Synthesis of Aldehydes And Ketones
1o
H2
C
[o]
R
2o
O
C
OH
OH
[o]
R
H
R
R'
R
C
O
R
C
C
R
C
CH
R
H
C
H
C
R
BH 3
R
OH
[o]= KCr07 , KCrO4 , NaCrO4
O
R
R'
H2O , H
R
Hg
CH
only , and not to the Carboxylic Acid we
should use PCC as the oxidating
Reagant
[o]
CH
R
To Oxidate of the 1oalcohol to Aldehyde
O
C
C
H2O , H
Hg
O3
O
H2
C
C
H2O2
R
H
C
R
CH3
CHBH 2
R
CH
R
R
NaOH
O
CH
O
O
H2
C
C
R
O
D
D2O
O
C
Zn
2R
CH3COOH
Reaction of carbon α
O
H
O
H
O
D
OH
D
D
H+
OD
OH
OH
OD
D
D+
D+
O
O
O
O
BH+ +
+B
Tautomerisation
Keto-Enol
The Ketone form
is the favorite
form
O
R
O
O
O
CH3I
LDA
O
CH3
+
O
O
H+
+ Cl2
or
B
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Cl
CH3
H+
or
B
R
OH
+OH
O
O
+OH
X
X
+ H+
fast
slow
+ X2
O
X
OH
+ H++X
+X
O
O
O
fast
fast
O
_ X
X
X2
X
+X
O
Oxygen replacement
18
OH2
O
+
18
O
18
HO
OH
- H2O
+ H218O
+ H2O
Reactions with Alcohols
OH
O
R OH
+ H+
H+ +
H
R O
O
R
R
O
O
R
HO
O
RR
OH
HO O
O
R
+ H+
R
H2O O
H2 O +
O
H+
+
R
R'
OH
O
O
OH
R
R'
Synthesis of Imine
O
R +
R
NH 3
R
R
R
NH3
R
H
+ H2N OH
R
hydroxylamine
R
N
R
R
+ H2N NH2
Hydrazine
R
R
NH 2
N
R
R
HN
O
R
N
H N NH
R + 2
Phenyl-hydrazine
R
R
Phenyl-hydrazone
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H+
R
R
NH2
NH2
NH
OH
+
H+
O
R
NH 2
Reactions whit Derivatives of Amonia
O
OH2
OH
O
R
H2O +
R
R
R
Grignard reagnt reaction
O
OMgX
"R
R
C
R'
"R
R
MgX
O
R
R' H2O,H
or OH-
C
-
"R
C
R'
"R
R
Li
R
OLi +
C
R'
X=Cl,Br,I
OH
"R
H2O
R'
H+
C
OH
"R
+
R
C
R'
OH
O
+ HCN
R
R
H
CH
P
CN
O
(C6H5)3P
HO
R
CN
+
+ K CN
R'
C
R
R'
Wittig Reaction
C4H9Li
THF
I
(C6H5)3P
+ H3C I
CH3
(C6H5)3P
R
R'
O
(C6H5)3P +
Triphenyl-phosphine
O
R
0c
Triphenyl-phosphine-oxide
CH2
R'
0
CH2
(C6H5)3P
(C6H5)3P
CH2
R'
O
CH2
(C6H5)3P
O
0
R -70 c R
R'
(C6H5)3P CH2
Betain
Aldol Condensetion
OH
O
2x
H3C
C
H
C
H3C
+ OH
C
H3 C
2)
C
O
O
C
C
H
O
H2C
H
H3C
H
Aldol
O
O
1)
O
NaOH
H2C
C
H
H2C
O
O
C
H
H3C
H
H2O
H+
H3C
H
OH
A)
B)
O
H3 C C H
H3C
OH
O
C H
(BA)
+
O
O
(AB)
CH 3
OH
O
C
H
O
O
C
H
+ (AA) +
(BB)
H
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H3C
H
We can get 4 different products:
A & B can react with themselves or
with each other- A attacks B or B
attacks A.
O
O
+
H3C
C
H
O
Benzaldehyde
Only 1 option
because there is
no H on C alpha
Cinnamaldehyde
O
O
B
H2 O
C
H3C
H2
O
C
CH 3
O
H3C OH
H+
CH2
O
B
O
CH 3
O
H
C
O
CH 3
CH 3
O
Not stable
O
in this kind of
reactions we
will get rings of
5 to 7 Carbons
depending on
the starting
material
CH 3
CH 3
HO
CH 3
H + MeLi
H
HO
O
H+
CH 3
O
CH3
CH 3
+
H
cis 10%
trans 90%
Oxidation
[O]
O
R
O
R
H
Aldehyd
O
R
O
[O]
Ketone
Acid
VI
OH [O]= KMnO4,CrO3,Cr ,H2O2,RCOOOH (peroxy acids)
R
R
Ester
[o] = CF3 COOOH (Peroxide)
Mechanism is in Bayer Viliger
OR
OH
Bayer Villiger
R
R + F3C
R
R
O
O
O
O
OH
O
O
OH
O
CF 3
OR + F3C
R
O
H+
Clemmensen
O
R
R'
Zn(Hg)
HCl
O
R
H2
C
R
R'
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O
OR + F3C
OH
Wolf Kishner
O
B =KOH, R-ONa, R-OK
N
NH 2NH 2
R
R'
NH2
R
KOH
B
R'
R
H2
C
NH
N
NH
R'
B
N
BH +
R
R
R'
N
R'
NH
R
R'
H
B
B -H+
H2
C
B +R
R'
N
H
C
R
R' + N
N
N
BH + R
R'
H
Carboxylic Acid
Synthesis
O
O
C
R
R
C
O
O
R
C
OH
O
1)
H3 C
C
N
O
H2O H+
R
R
Mg
Ether
RX
CO 2
RLi
3)
R
4)
H2
C
RCOO -Li+
C
CO 2
-
(RCOO )2Mg
[O]
H
2+
H+ H2O
Not as good as Mg
RCOOH
OH
[O]= KMnO4,KrCrO4
O
R
OH
Acid ==>Ester
O
O
OH
+
R
OH + HO
RCOOH
O
R
OH
+NH 3
O
H+ H2O
[O]
O
R
RMgX
OH
O
H2O OH
2)
C
+ NH4+
H
CH 3
R
O
CH 3
-H
+
R
H+
OH
OH
R
OH + HO
CH 3
R
H 3C
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O
- H2O
OH
OH
H
R
H3 C O
O
OH 2
CH 3
Reaction with Diazomethane
O
O
OH + H2C N N
Diazomethane
R
O
O + H3C
R
N
N
R
CH 3 +N 2
O
Reaction with Thionylchloride (SOCl 2 )
O
O
OH + SOCl2
R
R
-
O
SOCl + Cl + H
Acid ==> Acyl
+
Cl + SO2Cl + H
R
OH + NH3
R
O
PBr3
R
O
NH4+
2000c
O
H2O + R
Br + H3PO3 + HBr
R
O
O
Cl + H3PO4 + HCl
R
O
OH
+
Acid ==> Amide
O
PCl5
R
SO2 + HCl
O
NH2
Hunsdiecker Reaction (chain reaction)
O
O
OH + Ag2O
R
O
O Ag +
R
Br2
CCl4
R
O
Br + AgBr
O
O
R-Br + R
O
R
O
O
Br
R + CO 2
R
O
H2
C
R
O + Br
O
C
Br2
PBr3
OH
H
C
R
O
C
Br
H2O
H
C
R
Br
C
OH
Br
Esters synthesis
O
O
R
Cl + HO
B
R'
R
O
B =
R' + HCl
,R"3N
N
Esters
Reactions
H2O
H+or
OH -
O
R
O
R'
O
OH + R'
R
OH
Acid Mechanism
O
R
OH
+
O18 CH 3 + H
R
OH
O18
CH 3+ H2O
+
OH
O
R
OH +H3C
O18H
-H
R
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R
H2 O
O18 CH 3
OH
OH +H3C
18
O H
R
O18 CH 3
OH H
Different mechanisem for 3o Carbon on the Ester
O
R
OH
CH3
H
O18
C
CH 3
C
R
OH
CH3
+
O18
C
C
CH3
CH3
R
C
CH3
O18
C
CH3
CH3
CH3
Base mechanisem for the same reaction (better than the Acid mechanisem if there are no halogens)
R
O
O
O
C
R
CH3 + OH
O
C
O
CH3O +
CH 3
R
OH
because of
pKa differrences
OH
O
O
+
CH3OH +
R
H
OH
CH3OH +
R
O
Amide ==>Acid
O
O
H2O
NH2 OH
R
R
O
+ NH3
O
H+
R
OH
Acylhalide ==>Acid
O
O
R
Cl
H2O
OH + HCl
R
Acid <==>Acid Anhydride
O
O
O
OH +HO
R
R
R
H2 O
O
O
O
Acid Anhydride
R
H2O
2 R
Trans-esterification
O
O
O
O
+
OH
Et
O
OH
OH + R
C
O
Me
H
R
C
Me
R
O
R
O
O
R
NH2
+
O
R
OH
O
R'NH2
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R
R"2NH
NH2
O
R'
R"
O
O
NH3
Cl
R
O
Et
Acylhalide ==> Amides
O
NH3 R
O
+
O
succinic
acid
O
OH
N
R"
R
NH
Acid Derivatives
O
R
R'M
O
R
O
GX
OH
R
2 R'MGX
O
R'MGX
-70 C,FeCl 3
R
R 'M
GX
CH3
O
Intermdiate - Will react
with another Grignard
R' Reagant
R'
R
R'
R'
O
R'2 CuLi
R
Cl
R'
Reduction of Acid Derivatives
LAH
O
R
R
R
R
OH
CH2
R
OH
CH2
R
CH2 NR'2
R
OH
CH2
OH
O
R' OH
O OR'
NR'2
O
R
Cl
O
O
R
[H]
[H]
R
Cl
H
R
OH
Ch2
LiAlH(OR)3 H2/Pd/BaSO4
Quinoline (Poisoned Catalyzator)
O
R
O
H
OH
NaBH4
CH
R
R'
R
R'
LiAlH4 > LiBH4
All
Esters
N C
Ketones
HN CH Aldehydes
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> NaBH4 > LiAlH(OR)3
Ketones
Acyl Halides
Aldehydes
Not Acids
Alkyl Insert to Ester Alfa Carbon - Claisen Condensation
O
O
O
LDA
R
C
H2
O
R
CH3
C
H
O
CH3
R
C
H
O
CH3
RX
O
R'
R
O
O
OEt
H2 C
OEt
EtO
O
CH3
We Use the same RONa of the
Ester RO
O
O
EtONa
H3 C
C
H3 C
CH3
OEt
C
H2
OEt
O
EtO
O
H3C
C
H
O
H2
C
OEt
O
O
O
H
H3 C C C
OEt
H3 C
Explaination for the acidity of the H
C
O
H3 C
C
O
H2
C
OEt
pKa~11
O
H
C
OEt
Pyrolysis Elimination of Esters
O
formic acid
O
CH3
and
/or
HO
O
Chuger Reaction
HO
CS2
NaOH
NaS2 CO
CH3 I
H3 CS2 CO
~1800c
cyclohexanol
CH3 SH + COS +
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Conjugated Systems
H2
C
H
C
H3 C
H
C
H
H3 C
OH -H2O
C
H
H
C
CH2
CH 2
H3C
C
H
C
H
Allyl Cation
MgX
H
+
43%
57%
Acylic Bromination
(Radical Br picks the Allyl H)
NBS
O
N
+
(Allyl Shift
Product)
Br
Br
CCl4
+
hν
Br
peroxid
O
Grignard Reagent + Conjugated
MgCl
Cl
1)
H
C
H2 C
H
C
H2C
Mg
CH 2
CH2 Br
Ether
Attacks Itself
In Order to prevent this - Take low
Concentration Of the Allyl
2)
H2C
H
C
Mg
H
C
CH2Br
Ether
H2C
H2C
H
C
CH 2Br
H2
C
H
C
CH2MgBr
C
H2
H2 C
Additions
H
C
CH2
+ Br2
-150C
H2 C
Br
1,2 Addition
If the reaction is in 60 c there are
+
10% 1,2 addition and 90% 1,4 addition Br
H 2C
C
H
Br
1,4-E-dibromobutene
H
C
CH2
+ Br2
46%
3,4-dibromobutene
0
H
C
C
H
Br
C
H
H2C
CH2
H2 C
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1,4 Addition
H
C
CH2Br
C
H
H2 C
+
H
H2 C C
1,4
Br
CH 2Br
C
H
H
C
1,2
54%
CH2Br
Allenes
N - No. Of C atoms
Even - A-Chiral (2N)
Odd - Chiral (2N+1)
C
(Just A Possibility)
Conjugated sytems with Carbonyls
A) Synthesis
O
1)
2x
H3C
B
2x
H3 C
H3 C
H
C
C
CH3
3)
C
H
O
H+
O
Aldolyc Condensation
CH
H3 C
CH
2)
O
H
C
C
CH 3
H3 C
O
O
+
CH
O
H
C
C
H3 C
C
C
H
CH3
CH3
Aldolyc Condensation
4)
OH
H
C
MnO 2
Synthesis Of Dienes
1) Wittig
H2
C
H2C
C
H
Br_
H2
C
H2C
C
H
Br
MnO2 Only
Oxydates Allyl
R Alcohols
O
C
RHC
CHR
RHC
H
C
B
H 2C
C
C
H
PPh 3
PPh3
R
Synthesis from Alekne with Halide
R
H
C
C
C
H
H2 C
2)
Me
OH
OH
C
C
Me
Me
KHSO4
Me
Me
C
C
O
CH2
POCl3
Al
H2O
or
OH HO
Mg 2
O
O
Mg
Me
H2C
or
Me
H2
OH
Or
C
C
O
Or
Me
KHSO4
C
Me
Me
Me
Me
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Or
Al2O3
O
C
R'
R'
Reactions with conjugated Carbonyl
1) 1,2-Addition
O
H
C
R'
H
C
Br2
C
R
R'
H
C
H
C
Br
Br
O
C
R
2) 1,4-Addition
O
H
C
R'
H
C
H
+
HCN
R'
C R
Appearantly it looks like 1,2 - addition
+
But actually it's 1,4-Addition, because the H
Attacked on the carbonyl first
OH
OH
H
C
R'
H
C
C
R
1O
7
6
CN
R'
H
C
H
C
OH
1,2
C
R
R
H2
C
H
C
O
C
R
CN
-
OH
R'
H
C
H
C
C
Grignard Reagent
2
3
1, 4
Li
Cu
R2
5
+ RMgX
4
O
OH
1,
4
on
iti
dd
-A
RLi
(1,2 - Addition)
R
R
OH
R
O
R
It seems like ordinary 1,2-addition to
double bond but it's a delicate
reduction reagant for carbonyl
conjugated sytems
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R
Reduction ( "Hizur")
O
LiAlH4
(Carbonyl Only)
OH
1,2
Na
4
BH
OH
1,4
NaBH4
1,2
O
OH
NaBH4
41%
OH
59%
Birch Reduction
O
O
O
O
O
H2/Pd/C
+ Li
/
Li
(l)
o
3
3
NH
-3
OH
O
O
O
O Li
Li NH 3
H3O+
NH 3
Li
H
Ketene
H2C
C
O
O
O
H 2O
H3 C
OH
O
C
C
H3 C
O
CH3
Acetic Anhydride
O
o
H3C
CH3 700
Zn
Br Br
RHC
H2 C
+ CH
RHC
C
O
O
OH PBr 3
RH2 C
4
CHR'
O
NPh 3
CHR'
H
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O
Br2
Cl
R
Zn
Br
Br
C O
RHC
C
O
4+2 Cyclo-Addition - Diels Alder
CH
CH
O
O
Note: The reactions are at the
same time and threfore
Streo-specific - Cis And Trans
will be preserved - It's named
Cis Addition Or Syn Addition
Nomenclature of Bicyclic Molecules
7
Bicyclo[2.2.1]heptane
6
2
1
4
5
3
Count the number of Carbon atoms to the
lefr,right and above the "Bridge" Carbon (No.
1) until the next "Bridge" Carbon (No. 4).
e.g. the Number 1 in the name is for the
carbon above (No. 7)
O
O
C
+
O
O
O
Endo Rule - Endo is the favorite
Isomer and in low Temp. we'll
get only Endo
O
C
O
C O
C
O
-exo
-endo
CO2Me
CO 2Me
CO2Me
CO2Me
MeO2C
CO 2Me
Trans Double Bond
in a cyclic molecule
causes High cyclic Strain
A
A
A
A
A
Para-diA
ortho-di
meta-di-
A
By adding 1 more functional group (to ortho para or meta isomeres) we can
identify which isomere we had according to the number of new isomeres:
Ortho - 2(A3) , meta -3(A3) and para - 1(A3 ).
Cl
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Cl
This isomere conributes to the
Moment Dipole of the molecule
Reactions with Aromatic systems
H3 C
Radical Reacions
Cl
CH2
+Cl2
+Cl
Cl
ClH2 C
Cl2
Cl
Cl
hν
Cl
Cl
Nucleophilic Substitution
It appears that the reaction is SN 1 and not SN 2.
CH2CN
CH2 Cl
attacking the methyl
2Cl
Cl2
doesn't damage the
aromaticity.
There is a resonance of the radical
all over the aromatic cycle
hν
the reason is that Benzylic Carbocation is very
NaCN
stable - Even more than 3o Carbocation.
SN 1
X
X
Oxydation Reactions
CH3
CHO
[O]
COOH
[O]
[O] = CrO3
CH2 CH2 CH2CH3
COOH
[O]
O
_
CH2
...
COOH
Reduction Reactions ("Hizur") (Part 1)
R
R H2 /Rh/C
100 atm
o
T>100 C
HC
H2 C
CHR
H2 /Cat
R
R
CHR
Note: Without the rough conditions of
temp. and pressure, the aromatic cycle
isn't Reduced as it was in the example
above
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Reduction Reactions of Aromatic Systems (Part 2)
Hydrogenolysis
O
O
H2/Pd/C
OCH
+
H2
C
CH3
H
+
H3C
C
OH
toluene
R
O
C
R
OH
Birch Reduction
Na/NH3(l)
O
C
In Order to protect the Carboxyl
Group we create a Benzylic Ester that
we can Remove by simple
hydrogenolysis that doesn't require
any tough conditions
O
H2
C
De-Aromatization
1)Fe,HCl
EtOH
NO2
C
H2
OMe
OMe
Li/NH3(l)
OMe
H+
NH2
2) H2 / Ni
H2 O
OMe
Me
O
H
HO
H2O
EtOH
Not conjugated Intermediate is
less stable and therefore it's
turned intoa conjugated
O isomere
O
+
Benzene Reactions
Part 1
Halidation
+Br2
OH
-H
FeBr 3
+HBr
+ I2 +HNO3
Br
+
Br FeBr4
H
I
+NO + NO2+ HI
Br H
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Benzene Reactions (Part 2)
H
H O NO2 +H2 SO4
-
O NO2
H
H3O+ + NO2 + HSO4
+H2 SO4
( HSO4 -)
+ NO2
HNO3, H2SO4
H
NO2
NO2
Friedel - Crafts Reactions
Acylation
O
+
R C Cl
O
+ AlCl3
R C Cl
O
R C
AlCl3
R C O AlCl 4
O
R C
H
R C O
Important: The Reaction is good for
Benzene without any Functional groups
Or one that don't De-Activate it (See
effects of functional groups on Benzene)
Alkylation
Multi
Alkylation
AlCl3
+ CH3X
+
+
CH3
Example:
+ CH3 CH2 CH2Cl
AlCl3
CH3
CH3
CH3
CH3
H3C
CH3
CH3
C
H
n
n=1,2,3,4
CH3CH2 CH2Cl + AlCl3
CH3 CH2 CH2
o
1 Carbocation
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CH3 CHCH3
o
2 Carbocation
Multiple Electrophilic Aromatic Reactions
X
X
Y
Ortho
X:
Meta
X
X
Y
Meta
Y
Y
Para
O
O
O
O
- NO2 , C R , C OR , C NR , C OH , CN ,CF3
O H
,R C N
O
Ortho, Para R C O
De-Activation
Activation
, NH2 , -OR , -R
Activation ( Weak)
Hallogenes
Note: For 3rd Subtitution - If one of the 2 functional groups leads strongly to o,p we'll
get o,p ,Otherwise We'll get a mixture of Products
OCH3
OCH3
OCH3
OCH3
OCH3
X
X
X
X
Ortho
X
Meta
Para
The Resonance of the + charge can give a form with the + on the carbon attached to the
Electron Drawing group and stabilizes it. This can't happen on the Meta isomere and
therefore Para and Ortho are the favorite isomere.
The Activation is resolved by the Electron Donation to the Aromatic system.
CHO
CHO
CHO
CHO
CHO
X
X
Ortho
X
X
Meta
X
Meta
Para
The + charge will never be on the Electron Donating group in Meta Isomere, but on Ortho
and Para Isomere it can be. Therefore it De-stabilizes these 2 isomeres and the Meta
Isomere is the favorite one.
The De-Activation is resolved of the Electron drawing from the Aromatic system.
SO3
H2 SO4
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SO3 H
D2 O / D
+
D
O
O
H2 C
C
H2 C
C
O
+
C
AlCl3
CH2
H2 C
O
Zn ,
H
T o lue C l
ne
Note: Without the reduction of the
Ketone, we wouldn't get thisReaction,
due to the De-Activation
(CH2 )3
AlCl3
COOH
H2
C
SOCl2
C
CH2
H2 C
COOH
O
Cl
O
Aromaticity
Huckel Rule: Number of π Electrones: 4n+2
4n
Aromaticity
Anti-Aromatic
2K+
K
2-
THF
Aromaticity
(4n+2)
Diels-Alder
Dimerization
sp3
sp2
sp2
H
H H
B
Aromaticity: 6e-
-
Aromaticity
we took the e the p orbital
Aromaticity : 6e
A
H
BF4
H
Na+
(4n+2)
is now empty for resonance
-
A = HBF4
(4n+2)
-
H
C10H10Fe :
Fe
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The π Electrones are shared
with the Fe atom's empty
orbitals (also possible with other
Transition Metals such as Ni
and Co
Amines
pKa
Electron drawing Functional groups such as NO2 ,Cl,Br make the Amine to be
Less Base and to have a lower pKa
This is a good way to make rings with 3
CHCl3 + NaOH
CCl2
carbons. However th problem is that NaOH is
Cl
Cl
Chloroform
not soluable in Chloroform. And we
H H
CCl2 + R C C R
RHC CHR need the Amine.
R'
Synthesis Of Amines
a
+
R
N
OH
+
NaOH
4
R N R'''
1)
Ion Pair
0
+ (1 ) RNH2 + (HX) NH4 X
RX + NH3
R''
RNH2 +RX
(20 ) R2NH +HX
This is not a good way of Amine Synthesis
(30 ) R3N + HX
R2 NH + RX
0
+ because it creates a mixture of Amines
(4 ) R4 X
R3 N +RX
O
2)
O
O
CO2 H
KOH
H3O+
RX
RNH
+
2
N R
N K
NH
Or
CO2 H
O
O
Phtalimide
Reduction
A) RNO 2
RNH 2
C)
E)
NO2
NH2
O C
H
Aniline
[H]
N
NH3
O
R CR'
D)
RCH 2NH 2
NH
R CR'
+R''NH2
LAH
4) O
RC NHR'
O
R''
[H] = LAH , H2/Cat
RCH 2NH2
O
[H]
RC NH2
O
R CR'
NH2
SnCl 2
O C
H
RC
R'
NOH
RC R'
NH2
[H]
R CR'
NR'' [H]
R CR'
Schiff's Base
RCH2NHR'
R''
LAH
[H]
10
R CR'
30
R'
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O NH2OH NOH
RC R'
RC R' R CR'
H2 N
[H] = H2/Ni , LAH
NHR''
20
RH2 C N
RC N
OH-
[H] = Fe/HCl , SnCl2 , H2/Pd , H2 /Ni
[H]
NO2
B)
O
20
R
CH2 O
NH2
HCOOH
H
R N
RC NH2
+OH
+NH2NH2
Me
R
Cl2 Or Br2
5) O
RC NH2 NaOH
O
O
R'
O
HO
2
R N C O
Isocyanate OH
N
-
Amide
O
H
RC N R'
O
R N C O
H2O
O
RC Cl
+ R'NH2
R'
NaNO2 R'
OH
R
R
+
O N OH + H
(HNO2 )
Me
N
N Me NaNO
2
O
Nitrosoamine
H
O N O
H
O
N
H2O
RNH2 +CO2
O
+
O
RC N
NH
O
HNO2 + NaCl
NaNO2 + HCl
N O
+ H2O
R'
R'
R
NH
H
N N
O
R
H2 O
N Me This is an Exception because
normally 3o Amines DO NOT React
HCl
R'
R
R
Mechanism
Me
R N C O
Isocyanate
C
R'
R C
N
NH HCl
-
+Cl
RNH2 +CO32-
-
Carbamate
O
R C
RC N
+H 2O
6) Curtius
O
H2O
O
R
RC N N N
RC Cl + NaN3
Azide
1. H+
7) Schmidt
RNH2 +CO2 +N2
RCOOH + HN3
2. OH
Reactions
RC NCl
+Cl
O
C
RC NHCl
OH
Nitrene
O
O
-
+H2O
R
O
Cl2
RC NH
Hydrazone
Synthesis
Hoffmann Rearrangement
RNH2 +CO32-
-
N NH2
R'
N N O
R
with HNO2 , But this Amine has an
Aromatic group.
NO
NaNO2
CH3 CH2CH2 CH2NH2
N2 + CH3CH2 CH2CH2Cl + CH3 CH2CHCl-CH3 +
HCl
CH3CH2CH2 CH2OH + CH3CH2CHOH-CH3 +
1o
CH3 CH2CH2 CH2
2o
CH3 CH2CHCH3
We get the Nucleophilic Substitution,
H
C
H3 C
Elimination ,Of 1o and 2o Alkyls.
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C
CH3
H3 C
+
H
C
H
C
CH3
H
+
HONO
H
R N NO
R NH2
H
H
R N N OH
H
R N N OH
B
+
H
H
+
R + N2
N Cl
N
NH2
HCl
R
H
-
Slow
HONO
R N N OH
R N N O
R N N + H2 O
R N N
If the R Group of the
Amine is Aromatic
then the Diazonium
Salt is stable
N2 + Aromatic
R
Diazonium Salt
Oxidation Of Amines
1o R NH2 H2 O2
H2 O2
o R NH
2 2
R
H2 O2
3o
R' N
R''
O
HN C CH3
H
R N
OH
Hydroxylamine
R2 NOH
R
R'
N
Nitroxide
O
R''
O
O
Me
HN C CH3 HN C CH3
Me
N
Br
Br2
FeBr 3
Me
Br2
Br
+
Br
Br
R'
O
HN C CH3
R
N
77% NO2
In case we have a Cyclic Amine:
Ag2 O
R'
CH3 X
R
R''
NH2
23%
K2 CO3
Me
Br
Ch3 COOH
Less
HNO3
90%
O
HN C CH3
NO2 +
N
N
CH3 X
R''
OH - N(CH3 )3
I - N(CH3 )3
Ag2 O H
K2 CO3
H2 O
( E2 )
CH3 I
Excess
Hoffmann Degradation
K2 CO3
H
CH3 I
N I - H2 O
N
OH Excess H C CH
H3 C CH3
3
3
By the number of the double bonds after the
degradation we can determine: 1 - Chain Amine,
2 -Cyclic Amine , 3 - Bicyclic Amine
N
H
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N
H3 C
CH3 I
Excess
CH3
(CH3 )3N +
IH3 C CH3
CH3
Ag2
O
H2 O
N
CH3
CH3
CH
I
Excess
3
NH2
N(CH3 )3 Ag2O
K2 CO3
H2 O
We get the
Anti-Zeizev
Product
1%
99%
Cope Reaction
O
H2C N Me
0
Me ~160
RNH2 +
O
C
NR
C
+ (CH3 )2NOH
(For the following
Reaction)
O
OH
H+
N
H
N
+
H
N
OH
N
N
OH2
B
N
H
-H+
N OH
R
OH
R
N OH
2
R
H2 O X
1o RX
-
N
N
N
R
Enamine
OH
+
R
N
H
DiazoAlkanes Derivatives
CH2 N2 H C
2
N
N
O
RC
O
OH
O
RC
H2 C
N
+CH2 N2
CH2CH3
RC
+ OH -
N
N
Diazomethane
OCH3
+N2
H3 CHC
NO
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N
N
O
O
SOCl2
RC
OH
O
CH2 N2
RC
H
C
RC
Cl
This is a good way to
add Carbon atoms to
Carboxylic acids
N
N
R
O
OH
H2
C
R
R N
C
CH
H2 O
R
O
O
C
Carbene
H2
C
R
(Works in the same
mechanism)
R
O
OR'
H
C
C
O
R'OH
O
N
O
O
RC
Cl
H2 C
N
N
O
RC
Cl
H2 C
N
O
H2
C
RC
N
N
N
-
RC
H
C
N
N
+H+
+Cl
This is the mechanism of the Reaction above
Cu ( Powder)
+CH2 N2
O
+
N
N N Cl
N
H
C
C
Cu ( Powder)
OEt
COOEt
OH
Cl
+
H2 O
(Solvent)
Phenol
N N
NH2
NaNO2
Chlorobenzene
OH
Synthesis of only the
Phenol
H2 O
H2 SO4
H2 O
Phenol
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Sandmeyer Reaction
-
N N Br
-
N N Cl
Br
CuBr
Cl
CuCl
N N X
N N
COOH
CN
I
KI
H2 O
CuCN
+
-
H Or OH
F
N N BF4 -
NH2
N N
HBF4
H
H3PO2
NaNO2
H2 O
NaNO2
D3 PO R
R
2
NO2
D
This a good way of
synthesis that we can
remove the Diazonium
group when we want
NO2
NO2
R
NHCOCH3
NHCOCH3
Cl2
1)
FeCl3
2)
Cl
[H]
HNO3
O
O
H3 CC O CCH3
H2SO4
Cl
Cl
NO2
This is an example how we can use Sandmeyer Reaction in
order to insert 2 Halogene groups in desired positions.
In this case we put them in Ortho and without
this method they would have been Para
N N X
H3PO2
-
1 ) [H]
2 ) HONO
3)
NH2
CuI
NHCOCH3
HNO2
H2 O
+
Cl
Cl
I
I
Biaryls - Synthesis
NaOH
Cl
I
Cl
I
1) Gomberg - Bachman
-
N NCl
R
H
R
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X
2) Ulmann Electrone Drawing groups
(like NO2 ) or Simetric
materials help this Reaction
Cu
2
X= Br , Cl , I
Me
+
N
N
R
Me
N
N
R
Me
The Azo compounds are colorful and
absorb UV light in wavelengths close
to the Seen Light
N
N
H
Me
B
N
R
Me
N
N
Azo
Me
Coupling Reactions
Haloaromatic Compounds
Cl
OH
NaOH
H2 O
Cl
370
high pressure
NO2
Cl NO2
CH3 ONa
<1000C
0
F
OH
NaOH
H2O
CH3 OH
NO2
NO2
NHR
NO2
RNH2
NO2
OCH3
NO2
NO2
NO2
NO2 and CN are Electrone Withdrawing
groups that can help the reaction go.
NO2
Mechanism For Coupling Reactions
X
X
Nu
Nu
+X-
+ Nu
NO2
O
Cl
NaNH2
N
O
O
NH3
*
NH2
*
NH3
50 %
NH2
+
NH2
NH2
Cl
NaNH2
CH3
NaNH2
NH3
F
O
CH3
CH3
NH2
N
* -
14
C
*
50 %
The mechanism is
shown in the next page
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Mechanism for the coupling reaction from the previous page
X
H
*
X
H
H
H
+ NH2
( As a base)
H
NH2
H
H
H
*
H
+
H
H
Cl
NH2
*
H
H
H
+ NH3
H
H
H
H
NH2
NH2
*
H
+
H
H
H
Biphenylene
triphenylene
N N
+
COO
Ortho boom-boom
Benzyne
+
Cl
NO2
NO2
Cu
220
*
H
3
NH2 C5 H11ONO
0
O2 N
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H
NH2
NH3
2
H
H
H
H
This reaction ( was shown on the
previous page) also goes through the
benzyne intermediate
3700C
high pressure
COOH
*
Benzyne
H
OH
H
H
NH3
H
NaOH
H2 O
H
*
*
Ullmann
NO2
NO2
Cl
We get this product
O2 N
inspite of the steric effect.
The rings are, in avarage
COOH
at 900 One to the other
COOH HOOC
NO2
CO2 H
O2 N
HO2 C
Phenols
OH
O
Enol
In opposite to the regular Keto-Enol
Tautomerization , This Enol is the more stable form
due to the aromatic stabilization
Ketone
Phenols Synthesis
1)
Cl
+
O Na
NaOH
+
3700
O
High pressure
Long time
2)
H2 SO4
OH
SO3 H
NaOH
SO3
3)
N N HSO4
OH
H2 O
4)
H
H
+ H3 C C CH2
O OH
+
O2
H2 SO4
OH
+
H3 C
cumene
The Mechanism will be shown in the next page
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O
C
CH3
Mechanism for the synthesis from the previous page
O O H
H
O OH
H
+
O
O
O
H2 O
+ H2 O
H
H
O
HO
+
OH2
HO
O
O
Hemiketale
+H
+
OH
Reactions
O
O
Br
Br2
OH
Br
Br
Br
Br
Br
CO2
~800
Pressure
Kolbe Reaction
OH
Br
O
CO2
H
+
OH
O CCH3
COOH Acetic Anhydrid
Salicylic Acid
COOH
Acetylsalicylic acid
( Aspirin )
OH
O
0
~240
Kineticly we get the Ortho product
CO2
Pressure
( shown above) but Themodynamicly we
COOH
get the Para product and if we heat the
0
reaction to 240 , we'll get the Para Product
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Reimer - Thiemann
O
OH
O
CH
NaOH
Salicylaldehyde
+ CHCl3
Mechanism
CCl3 - + H2O
CHCl3 + OH -
Cl - + CCl2
Carbene
CCl3
O
O
Cl
C
Cl
O
C
Cl
Cl
CH
Cl
Cl
O
O
O
+ H2 CO
O
CH2 OH
CH2OH
+
CH2 OH
CH2 OH
O
CH2 OH
Polymer
CH2 CH OH
2
O
O
OH
O
[H]
[O]
OH
[O]
O
p-benzoquinone
O
O
O
O
O
O
O
OH
+
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Diels-Alder
Polycycyclic Aromatic Hydrocarbons
0
1.397 A
Biphenyl
p-Terphenyl
O
The Naphtalene has a
similar structure as the
o-quinone 's structure
O
Naphthalene
o-quinone
Anthracene
Naphthacene
Phenanthrene
Pyrene
As we enlarge the system, the number of the
Hydrogen atoms is getting smaller, compared
to the number of the carbon atoms
Coronene
The more we add rings, we'll get a spiral
structure and the molecule will be Chiral. This
systems are called Helicenes. Their optical
activity is different than the usual
H H
H
H
Synthesis of Polycyclic Aromatic Hydrocarbons
O
R
AlCl3
+
R
O
NH2 NH2 R
HF
O
O
HO2 C
PPA = Polyphosphoric acid
H3 PO4 + P 2 O5
KOH
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PPA
HO2 C
O
[H]
Pd/C,
Or
R
Se,
R
O
O
O
H2 SO4
+
O
HOOC
O
Electropilic Substitution
O
In the Ortho
position there is
a steric effect
Br
because of the
hydrogenes
H H
Br2
FeBr 3
H H
anthracene
NO2
HNO3
NO2
H2 SO4
E+
H
E
1α-
Kineticly
Favorable
H
E
2βH
Thermodynamicly
Favorable
E
H
E
There are 2 Resonance forms that have one benzene ring
H
E
H
E
H
H
E
H
E
E
There is only 1 resonanace form with one benzene ring
Reactions
Sulfunations
1)
O3
/S
4
SO
H 2 icate ns
l
De ditio
con
H2 SO4 /SO3 ,
2) Friedle-Crafts
O
SO3H
α
β
SO3 H
O
O
Cl
O
AlCl3 , CS2
AlCl3
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Cl
NO2
Guidance in Naphtalene
SO3 H
1) Meta guiding group :
α position on the other ring
2) Ortho,Para guiding group
in No.1 Position:
Para is favorable because it's α
OR
Primary
3) Ortho,Para guiding group
in No.2 Position:
OR
Difunctional Compounds
Examples
1,2 - Diols
Synthesis
C
R
Mg
O
KM
nO
Or
4
Os
O
R
C
R'
OH OH
R
C
R'
O
OH
RC O
C
R'
OH
OH
2 R
R'
OH OH
OH
OH
C
R
R'
H2 O
O
H
OH
+
OH
OH
OH
OH
4
pinacol
OH
+ OsO4
O
Os
O
O Na2 SO3
O
Or
H2 O2
We can see that with OsO 4 and KMnO4 we
get the Cis Product. This happens due the
formation of the ring.
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OH OH
R
R'
C
C
R
R'
CH3
H3 C
OH
H3 C
OH
+
H
CH3
Pinacole Rearrangement
pinacol
+
H
CH3
H3 C
OH2 -H2 O H3 C
H3 C
OH
H3 C
H3 C
CH3
CH3
CH3
H3 C
OH
OH
CH3
CH3
pinacolone
CH3
CH3
CH3
CH3
H3 C
CH3
OH
CH3
O
CH3
O
1)Mg
Benzene
2) H2 O
OH H
OH
+
-H2 O
OH2
OH
OH
OH
+ H+
Bicyclic compounds having one
carbon common to both rings are
spiro compounds
O
Spiro[5.6]dodecane
Oxidation
OH OH
R CH CH
R MeOH
OH
Me
HIO4
OH
HIO4
MeOH
2
HO HO
O
R
RC H
Aldehyde
C
C
R
R
O
C
H
R
HIO4
R
2
C
R
O
Ketone
Another Reagent is Lead tetracetate
Pb(CH3 COO)4
O
Me
HIO4 is specific to 1,2 Diols and
doesn't oxidate other groups
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Diacids
O
COOH
C OEt
CH2
H2 C
B
O
O
C OEt
C OEt
HC
HC
R
C OEt
C OEt
C OEt
Malonic Acid
O
O
O
Diethyl Malonate
pKa ~ 10
O
O
0
C OEt
(1 R is better)
C OEt H O/H+
2
RX
HC
R CH
C OEt
O
C O
C
H
O
R CH
C O
OH
O
C OEt NaOEt
R CH
COOH
HO
This is Specific for β Diacids Only
CHR
RCH2 COOH
C
The overall process is a way to add 2
carbons to the chain RCH2 COOH
RX
OH
O
C OEt
R C
+CO2
Decarboxylation
O
O
RH2C C OH
R CH
C OEt
O
O
COOH
R'X
R
R'
C
COOEt
+
H2 O/H R
CH COOH
COOEt
C OEt
R'
C OEt
O
O
1) NaOEt
O
O
2)ClCH2 CH2CH2 Br
C OEt NaOEt
C OEt
ClH 2CH 2CH2C
ClH2CH2 CH2C CH
H2C
C OEt
C OEt
O
O
H2 O/H+
O
C OEt
C
C OEt
O
COOEt
COOH
COOEt
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Knnovenagel Reaction
O
RC H
+ HC
2
H2 C
O
H2 O/H+
C OEt
N
H
C OEt
O
RHC
+ HC
2
O
RHC
C
H
C
C
OH
C OEt
O
O
C OH
O
RC H
Synthesis of Conjugated Carboxylic Acids
O
C OEt
O
C OH
N
C OH
O
RHC
O
RHC
C
H
C
C
OH
C OH
O
O
C OH RCHO
COOEt
RHC
CN
Ethylcyanoacetate
C
C
N
Michael Addition
O
C OEt
O
1,4-addition
O
+ HC
C OEt
O
COOEt
HC
α,β-unsaturated
Ketone
COOEt
O
O
H
H
O B
O
H
O
O
O
O
O
ETONa
EtOH
O
O
H3 C
H
C O ETONa
H2 C C C CH3 O
O
CH2
EtOH
CH2
This is a good
way for
Cyclization
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O
O
Dieckmann Reaction
O
O
O
EtO Na
EtO C CH2 CH2CH2CH2 C OEt
O
EtO C CH2CH2 CH2CH C OEt
EtOH
We can Synthesis 5or 6 carbons
rings.For larger rings we need low
concentrations to prevent Dimerization
O
O EtO
+
H
EtOOC
O
O
O
EtO
EtO
EtO C
EtO
O
C
EtOOC
H
Acyloinic Cyclization (Here we don't need low concentrations)
O
C OEt
H2 C
ONa
(CH2)n
Xylene
C OEt
O
(CH2 )n
+
H
C
ONa
Mechanism
O
C OEt
C O
H2 O
C
Na ,
n
CH3
+ Na
O Na
+
C OEt
+ Na+
C OEt
O
C OEt
O
o,p,m Xylene
Acyloin
O
C OEt
C OH
CH3
Acyloin
C OEt
O
Hydroxyacids
OH
RHC CH2COOH RHC CH2 CH2COOH RHC CH2 CH2CH2COOH
RHC COOH
α−
OH
β−
OH
γ−hydroxyacid
+
RHC CH2 CH2COOH H
OH
γ−hydroxyacid
β
α
δ γ
RHC CH2CH2 CH2COOH
OH
δ−hydroxyacid
OH
δ−hydroxyacid
RHC CH2 CH2 CH2COOH
R
O
O
γ-Lactone
R
OH
δ−hydroxyacid
O
OH
Cyclic "Hemiacetal"
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R
O
O
δ−Lactone
Amino-Acids , Peptides , Proteins
R
H2 N CH
R
COOH H2 N CH
H3 N
COOH
α-Amino acids
H3 N
H2
C
H3 N
O
C
H2
C
H2 N
O
H2
C
H3 N
C
pKa = 9.8
C
O
O
O
O
*
R
pKa=2.35
C
RHC COOH
COO
COO
NH2
H3 N C H H3 N
H
RHC COOH
OH + NH2
OH
O
O
ninhydrin
R
1) RHC COOH
Br
2)
O
C OEt
H2 C
NH3
RHC COOH
NH2
+
O
C OEt
+ Br2
C OEt
O
RHC COOH
NH
+
RCHO
CO2
3H2 O
HO
Double substitution by
the NH 3
RHC COOH
O
O
C OEt
N CH
C OEt
O
O
O
+
N K
Br CH
C OEt
O
Gives Purple
color
N
H
O
Synthesis
O
O
OH
SL-
C
Zwitterion
O
H2
C
O
H2
C
O
In order to Synthesis all kinds of Amino
Acids , all we need is to use another RX
R
H2 N CH
3) O
+ NH3 + HCN
RC H
H2 O
COOH
H+
+
RHC CN H2 O/H RHC COOH
NH2
NH2
O
RHC COOH
MeOH
RHC COOH
+
HN C R' R'C Cl
H
NH2
Amide
Base
O
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RX
R O
C OEt
N C
C OEt
O
O
O
Shtraker Reaction
RHC COOMe
NH2
Esetrification
O
RC OH + R'NH2
O
H N
RC NHR' 2
SOCl2
O
RC NHR'
O
R'NH2
RC Cl
H
H
SOCl2
C COOH
H2 N C COCl
R
R
With Amino Acids
we get
polymerization
Polymer
Protection for Amino Acids
Protection For The Amino side
1) Cbz
CH2OH
O
Amino
H2 C O C Cl Acid
COCl2
O
CH3
O H R
H2 /Pd
+CO2
CH2O C NCH COH
+
H
CarbobenzoxyCbz2) Boc
H3 N
H
C COOH
R
CN
O
Ph
N O
O
Boc On
+ R
H2 N CH COOH
O
CF3 CO2H CO2 H
H H
+
O C N C COOH
Amino
R
Or HCl
Acid
+
Protection For The Carboxylic side
For the protection we create the Ester ( Me,Et or Ph) with the Carboxylic
Acid and when we want to remove it there are 2 options:
We can use this method
1)
OH
H
to remove Et Ester and
H
+
MeOH
H
N
COOH
C
2
H2N C COOMe
we'll get EtOH
R
R
2) OBnO
H
OH
H
N
C COOH
H2 /Pd 2
H
H
H2 N C COOH
H2N C C OCH2Ph
R
R
R
SO3 H
+
+ CO2
The OBn Protection doesn't fall in
Hydrolysis , which could be very useful
N
C
N
DCC- A catlyst for the condensation of 2
Amino Acids into a Peptide
dicyclohexylcarbodiimide
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O
R
H
CH2O C N CH COOH
+
R
Good for Dipeptides ONLY
R O
H2 N
CHC
Merrifield
H
CH2 O
HC CH2
Washing
1)H
+
2) HF
Peptide
CH
ClCH2 Pol
O R O
R'
O R O H+
R'
R'
Boc NHCHCOOH Filtration H2 NCHCNHCHC
Boc NHCHCNHCHC
Washing
DCC
Pol C O
Pol C O
H2
H2
In order to add more A.Acids we need to
add another A. Acid with Boc.It's Important
to wash and Filter after every time.
Sanger Method
R O
NHCHCOOCH2
n
Cl
Cl
R'
H
H2N C COO CH2 Pol Boc NHCHCOOH
R
DCC
H
H2 N C COO Na + ClCH2 Pol
R
+
Filtration H
H2 H
C C
H
C
C
ZnCl2
R'
Cbz NHCHC
+
NHCHCOOH
+
R O
H2 /Pd
R'
HC CH2
H H2 H
C C C
DCC
H
H2N C COOCH2
R' O
F
R'' O
H2 NCHC HNCHC HNCHC
R'' O
H2 O ,
HNCHC HNCHC HNCHC
NO2
H Or
OH -
R O
NO2
NO2
R' O
+
NO2
R O
Sanger Method can determine what
kind of A.Acid is in the N-Terminal
part of the Peptide
HNCHCOH
NO2
NO2
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R'' O
H2 NCHC
+ R' O
HNCHCOH
Edman Method
R O
R' O
Phenylisothiocyonate
R O
R' O
R'' O
S
H
N CH HNCHC HNCHC HNCHC
N C S
R'' O
H2NCHC HNCHC HNCHC
n(Amino Acid)
H+
n-1(Amino Acid)
S
We can repeat it many times and
determine the Sequence of the
Amino Acids in the Peptide
C
R' O
R'' O
N NH +
HNCHC HNCHC
C CH2
O
Thiodantoin
Carbohydrates - Cn(H 2O)n
1) Monosacharides
2)Oligosacharides - 2-7 monomers
3)Polysacharides -7+ monomers
OH
OH
n
*
HO
HO
O
*
H
2
1 CHO
* OH
HO
2
O
Ketose
2
H
* H
OH
D(+)-Glyceraldehyde
OH
3
CHO
1 CHO
3
HO
3
4
H
L(-)-Glyceraldehyde
5
H
We determine L or D by the Chiral
Carbon with the highest number - If it's
on the Right -D , and if it's on the Left -L
HO
OH
H
2
* OH
OH
1CHO
1) H 2 OH
H 3 * OH
4
OH
3
2) HO
H
1 CHO
2
3
4
H
*
OH
OH
HCN/OH -
1
1
HO
H
* OH
HO
* H
CH2OH
6 CH2OH
D-Glucose
CN
+
H
OH HO
H H3O
+
3
3
H * OH
H * OH
4
OH
4
OH
Mixture of Diastreomeres
2
2
CHO
1
Separation H
2
H
3
4
*
1
L-Glucose
2
OH
OH
3
+
H
4
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*
COOH
H
2
H
3
4
1
OH
* OH
+
COOH
HO
2
H
3
OH
H
* OH
4
OH
~1000
CHO
OH HO
OH
OH
1
CN
H
H
H
Fischer - Kiliani Synthesis
1 CHO
OH
H
Aldose
1 CHO
n
O
O
1
1
H Na(Hg) H
2
OH
OH
3
H
4
OH
*
+ HO
2
H
3
OH
H * OH
4
O
O
Mixture of Lactones
1)
H
1 CHO
2
H
OH
HNO3
H
3 *
H
HO
H
1 CHO
2
H
3 *
4
OH
3 *
OH
0
Alcohols and
HNO
oxidates
only
1
3
4CO H
2
Meso-tartaric Acid Aldehydes to Carboxylic Acids.
[α]D = 0
This helps the separation process,
OH
4
OH
Erythrose
2)
1 CO2H
2
HO
HNO3
H
OH
OH
Threose
1 CO2H
2
because here we get a Chiral
H
product and an a-Chiral product.
3 *
OH
4 CO H
2
[α]D = 0
Identifying Glucose and Manose
CHO
H
HO
CO2H
OH
H
H
OH
HO
H
H
H
OH
H
OH
HNO3 HO
H
H
OH
H
OH
H
HNO3
CH2OH
CHO
CH2OH
CO2H
OH
H
H
1800 HO
H
OH
H
OH
HO
CHO
D-Glucose
HO
OH
H
CH2OH
L-Gulose
If we have L-Gulose we can oxidate it and then if we oxidate Glucose and
Mannose -Only Glucose will give the same product as the L-Gulose
If we Crystalize Glucose from an H2O solution we'll get 2 Different materials:
[α]D = 520
1) [α]D = 1120 Equlibration Time
Equlibration Time
[α]D = 520
2) [α]D = 190
Appearantly these 2 materials are shifting from one form to the other until they
reach Equlibrium and we get the 520. This lead to the conclusion that glucose is
mostly in a Cyclic Form:
O C H
H
OH
HO
H
H
OH
H
OH
CH2OH
1
HO C H
H
OH
HO
H
H
OH
H
O
CH2OH
1
+
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H C OH
H
OH
HO
H
H
OH
H
O
CH2OH
We get a mixture of
these 2 Diastreomeres
Haworth Projections
5
O H
4
α- Pyranose
1
3
5
2
OH
O OH
4
1
3
2
Carbon No. 1 is Named The Anomeric Carbon
H
O
4
O
3
1
2
β- Pyranose
4
1
3
H
O
OH
OH
O
Pyran
α- Furanose
2
H
Furane
β- Furanose
Transformation from Fischer ro Haworth Projection
HO
H
HO
H
H
1
HO C H
Modification H 2 OH
3
HO
H
4
H
OH
5
O
H
HOH2C
6
OH
O
5
6
OH
1
C H
2 OH
3
H
4
OH
Above the ring
O
OH
OH
OH
OH
All the groups Right of the
middle are Below the ring and
the Left ones are above the ring
Below the Ring
Reactions
CO2H
Oxidation
CHO
Br2
Oxidates the
Aldehyde Only
CHOH
CO2H
CHOH
and 10 Alcohol
CHOH
H2O
CHOH
CH2OH
CHOH
CH2OH
HNO 3
CHO
CH2OH
CHOH
NaBH4
CHOH
Or H2/Pd
CHOH
CH2OH
Oxidates Both Aldehyde
CHOH
CH2OH
CHOH
CO2H
Reduction
These Reagants reduce the Aldehyde to Alcohol.
If we use HNO3 or H2/Pd or NaBH4 on Allose
or Galactose we would get Meso form and
lose Chirality
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CHO
NHNH2
NNH
HC
CHOH
CHOH
CHOH
CH2OH
Phenylhydrazone
CHOH
CH2OH
NHNH2
HC
NNH
C
Excess
This reaction occurs only on carbons 1
& 2.It helps us to identify the
Carbohydrates that have a difference
only in carbon No. 2, because they'll
give the same Osazone
NNH
Osazone
CHOH
CH2OH
Ruff Reaction
COO
COOH
CHO
CHOH Br2
H2O
CHOH
CaCO3
H2O2
CHO
+CO2
Fe
CHOH
CH2OH
This reaction
removes one
carbon from the
aldehyde end
2
Glucose
OH
These Ketals
don't change
O
O
HO
H HO
HO
between the α,
OMe
HO
OH
β Mutarotations
OH
OMe
and therefore
α-glycoside
β-glycoside
we can seperate
them
MeOH
H+
OH
OH
HO
HO
Ca
CHOH
+2
OH
O
OH 1
O
H
4
HO
HO
HO
OH
O
H
OH
OMe
Satrch (‫)עמילן‬
Polymer of β(1-4) Glucose
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OH
O
OH
O
O
HO
OMe
OH
Cellulose (‫)תאית‬
Polymer of a(1-4) Glucose