40th Austrian Chemistry Olympiad
National Competition
O
H
1. Br2
h
1.+ CH3MgBr
L
K
+
2. H2O/H+
2. Base (-HBr)
1. Base
B
H
C
2.
CN
Br
A
O
O
OH
H
I
Solution
H2SO4, H2O, 
H
1. Ph3P=CH2
Grandisol
CrO3
Ph3P=CH2
Grandisol
M
1. Base
1. O
2. H2.2ON
MCPBA
G
KOH/H O
2. BH3/THF2
(C14H26O3)

O
NH2N
C
F
N
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
Task 1
15 points
Titanium and other metals
A. From the ore to the element
1.1. Calculate the empirical formula of titanite.
formula: CaTiSiO5
calculations:
Ti:
24.42
47.87
= 0.5101
Ca:
20.44
40.08
= 0.5100
Si:
14.33
28.09
= 0.5101
O:
40.81
16.00
= 2.5506
proportion 1:1:1:5
1.2. Write down balanced equations for the reactions in the scheme and derive from them an overall
equation for the depicted process.
A: FeTiO3 + C  Fe + CO + TiO2
B: TiO2 + 2 C + 2 Cl2  TiCl4 + 2 CO
C: TiCl4 + 2 Mg  Ti + 2 MgCl2
D: MgCl2  Mg + Cl2
overall: FeTiO3 + 3 C  Fe + 3 CO + Ti
1.3. 1000 kg of the reaction mixture according to the overall equation were used. Which maximum
mass of titanium may be generated starting from these 1000 kg?
mass of Ti: 255 kg Ti
calculations:
M(FeTiO3) = 151.72 g/mol
n·151,72 + 3n·12,01 = 106 g therefore n = 5326 mol
m(Ti) = 5326·47.87 = 255 kg Ti
1.4. Which process is used to execute reaction D?
electrolyis, fused-salt electrolysis
1.5. Complete the equation for the synthesis of barium titanate(IV) from TiO2:
BaCO3 + TiO2  BaTiO3 + CO2
1
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
1.6. Write down the complete equation for the radioactive decay of 45Ti.
45
22Ti
→
45
21Sc
+ 𝑒 + + 𝜈𝑒
1.7. What was the specific activity of the sample put aside after 3.0 hours?
specific activity after 3.0 hours: 33.54 Bq/g
calculations:
ln 2
𝜆=𝑡
1⁄
2
ln 2
= 3.078 h = 0.2252 h−1
𝐴(𝑡) = 65.91 ⋅ 𝑒 −0.2252⋅3 = 33.54
1.8. Calculate the overall yield of Ti of the reaction sequence (A,B,C,D) based on the starting amount
of FeTiO3.
yield in %: 80.0 %
calculations:
total activity at the beginning: 82.6 g · 65.91 = 5681.4 Bq
After some time – activity after 210 min (3.5h) 𝐴(𝑡) = 5681.4 ⋅ 𝑒 −0.2252⋅3.5 = 2584.4 Bq
2068
yield: 2584.4 = 0.8002
B. Other Ti-compounds – crystal lattice
1.9. Write down the electron configuration of Ti3+.
1s2 2s2 2p6 3s2 3p6 3d1
1.10. Calculate the ionic radius of N3− using the Kapustinskii-equation. Show your calculations.
(d* = 34.5 pm; κ = 1.21·105 kJ·pm/mol)
Ionic radius of nitride N3−: 1.64Å
calculations:
𝑁𝑖𝑜𝑛 |𝑧− 𝑧+ |𝜅
𝑑∗
(1
−
)
𝑟0
𝑟0
𝑁 |𝑧 𝑧 |𝜅
2⋅3⋅3⋅1.21⋅105 l
𝐵 = − 𝑖𝑜𝑛𝛥𝐻− + = − −8033
= 2.711 ⋅ 102 𝑝𝑚
1
1
𝑑∗
𝑟2
= −
⇒ 0 = 𝑟0 − 𝑑∗ ⇒ 𝑟02 − 𝐵𝑟0 + 𝑑∗ 𝐵 = 0
𝐵
𝑟0
𝑟0
𝐵
𝛥𝐻 = −
𝑟0 =
𝐵
2
𝐵2
+ √ 4 − 𝑑∗ 𝐵
r0 = 2.31Å = rTi + rN
therefore rN = 1.64Å
2
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
1.11. Calculate the lengths of x and y using this information.
x = 3,562 Å
y = 3,548 Å
calculations:
Looking at the front plane: ∠(𝑇𝑖 − 𝑇𝑖 − 𝐶) =
and
𝑥
2
180−118.4
2
= 30.8°
= (𝐶 − 𝑇𝑖) ⋅ cos 30.8° = 2.074 Å ⋅ 0.85896 = 1.781Å
⇒ 𝑥 = 3.562Å
𝑦 = 2 ⋅ (𝐶 − 𝑇𝑖) ⋅ sin 30.8° + (𝐶 − 𝐶) = 2.124 + 1.424 = 3.548 Å
C. Complexes of chromium and cobalt
1.12.
Write down the missing name or formula respectively for each of the compounds (a) to (e).
(a) pentaamminbromidocobalt(III)-sulfate
(b) pentaamminsulfatocobalt(III)-bromide
(c) [CrCl2(OH2)4]Cl · 2 H2O
(d) [CrCl(OH2)5]Cl2 · H2O
(e) potassium hexafluoridocobaltat(III)
1.13. Assign (a) to (e) with the roman numbers I to V.
I:
(a)
II: (e)
III: (c)
IV: (b)
V: (d)
Give your reasoning for the classification using either calculations or chemical equations.
Vessels III, IV and V must contain bromide, and twice chloride respectively.
Vessel I: SO42- + Ba2+  BaSO4
𝑚
3.28
M(c) = M(d) = 266.47 g/mol 𝑛 = 𝑀 = 266.47 = 0.123 mol ⇒ corresponds to 1.76 g AgCl
 vessel III contains (c)
Mass loss by drying  crystal water  have only (c) and (d)  vessel V must (d).
molar mass 290.23 g/mol  (e)
1.14. How do you call the constitution isomerism of (a) and (b)?
ionisation isomersm
1.15. How many diastereomeres may occur in the case of the cation of (c)? Give also the prefixes
which you use in naming them.
two -- cis and trans
3
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
1.16.
Calculate ΔO in kJ/mol.
ΔO = 156.7 kJ mol-1
calculation:
𝑐
𝐸 = ℎ 𝜆 = ℎ𝑐𝜈̃ = 6.6262 ⋅ 10−34 ⋅ 2.9979 ⋅ 108 ⋅ 13100 ⋅ 100 = 2.6022 ⋅ 10−19 J
E = 2.6022 ⋅ 10−16 ⋅ 6.0221 ⋅ 1023 = 156.7 kJ mol-1
1.17. Sketch a d-orbital scheme for the complex in (e)
1.18. Calculate the magnetic moment in units of μB.
μ = 4.90𝜇𝐵
calculation:
𝜇 = √𝑛(𝑛 + 2)𝜇𝐵 = √4(6)𝜇𝐵 = 4.90𝜇𝐵
4
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
Task 2
13 points
Some Equilibria
A. Electrochemical Equilibria – Alkaline Cell
2.1. Write down the reaction equations for the generation of MnOOH and Mn(OH)3 respectively,
starting from Mn2O3 using water as reaction partner.
Mn2O3 + H2O ⇌ 2 MnOOH
Mn2O3 + 3 H2O ⇌ 2 Mn(OH)3
2.2. Calculate the free standard reaction enthalpy for the reaction Mn2O3 → Mn(OH)3.
ΔG0= 77 kJ
calculation:
2 Mn(OH)3 + 2e- ⇌ 2 Mn(OH)2 + 2 OHMn2O3 + 3 H2O + 2e- ⇌ 2 Mn(OH)2 + 2 OHMn2O3 + 3 H2O ⇌ 2 Mn(OH)3
E03 = 0.15 V
E04 = -0.25 V
E07 = E04 – E03 = -0.40 V
ΔG07 = -z·F·E07 = 77 k
2.3. Calculate the free standard reaction enthalpy for the reaction Mn2O3 → MnOOH.
ΔG0= 5.8 kJ
calculation:
2 MnOOH + H2O + Zn ⇌ 2 Mn(OH)2 + ZnO
Zn + 2 OH- ⇌ ZnO + H2O + 2e2 MnOOH + 2 H2O + 2e- ⇌ 2 Mn(OH)2 + 2 OHMn2O3 + 3 H2O + 2e- ⇌ 2 Mn(OH)2 + 2 OHMn2O3 + H2O ⇌ 2 MnOOH
E02 = 1.06 V
E01 = 1.28 V
E08 = E02 – E01 = -0.22 V
E04 = -0.25 V
E09 = E04 – E08 = -0.03 V
ΔG09 = -z·F·E09 = 5.8 kJ
2.4. Calculate the solubility product of Mn(OH)2.
KL = 1.42·10-13
calculation:
Mn2+ + 2e- ⇌ Mn
Mn(OH)2 + 2e- ⇌ Mn + 2 OHMn(OH)2 ⇌ Mn2+ + 2 OH-
E05 = -1.18 V
E06 = -1.56 V
ΔG05 = -z·F·E05 =228 kJ
ΔG06 = -z·F·E06 =301 kJ
ΔG010 = ΔG06 - ΔG05 = 73.3 kJ
𝛥𝐺0
10
𝐾𝐿 = 𝑒 −𝑅·𝑇 = 1.42 · 10−13
5
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
B. The coral reef
2.5. Calculate the minimum concentration of carbonate ions in sea water in order to build the lime
frame.
[𝐶𝑂32− ] = 8.3·10-7 mol/L
calculation:
[𝐶𝑎2+ ] =
0.42
40.08
= 0.0105 𝑚𝑜𝑙/𝐿
[Ca2+]·[CO32-] = 8.7·10-9
[𝐶𝑂32− ] =
⟶
8.7·10−9
[𝐶𝑎 2+ ]
=
8.7·10−9
0.0105
= 8.3 · 10−7 𝑚𝑜𝑙/𝐿
2.6. Calculate the highest pH, where the lime frame dissolves.
pH = 6.92
calculation:
[𝐶𝑂32− ]·[𝐻3 𝑂 + ]
[𝐻𝐶𝑂3− ]
= 5.6 · 10−11
⟶
[𝐻𝐶𝑂3− ] =
[𝐶𝑂32− ]·[𝐻3 𝑂 + ]
5.6·10−11
=
8.302·10−7
5.6·10−11
[𝐻2 𝐶𝑂3 ] + [𝐻𝐶𝑂3− ] + [𝐶𝑂32− ] = 2.3 · 10−3 𝑚𝑜𝑙/𝐿
· [𝐻3 𝑂+ ] = 14826 · [𝐻3 𝑂+ ]
⟶
[𝐻2 𝐶𝑂3 ] = 2.3 · 10−3 − 14826 · [𝐻3 𝑂+ ] − 8.302 · 10−7
[𝐻𝐶𝑂3− ]·[𝐻3 𝑂 + ]
[𝐻2 𝐶𝑂3 ]
=
14826·[𝐻3 𝑂 + ]2
2.299·10−3 −14826·[𝐻3 𝑂 + ]
= 4.3 · 10−7
⟶
[H3O+] = 1.21·10-7
⟶
pH = 6.92
C. A mixture of acids
2.7. Calculate the pH of this solution. Neglect the autoprotolysis of water.
pH = 2.88
calculation:
𝑝𝐻 = 0.5 · (4.76 − 𝑙𝑜𝑔(0.1)) = 2.88
2.8.
Give for the concentrations of all particles in this mixture numbers or terms with variables,
whereby you should show reasonable assumptions about negligible species.
[HAc] ≈ 0.05
[Ac-] = x
[H2SO4] ≈ 0
[HSO4-] = a
[SO42-] = b
[H3O+] = 10-(2.88-0.300) = 10-2.58 = 0.00263 mol/L
6
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
2.9. Calculate the concentration of the diluted sulphuric acid, which is added to the acetic acid
solution.
c = 0.00251 mol/L
calculation:
𝑥·0.00263
0.05
= 10−4.76
⟶
x = 0.0003303 mol/L
[H3O+] = x + a + 2b
⟶
0.00263 = 0.0003303 + a + 2b
𝑏·0.00263
𝑎
⟶
𝑏=
⟶
0.00263 = 0.0003303 + a + 2·4.571·a
⟶
a = 0.000227 mol/L
⟶
b = 0.00104 mol/L
= 10−1.92
10−1.92 ·𝑎
0.00263
= 4.571 · 𝑎
𝑐𝐻𝑡𝑜𝑡𝑎𝑙
= 𝑎 + 𝑏 = 0.001253 𝑚𝑜𝑙/𝐿
2 𝑆𝑂4
Before dilution:
c = 0.00251 mol/L
2.10. Give reasoning for your assumptions in 2.8.
[HAc] = 0.05 – x
[𝐻2 𝑆𝑂4 ] =
x = 0.0003303 mol/L
⟶
assumption [HAc] ≈ 0.05 true
[𝐻𝑆𝑂4− ] · [𝐻3 𝑂+ ] 0.0002237 · 0.002613
=
= 5.85 · 10−9 𝑚𝑜𝑙/𝐿
𝐾𝐴1
102
⟶
assumption [H2SO4] ≈ 0 true
7
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
Task 3
15 points
Kinetics and Thermodynamics
A. Two kinetic problems
3.1. Write down the differential rate law for reaction (1) and give proof by a short calculation.
𝑣 = 𝑘 ∙ [𝑡 − 𝐵𝑢𝐵𝑟]
rate law:
calculation:
𝑡 − 𝐵𝑢𝐵𝑟: 𝑉 ∝ 𝑚 ∝ 𝑛 ∝ 𝑐; 𝑉(𝐴𝑔𝑁𝑂3 ) ∝ 𝑣
𝑡 − 𝐵𝑢𝐵𝑟(𝐸𝑥𝑝1 − 𝐸𝑥𝑝2): 𝑐(2): 𝑐(1) = 1.5: 1 𝑎𝑛𝑑 𝑉(2): 𝑉(1) = 1,5: 1 ⇒ 1st order
𝐻2 𝑂(𝐸𝑥𝑝3 − 𝐸𝑥𝑝2): 𝑐(3): 𝑐(2) = 1: 0.75 𝑎𝑛𝑑 𝑉(3): 𝑉(2) = 1: 1 ⇒ zero order
3.2. Calculate the initial concentration of t-butyl bromide.
exp1: c0 (t-BuBr) = 0.212 mol/L
exp2: c0 (t-BuBr) = 0.318 mol/L
calculation for one example:
𝑚 = 𝜌 ∙ 𝑉 = 1.20 ∙ 1.21 = 1.452 g
𝑛=
𝑐=
𝑚
1.452
=
= 0.0106 mol
𝑀
137
𝑛
0.0106
= 50 = 0.212 mol/L
𝑉
3.3. Calculate the concentration of t-butyl bromide after one hour for experiment (1) and (2).
ct (1) = 0.201 mol/L
ct (2) = 0,302 mol/L
calculation for one example:
𝑛(𝐴𝑔+ ) = 𝑛(𝐵𝑟 − ) = 𝑉(𝐴𝑔+ ) ∙ 𝑐(𝐴𝑔+ ) = 10.6 ∙ 0.05 = 0.530 mmol
𝑛𝑡 (𝑡 − 𝐵𝑢𝐵𝑟) = 𝑛0 (𝑡 − 𝐵𝑢𝐵𝑟) − 𝑛(𝐵𝑟 − ) = 10.6 − 0.53 = 10.07 mmol
𝑐𝑡 =
𝑛𝑡
𝑉
=
10.07
50
= 0.2014 mol/L
8
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
3.4. Calculate the rates in experiment 1 and 2 using the differential rate law, and thereby find also a
mean value for the rate constant.
v(1) = 0.0106 mol/L·h
v(2) = 0,0159 mol/L·h
k = 5.00·10-2 h-1
calculation:
𝑣=
𝑐0 (𝑡 − 𝐵𝑢𝐵𝑟) − 𝑐𝑡 (𝑡 − 𝐵𝑢𝐵𝑟) 0.212 − 0.2014
=
= 0.0106 mol/L ∙ h
∆𝑡
1
𝑣
0.0106
𝑘=
=
= 5.00 · 10−2 h−1
𝑐0 (𝑡 − 𝐵𝑢𝐵𝑟)
0.212
Concerning reaction (2):
3.5. Which reaction order do you suppose based on the chemical system in question?
reaction order: 2nd order
3.6. Calculate the initial concentrations of the starting products.
c0(methyl bromide) = 1.00 mol/L
c0(sodium methylate) = 1,00 mol/L
calculation:
𝑚
94.9
𝑚
54.0
𝑐(𝐶𝐻3 𝐵𝑟) = 𝑀∙𝑉 = 94.9∙1 = 1.00 mol/L
𝑐(𝐶𝐻3 𝑂− ) = 𝑀∙𝑉 = 54.0∙1 = 1.00 mol/L
3.7. Calculate the concentrations of methyl bromide and sodium methylate respectively after 30, 60,
90, 120 and 150 minutes.
ct(30) = 0.984 mol/L
ct(60) = 0.968 mol/L
ct(120) = 0.938 mol/L
ct(150) = 0.923 mol/L
calculation for one example:
𝑐𝑡 =
𝑉(𝑚𝑖𝑥𝑡𝑢𝑟𝑒)−𝑐(𝐴𝑔+ )∙[𝑉(𝐴𝑔+ )−𝑉(𝑆𝐶𝑁 − )]
𝑉(𝑚𝑖𝑥𝑡𝑢𝑟𝑒)
50−0.1∙[50−𝑉(𝑆𝐶𝑁 − )]
45+0.1∙𝑉(𝑆𝐶𝑁 − )
=
50
50
45+0.1∙33.9
𝑐𝑡 (60) =
= 0.9678 mol/L
50
𝑐𝑡 =
9
ct(90) = 0.953 mol/L
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
3.8. Proof your assumption in part 2.7. by a calculation and find a mean value of the rate constant.
k (30) = 5.56·10-4 L/mol·min
k (60) = 5.55·10-4 L/mol·min
k (90) = 5.53·10-4 L/mol·min
k (120) = 5.53·10-4 L/mol·min
k (150) = 5.55·10-4 L/mol·min
kM = 5.54·10-4 L/mol·min
calculation for one example:
𝑘=
1
𝑡
1
𝑐𝑡
1
𝑐0
∙( − )
1
1
𝑘(60) = 60 ∙ (0.9678 − 1) = 5.55 ∙ 10−4 L/mol ∙ min
k remains constant ⇒ 2nd order
10
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
B. Four (extraordinary?) compounds
3.9. What is element X?
X: Fluorine, F2
3.10.Give the formulae of A, B, C, and D. Show your calculations, if this is possible.
A: CF4
B: PF3
C: XeF4
D: SF4
Calculation:
𝑚
A: 𝑓𝑟𝑜𝑚 𝑝𝑉 = 𝑛𝑅𝑇 𝑎𝑛𝑑 𝑛 = 𝑀 𝑎𝑛𝑑 𝜌 =
𝑀(𝐴) =
3570∙8.314∙300
101300
𝑚
𝑉
𝑤𝑒 ℎ𝑎𝑣𝑒: 𝑀 =
𝜌∙𝑅∙𝑇
𝑝
= 87.9 g/mol
3 fluorine: M(element) = 87.9 − 57 = 30.9 ⇒ P
4 fluorine: M(element) = 87.9 − 76 = 11.9⇒ C
PF3 or CF4, the description of properties and the fact that B obviously is PF3, lead to CF4.
C: 207 - 3·19 = 150 g/mol, could be M of samarium, SmF3 possible, but not a molecule!
207-4·19 = 131, is M of Xe, therefore XeF4
3∙19
D: 𝑀(𝐷) = 0.704 = 81.0 g/mol 𝑀(𝑒𝑙𝑒𝑚𝑒𝑛𝑡) = 81.0 − 57 = 24.0 g/mol could be Mg, but MgF3 does not
exist
4∙19
g
𝑀(𝐷) = 0.704 = 108 g/mol . 𝑀(𝑒𝑙𝑒𝑚𝑒𝑛𝑡) = 108 − 76 = 32 mol, sulphr, SF4 makes sense
3.11.At first sight, A and C have very similar fornulae. In fact, their three dimensional structures are
quite different. Sketch the three dimensional structures of both molecules using the VSEPRtheory, and find for both structures the point groups.
TD
3.12. Write down a balanced equation for this synthesis.
P4 + 6 H2 ⇄ 4 PH3
11
D4h
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
3.13.Calculate the standard reaction enthalpy of process 3.12., and consequently the standard
formation enthalpy of phosphane at 298 K.
𝑂
∆𝐵 𝐻298
= 1.25 kJ/mol
Calculation:
The equation in 3.6. follows from: (a) + (b) + 3×(c) - 4×(d)
-2984 – 415 - 3·572 + 4·1280 = 5.0 kJ for 4 mol ⇒ 1.25 kJ/mol
3.14.Calculate the equilibrium constant of the phosphane synthesis at 300°C.
K = 4,9·10-9
Calculation:
∆𝑋 = ∑𝑖 𝑋𝑖 (𝑒𝑛𝑑) − ∑𝑖 𝑋𝑖 (𝑠𝑡𝑎𝑟𝑡)
∆𝐻(𝑇2 ) = ∆𝐻(𝑇1 ) + ∆𝐶𝑃 ∙ ∆𝑇
∆𝑆(𝑇2 ) = ∆𝑆(𝑇1 ) + ∆𝐶𝑃 ∙ 𝑙𝑛
𝑇2
𝑇1
∆𝐺 = ∆𝐻 − 𝑇 ∙ ∆𝑆
∆𝐶𝑃 = 4 ∙ 37 − 6 ∙ 29 − 21 = −47 J/K
𝑂
∆𝑅 𝐻298
= 5 − 59 = −54 kJ
𝑂
∆𝑅 𝐻573
= −54 + (573 − 298) ∙ (−0.047) = −66.925 kJ
𝑂
∆𝑅 𝑆573
= −226 + (−47) ∙ 𝑙𝑛
𝑂
∆𝑅 𝑆298
= 4 ∙ 210 − 6 ∙ 131 − 280 = −226 J
𝑂
∆𝑅 𝐺298
= −66925 − 573 ∙ (−256.73) = 80181 J
∆𝐺𝑂
80181
𝐾 = 𝑒 − 𝑅𝑇 = 𝑒 −8.314∙573 = 4.9 ∙ 10−9
12
573
298
= −256.73 J
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
Task 4
17points
Nitrogen-Heterocycles
Cimetidin, an antagonist for histamine receptors
4.1. Draw the constitutional formulae of A, B, D, E, and F.
N
NH2
HS
OH
N
H
H
N
B
F
A
S
N
S
N
H
N
H
N
N
S
N
H
S
N
H
N
H
CN
E
D
4.2. Which type of reaction is meant by the conversion from C to E, and from E to cimetidine (Look at
the –CN-group!)
1,4-nucleophilic addition to a conjugated system, then elimination
4.3. Draw the mechanism of the reaction C→E. Use for C R-NH2, and arrows to show the positions of
the respective attacks.
S
S
R-NH2
- H+
N
R
H S
S
N
R
- CH3S-
N
N
N
13
-
H
N
N
N
N
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
B: Ketorolac, an inflammation inhibitor
4.4. Draw the structural formulae of U, V, X, and Y.
N
N
O
O
COOH
U
COOCH3
V
N
I
N
O
COOCH3
O
COOCH3
Y
X
4.5. What is the mechanism of the reaction X →Y? Think about AIBN.
radical cyclisation
Ketorolac appears as racemic mixture, whereby only the (S)-enantiomere has a medical effect.
4.6. Draw the structure of the (S)-enantiomere of ketorolac.
COOH
N
O
C. Norflurazon and Diclomezin, two herbicides belonging to the pyridazinon-type
4.7. Draw the structural formulae of G, H, and J.
+
N2 Cl-
NH-NH2
CH3NH2
CF3
CF3
G
J
H
4.8. Write down the formula of the reaction product of a carboxylic acid RCOOH with hydrazine.
How do you call this class of substances?
formula:
name:
RCONHNH2
acid hydrazides, hydrazides
14
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
4.9. Write down the formula of the reaction product of an aldehyde RCHO with hydrazine.
How do you call this class of substances?
formula:
name:
RCH=NNH2
hydrazone
4.10. Allocate the two N-atoms in compound I to the classes of compounds from above.
Hydrazon
CF3
N
N
Cl
O
Cl
Hydrazid
In reaction H →I, mucochloro acid is involved. Some sub-tasks for this issue:
4.11. Draw the possible stereo isomers of mucochloro acid.
H
Cl
Cl
O
O
O
H
Cl
Cl
O
OH
OH
(E)
(Z)
4.12. Write down a balanced equation for this reaction.
C5H4O2 + 5 Cl2 + 3 H2O → C4H2Cl2O3 + CO2 + 8 HCl
4.13. Mucochloro acid may also occur in a tautomeric cyclic form K.
Draw the structural formulae of K.
Cl
Cl
HO
H
O
O
K
4.14. Write down the molecular formula for the particle which causes the peak at 126 m/e.
[C3H235Cl37ClO]+
15
40th Austrian Chemistry Olympiad
National Competition - Deutschlandsberg
Theoretical Part - solution
June 12th, 2014
4.15. Draw the structure formulae of the compounds L, M, N, and O.
COOH
O
O
O
O
L
M
COOH
Cl
Cl
O
O
Cl
N N
H
Cl
O
N
16