www.pianetachimica.it
47 IChO 2015 Baku – Azerbaijan Soluzioni preliminari dei problemi preparatori 1 Problema 3 Essential ozone
Some hydrocarbon C10H16 participates in the transformations given in Scheme 1.
1. Determine the structural formulae of the hydrocarbon C10H16 and the molecules A–D accounting for the fact that compounds C and D are isomers of the initial hydrocarbon; the ozonolysis of C followed by the treatment of the reaction mixture with alkaline H2O2 produces a single product while the same transformations of D afford two compounds.
Solution
O
O
OH
O O
C10H16
1) O3 2) Zn/H+
EtONa EtOH
A
EtONa EtOH
OH
OH C10H16
1) O3 2) NaBH4
H2SO4
B C D
+
Some other hydrocarbon E (ωC = 90.6%) under ozonolysis (1. O3, CH2Cl2, –78 oC; 2. Me2S) forms three carbonyl compounds – F (C2H2O2), G (C3H4O2), and H (C4H6O2) in a ratio of 3:2:1. Initial hydrocarbon E doesn’t decolorize bromine water.
2. Write down the structural formulae of hydrocarbon E and products of its ozonolysis F–H.
Solution
Hydrocarbon E has 90,6% of C, 9.4% of H. Then its empirical formula is C4H5. Hydrocarbon E is aromatic because it doesn’t decolorize bromine water, so it is C8H10 ortho-xilene.
C C O O
H H
C C O O
H H CH3
C C C H3
O 1) O3 O
2) Me2S
E F F H
C C O O
H H
C C O O
CH3
H CH3
C C
O O
H 1) O3
2) Me2S
E F G G
www.pianetachimica.it
47 IChO 2015 Baku – Azerbaijan Soluzioni preliminari dei problemi preparatori 2 Hydrocarbon I having center of symmetry was used as an initial material in the total synthesis of pentalenene (Scheme 2):
The ozonolysis of hydrocarbon I furnishes a single compound P or Q depending on the treatment of the ozonolysis product. Under treatment with I2 and NaOH, compound Q forms a yellow precipitate containing 96.7% of iodine. Under basic conditions compound Q is transformed into compound R containing 4 types of hydrogen atoms (4 signals in 1H NMR spectrum with integral intensity of signals 1:1:2:2). Molecular formula of R is C5H6O. Molecule of compound N has bicyclic framework containing R as a fragment. Molecule of O consists of three rings.
3. Descript the scheme of the synthesis of pentalenene.
Solution
P and Q differ for a single oxygen atom then Q is ketone and aldehyde and Q is ketone and acid.
Q is a methylketone because it gives the iodoform reaction. Q and P are then:
O
OH O
O
H O
P Q
O
H O
O OH O
Q
OH−
OH−
R
Assembling P and Q we obtain I. There are two possibilities:
But only the first one is correct because it has a center of symmetry.
Then the synthesis of pentalenene is as follows:
www.pianetachimica.it
47 IChO 2015 Baku – Azerbaijan Soluzioni preliminari dei problemi preparatori 3
OH O
1) 9-BBN 2) H2O2/OH−
PCC 1) LiN(SiMe3)2
2) CH2=CHCH2Br
I J K
O O
O
O
OH O2 PdCl2 CuCl
L
NaH
M
THF
NaH THF
O O O
BF3-OEt2 HCOOH
N
+
Al(CH3)3
O pentalenene
Soluzione proposta da
Mauro Tonellato - ITI Marconi - Padova
