1
EXAM OF SCIENTIFIC CULTURE
CHEMISTRY
PROBLEM 1:
Theophylline (TH), the structure of which is presented below, is a bronchial-dilator used for the
treatment of asthma.
O
H3 C
N1
O
H
N7
N3
N9
CH3
1.1 Structural study and acid-base properties
1.1.1 Indicate for the nitrogen atoms 7 and 9 their hybridization state, the nature of the orbital
containing the non-bonding electron lone pair, and specify whether this orbital is in the same plane as
the molecule or out of the plane.
1.1.2 Write another form of the theophylline which allows identifying its aromatic character and
specify the number of electrons involved in the aromaticity.
Theophylline (TH) behaves like a weak acid. The acid-base couple (TH/T-) is characterized by a
pKa1 of 8.4.
1.1.3 Present the structure of the conjugated base of theophylline (T-) and write another form
showing the most important delocalisation of the electrons.
Theophylline is administered as an ammonium salt of theophylline, the cation is symbolized by
RNH3+. The pKa2 value is given: pKa2 (RNH3+/RNH2) = 9.6.
1.1.4 In water, theophylline is less soluble than its ammonium salt. Explain.
1.1.5 What is the pH of a solution of the ammonium salt of theophylline at a concentration of 10-1
mol.L-1?
Theophylline is also a ligand for metal ions, such as copper. In the presence of a salen ligand a neutral
complex is formed with copper in a +II oxidation state:
CH3
O
N
O
H3C
N
N
NH2
H2N
N
OH
salen
7N
Cu
N
O
complex
1.1.6 Write the electronic configuration of the Cu2+ ion (Cu: Z = 29).
1.1.7 How many electrons can be found in the valence shell of copper in this complex?
2
1.1.8 Even though the 18 electron rule is not verified, such a complex is stable. Propose an
explanation.
1.2 Oxidation of theophylline
The oxidation of theophylline (TH) by dissolved dioxygen in the presence of the enzyme xanthine
oxidase, results in the formation of 1,3-dimethyluric acid (UH) and hydrogen peroxide (H2O2). This
reaction was studied at 300 K and implies the couple UH / TH. The redox equation is given below:
O
O
H 3C
H 3C
N
N
2 e-
+
+
N
N
N
N
2 H+
+
O
O
H
H
O
N
H
CH3
H2O
N
CH3
1,3-dimethylurique acid (UH)
Theophylline (TH)
–
1.2.1 1,3-dimethyluric acid (UH) is a weak acid with a pKa (UH/U ) of 5.4. Which form of this acid
is predominant at physiological pH?
–
–
1.2.2 Write the redox equation of the couple (U /TH) and explicit the potential E1 of the couple (U
–
/TH) as a function of its standard potential E1° and the concentrations of [H+], [U ] and [TH]. How
does this expression change when E1°’ is introduced into the equation? E1°’ corresponds to the
–
standard biological potential of the couple (U /TH) with all the concentrations equal to their standard
value except for the proton concentration which equals to 10-7 M (so that the pH is fixed at 7).
1.2.3 Write the equation of the redox couple (O2/H2O2) and explicit the potential E2 of the couple
(O2/H2O2) as a function of E2°’ (standard biological potential of the couple O2/H2O2) and the
concentrations of [O2] and [H2O2].
1.2.4 Write the reaction that occurs spontaneously under these standard biological conditions beween
–
the couples (U /TH) and (O2/H2O2) knowing that E1°’ = - 0,36 V and E2°’ = 0,3 V. Justify you
answer.
1.3 Ribofuranosyl theophylline
Ribofuranosyl theophylline 5 was synthesized in order to evaluate its antiviral properties. This
synthesis was carried out starting from D-ribose 1 and 4,5-diamino-1,3-diméthyluracile 2 :
CH3
CH3
HOCH2
O
OH
1
(a)
HO
H2N
4
H2N
5
N
H 2N
O
+
4'
HO
CH3
3'
+
N
2'
HO
O
1
O
1'
5'
N
OH
N
OH
N
OH
2
H2O
CH3
O
3
CH3
O
(b)
3
+
N
OC2H5
O
OC2H5
HOCH 2
O
N
CH3
O
HO
OH
+
N
N
4
O
5
2 C2H5OH
+
CH3CO2H
3
D-ribose 1
exists in aqueous solution as a cyclic form and also in very small quantities as an open form.
1.3.1 Write the mechanism of the reaction that transforms the cyclic form into the open form in an
acidic medium.
1.3.2 Treatment of D-ribose 1 by 4,5-diamino-1,3-diméthyluracile 2 results in the formation of 3
according to reaction (a). Write the mechanism of the reaction (a), catalyzed by a weak acid AH,
which transforms the open form of D-ribose into compound 3.
1.3.3 The amine function of 2 carried by carbon 4 is less nucleophilic than the one carried by carbon
5. Represent another mesomeric form of 2 which allows justifying this difference in nucleophilicity.
When compound 3 is treated with diethoxymethyl acetate 4, ribofuranosyl théophylline 5 is formed
according to reaction (b). This transformation can be interpreted under acidic catalytic conditions as a
succession of four steps, denoted as (i), (ii), (iii) and (iv). For these steps, compounds 3 and 5, and the
intermediates are represented by a simplified formula.
1.3.4 Propose a mechanism for step (i) below:
O
(i)
H
OC2H5
C
+
CH
H 3C
O
CH3-CO2H
H+
+
C2H5
O
OC2H5
C
A
4
OC2H5
1.3.5 Specify the electrophilic or nucleophilic behavior of the intermediate A and propose a
mechanism for step (ii) below:
C2H 5O
(ii)
C2H5
O
H
N
H2N
H
C
C2H5O
+
OC2H5
N
HO
A
+
H+
N
HO
3
B
1.3.6 Propose a mechanism for the following step (iii) and represent the monocyclic reaction
intermediate C:
C2H5O
H
N
H
N
C2H5O
(iii)
+
H
+
C2H5OH
+ C
N
N
HO
+
C2H5O
O
B
D
1.3.7 What is the reaction type of the following step (iv)?
H
N
(iv)
N
H+
+
C2H5O
N
N
O
O
D
5
C2H5OH
H+
+
C2H5OH
4
ÉPREUVE DE CULTURE SCIENTIFIQUE
CHIMIE
PROBLEM 2 :
This exercise examines part of the vernolepin synthesis reported in 1976 by the Danishefsky’s group.
2.1 In a first step, one considers the following sequence of reactions:
2.1.1 What is the name of the reaction leading to the formation of the bicyclic molecule 3? Indicate in
the formula of the product 3 which new σ C-C bonds have been formed during this step.
2.1.2 The following diagram displays the overlap of the molecular orbitals, which are involved in the
formation of the product 3. Make an interpretation of the diagram. Indicate which overlap drives the
cyclization reaction according to respective values of the energy differences ∆E1 and ∆E2.
2.1.3 Solubilized in an acidic aqueous solvent, the compound 3 yields the ketoester 4. Propose a
mechanism to account for this conversion. Which side-products are formed during this conversion?
2.2 Two subsequent transformations are performed:
2.2.1 Draw the structure of the product 5 associated to the formula C11H12O3.
2.2.2 The conversion of the compound 5 into the product 6 is performed under kinetic control. What is
the corresponding information? What is the other regime of control which is often used in organic
chemistry?
5
2.2.3 Propose a mechanism for the formation of the product 6 from the compound 5. Which explanation
could you propose to account for the formation of only one stereoisomer during this step?
2.3 The compound 6 is treated by the 1,8-diazabicycloundec-7-ene (DBU), which acts as a base:
2.3.1 What is the most basic site in the DBU molecule? Justify.
2.3.2 What is a priori the mechanism involved in the conversion of the halogenated derivative 6 into the
ethylenic derivative 7? Justify your answer and write the corresponding mechanism.
2.4 Two oxidation steps follow:
Which reagents would you propose to achieve both conversions from 7 to 8, and from 8 to 9 ?
2.5 The carboxylic acid 9 is treated by diazomethane CH2N2 to yield the methyl ester 10:
2.5.1 Draw the Lewis structure of diazomethane.
2.5.2 Propose a mechanism to account for the conversion of 9 into 10.
2.6 The compound 10 is eventually converted into the aldehyde 11 by a series of transformations which
will not be examined. Part of the investigated synthesis is finally achieved by the formation of the
ethylenic derivative 12 after reaction of 11 with the reagent Ph3PCH2.
2.6.1 Draw the Lewis structure of the reagent Ph3PCH2.
2.6.2 Propose a mechanism accounting for the conversion of 11 into 12.
6
EXAM OF SCIENTIFIC CULTURE
CHEMISTRY
PROBLEM 3:
3.1. The glyphosate (GP), which is the main component in Roundup®, and 2Fluorophosphoenolpyruvate (FPEP) are both molecules that block the biosynthesis of aromatic amino
acids.
pKa4 = 10,6
O-PO3H2
H
HO2C
N
H
1
PO3H2
pKa3 = 2,2
CO2H
F
GP
pKa1 = 0,7
pKa2 = 5,9
FPEP
3.1.1 What is the state of protonation of GP at physiological pH (pH = 7)?
3.1.2 Write the Lewis structures of the CO2H and PO3H2 groups.
3.1.3 Does the presence of the nitrogen atom in GP have an influence on the acidity of the functional
group PO3H2 (compared to CH3-PO3H2) and on the functional group CO2H (compared to CH3CO2H)?
Explain your reasoning.
3.1.4 Which is the configuration, E or Z, of the molecule FPEP?
3.1.5 During the inhibition FPEP is protonated and forms a carbocationic intermediate at carbone 1.
Represent this carbocation and compare its stability to the one of the same carbocation formed from
O-PO3H2
PEP :
H2C
PEP
CO2H
Glyphosate (GP) is an inhibitor of the enzyme 5-enolpyruvoyl-shikimate-3-phosphate synthetase
(denoted E). Thermodynamic values, such as the dissociation constant Kd’ and the standard biological
enthalpie ∆rH°’ were determined for the dissociation reaction of the enzyme-inhibitor complex [E-GP]
and the enzyme-substrate-inhibitor complex [E-S-GP]. Kd’ and ∆rH°’ correspond to the dissociation
constant and the reaction enthalpy with all the concentrations equal to their standard value except for
the proton concentration which equals to 10-7 M (so that the pH is fixed at 7).
(1)
[E-GP]
E + GP
Kd1’ = 2.10-2
∆rH°’1 = 55 kJ.mol-1
(2)
[E-S-GP]
[E-S] + GP
Kd2’ = 3.10-7
3.1.6 Does a comparison of the values Kd1’ and Kd2’ allow to deduce whether GP is more strongly
fixed to the enzyme in the absence or in the presence of substrate S?
3.1.7 Does the value of ∆rH°’1 give any information about the interactions between the active site of
the enzyme and the inhibitor? Explain your reasoning.
7
3.2. We are now interested in the synthesis of a precursor of the inhibitor FPEP starting from ethyl
difluoroacetate CHF2-CO2H and diethyl oxalate EtO2C-CO2Et. The first step is a Claisen condensation
in ethanol in the presence of sodium ethanolate CH3-CH2-O−, Na+ (EtO−, Na+). The reaction equation
is shown below:
EtO, Na
CHF2 -CO 2Et + EtO 2C-CO 2 Et
EtO 2 C-CF 2-CO-CO 2 Et + EtOH
3.2.1 Propose a mechanism for this reaction.
3.2.2 Would the same diester be obtained if sodium ethanolate would be replaced by sodium
methanolate (CH3-O−, Na+)? Justify your answer.
The second step of the synthesis is the acidic hydrolysis of the diester in water according to the
reaction below:
HCl
EtO 2C-CF2 -CO-CO 2Et
H 2O
HO 2C-CF 2-CO-CO 2H + 2 EtOH
We will consider the hydrolysis reaction of a monoester RCO2Et under the same conditions. The
constant k corresponds to the apparent rate constant of the water addition step to the protonated ester
R-(CO2H)+Et.
3.2.3 Knowing that the rate determining step is the addition of the nucleophile, write the rate law v of
the reaction as a function of k and Ka, where Ka is the acidity constant of the couple R-(CO2H)+Et/RCO2Et.
Water can react with the diacid 1 to form a hydrate according to the reaction (1):
F
O
F
F
HO
OH
+ H 2O
F
O
HO
OH
1
O
O
HO
O
(1)
OH
In the general case, represented by the hydration of acetone H3C-CO-CH3 in the reaction equation (2),
the equilibrium is in favour of the ketone:
O
HO
OH
+ H2O
H3C
CH 3
(2)
H 3C
CH 3
3.2.4 What is the influence of the inductive effects of fluoride on the reaction rate of the water
addition to the carbonyl of 1 compared to the one of acetone? Explain your reasoning.
The diacid 1 is then decarboxylated in basic medium. The reaction mixture is then again acidified in
order to form compound 2.
Since the ketone is in equilibrium with the hydrated form (see question 3.2.4), the decarboxylation can
be envisaged either starting from the hydrated molecule (mechanism A) or starting from the ketone
(mechanism B).
8
mécanisme A
COO
F
+ OH
OH
F
COO
F
O
COO
O
OH
F
F
H
O
+ CO2
O
F
O
H
O
H
O
H
+ OH
F2HC
COOH
F2HC
2
COO
mécanisme B
COO
F
O
O
F
F
CO2
F2HC
F
COO
O
O
+
COO
H3 O
O
O
H
+
OH
H
O
+ H2 O
F2HC
COOH
2
3.2.5 The first step of the mechanism A is an anti unimolecular elimination. Would the mechanism A
always be possible independent of the conformation of the hydrated form of compound 1? Justify
briefly.
3.2.6 Would a decarboxylation also be possible for the second carboxylate function CO2-? Justify.
The compound F2HC-CO-CO2H, 2, is then placed into methanol, CH3-OH, in the presence of a strong
acid. Three products are obtained under these conditions:
O
HO
F2HC
COOCH3
3
OCH3
F2HC
COOCH3
4
H3CO
OCH3
F2HC
COOCH3
5
3.2.7 Is the solution obtained in the end containing 3, 4 and 5 optically active? Justify your answer.
3.2.8 Would the formation of these three products be equally observed if the reaction were performed
in basic medium? Justify your answer.