 x( Al ? Al ) = 2( Al ? Cl ) cos(39.5o ) = 2 × 221 pm × 772 = 325.63 pm 2

2.1 n = 7+ 2.2 The Aloctahedral/Altetrahedral is estimated ~ 12/1. The center Al atom at number 7 is tetrahedral; the other atoms are octahedral. 2.3 and 2.4 The Al-Keggin cation structure is composed of one Altetrahedral cation surrounded by four oxygens. This Al atom is located centrally and caged by 12 octahedral AlO6-units linked to one another by the neighboring oxygen atoms. There are a total of 24 bridging oxygen atoms that link the 12 adjacent atoms. The cations centered in the 12 octahedra are arranged on a sphere almost equidistant from each other. The formula can be expressed as (AlO4Al12(OH)24(H2O)12)7+.

(with the permission of Aleksandar Kondinski, Jacobs University) 2.5 13AlCl3 + 32 NaOH+ 8 H2O = [(AlO4Al12(OH)24(H2O)12)7+]Cl7 + 32NaCl
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3.1 13Al [(H2O)6]Cl3 + 16(NH4)2CO3 + 24H2O = [(AlO4Al12(OH)24(H2O)12)7+]Cl7 + 32 NH4Cl + 16 CO2 ↑ + 54 H2O 3.2 Volume of a ball = (4/3) πr3 = (4/3)× 3.14 × (0.542)3 = 0.667 cm3. Volume of 3 balls = 3 × 0.667 = 2.00 cm3. Inner volume of crucible = 15 cm3 Volume of gas = 15 – 2 = 13 cm3. PV = nRT → n = 1×13×10-3/ 0.082 × 298 = 5.32 × 10-4 mol. After reaction:

∑n

gases

= nbefore + nCO2

PV = ngasesRT → n = (2.50 atm) ×(13.10-3 L) / (0.082 L.atm.mol-1.K-1) × (298 K) = 1.33 × 10-3 mol
nCO2 = nafter ? nbefore = 1.33 × 10 ?3 ? 5.23 × 10 ?4 = 8 × 10 ?4 mol

Molar number of Al13-Keggin cation = 8×10-4 /16 = 5×10-5 mol. Number of Al13-Keggin cations = 5×10-5 mol × 6.023×1023 = 3×1019 ionic molecules

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Problem 16. Safrole
1.

Reaction 1 2 3
2.

Balanced equation K[PtCl3C2H4] + C10H10O2 → K[PtCl3(C10H10O2)] + C2H4 2 K[PtCl3(C10H10O2)] → [Pt2Cl2(C10H9O2)2] + 2 KCl + 2 HCl [Pt2Cl2(C10H9O2)2] + 2 C5H5N → 2 [PtCl(C10H9O2)(C5H5N)]

in A From the IR data From the 1H NMR data
3. A

in B C9 and C10 bond with Pt safrole lost H5, C5 bonds with Pt
B

in C C9 and C10 bond with Pt safrole lost H5, C5 bonds with Pt
C

C9 and C10 bond with Pt safrole coordinated with Pt

4.

Reaction Driving force 1 K[PtCl3C2H4] + C10H10O2 → K[PtCl3(C10H10O2)] + C2H4 ↑ 132

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Preparatory Problem Solutions

Ethylene (C2H4, gas) is more volatile than safrole (C10H10O2, liquid). 2 3 2 K[PtCl3(C10H10O2)] → [Pt2Cl2(C10H9O2)2] + 2 KCl + 2 HCl The chelate complex [Pt2Cl2(C10H9O2)2] is more stable. [Pt2Cl2(C10H9O2)2] + 2 C5H5N → 2 [PtCl(C10H9O2)(C5H5N)] In the dinuclear complex [Pt2Cl2(C10H9O2)2], two bridging Cl weakly bond with Pt but in [PtCl(C10H9O2)(C5H5N)] the ligand C5H5N strongly bonds with Pt.
5. This reaction was controlled by steric effects rather than the trans effect.

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Problem 17. Imidazole
1.

Structure Imidazole (C3H4N2) Imidazol-1-ide anion (C3H3N2) Imidazolium cation (C3H5N2) Oxazole (C3H3NO) Thiazole (C3H3NS)
N S NH N H

aromatic or not aromatic

aromatic

aromatic aromatic aromatic

2.

Melting point

Imidazole > Thiazole > Oxazole Imidazole is the first because of intermolecular hydrogen bonding. Thiazole is placed before oxazole because thiazole’s molecular mass and polarizability are lager than those of oxazole. Imidazole > Thiazole > Oxazole Imidazole is the first because of intermolecular hydrogen bonding. Thiazole is placed before oxazole because thiazole’s molecular mass and polarizability are lager than those of oxazole. 134

Justification

Boiling point

Justification

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Preparatory Problem Solutions

Equation

for

the ionization

Kb Justification

Imidazole > Thiazole > Oxazole Conjugate acid of imidazole is symmetrical delocalized, forms stronger hydrogen bonding with water, i.e. more stable, thus imidazole more basic than oxadiazole and thiazole. Atom O is more electronegative than N and S, it decreased electron density at N of oxazole, decreased stability of oxazole’s conjugate acid making oxazole less basic than thiazole.

4.

Reaction mechanism:

Explanation: Atom N-3 (N at 3-position) is strong nucleophile; The positive charge is delocalized; The imidazole is good leaving group.

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Preparatory Problem Solutions

6.

The pair of electrons from N-1 and The

pair

of

electrons

from

N-1

four electrons of the remaining four conjugate with C=O decreasing the atoms form a sextet of π-electron of bond order of C=O, hence decrease its aromatic system. They do not IR stretching frequency. conjugate with C=O, thus do not affect the bond order of C=O.
7.

4 C3H4N2 + COCl2 → (C3H3N2)2CO + 2 [C3H5N2]Cl

(1)

2 mol of imidazole react with 1 mol of phosgene to form 1 mol of CDI and 2 mol of HCl; the other 2 mol of imidazole are used to react with the HCl. 2 C3H4N2 + COCl2 + 2 NaOH → (C3H3N2)2CO + 2 NaCl + 2 H2O (2) In imidazolyl groups of CDI the pair of electrons from N-1 and four electrons of the remaining four atoms form a sextet of π-electron of aromatic system. They do not conjugate with C=O. Two electron-withdrawing imidazolyl groups make C=O more active, the imidazole is good leaving group, hence CDI readily reacts with water from reaction (2): (C3H3N2)2CO + H2O → 2 C3H4N2 + CO2 (3)

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Preparatory Problem Solutions

8.1 R = CH3(NH2)CH

8.2

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Problem 18. Small heterocycles
1.

Scheme 1:

Scheme 2:
Bn Bn

O F3C

O OEt

NH O
BnNH2 AcOH

NaBH4

NH F3C I OH

F3C H

OEt EtOH THF Bn

(C13H14NO2F3) Bn SOCl2 CH2Cl2

NH F3C K Cl

LiHMDS

N CF3

L

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2. Reaction mechanism for the transformation from B to C:

3. Reaction mechanism for the transformation from F to G:
Ts N OTs F3C
-

Ts N F3C
-

OTs

Ts N F3C OPh

O-Ph

OPh

Problem 19. Vitamin H
1. The chloride acid was reacted with the amino group of the “bis(L-cystein)” (A) to give

amide (B). Zn powder in acetic acid solution reduced the S-S bond of (B) to give an intermediate containing a thiol (-SH) group. Under normal condition, the -SH group added spontaneously to the terminal alkyne group to yield (C) with a ten-membered ring, of which the newly formed C=C double bonds had (Z) configuration.

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2. Diisobutyl aluminium hidride (DIBAL) partially reduced ester (C) into aldehyde (D)

which was condensed with benzylhydroxylamine to give nitrone (E) with (E) configuration. In the intramolecular [4+2] cyclization reaction of (E) (note that ‘4’ and
‘2’ are the numbers of π electron of the nitrone and the double bond involved in the cyclization, respectively), the configuration of the double bonds C=C and C=N remained

unchanged. The resulting compound (F) had three new chiral centers, two of which were (3aS, 4R). They were the configurations of the corresponding C3 and C4 in the skeleton of (D)-(+)-Biotin. The third chiral carbon which was attached to the oxygen atom had an (R) configuration.

3. (F) was reduced by Zn in acetic acid to give (G) containing one –OH group and a

second-order amino group. The amino group was reacted with chlorofomate in the

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presence of Na2CO3 in THF solution produce (H). Under basic condition, the tenmembered ring of (H) was opened to give δ-hydroxy acid (I). The configuration (6aR) of (I) resulted from the (R) configuration of the chiral carbon in L-cysteine. Therefore, Lcystein is chosen as the starting material for the synthesis.

4. The sulfur atom caused an anchimeric effect by which the configuration of the carbon

attached to the –OH in compound (Ι) remained unchanged as this –OH group was replaced by the halogen atom to yield (K). The halogen atom was then replaced when (K) was reduced with NaBH4 in which the “pentanoic acid” branch of (D)-(+)-biotin was formed. The hydrolysis of ester (M) in the aqueous solution of HBr, followed by the removal of the benzyl group resulted in the formation of the target molecule, (D)-(+)Biotin.

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Problem 20. No perfume without jasmine

1. In a [4+2] cycloaddition reaction (Diels-Alder reaction), the configuration of the

dienophine (B) remained unchanged: the two ester groups –COOLac of compound (C) were placed in different sites in comparison to the six-membered ring. The hydrolysis of these two ester groups in LiOH solution gave the two corresponding
trans carboxyl groups.

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2. When dicarboxylic acid (D) was treated with I2/KI, it was transformed into γ-

iodolactone (E) of the endo –COOH. This lactone then underwent a decarboxylation-cyclization in basic solution step to give lactone (F) containing a three-membered ring. The secondary –OH group which resulted from the hydrolysis of (F) was oxidized by NaIO4 to form a carbonyl group.

3. The addition-ring opening step of the cyclopropane ring with HI oriented by the

(-C) conjugation effect of the carbonyl resulted in the formation of γ-iodo acid (H) containing only five-membered rings. The reductive elimination of the iodine atom by Zn in acetic acid produced ketoacid (I) which underwent a Bayer-Viiliger oxidation to yield lactone (K) (the main product) with one carboxyl group. The carboxyl group was treated under Rosenmund reduction condition in which lactone (L) with an aldehyde functional group was obtained. Vinyl ether (Μ) was separated from the Wittig reaction between the aldehyde and the ylide Ph3P=CHOMe.

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4. Methyl vinyl ether (M) was hydrolyzed in acidic medium to give lactone (N)

containing an aldehyde functional group. From the Wittig reaction between (N) and the ylide Ph3P=CHCH2CH3, lactone (O) with a cis carbon-carbon double bond was separated. The hydrolysis of the lactone (O) followed by treatment with diazomethane produced ester (P) which was oxidized with pyridine dichromate to give the target compound (Z)-(3R,7S)-methyl jasmonate.

methyl jasmonate

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Problem 21. Vietnamese cinnamon
1. Give the structure for A, B, C. A B C

2. Assign 1H NMR signals in first spectrum to appropriate proton groups of C. C

9.0 ppm, s
He

7.4-7.3 ppm, m
C6H5

5.5 ppm, s
Hd

4.2 ppm, t
Hc

2.8 ppm, dd
Ha

2.6 ppm, dd
Hb

3. Propose a reaction mechanism for the formation of C from B.

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4. Among four given below structures, select one for D and give the reasons for

your selection. The structure for D Reasons for your selection.
D3 derives from C and corresponds with the given 1H NMR

spectrum.
D1 and D4 do not derive from C. D1 does not correspond with second 1H NMR spectrum (For

example in the spectrum there are not two ethylenic protons).
D2 and D4 seem corresponding to the given 1H NMR

spectrum, but D2 contains three members and D4 contains four members cycles, which cannot exist after reflux for 12 h.
5. Assign 1H NMR signals in second spectrum to appropriate proton groups of D. D

9.6 ppm, s

8.3 d

7.6 d

7.47.3 ppm, m

3.0

2.9 dd

2.6 ppm, dd

ppm, ppm,

ppm, t ppm,

Hd

2 Hf

2 He

C6H5

Hc

Ha

Hb

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Problem 22. Cinnamic acid
1.

1.1 Reaction mechanism for the isomerization

1.2
Ph HOOC H H H H

COOH

Ph

α-truxillic acid

1.3

β-truxinic acid

Has an enantiomer

Has an enantiomer

Has an enantiomer

Has an enantiomer

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1.4 The packing arrangement of α-type of trans-cinnamic acid which leaded to the formation of α-truxillic acid

The packing arrangement of β-type of trans-cinnamic acid which leaded to the formation of β-truxinic acid

1.5 In solution all molecules of cinnamic acid were solvated and randomly arranged.
2.

2.1
A B C

Optically active since A Optically active since B has Optically inactive since has four and asymmetric three asymmetric carbons C has has symmetrical carbons no and has no symmetrical plane although has two

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symmetrical plane and no plane and no symmetrical asymmetric carbons. symmetrical center. 2.2 Optical active since each bicyclooctane moiety has four asymmetric carbons and their configurations are exactly the same as in A thus α-truxilline has no symmetrical plane and no symmetrical center. center.

Problem 23. Tris(trimethylsilyl)silane and azobisisobutyronitrile
The products from the each reaction: 2.1 A, B and C from the compound (I):

(I )

A

B

C,

2.2 D from the compound (II): TTM

(II )

D

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2.3 E from the compound (III):

(II )

D

(III)

E
E

2.4 I from the compound (IV) through radicals F and G:
Bn N Bn O TTMSS / AIBN MeO I N3 Benzene, MeO N O MeO N N=N=N F Bn N O MeO NMe J (+/-)-Horsfiline MeO I NH Bn N O G + H+ - N2 Bn N O N=N

.

.

(IV)

Problem 24. (-)-Menthol from (+)-δ-3-Carene
Catalytic isomerisation of δ-3-Carene provides (+)-δ-2-Carene (A) which then was pyrolysed to cleave the cyclopropane ring forming diene (B):

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CH3 CH3

Preparatory Problem Solutions
CH3

S
R

H3C

CH3

H3C

CH3 A B

1S, 6R-(+)-delta-3-Carene

Treatment of the unconjugated diene (2,8-menthadiene, B) with HCl to give C and then, dehydrochlorination led to a conjugated diene ((+)-2,4(8)-p-menthadiene, D). Treatment of (+)-2,4(8)-p-menthadiene with hydrogen chloride affords 8-chloro-3p-menthene (E):
CH3 HCl CH3 HOCH3 HCl CH3

Cl B C D E

Cl

E reacted with sodium acetate and acetic acid to give mixed (cis/trans) pulegol

esters (F) via allylic displacements. Hydrolysis F affords (-)-cis and (+)-transpulegol (G). Reduction of either pulegol isomer provides menthol isomers which can be readily equilibrated to predominently (-)-menthol.
CH3 AcONa E AcOH AcO HOHO CH3 H2 HO CH3

F

G

(-)-Menthol

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46th International Chemistry Olympiad Hanoi, Vietnam – 2014 Problem 25. Cefalotin 1. Synthetic scheme:
HO2C H H2N HO2C acetone SH HN Me B MeO2C MeO2C N NHCO2Me MeO2C CO2Me
tBuO tBuOH

Preparatory Problem Solutions

HO2C S Me
tBuO

MeO2C CH2N2 S Me
tBuO

COCl2

N

N

S Me

O Me C

O Me D

N N

MeO2C Pb(OAc) 4 NaOAc, MeOH
tBuO

OH S Me

N

S Me E

N

O Me

O Me F O

MeO2C MeOSO2Cl NaN3
tBuO

N3 S Me Al(Hg)

MeO2C
tBuO

NH2 S Me Al(OiBu)3
tBuO

NH N S Me

N

N

O Me G

O Me H Cl3CH2CO2C O N

O Me I

OHC CHCO2CH2CCl3 OHC

OH O CF3CO2H O N H 2N

CO2CH2CCl3 CHO S L CO2CH2CCl3 CH2OH S

tBuO

N

S Me K CO2CH2CCl3 CHO B 2H 6 S S O N H N

O Me

O S CH2COCl S O N H M N

O N

O Ac2O S O N H O N

CO2CH2CCl3 CH2OAc Pyridine S S O

O N N H P

CO2CH2Cl3 CH2OAc S

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2. Reaction mechanism for the transformation from K to L

3. There are two asymmetric carbon atoms on cefalotin, so we should expect to

have four optical isomers.
O O S N H N CO2H CH2OAc O S S CO2H CH2OAc O S S N H N H O N CO2H CH2OAc S

O O S N H N

O N

CO2H CH2OAc S

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Problem 26. Heterocyclic compounds
1. Synthesis of A from levulic acid:

The purpose of 1.3-dioxolane formation was to protect the carbonyl group from the reaction with NH2NH2. TsOH preparation from toluene:

2. Synthesis of B from A:

B

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The reaction of A and PhNCS is nucleophilic addition of the –NHNH2 group to the –NCS group. PhNCS preparation from aniline:
PhNH2 + CS2 + NH4OH PhNHCS2NH4 + Pb(NO3)2 PhNHCSNH4 + H2O S PhNCS + NH4NO3 + HNO3 + PbS

3. Synthesis of C from levulic acid:

3-O2NC6H4SO3Na is a dehydrogenation (or oxidation) reagent to convert dihydropyridazine into pyridazine ring. 3-O2NC6H4SO3Na preparation from benzene:

1

H-NMR spectrum:

Before the reaction, the heterocyclic should provide 2 resonance signals with 2H intensity in the strong field of the two CH2 groups (experimental: 2.57 ppm and 46th IChO Preparatory Problem Solutions, Hanoi, Vietnam, July 2014 155

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2.83 ppm). After the reaction, these 2 signals should disappear, and two new signals in with 1H intensity the weak field of the two CH groups should appear (experimental: 7.01 ppm and 7.64 ppm).
4. Reaction mechanism of R-CONHNH2 with PhCHO:

First, the hydrazide -CONHNH2 group performed nucleophilic addition to the C=O group of benzaldehyde, then the dehydration step occurred:

- H2O

The electron-withdrawing group –NO2 facilitates the reaction, while the electrondonating group –NMe2 retards the reaction.

4-Me2NC6H4CH=O

<

C6H5CH=O

< 4-NO2C6H4CH=O

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Problem 27. Lotus
1. Structural formulae and the reaction conditions for (a3) and (b3):
MeO MeO A1 MeO MeO N COOMe NO2 MeO MeO A2 MeO MeO N COOMe NH2 Br B1 NO2 Br B2 NO2

Br X1 X2

(a3) MeOCOCl, THF (b3) 1. NaOH/EtOH 2. H2SO4

2. Reaction mechanism: a. A1 from 3,4-dimethoxibenzaldehyde: aldol condensation, the water elimination,

(crotonation).
H MeO MeO O H+ MeO MeO H OH
CH3NO2 -H
+

OH MeO MeO NO2 - H2O A1

b. Reaction mechanism for the formation of X1: Addition-cyclization as Pictet-

Spengler reaction mechanism.

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M eO M eO H NHCOOM e O Br M eO M eO N HO COOM e

Preparatory Problem Solutions
M eO xt, to - H2O M eO N+ COOM e

Br M eO M eO
+

Br

N H

COOM e

xt, to
-H
+

X1

Br

3. Structural formule for Y1a, Y1b, Y2 :
MeO MeO NMe2 MeO MeO NMe2

MeO MeO NMe CHO

Y1a

Y1b

Y2

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Problem 28. NMR Spectroscopy
1. A molecule can undergo fluxional process by interchanging two or more sites. If

the rate of exchange is faster than the NMR time scale, the two different groups will appear at an average shift. As temperature decreases the rate becomes lower and separate shift can be obtained. Rapid equilibration at room temperature between chair conformations leads to one peak. As one lowers the temperature, the interconversion is slowed down until, at temperatures below -66.7 °C, peaks due to the axial and equatorial hydrogens are observed. Axial and equatorial hydrogens have different chemical shifts under these conditions.
Ha He k
1

He Ha

k?1

k at coalescence (at -61 oC): kc = π?ν/√2

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2. The t-butyl-substituted rings are conformationally locked. The hydrogen at C1

has different chemical shifts, depending upon whether it is axial or equatorial. 4Bromocyclohexanes are conformationally mobile. No difference between axial and equatorial hydrogens is observed until the rate of chair–chair interconversion is decreased by lowering the temperature.

Problem 29. Infrared Spectroscopy (IR)
1. Resonance (conjugation) effect: the amino group pushes electron density into

the ring and into the carbonyl group resulting in a lower frequency carbonyl group (more single bond character). A nitro group withdraws electrons resulting in higher frequency carbonyl absorption (more double bond character).
2. Conjugation of a C=C double bond with either a carbonyl group or another

double bond provides the multiple bond with more single-bond character (through resonance, as the following example shows), a lower force constant K, and thus a lower frequency of vibration. For example, the vinyl double bond in styrene gives absorption band at 1630 cm?1. Esters show a very strong band for the C=O group that appears in the range of 1750–1735 cm?1 for simple aliphatic esters. The C=O band is shifted to lower frequencies when it is conjugated to a C=C or phenyl group.

(Hint: ν

=

1 2πc

K

?

, ?: reduced mas ?=m1m2/(m1+m2), c: speed of light).

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Therefore,
Spectrum A: Methyl acrylate. The absorption band appears at 1726 cm?1 belong to

the C=O group that conjugates to double carbon-carbon double bond. Similarly, the C=C bond in this molecule has the absorption bond at 1639 cm?1 due to the stretching vibration.
Spectrum B: Allyl acetate. The stretching vibrations of C=O and C=C double

bonds appear at the normal positions for these vabrations, at 1743 and 1650 cm?1, respectively. There are only the separated C=C and C=O double bonds in vinyl propionate and allyl acetate, so the stretching bands appear at the normal positions.

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PART 2. PRACTICAL PROBLEMS
Problem 30. Condensation between vanillin and benzylamine
2. Mechanism

3. Na2SO4 is a water-adsorbing substance. It removes water preventing H2O attack

to the product (imine).

Problem 31. Synthesis of eugenoxy acetic acid
1. Reactions in steps 1a, 1b and 2:

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3. The reaction in step 2 follows a SN2 substitution:

4. In alkaline media, eugenol is transformed into eugenolate which has adequate

nucleophilicity to replace the chlorine atom of monochloroacetic acid. An excess amount of alkaline, however, should not be used as the hydroxide ions can compete with the eugenolate ions to form the hydroxide derivative of acetic acid.
5. The carboxyl group, on the one hand, provides for an (-I) effect to increase the

positive charge density at the alpha carbon, facilitating the attack of nucleophiles. On the other hand, the carboxyl can delocalize the negative charges appearing in the transition state of the SN2 reaction.

6. Crystallized water in the product re-crystallized from hot water lowers down its

melting point. Re-crystallization of the product from dry benzene helps eliminate water and increase the melting point. Titration or TGA can be used to determine the amount of water crystallized in the product

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7. In basic condition, eugenoxyacetic acid is subjected to isomerization in which

the terminal C=C double bond is moved in and conjugated with the benzene ring to yield isoeugenoxyacetic acid as shown in the scheme below:

Isoeugenoxyacetic can exist in form of two configuration isomers (Z) and (E):

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Problem 32. Complexometric titration of iron, aluminum, and magnetism in the aqueous solution
1. The chemical equations of the reactions used in the titration:

1.1 Titration of Fe3+ The stability constant of FeY- is much larger (= 1025.1) >> The constants of complex of Al3+ ( 10 16.13 ) and of Mg2+( 10 8.7) therefore in the solution with pH of 2 only ion Fe3+ totally titrated: Fe3+ + Na2H2Y → FeY- + 2Na+ + 2H+ 1.2 In the solution with pH of 4.7 only ions Fe3+ and Al3+ are totally titrated: Fe3+ + Na2H2Y ( extra ) → FeY- + Na2H2Y Al3+ + Na2H2Y (extra ) → AlY- + Na2H2Y 1.3 Separate and titration of Mg2+: In the buffer NH3 + NH4+ (pH = 9.2) only Al(OH)3 and Fe(OH)3 are precipitated and total Mg2+ ions are existed in the solution. After the filtration of Al(OH)3 and Fe(OH)3 we can titrate Mg2+ in the filtrate: Mg2+ + Na2H2Y
→ MgY2- + 2Na+ + 2H+

2. The formulae for calculation of ion concentrations (in mol / L)

CFe(III) = V1 0.05 / 25.0 CAl(III) = [( 50.0 – V2 – V1) 0.05] / 25.0 CMg(II) = V3 0.05 / 25.0

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Problem 33. Determination of zinc and lead in zinc oxide powder
1.

1.1 ZnO(s) + H2SO4 (aq) → ZnSO4(aq) + H2O(l) PbO(s) + 2 HNO3 (aq) → Pb(NO3)2 (aq) + H2O(l) Pb2+ + SO42- → 1.2 PbSO4 (s) + 4 NH4CH3COO(aq) → Pb(CH3COO)4(NH4)2 (aq) + (NH4)2SO4 1.3 2Pb(CH3COO)4(NH4)2 1.4 2PbCrO4(s) + 4NaCl(aq) + 4HCl(aq) → 2Na2PbCl4aq) + H2Cr2O7(aq) + H2O(l) 1.5 Cr2O72- + 6 Fe2+ I2 + 2 Na2S2O3 (aq) →
2.
(aq)

PbSO4 (white ppt.)

+ K2Cr2O7

(aq)

+

H2O(l) →

2PbCrO4

(yellow ppt.)

+

2KCH3COO(aq) + 4NH4CH3COO(aq) + 2 CH3COOH(aq)

+ 14H+ 2Cr3+



2Cr3+ 3 I2 +

+

6Fe3+

+

7H2O

H2Cr2O7 + I- + 12H+ →

+

7H2O

2 NaI(aq) + Na2S4O6 (aq)

Mass of zinc oxide powder = a (g). The volume of standard solution is recorded in mL.

%Zn = (VEDTA × CEDTA ) ×10 ×100
1 %Pb = × 207.02 × (VFe2+ × CFe2+ ) ×100 3

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Preparatory Problem Solutions

The half reactions

Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O I2 + 2e → 2 IS4O6 2- + 2e → 2S2O322

E° = 1.33 V

EI0 /2 I ? = +0.54 V;
0 ES = +0.08 V O 2? / S O 2?
4 6 2 3

As K2Cr2O7 is a strong oxidant, it can oxidize S2O32- to form S4O62- and SO42-. The reactions are not stoichiometry.
4.

4.1 Pb(OAc)2(aq) + K2CrO4(aq) → PbCrO4(s) + 2KOAc(aq) Then PbCrO4(s) ? Pb2+ + CrO42– [Pb2+] = 0.1×10-5/0.12= 8.3×10-6 mol/L [CrO42-] = 0.02×1.0×10-3/0.12= 1.7×10-4 mol/L Therefore Q = 8.3x10-6 × 1.7×10-4= 1.4×10-9 > Ksp . So a precipitate will occur. 4.2 Since [Pb2+] = 8.3 × 10-6 and [CrO42-] = 1.7 × 10-4 and there is a 1:1 stoichiometry, Pb2+ is completely reacted. PbCrO4(s) ? I. (after ppt.) C. E. Pb2+ 8.3 × 10-6 x x + CrO42– 1.7 × 10-4 -8.3 × 10-6 x 1.7 × 10-4 +x = 1.7 × 10-4

Ksp = [x][1.7 × 10–4 + x]= 1.8×10-14 Solving for x gives x= 1.1×10-10, so the concentration of Pb2+ remaining in solution is very small.

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Problem 34. Preparation of copper(II) acetylacetonate
1. Cu(acac)2 2. There are four main equilibria involved the complex formation:

Hacac (aq) + H2O H3O+ (aq) + OH- (aq) Cu2+ (aq) + 2acac- (aq) Cu2+ (aq) + 2OH- (aq)

H3O+ (aq) + acac- (aq) 2H2O (l) [Cu(acac)2] (s) Cu(OH)2 (s)

At a low pH, acac– is not sufficiently concentrated to precipitate complex or to form the complex with a high yield. On the contrary, at high pH regions, Cu(OH)2 can be competitively precipitated and an impure product can be obtained.
3. Square planar complex with two six-membered chelate rings.

Problem 35. Kinetic analysis of the hydrolysis of aspirin
1.

The theoretically obtained amount of aspirin is:
n(salicylic acid) = 2.00g/138.1 = 0.0145 mol m(aspirin) = 0.0145 mol × 180.2 = 2.6129 g

Experimentally, the amount of aspirin obtained is 2.0132 g
→ The yield of the reaction is: (2.0132/2.6129) ×100 = 77.04 %

2. Magnesium hydroxide, magnesium carbonate and aluminum glycinate, when

mixed into the formulation of the aspirin will reduce the irritation. 168

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3.

Ignoring the volume change upon mixing and supposing that aspirin occupies only negligible volume: In the 5×10-4 M solution of acetylsalicylic acid, the concentration of NaOH = (5.0 × 10-3 mol L-1× 40 mL)/50 mL – 5.0 × 10-3 mol L-1 = 3.5 × 10-3 mol L-1
4. Determine the order with respect to the concentration of aspirin and the pseudo -

order rate constant of the reaction. The UV-Vis absorption obtained in the experiments is given below: Time/minute Absorbance A
1.2 1.0 0.8 [A - A] 0.6 0.4 0.2 0.0 0 10 20 30 40 50 60 70 t /minutes

5 0.549

10 0.829

20 1.178

30 1.389

40 1.506

50 1.569

60 1.602



1.653

0.5 0.0 -0.5 ln(A - A)
8

-1.0 -1.5 -2.0 -2.5 -3.0 -3.5 0 10 20 30 40 50 60 70 t /minutes
y = -0.056x + 0.372 R2 = 0.999

8

Figure 1. Plot of (A∞ - A) versus time

Figure 2. Plot of ln(A∞ - A) vs. t

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3 .5 3 .0 ln[(A - A )/(A - A )] 2 .5 2 .0 1 .5 1 .0 0 .5 0 .0
t2

Preparatory Problem Solutions

20 [1/(A - A) - 1/A ]
8

15

8 t1

y = 0 .0 5 6 x R 2 = 0 .9 9 9

8

10

8

5

0 0 10 20 30 40 50 60 70 t /m in u t e s

0

10

20

30 (t 2 - t 1 ) /m in u t e s

40

50

60

Figure 3. Plot of ? ?

?

1 1 ? ? ? ? vs. t A ? A A ∞ ? ? ∞

Figure 4. Plot of ln

A ∞ ? A t1 A∞ ? A t2

vs. (t2-t1)

5. Experimental results showed that the reaction obeyed the pseudo-first-order rate

law (figure 2 and 4), but not the second-order (figure 3). Based on the equation ln
A ∞ ? A t1 A∞ ? A t2

= kt and the results in figure 4, as well as in figure 2, we can easily

calculate kobs = 0.056 min.-1. Hence, the half-life is 17.85 minutes and the reaction time is 3.36 times greater than the half-life.
6. From the obtained experimental results the reaction is first order with respect to

both [asp] and [OH-], therefore the rate law may be given as Rate = k [asp]1[OH-]1. According to this mechanism, in step 1, the hydroxide nucleophile attacks at the electrophilic C of the ester C=O, breaking the σ bond and creating the tetrahedral
intermediate. In step 2, the intermediate collapses, reforming the C=O and, the

last step (step 3) is an acid-base reaction which takes place very fast, a very rapid equilibrium. Hence, it is not the rate-determining step of the reaction. Let denote the tetrahedral intermediate as I and (2-HOC6H4COO-), a product in step (2) as P. The rate of formation of product may be given as 170

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Rate = d [ P] = k2 [ I ] dt

Preparatory Problem Solutions

With regard to the stability of the intermediate I, the two possibilities, which may be considered, are i) If k-1
k2, means the rate of reconversion of I into asp and OH- is significantly

greater than the rate with which it undergoes to give the P, the concentration of intermediate I, [I] can be calculated by considering equilibrium (1) alone [I] = K[asp][OH-] Where equilibrium constant K = k1/k-1 and, therefore Rate of the reaction = k2K[asp][OH-] which is matching well with experimental results. ii) If intermediate complex I is much less stable species, means the rate of its conversion to product (step 2) is not small compared with the reverse rate in step 1. In this case, the concentration of I must be calculated by using steady-state treatment. By applying the steady state with respect to [I], we get
k1[asp ][OH ? ] = k?1[ I ] + k2 [ I ]

or and, therefore

[I ] =

k1[asp ][OH ? ] k ?1 + k 2

k1k2 [asp ][OH ? ] Rate of reaction = k ?1 + k 2

However, when k-1
k2.

k2, the rate law becomes same as case 1. Thus, the steady

state treatment is the general one, and reduces to the equilibrium treatment when k-1 Conclusion: if the equilibrium in step 1 is controlled throughout the reaction process (step 1 is very fast and represents rapid pre-equilibrium to the rate), the 46th IChO Preparatory Problem Solutions, Hanoi, Vietnam, July 2014 171

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given mechanism agrees with the rate law and in this case the step 2 is ratedetermining step of the reaction. The obtained rate laws clearly show the rate is dependent on the NaOH concentration. For a given concentration of NaOH, we may write Rate of reaction = kobs [asp] Where kobs =
k1k2 [OH ? ] or k2K[OH-] (when k-1 k ?1 + k 2

k2)

Constant kobs is proportional to [OH-] and is known as catalytic coefficient for the catalyst.

Problem 36. Complex formation of ferric ion and salicylic acid
1. n = 1, thus the empirical formula is Fe3+(H2Sal) 2.

2.1 The chemical equation: H2Sal + Fe3+ 2.2 K (1) = K f × K a1 × K a 2 [Fe(Sal)]+ + 2 H+
[ Fe( Sal )] × [ H + ]2 = [ H 2 Sal ] × [ Fe3+ ]

(1)
K eq × [ H + ]2 K a1 × K a 2

or

Kf =

[H+] need be calculated from initial concentration 0.0025 M and dissociation concentration during complex formation: [H+]eq = 0.0025 + 2 × [Fe(Sal)]+ 2.3 The average value is about 1.4 × 1016

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2.4 The average Kf value is not the same with literature values which vary from
2.2×1016 to 2.7×1016 (ref. 2) due to the simplifications of the equilibrium as

mentioned, and also using concentrations instead of activities in the Kf equation.
[1]. D. R. Lide. CRC Handbook of Chemistry and Physics (84th Ed). CRC Press, 2003, pp. 1247. [2] Z. L. Ernst; J. Menashi. Complex formation between the Fe3+ ion and some substituted phenols. Part 1. Spectrophotometric determination of the stability constant of ferric salicylate. Trans. Faraday Soc., 1963, 59, 1794-1802.

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