Transcript Problema 3 Essential ozone Some hydrocarbon

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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
OH
1) O3
EtONa
EtONa
2) Zn/H+
EtOH
EtOH
O
C10H16
O
O
A
OH
1) O3
H2SO4
+
2) NaBH4
OH
C10H16
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.
1) O3
2) Me2S
O
O
C
O
C
H
2) Me2S
O
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CH3
H3C
O
CH3
G
O
C
C
H
C
H
O
C
H
F
E
C
H
O
C
H
O
F
O
C
O
C
H
F
1) O3
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C
H
E
O
C
H
CH3
G
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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
O
H
OH
O
O
Q
P
O
−
OH
O
OH
−
OH
O
H
O
Q
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:
47 IChO 2015
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Soluzioni preliminari dei problemi preparatori
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OH
1) 9-BBN
O
PCC
−
2) H2O2/OH
1) LiN(SiMe3)2
2) CH2=CHCH2Br
J
I
K
O
O
O
O
O2 PdCl2 CuCl
NaH
NaH
OH
THF
L
THF
M
O
O
O
Al(CH3)3
BF3-OEt2
HCOOH
+
N
O
pentalenene
Soluzione proposta da
Mauro Tonellato - ITI Marconi - Padova
47 IChO 2015
Baku – Azerbaijan
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