synthetical methods for the preparation of hetarenes

Transcription

MAIN SYNTHETICAL METHODS FOR
THE PREPARATION OF HETARENES
1. Ring Synthesis Strategy
1.1. Hydrolytic disconnection
1.2. Redox disconnection
1.3. The mains precursors
1.3.1. Precursors as nucleophiles
1.3.2. Precursors as electrophiles
1.3.3. Precursors as both electrophiles and nucleophiles
2. Substituent modification
3. Nomenclature: I.U.P.A.C., general rules
References:
1. Alan R. Katritzky, A. F. Pozharskii Handbook of Heterocyclic Chemistry 2nd Edition, Pergamon 2000
2. Alan R. Katritzky Short Course in Heterocyclic Chemistry for Ph. D. Students, University of Florida
1996/1997
3. David T. Davis Chimie des Hétérocycles Aromatiques De Boeck Université 1997, Oxford University Press
1992
4. René Milcent Chimie Organique Hétérocyclique EDP Sciences 2003, www.edpsciences.org
5. Jonathan Clayden, Nick Greeves, Stuart Warren, Peter Wothers Organic Chemistry, De Boeck Diffusion
s.a., 2003, Oxford University Press 2001, www.deboeck.com
Modifications (improvements, additions, corrections, up to dates etc.) are subjected to no notice.
Mircea Darabantu, MASTER DIA. I N T R O
D-1
MAIN SYNTHETICAL METHODS FOR
THE PREPARATION OF HETARENES
D e f i n i t i o n : the term h e t a r e n e s designes all heterocyclic compounds possessing aromatic
character according to Hückel rule.
Synthesis
Ring
Synthesis
Combination of
the Two
Substituent
Modification
1. Ring Synthesis Strategy : the following three factors increase the importance
of the ring synthesis
a) Fused Ring vs. Mono cyclic
b) Five vs. Six memebered Ring
c) More Hetero atoms
N
pyridine
N
quinoline
N
acridine
)
Ii it is easier to synthesise a
second ring onto a first one
than
to
synhtesise
a
monocyclic compound.
X
pyrrole X=NH
thiophene X=S
furane
X=O
X
indole
benzothiophene
benzofurane
N
N
pyrimidine
N
N
H
imidazole
N
N
quinazoline
N
N
H
benzimidazole
X
carbazole
dibenzothiophene
dibenzofurane
N
N
The more heteroatoms one
has in the ring, the more
methods of synthesis they
are.
N
N
pteridine
N
N
N
H
purine
N
))
)))
Substitution reactions tend
to be easier on the whole
when
one
has
fewer
heteroatoms in the ring.
D-2
Baldwin’s Rules for 3 to 7 Membered Ring Closure
J. E. Baldwin J. Chem. Soc., Chem. Commun. 1976, 734
-
Z: sp3 Exo - Tet
Z
X
X
Z
X
Z
Exo-Tet: From 3 to 7 membered rings
-
Exo-Trig: From 3 to 7 membered rings
Z: sp2 Exo - Trig
Z
X
& FAVOURED
Y
Y
& FAVOURED
Y-
Y
Exo-Dig: From 3 to 4 membered rings
-
Z
X
&
Z: sp Exo - Dig
Z
X
Y
-
Y: sp3 Endo - Tet
X
Z
X
Y: sp2 Endo - Trig
Z
Z
Endo-Trig: From 3 to 5 membered rings
-
& DISFAVOURED
Endo-Trig: From 6 to 7 membered rings
&FAVOURED
Endo - Dig
Z
X
Z
-
Endo-Dig: From 3 to 7 membered rings
& FAVOURED
Y
Y
i)
ii)
iii)
iv)
& FAVOURED
Y
Y: sp
X
Endo-Tet: From 5 to 6 membered rings
Z
X
Y
-
& FAVOURED
Y
Y
- X
DISFAVOURED
Exo-Dig: From 5 to 7 membered rings
Y-
all the above rules have empirical support only.
disfavoured does not mean impossible but more difficult to realise.
the basic support of the above rules is stereochemical (bond lenghts and bond angles).
a lot of cases (before and after 1976 are in substantial accord with these rules).
(Y)
HO
(Z)
(Z=Y)
CH=O
OH
HO
Exo-Trig 6
OH
OH (X)
CH2OH
D-glucose
Endo-Trig
Ring Closure
5 Membered ring
OH
HO
O (X)
(X)HO
N
Z
OH
Ar Y
CH2OH
Pyranose
!
(Z)
O
NH
(X) (Y)
Ar
chain
ring
tautomerism
D. E. Bergmann, Chem. Rev. 1953, 53, 309-353
L. Lázár, F. Fülöp, Eur. J. Org. Chem. 2003, 3025-3042
D-2a
Examples : exo-cyclisations according to stereoelectronic requirements in the
transition state
.
2
3
2
Br
1
+ Br-
..
4
5
-
3
5
six membered
6-exo-tet cyclisation
(Intramolecular SN2)
FAVOURED
Stereoelectronically
2
1
H2N
3
4
six membered
6-exo-trig cyclisation
(Intramolecular RA)
FAVOURED
Br
N
H
aziridine
H
2
1O
5
O
O
OH
O
five membered
(tetrahydrofuran-2-one)
G e n e r a l i s a t i o n:
lone Y ..
pair
(donor)
X
σ∗
empty
(acceptor)
n-exo-tet cyclisation
k=0-4
FAVOURED
five membered
4
FAVOURED
( )k
5
5-exo-dig cyclisation
(Intramolecular RA)
FAVOURED
(three membered)
3-exo-tet cyclisation
FAVOURED
.
3
6
3
2
1
.
4
6
.
1
( )k
lone Y ..
pair
(donor)
X
π∗
empty
(acceptor)
n-exo-trig cyclisation
k=0–4
FAVOURED
exo-dig cyclisations are favoured for five to
seven membered rings
D-2b
Examples: endo-cyclisations according to stereoelectronic requirements in the
transition state
NOTE: only 6- and 7-endo-tet cyclisation are favoured
all 3 – 7-endo-dig cyclisations are favoured
almost all 3 – 7-endo-tet cyclisations are less favoured
Plausible but…
+
+
O
O
3
4
2
1NH
..
2
+
O
H
OEt
OEt
HN
5
π∗ empty
N ..
(acceptor)
H lone
pair
(donor)
5-endo-trig cyclisation
DISFAVOURED
(Intramolecular 1,4-addition)
lone
pair
H (donor)
H
N ..
O
OEt
OEt
π∗ empty
(acceptor)
Inappropriate
orientation
Too far...
This is the real:
3
4
2
5
1NH
..
2
EtO
HN
O
FAVOURED
O
A 6-endo-trig cyclisation: favoured, occuring in 89 % yield
O
MeO2C
MeOOC
O
Base
MeO2C
∗
π
..
2
O
O
O 3
4
MeO2C O
O
1
6
O
5
H
Appropriate
orientation
trans stereochemistry
with respect to
cyclanone ring
A 5-endo-dig cyclisation: favoured
π∗ empty (acceptor)
inapropriate orientation
O
O
Base
OH
i) 5-endo-dig cyclisation
FAVOURED
+
ii) H
Ar
2
1O -
3
4
O
5
Ar
O
O
_
O ..
lone
pair
(donor)
π∗ empty (acceptor)
apropriate orientation
D-3
1.1. Hydrolytic disconnection
- hydrolytic disconnections of the double bonds are very useful since most of the ring closures to afford
heterocyclic systems are simple condensation
H+ / -H2O
i) Nucleophilic addition
ii) Elimination
R1
R1
N R3 + H2O c o n d e n s a t io n
O + H2N-R3
R2
Imine
Schiff base
R2
Example 1:
originates
from...
R1
R1
N R3
good
O + H2N-R3
R2
R2
Target
Compound
R1
precursors
bad
NH + HO-R3
R2
o p t i o n !!
o p t i o n !!
precursors
R1
OR3
R2
NH2
e.g.
Example 2:
hydrolytic disconnection
EWG
EWG
O +
precursors
H2C
(EWG)
(EWG)
EWG: Electro(no) Withdrawing Group
CO, COOR, CN, etc.
Example 3:
N
heterocyclic compound
seen as a cyclic imine
R
N
heterocyclic compound
seen as a cyclic amine
precursor: good option !!
R
R
O
NH2
R
NH OH
R
H2N
- NH3
O
? bad option
D-4
Example 4:
Imine or enamine ? It doesn’t matter…
H
H
sp2
R
sp3
N
H
(masked)
enamine
R
H
sp2
R
OH
H
H
R
sp3
N
H
(masked)
enamine
NH2
H
H
R
O
NH2
N
imine
Example 5:
Retrosynthesis of pyrroles seen as hydrolytic disconnection:
basic
basic
center +H+ +H+ center
4
3
H3C
2 CH3
N1
O
H
H
2,5-dimethylpyrrole
seen as
acid
double enamine
center -H+
enol
H3C
CH3
NH H
5
H3C
H3C
CH3
CH3
O O
NH3
O N
H
acid
-H+ center
enamine
Example 6: the importance of the formal charges
Rings containing Nitrogen
+1
+1
+1
H -1
H
H
H
-2
+1
R
-3
N
H
+1
+1
+2
+1
R
OH
-2
+1
+1
R
NH2
O
+1 -3 2 x (+1)
-2
Rings containing Sulphur
the best precursor
+1
+1
+1
H -1
H
NH2 2 x (+1)
-3
H
H
-2
+1
R
-2
S
+1
+1
+2
+1
R
OH
-2
+1
O
+1 -2
-2
Rings containing Oxygen
SH
+1
-2
+1
H -1
H
best X
electrophile
H
-2
+1
R
-2
O
+1
+1
R
+1
OH
-2
OH
+1 -2
+2
+1
R
+1
O
-2
H
&
R
+1
+1
H
+1
R
SH
H
+1
OH
-2
Y
%
best
nucleophile
D-5
To be kept in mind:
1. If a hydrolytic disconnection appears suitable, the best pair nucleophile-electrophile sould be
considered
2. All cyclisations (or cyclocondensations) involve classic tautomerism: keto-enolic, imino-enamine, etc.
3. During cyclocondensation (or retrosynthetic hydrolytic disconnection) no global redox process,
involving the whole molecule occurs.
1.2. Redox disconnection
-this methodology provides information about the general strategy to be used to access the target
compound: is there any redox step ?
Example 1:
- a five membered hetarene:
it looks like
"organic"
N
H
i) cut (remove) the bonds
between the organic
and inorganic parts
ii) add the formal charges +1
O
add two functional
groups at both ends
of the "cation"
+1 to "neutralise" charges
-1
HO ]
N -1-3
H
O
-1
[OH
+1
-3
+1
3x(+1)
NH3
precursors
it looks like
"inorganic"
it sounds like
ammonia
Obs: the structure of the target compound (bond connection and aromaticity) is automatically issued by
preserving the formal charges of each of the involved (hetero)atoms: no redox step in the synthesis
Example 2:
- a six membered hetarene
redox
disconnection
+2
H3C
N
CH3
h
H3C +2 +1 CH3
N3-
H3C
+1
CH3
-2
-2 O OH
+1
-3 NH
3 3x(+1)
???
1
2[H]
H3C
3
H3C
2
O
7
5
4
CH3
6
O
CH3
O O
2,6-heptanedione
Z-1,4-diketone
currently nonavailable
commercial product
difficult to obtain
is the above disconnection useful for a chemist ? Why ?
a) Because it provides rapidly the type of precursors
b) Because it provides rapidly the most convenient precursors.
c) Because it infers that if, for reason of availability, a redox step is revealed (e.g. reduction) by the
redox disconnection, in the synthesis of the target molecule a redox step must be accomplished: e.g.
oxidation.
☺
D-6
The direct synthesis:
+ NH3
-H2O
H3C
OO
CH3
H3C
O
CH3 H3C
NH
-H2O
CH3
CH3 -H2O H3C 6 N 2 CH3
N
H3C
NH2
-H:- anticipated oxidation step
H
4
3
+ [O]
5
H
H3C
N
OH H
CH3
O
1
H
2,6-dimethylpyridine
-H+
1.3. The main precursors
1.3.1. Precursors as Nucleophiles: this is the traditional route to bring heteroatoms (N, S, O) in the target
compound
Example 1: b r i n g i n g one heteroatom in the target compound:
NH3
+
4
H2S
3
4
2
5
+
H2 O
3
2
5
N1
H
+
4
2
5
S
3
O
1
1
Example 2: b r i n g i n g two heteroatoms in the target compound:
H2N-NH2
HO-NH2
hydrazine
hydroxylamine
+
+
4
3
4
N2
5
5
N1
H
pyrazole
S
+
4
NH2
isoxazole
N3
5
S 2 NH2
1
NH2
S
R
4
N3
5
S 2 R
1
2-substituted thiazoles
amidines
NH2
O
+
+
+
N2
O1
Example 3: b r i n g i n g three atoms in the target compound:
thiourea
thioamides
urea
NH2
3
NH2
NH2
HN
6
HN 1
guanidine
4
5
5
4
6
O 2 N3
2-pyrimidone
NH2
R
HN
+
4
N3
+
5
2
N1
NH2
R
6
N3
2
N1
2-substituted pyrimidines
NH2
D-7
Example 4: b r i n g i n g three, four and five atoms in the target compound: benzoderivatives
NH2
NH2
5
+
4
7
8
N
+
8
3
7
2
6
6
NH2
NH2
N
N
5
1
Quinoline
5
1
2
6
3
7
4
4
3
N2
8
Quinoxaline
+
1
Isoquinoline
1.3.2. Precursors as Electrophiles:
Example 1: b r i n g i n g one atom:
δO
δ+
R
LG
NH2
)
4
O
R
6
NH2 Cl
)
R1
O
Cl
N1
H
2-substituted
benzimidazoles
7a
R1
:NH2
4
5
:S
R2
R
2
7
LG: leaving group as LG:e.g. Cl as Cl-; R'O as R'OH etc.
Example 2: b r i n g i n g two atoms:
δδO
O
δ+
δ+
δ+
δ+
R
OR
δδ
LG
LG
N3
3a
5
1S
N3
2
R2
2,4-disubstituted
thiazoles
Example 3: b r i n g i n g three atoms:
δO
Oδ
δ+
δ+
R
R
1,3-dicarbonylic
compounds
δ-
O
δ+
RO
Oδ
δ+
OR
malonic esters
O
)
O
4
R
5
1X
R
R
HX
NH2
N2
O
OR
O
R
3,5-disubstituted
X = NH pyrazoles
X = O isoxazoles
O
)
3
OR HN
NH2
R'
5
4
N3
O 6 N 1 2 R'
H
2-substituted-pyrimidin-4,6-dione
D-8
O
δO
δ+
R
R
R
)
δ+
acrylic derivatives
5
4
3
6
7
NH2
2
N
8
1
4-substituted-quinoline
Example 4: b r i n g i n g four atoms:
R
δ+
δ+
)
R
O O
δ- δ1,4-diketones
R
δ+
δ+
3
4
R
R
5
O O
X 2
R
1
H2X (X = NH, S)
Note: electrophilic fragments usually do not bring heteroatoms in the target compound
1.3.3. Precursors as both Electrophiles and Nucleophiles:
Example 1: b r i n g i n g two atoms:
δas Nu:O
δ+
R
LG
as E+
R2
)
R2 4
NH2
R3
O
R3
R1
O
δ+
O
δ-
as Nu:LG
LG
as E+
HO
LG
1
O 2 R
1
H
O
LG
5
2,4,5-trisubstituted-oxazoles
LG: leaving group as LG:e.g. Cl as Cl-
O
N3
LG
O
δ-
O
)
OH
HO
5
OR
δ+
OR
4
3
6
OR
2
7
8
O
1
O
3-substituted-coumarines
LG: OR, Halo
Example 2: b r i n g i n g three atoms, see Example 1
as E+
R3
δ+
R2
δO
NH2
as Nu:-
an α-aminoketone
D-8a
Example 3: acid chlorides as good electrophiles to generate good nucleophiles.
The target compound:
O
5
R1 4
R1 =/ R2 =/ H
O1
3N
2
R2
Retrosynthetic analysis:
O
5
R1 4
O1
3N
O
O
2
hydrolytic disconnection R1
N-3-C-2 and C-2-O-1
N
OH
O
R2
R2
poor ability as Leaving Group
weak nucleophile in this environmen
O
O
HN
R2
Cl
OH
R1
O
OH
NH2
OH
NH2
R2
O
R1
R1
O
better LG than HO
C=O more electrophilic
C
R2
R1
O
O
HN
OH
R2
weak
electrophile
Solution of the problem:
good LG
R2
O
O
O
OH
R1
Cl
HN
C
R2
O
R2
O
R1
Δ excess
N
H
O
C
R2
O Nucleophile
O
_
R2
O
-H+
O
R1
O
N
Note: obsolete cyclisation of α-aminoacids as double protected N-, COO- form (oxazoline)
R2
D-9
2. Substituent modification
-
the below three main types of substituent modification are of general interest
i) nucleophilic displacement:
π
X
π
+ Nu:-
+ LG:Nu
X
LG
ii) electrophilic displacement:
E
H
π
π
+ E+
+ H+
X
X
iii) electrophilic displacement via metallation:
π
X
H
+ R-Mn
- RH
π
X
Mn
π
+ E+
- Mn+
Mn: LiI, MgII etc
X
E
3. Nomenclature: IUPAC general rules
-
important trivial names are given at the beginning of each chapter: e.g. pyridine instead of azabenzene
the type of heteroatom present in the ring is indicated by the below prefixes, in decreasing order of
citation (nomenclature as a):
O oxa > S thia > Se selena > N aza > P phospha > As arsa > Si sila > B bora etc.
Note: the final a is however elided before a vowel e.g. azaole → azole
-
two or more identical heteroatoms are indicated by dioxa, triaza, etc.
If different, the prefixes should be combined according to the above order: e.g. thia-aza-ole → thiazole;
e.g. oxa-diaza-ole → oxadiazole; e.g. oxa-thia-ane → oxathiane.
numbering of the positions starts at an oxygen, sulphur or nitrogen (in decreasing order of preference)
and continues in such way that the heteroatoms are assigned the lowest possible numbers.
other things being equal, numbering starts at a substituted rather than at a multiply bonded nitrogen
atom.
4
5
N
3
2
5
4
N1
2
N3
N1
CH3
CH3
1-methyl-1,3-diazole 3-methyl-1,3-diazole
1-methylimidazole 3-methylimidazole
MAXIMUM UNSATURATION is defined as THE MAXIMUM POSSIBLE NON CUMULATIVE
DOUBLE BONDS BY CONSIDERING:
O- two valencies
S – two valencies
N – three valencies
D-10
The Hantzsch-Widman Nomenclature for Heterocyclic Compounds
Rings w i t h Nitrogen
Ring Maximum Unsaturation
Size
3
3
2
2
Eng.
3
-irine
N
N1
1
H
Rom.
1-azirine 2-azirine
-irină
3
2
Eng.
4
-ete
4 N
-
Eng.
-etine
1
azete
Rom.
-etă
Saturated
One Double Bond
Rom.
-etină
-
3
2
3
2
4
N1
4
NH
1
1-azetine 2-azetine
3
4
2
NH
O1
3
4
2
N
O1
Eng.
-iridine
Rom.
-iridină
Eng.
-etidine
Rom.
-etidină
3
O1
3
2
N1
H
N2
H
aziridine
oxaziridine
3
2
3
O2
4
NH
4
NH
1
1
azetidine 1,2-oxazetidine
2H-1,2-oxazetine
4H-1,2-oxazetine
5
Eng.
-ole
Rom.
-ol
6
N3
4
5
4
6
8
2
1,3oxazole
Eng.
-oline
Rom.
-olină
-
N3
5
2
N
1
5
N3
4
5
S1
2
4
5
N
N4
-
-
-
-
-
3
2
O1
5
4
6
7
3
Rom.
-ocină
2
N
8
1
azocine
9
Rom.
onină
5
6
Eng.
-onine
4
7
8
1
N 2
H
1H-azonine
9
2
4,5dihydropyrimidine
-
1,4-oxazepine
Eng.
-ocine
N3
1
6
7
2
Δ -thiazoline
6
1,3-diazine
pyrimidine
Eng.
-epine
Rom.
-epină
O1
N1
H
1,3diazole
imidazole
Rom.
-ină
7
5
2
Eng.
-ine
N3
4
3
Eng.
-olidine
3
4
NH
5
Rom.
-olidină
O1
2
1,3-oxazolidine
Nomenclature is made by using the
prefix perhydro (total hydrogenated)
which precedes the name of the
corresponding non saturated structure
4
6
3
NH
5
1
2
N
H
perhydro-1,3-diazine
D-11
The Hantzsch-Widman Nomenclature for Heterocyclic Compounds
Rings w i t h o u t Nitrogen
Ring Maximum Unsaturation
Size
3
3
Eng.
2
2
3
-irene
O
S
4
5
Rom.
irenă
Eng.
-ete
Rom.
-etă
1
oxirene
thiirene
4
3
O1 4
oxete 1,2-oxathiete
4
Eng.
-ole
4
3
5
2
O1
oxole
furane
3
5
2
S1
thiole
thiophene
Rom.
-olen
-
5
3
6
Rom.
-ină
2
S1
H
Eng.
-irane
Rom.
-iran
Eng.
-etane
Rom.
-etan
-
Eng.
-etene
Rom.
-etenă
Eng.
-olene
4
Eng.
-in
-
-
S2
O1
3
2
Rom.
-ol
6
1
Saturated
One Double Bond
4
5
2
3
O1
4
2 5
3
3
S1
2
Δ -oxolene Δ -thiolene
Eng.
-olane
Rom.
-olan
3
3
2
2
O
S
1
1
oxirane
thiirane
4
3
4
3
O1 2 S O1
oxetane
1,2-oxathietane
2
4
5
1O
4
3
O3 5
S1
2
1,3-dioxolane
thiolane
5
Eng.
-ane
Rom.
-an
6
4
1O
O3
2
1,3-dioxane
thiin
7
4
5
Eng.
-epin
6
Eng.
-ocin
Rom.
ocină
5
7
4
5
3
6
2
7
3
S1
8
1
4H-oxocin
Rom.
onină
Rom.
-epan
-
O
Eng.
-ocane
Rom.
-ocan
2
5
4
6
O
3
7
S1
2
thiepane-3-one
4
5
6
3
7
2
S1
8
thiocane
2H-thiocin
5
6
4
7
8
3
9
Eng.
-epane
1,3-dioxepin
4
Eng.
-onin
2
1
6
8
9
O
1
oxepin
8
7
2
O
O3
6
3
7
Rom.
epină
4
5
O
1
oxonin
2
5
6
Eng.
-onane
Rom.
-onan
2
4
7
8
3
9
O
1
2
oxonane
D-12
indicated hydrogen: H in compounds possessing maximum unsaturation, if the double bonds can be
arranged in more than one way, their positions are defined by indicating the nitrogen or carbon
atoms which are not multiply bonded and consequently carry an “extra” hydrogen atom: 1H, 2H,
etc.
in partially saturated compounds, the position of the hydrogen atoms can be indicated by the
prefixes dihydro if convenient suffix is not available
in partially saturated compounds, the position of the double bond can be indicated by the symbol Δa,b
which indicates that the double bond is between the atoms numbered “a” and “b”
-
-
Example 1:
4
5
5
2
4
3
4
3
5
2
2
N1
N1
N1
H
1H-pyrrole
3
3H-pyrrole
2H-pyrrole
Example 2:
4
4
5
6
S
5
3
5
2
6
2
6
1
thiane (I.U.P.A.C.)
4
4
3
S1
H
1H-thiin
S1
3
5
2
6
2H-thiin
S1
4
3
5
2
6
4H-thiin
3,4,5,6-tetrahydro-2H-thiin
S1
4
3
5
2
6
3,4-dihydro-2H-thiin
with suffix according to IUPAC: in
3
S1
no suffix available according to IUPAC
Note: the number of the indicated hydrogen is the lowest possible
Example 3:
4
3
5S
6
NH
S 2
1
4
5S
6
3
N
S 2
2H,6H-1,2,5-dithiazine → max. unsaturation possible is 1 double bond: suffix i n e
→ order of citation: S, S > N
→ supplementary hydrogen at N-2 and C-6: 2H,6H
4H,6H-1,2,5-dithiazine → max. unsaturation possible is 1 double bond: suffix i n e
→ order of citation: S, S > N
→ supplementary hydrogen at C-4 and C-6: 4H,6H
1
Example 4:
4
5
3
N
6
N
S 2
1
4
5
3
N
6
N
S 2
1
2
6H-1,2,5-thiadiazine → max. unsaturation possible is 2 double bonds: suffix i n e
→ order of citation: S > N, N
→ supplementary hydrogen at C-6: 6H
4H-1,2,5-thiadiazine → max. unsaturation possible is 2 double bonds: suffix i n e
→ order of citation: S > N, N
→ supplementary hydrogen at C-4: 4H

synthetical methods for the preparation of hetarenes

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