Organic chemistry : third stage/ College of science ...§لكيمياء...
Transcript of Organic chemistry : third stage/ College of science ...§لكيمياء...
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A-heeti
Ref..Morrison and Boyd
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α,β-Unsaturated Carbonyl Compounds
C OCC conjugated system
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CH2=CHCH=Oacrolein
CH2=CHCOOH
acrylic acid
H2C CHC N
acrylonitrile
H2C CCH3
COOH
methacrylic acid
CH3CH=CHCH=O
crotonaldehyde
O
mesityl oxide
CH
CH
CH=O
cinnamaldehyde
CH
CH
CO
benzalacetophenone chalcone
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CC
C
CH
HO
O
OH
OH
maleic acid
CC
C
HC
HO
OH
OHO
fumaric acid
O
O
O
maleic anhydride
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α,β-Unsaturated Carbonyl Compounds, reactions:
1. electrophilic addition
deactivated
2. nucleophilic addition
activated
3. Michael Addition
4. Diels-Alder Reaction
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1. electrophilic addition
The carbonyl group is an electron withdrawing group when conjugated with a double bond. This decreases the electron density and deactivates the double bond to electrophilic addition.
C C C O + HZ C C C OHZ
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H2C CH
CH
OHCl
CH2CH2CHOCl
O CH3OH
H2SO4
O
OCH3
H2C CH
COOHH2O,H+
H2CH2C COOH
OH
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C C C O + H C C C OH
electrophilic addition mechanism:
C C C OH
H
C C C O C C C OH H
more stable
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C C C O + H C C C OH
electrophilic addition mechanism:
C C C OH
Z
C C C OH
C C C OHZ
Zunstable
enol
C C C OZ H
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2. nucleophilic addition
The carbonyl group is an electron withdrawing group when conjugated with a double bond. This decreases the electron density and activates the double bond to nucleophilic addition.
C C C O + Z- C C C OHZ
H2O
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CH
CH
COOHNH2OH
CHCH2COOHNHOH
O O
NHCH3
CH
CH
COOEtNaCN,aq. H
CH2C COOEt
CNH3C H3C
CH3NH2
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C C C O C C C O
nucleophilic addition mechanism:
C C C O C C C OZ Z
more stable
+ Z
Z
Z
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C C C O
nucleophilic addition mechanism:
C C C O C C C OZ Z
ZH
C C C O C C C OHZ ZH
keto-enol tautomers
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3. Michael Addition.
Carbanions as the nucleophiles in nucleophilic addition to α,β-usaturated carbonyls. (The enolate anion must be in reasonably high concentration for the Michael Addition to take place. Such enolates can be obtained from removal of alpha-hydrogens that are next to two electron withdrawing groups.)
CH
CH
CO
+COOEtCH2COOEt
piperidine HC
H2C C
O
CHEtOOC COOEt
benzalacetophenone diethylamalonate
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H2C CCH3
COOEt +COOEtCH2C N
OEtH2C C
H
CH3COOEt
CHCOOEtCN
HC C
HCO
CH3 +COOEtCH2CO
CH3
base HC
H2C C
OCH3
CHC COOEtO
H3C
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G+
G
mechanism: (concerted)
G G
4. Diels-Alder reaction: diene + dienophile cyclohexene
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+CH=O
O
1,3-butadiene acrolein
+ O
O
O
O
O
O1,3-butadiene maleic anhydride
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+
1,3-butadiene
O
O
O
Op-benzoquinone
O
O
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+ O
O
OO
O
O
endo is preferred over the exo productany unsaturated groups in the dienophiletend to lie near the developing double bondin the diene rather than farther away (exo).
O
O
O
endo
exo
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α,β-Unsaturated Carbonyl Compounds, reactions:
1. electrophilic addition
deactivated
2. nucleophilic addition
activated
3. Michael Addition
4. Diels-Alder Reaction
C C C O + HZ C C C OZ H
+CO
CO
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A-heeti
Ref..Morrison and Boyd
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Amines, reactionsAmines are similar to ammonia in their reactions.
Like ammonia, amines are basic.
Like ammonia, amines are nucleophilic and react with alkyl halides, acid chlorides, and carbonyl compounds.
The aromatic amines are highly reactive in electrophilic aromatic substitution.
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Amine, reactions:
1. As bases
2. Alkylation
3. Reductive amination
4. Conversion into amides
5. EAS
6. Hofmann elimination from quarternary ammonium salts
7. Reactions with nitrous acid
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1. As bases
a) with acids
b) relative base strength
c) Kb
d) effect of groups on base strength
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with acids
NH2 + HCl NH3+Cl-
(CH3CH2)2NH + CH3COOH (CH3CH2)2NH2+, -OOCCH3
anilinium chloride
diethylammonium acetate
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relative base strength
RNH2 > NH3 > ArNH2
Kb ionization of the base in water
:Base + H2O H:Base+ + OH-
Kb = [ H:Base+ ] [ OH- ] / [ :Base ]
Kb
aliphatic amines 10-3 – 10-4
ammonia 1.8 x 10-5
anilines 10-9 or less
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Why are aliphatic amines more basic than ammonia?
NH3 + H2O NH4+ + OH-
R-NH2 + H2O R-NH3+ + OH-
The alkyl group, -R, is an electron donating group. The donation of electrons helps to stabilize the ammonium ion by decreasing the positive charge, lowering the ΔH, shifting the ionization farther to the right and increasing the basicity.
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Why are aromatic amines less basic than aliphatic amines?
R-NH2 + H2O R-NH3+ + OH-
NH2
+ H2O
NH3
+ OH
NH2 NH2 NH3 NH3
NH2 NH2 NH2 resonance stabilization of the free base, increases the ΔH, shifts the ionization to the left, decreasing base strength.
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Effect of substituent groups on base strength:
NH2
+ H2O
NH3
+ OH
G G
Electron donating groups will stabilize the anilinium ion, decreasing the ΔH, shifting the ionization farther to the right and making the compound a stronger base.
Electron withdrawing groups destabilize the anilinium ion, increasing the ΔH, shifting the ionization towards the reactants, making the compound a weaker base.
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Common substituent groups:
-NH2, -NHR, -NR2-OH-OR-NHCOCH3 electron donating-C6H5 groups-R-H-X-CHO, -COR-SO3H electron withdrawing-COOH, -COOR groups-CN-NR3
+
-NO2
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Number the following in decreasing order of base strength (let #1 = most basic, etc.
NH3
NH2 NH2 NH2 NH2
NO2 OCH3
4 1 5 3 2
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2. Alkylation (ammonolysis of alkyl halides)
RNH2R-X R2NH
R-XR3N
R-XR4N+X-
1o 2o 3o 4o salt
SN2: R-X must be 1o or CH3
CH3CH2CH2CH2BrNH3
CH3CH2CH2CH2NH2n-butylamine
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CH3CH2CH2NH2CH3Cl
CH3CH2CH2NHCH3n-propylamine methyl-n-propylamine
NH22 CH3CH2Br
NEt
Et
aniline N,N-diethylaniline
H2C NH2
benzylamine
(xs) CH3I H2C N
CH3CH3
CH3 Ibenzyltrimethylammonium iodide
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3. Reductive amination
C OH2, Ni
or NaBH3CNCH NHR+ RNH2
C OH2, Ni
or NaBH3CNCH NR2+ R2NH 3o amine
2o amine
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CCH2CH3
O
propiophenone
+ CH3CH2NH2NaBH3CN
CHCH2CH3
NHCH2CH3
1-(N-ethylamino)-1-phenylpropane
O
cyclohexanone
CH3NH2, H2/Ni NHCH3
cyclohexylmethylamine
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4. Conversion into amides
R-NH2 + RCOCl RCONHR + HCl
1o N-subst. amide
R2NH + RCOCl RCONR2 + HCl
2o N,N-disubst. amide
R3N + RCOCl NR
3o
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NH2 + (CH3CO)2OHN C CH3
O
N-phenylacetamide
CO
Cl(CH3CH2)2NH + C
O
N CH2CH3CH2CH3
N.N-diethyl-m-toluamide
N CH3CH3
+ CH3CO
ClNR
H3C H3C
DEET
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Conversion into sulfonamides
R-NH2 + ArSO2Cl ArSO2NHR + HCl
1o N-subst.sulfonamide
R2NH + ArSO2Cl ArSO2NR2 + HCl
2o N,N-disubst.sufonamide
R3N + ArSO2Cl NR
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Schotten-Baumann technique: reactions of aromatic acid chlorides are sped up by the addition of base.
R-NH2 + ArSO2Cl + KOH ArSO2NHR
1o acidic
ArSO2NR
water soluble salt
R2NH + ArSO2Cl + KOH ArSO2NR2 + HCl
2o N,N-disubst.sufonamide
water insoluble
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Hinsberg Test:unknown amine + benzenesulfonyl chloride, KOH (aq)
Reacts to produce a clear solution and then gives a ppt upon acidification primary amine.
Reacts to produce a ppt secondary amine.
Doesn’t react tertiary amine.
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NH2 N
N CH3CH3
+ SO2Cl SO
O
KOH
(CH3CH2)2NH SO2ClKOH
+ SO2ClKOH
+ SO
ON
CH2CH3
CH2CH3
NR
water sol.
ppt
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sulfanilamide “magic bullet” antibiotic
NH2
SO2NH2
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N
N N
NOH
H2N
H2C
HN C
O HN CH
COOH
CH2CH2COOH
folic acid
H2N COOH
p-aminobenzoic acd
H2N SO2NH2
sulfanilamide
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5. EAS
-NH2, -NHR, -NR2 are powerful activating groups and ortho/para directors
a) nitration
b) sulfonation
c) halogenation
d) Friedel-Crafts alkylation
e) Friedel-Crafts acylation
f) coupling with diazonium salts
g) nitrosation
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a) nitration
NH2
HNO3
H2SO4
TAR!
(CH3CO)2O
NHCOCH3
HNO3
H2SO4
NHCOCH3
NO2
+ ortho-
H2O,OH-
NH2
NO2
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b) sulfonation
NH2
+ H2SO4
NH3
SO3
cold H2SO4
NH3 HSO4
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c) halogenation
NH2
+ Br2, aq.
NH2Br Br
Brno catalyst neededuse polar solvent
Br2,Fe
Br
HNO3
H2SO4
Br
NO2
+ ortho-
H2/Ni
Br
NH2
polyhalogenation!
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NH2
Cl2 (aq.)NH2
CH3
Cl
ClCH3
o-toluidine
bright yellow!
Swimming pool test kit for chlorine:
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e) Friedel-Crafts alkylation
NR with –NH2, -NHR, -NR2
NH2CH3
+ CH3CH2Br, AlCl3 NR
Do not confuse the above with the alkylation reaction:
NH2CH3
+ CH3CH2Br
NHCH2CH3CH3
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f) Friedel-Crafts acylation
NR with –NH2, -NHR, -NR2
NH2CH3
+ NR
Do not confuse the above with the formation of amides:
NH2CH3
NHCCH3CH3
H3C CO
Cl
AlCl3
+ H3C CO
Cl
O
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g) nitrosation
NH3C CH3
NaNO2, HCl
NH3C CH3
NO
The ring is sufficiently activated towards EAS to reactwith the weak electrophile NO+
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h) coupling with diazonium salts azo dyes
NH2CH3
+
N2 Cl
benzenediazoniumchloride
CH3
NH2
NN
an azo dye
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6. Hofmann elimination from quarternary hydroxides
step 1, exhaustive methylation 4o salt
step 2, reaction with Ag2O 4o hydroxide + AgX
step 3, heat to eliminate alkene(s) + R3N
CH3CH2CH2CH2
(xs) CH3ICH3CH2CH2CH2NH2 N
CH3
CH3
CH3 I-
CH3CH2CH2CH2 NCH3
CH3
CH3 I-Ag2O
CH3CH2CH2CH2 NCH3
CH3
CH3 OH- + AgI
CH3CH2CH2CH2 NCH3
CH3
CH3 OH CH3CH2CH=CH2 + (CH3)3N
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CH3CH2CHCH3NH2
+ (xs) CH3I CH3CH2CHCH3NCH3
CH3H3C I-
CH3CH2CHCH3NCH3
CH3H3C I-Ag2O CH3CH2CHCH3
NCH3
CH3H3C OH + AgI
CH3CH2CHCH3NCH3
CH3H3C OH CH3CH2CH=CH2 + CH3CH=CHCH3
+ (CH3)3Nchief product
Hofmann orientation
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7. Reactions with nitrous acid
NH2 + HONO N N diazonium salt
R-NH2 + HONO N2 + mixture of alchols & alkenes
primary amines
secondary amines
HN R + HONO N R
N O
N-nitrosamine
tertiary amines
N RR
+ HONO N RR
NO
p-nitrosocompound
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note: 90% of all tested nitrosamines are carcinogenic in man. Many nitrosamine cancers are organ specific. For example, dimethylnitrosamine causes liver cancer while the nitrosamines in tobacco smoke cause lung cancer.
Sodium nitrite (“cure”) is used as a preservative in meats such as bacon, bologna, hot dogs, etc. to kill the organism responsible for botulism poisoning. In the stomach, the nitrous acid produced from sodium nitrite can react with secondary and tertiary amines to form nitrosamines. To reduce the formation of nitrosamines, ascorbic acid (Vitamin C) is now added to foods cured with sodium nitrite.
Nitrosamines are also found in beer!
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Amines, reactionsAmines are similar to ammonia in their reactions.
Like ammonia, amines are basic.
Like ammonia, amines are nucleophilic and react with alkyl halides, acid chlorides, and carbonyl compounds.
The aromatic amines are highly reactive in electrophilic aromatic substitution.
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Amine, reactions:
1. As bases
2. Alkylation
3. Reductive amination
4. Conversion into amides
5. EAS
6. Hofmann elimination from quarternary ammonium salts
7. Reactions with nitrous acid
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A-heeti
Ref..Morrison and Boyd
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Aryl Halides Ar-X
Organic compounds with a halogen atom attached to an aromatic carbon are very different from those compounds where the halogen is attached to an aliphatic compound. While the aliphatic compounds readily undergo nucleophilic substitution and elimination reactions, the aromatic compounds resist nucleophilic substitution, only reacting under severe conditions or when strongly electron withdrawing groups are present ortho/para to the halogen.
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Aryl halides, syntheses:
1. From diazonium salts
Ar-N2+ + CuCl Ar-Cl
Ar-N2+ + CuBr Ar-Br
Ar-N2+ + KI Ar-I
Ar-N2+ + HBF4 Ar-F
2. Halogenation
Ar-H + X2, Lewis acid Ar-X + HX
X2 = Cl2, Br2
Lewis acid = FeCl3, AlCl3, BF3, Fe…
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reactions of alkyl halides Ar-X
1. SN2 NR
2. E2 NR
3. organo metallic compounds similar
4. reduction similar
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C C X
X
aryl halide
vinyl halide
Ag+
-OH
-OR
NH3
-CN
ArH
AlCl3
NO REACTION
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Bond Lengths (Å)
C—Cl C—Br
CH3—X 1.77 1.91
C2H5—X 1.77 1.91 sp3
(CH3)3C—X 1.80 1.92
CH2=CH—X 1.69 1.86
C6H5—X 1.69 1.86sp2
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In aryl halides, the carbon to which the halogen is attached is sp2 hybrizided. The bond is stronger and shorter than the carbon-halogen bond in aliphatic compounds where the carbon is sp3 hybridized. Hence it is more difficult to break this bond and aryl halides resist the typical nucleophilic substitution reactions of alkyl halides.
The same is true of vinyl halides where the carbon is also sp2 hybridized and not prone to nucleophilic substitution.
In a manner analogous to the phenols & alcohols, we have the same functional group in the two families, aryl halides and alkyl halides, but very different chemistries.
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Aryl halides, reactions:
1. Formation of Grignard reagent
2. EAS
3. Nucleophilic aromatic substitution (bimolecular displacement)
(Ar must contain strongly electron withdrawing groups ortho and/or para to X)
4. Nucleophilic aromatic substitution (elimination-addition)
(Ring not activated to bimolecular displacement)
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1) Grignard reagent
Br
Cl
Mg
Mg
anhyd. Et2O
THF
MgBr
MgCl
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2) EAS The –X group is electron-withdrawing and deactivating in EAS, but is an ortho/para director.
BrHNO3, H2SO4
H2SO4,SO3
Br2,Fe
CH3CH2-Br, AlCl3
+
+
+
+
Br Br
Br
Br Br
Br
NO2
SO3H
Br
CH2CH3
Br
NO2
Br SO3H
Br CH2CH3
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3) Nucleophilic aromatic substitution (bimolecular displacement)
Ar must contain strongly electron withdrawing groups ortho and/or para to the X.
ClNO2
NO2
+ NH3
NH2NO2
NO2
BrNO2
NO2
+ NaOCH3
OCH3NO2
NO2
O2N O2N
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Cl
NO2
OH
NO2
ClNO2
NO2
OHNO2
NO2
O2N O2N
Cl
+ NaOH NR
350oC, 4500 psi H+OH
15% NaOH, 160oC H+
warm water
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Cl
NO2
NH2
NO2
ClNO2
NO2
NH3NO2
NO2
O2N O2N
Cl NH2
NH3, 170oC
NH3, room temp.
NH3, Cu2O, 200oC, 900 psi
NO2 NO2
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bimolecular displacement (nucleophilic aromatic substitution)
mechanism:
1) + :ZXX
ZRDS
X
Z2) Z + :X
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X
Z
Z
X
Z
X
Z
X
Z
X
Z
X
G
G
If G is an electron withdrawing group in the ortho andpara positions, it will stabilize the intermediate anion.
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evidence for the bimolecular displacement mechanism:
no element effect : Ar-I Ar-Br Ar-Cl Ar-F
(the C—X bond is not broken in the RDS)
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4) Elimination-Addition, nucleophilic aromatic substitution.
When the ring is not activated to the bimolecular displacement and the nucleophile is an extremely good one.
Br
+ NaNH2, NH3
NH2
F
+
Li
LiH2O
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Elimination-Addition mechanism (nucleophilic aromatic substitution)
1)
XH
+ :NH2
X
+ NH3
2)
X
+ :X
benzyne
3) + :NH2
NH2
NH24) + NH3
NH2
H
+ :NH2
elimination
addition
:
:
:
:
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While the concept of “benzyne” may appear to be strange, there is much evidence that this mechanism is correct.
Cl
*
* = 14C
NaNH2
NH3
NH2
* *+NH2
47% 53%
OCH3
BrH3C NaNH2
NH3
NR
ClD
+ :NH2
ClH
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+
benzyne intermediate has been trapped in a Diels-Alder condensation:
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A-heeti
Ref..Morrison and Boyd
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Carbanions
|— C: –
|
The conjugate bases of weak acids,strong bases, excellent nucleophiles.
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1. Alpha-halogenation of ketones
CCH
O+ X2
OH- or H+
CCX
O+ HX
X2 = Cl2, Br2, I2
-haloketone
H3CC
CH3
O+ Br2, NaOH H3C
CCH2Br
O+ NaBr
acetone -bromoacetone
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O
+ Cl2, H+
OCl
+ HCl
2-chlorocyclohexanone
C CH3
O+ Br2, NaOH C CH2Br + NaBr
O
-bromoacetophenone
cyclohexanone
acetophenone
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Alpha-hydrogens: 1o > 2o > 3o
CH3CH2CH2CCH3
O
2-pentanone
+ Br2, NaOH CH3CH2CH2CCH2Br + NaBr
O
1-bromo-2-pentanone
Hydrogens that are alpha to a carbonyl group are weakly acidic:
H3CC
CH3
O
H3CC
CH2
O+ OH + H2O
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RC
CH2
O
RC
CH2
O
"enolate" anion
Hydrogens that are alpha to a carbonyl are weaklyacidic due to resonance stabilization of the carbanion.
The enolate anion is a strong base and a good nucleophile
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Mechanism for base promoted alpha-bromination of acetone:
H3CC
CH3
O
H3CC
CH2
O+ OH + H2O
RDS
H3CC
CH2
O+ Br Br
H3CC
CH2Br
O+ Br
1)
2)
Rate = k [acetone] [base]
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Mechanism for acid catalyzed halogenation of ketones. Enolization.
H3CC
CH3
O
H3CC
CH3
OH+ H+
H3CC
CH3
OH+ :B
H3CC
CH2
OH+ H:B
H3CC
CH2
OH+ Br Br
H3CC
CH2Br
OH+ :Br
H3CC
CH2Br
OH
H3CC
CH2Br
O+ H
“enol”
1)
2)
3)
4)
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RC
CH3
O
Oxidation of "methyl" ketones. Iodoform test.
+ (xs) OI R CO
O+ CHI3
NaOH + I2
RC
CH2I
O
RC
CHI2
O
RC
CI3
O+ OH
R C CI3
O
OH
goodleavinggroup
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Carbanions. The conjugate bases of weak acids; strong bases, good nucleophiles.
1. enolate anions
2. organometallic compounds
3. ylides
4. cyanide
5. acetylides
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Aldehydes and ketones: nucleophilic addition
Esters and acid chlorides: nucleophilic acyl substitution
Alkyl halides: SN2
CO
+ YZ COY
Z
CW
O+ Z C
Z
O+ W
R X + Z R Z + X
Carbanions as the nucleophiles in the above reactions.
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2. Carbanions as the nucleophiles in nucleophilic addition to aldehydes and ketones:
a) aldol condensation
“crossed” aldol condensation
b) aldol related reactions (see problem 21.18 on page 811)
c) addition of Grignard reagents
d) Wittig reaction
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Carbanions as the nucleophiles in nucleophilic addition to aldehydes and ketones:
c) addition of Grignard reagents
Grignard reagents are examples of organo metallic carbanions.
CO
+ RMgX COMgX
R
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a) Aldol condensation. The reaction of an aldehyde or ketone with dilute base or acid to form a beta-hydroxycarbonyl product.
CH3CH=Odil. NaOH
CH3CHCH2CH OOH
acetaldehyde 3-hydroxybutanal
CH3CCH3
O dil. NaOHCH3CCH2CCH3
OOH
CH3acetone4-hydroxy-4-methyl-2-pentanone
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CH3CH=Odil. NaOH
CH3CHCH2CH OOH
acetaldehyde 3-hydroxybutanal
OH
CH2CH=O CH3CH+ O CH3CHCH2CH OO
+ H2O
+ H2O
nucleophilic addition by enolate ion.
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H3CC
CH3
O
OH
H3CC
CH2
O
H3CC
CH3
O
H3CCO
CH2
CO
CH3
CH3
+ H2O
+ H2O
H3CCO
CH2
COH
CH3
CH3dil. NaOH
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CH3CH2CH=O + dil. NaOH CH3CH2CHCH2CH2CHOH
O
CH3CHCH Oalpha!
CH3CH2CH CH3CH2CHCHCHOH
CH3
OO
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O
dil. OH-
O
OH
OH
O
O
O
O + HOH
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dil. H+
O
+ H2O
O
With dilute acid the final product is the α,β-unsaturated carbonyl compound!
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CH2CH O
phenylacetaldehyde
dil NaOHCH2 C
HCH
OHCH=O
dilute H+
CH2 CH
C CH=O
note: double bond is conjugatedwith the carbonyl group!
+ H2O
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NB: An aldehyde without alpha-hydrogens undergoes the Cannizzaro reaction with conc. base.
CHO
benzaldehyde
conc. NaOH COO- CH2OH+
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Crossed aldol condensation:
If you react two aldehydes or ketones together in an aldol condensation, you will get four products. However, if one of the reactants doesn’t have any alpha hydrogens it can be condensed with another compound that does have alpha hydrogens to give only one organic product in a “crossed” aldol.
CH3CH2CH + H2C OO CH3CHCH2 OHCH ONaOH
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N.B. If the product of the aldol condensation under basic conditions is a “benzyl” alcohol, then it will spontaneouslydehydrate to the α,β-unsaturated carbonyl.
CH=O + CH3CH2CH2CH=Odil OH-
CH=CCH=OCH2CH3
CHCHCH=OOH
CH2CH3
-H2O
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A crossed aldol can also be done between an aldehyde and a ketone to yield one product. The enolate carbanion from the ketone adds to the more reactive aldehyde.
C CH3
O
acetophenone
+ CH3CH=O
acetaldehyde
dil OH-CCH2
OCH
OHCH3
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b) Aldol related reactions: (see problem 21.18 page 811 of your textbook).
CH=O + CH3NO2KOH
CH=CHNO2 + H2O
CH2NO2
CH=O + CH2C NNaOEt
CH=C CN
CHC N
+ H2O
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Perkin condensation
CH=O + (CH3CO)2OCH3COONa
CH=CHCOOH
H2C CO
OCCH3
O
CHOH
CH2 CO
OCCH3
O
+ H2OHC C
HC
O
OCCH3
O
hydrolysis ofanhydride
+ CH3COOH
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d) Wittig reaction (synthesis of alkenes)
1975 Nobel Prize in Chemistry to Georg Wittig
C O + Ph3P=C R'
R
ylide
CO
C R'R
PPh3
C CR
R' + Ph3PO
CH2CH=O + Ph3P=CH2 CH2CH=CH2 + Ph3PO
Ph = phenyl
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CO
C R'R
PPh3
C CR
R' + Ph3PO
PPh
PhPh
CR
R' CO
ylide
nuclephilic addition by ylide carbanion, followed by loss of Ph3PO (triphenylphosphine oxide)
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O + Ph3PCHCH=CH2 CHCH CH2 + Ph3PO
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3. Carbanions as the nucleophiles in nucleophilic acyl substitution of esters and acid chlorides.
a) Claisen condensation
a reaction of esters that have alpha-hydrogens in basic solution to condense into beta-keto esters
CH3COOEtethyl acetate
NaOEtCH3CCH2COOEt
O+ EtOH
ethyl acetoacetate
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CH3COOEtNaOEt
CH3CCH2COOEtO
+ EtOH
CH3 COEt
OCH3 C OEt
O
CH2COOEt
nucleophilic acyl substitution by enolate anion
OEt
CH2COOEt
Mechanism for the Claisen condensation:
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ethyl propionate
CH3CH2CCHCOOEt
CH3
O
ethyl 2-methyl-3-oxopentanoate
OEtCH3CH2COOEt
OEt
CH3CHCOOEt CH3CH2CO
OEt CH3CH2CO
OEtCHCOOEtCH3
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CH2COOEt
NaOEtCH2C
OCHCOOEt
ethyl phenylacetate
CHCOOEt CH2CO
OEtCH2C
O
CHCOOEtOEt
OEt
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Crossed Claisen condensation:
COOEt + CH3COOEtNaOEt
CO
CH2COOEt
ethyl benzoate
HCOOEt + CH3CH2COOEt
ethyl formate
H CO
CHCOOEtCH3
OEt
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COOEt
COOEtCH3CH2COOEt
OC2H5+ C
O
CO
OEtCHCOOEtCH3
COOEt
COOEt2 CH3CH2COOEt
NaOC2H5+ C
O
CO
CHCOOEtCH3
CHCOOEtCH3
ethyl oxalate
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EtOCOEtethyl carbonate
+
COOEt
CH2COOEt
ethyl malonate
NaOEtC CHO COOEt
COOEtEtO
CH3CH2COOEt
ethyl propionate
+O
cyclohexanone
NaOEtCH3CH2C
OO
enolate from ketone in nucleophilic acyl substitution on ester
O
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b) Coupling of lithium dialkyl cuprate with acid chloride
R CCl
O+ R'2CuLi R C
R'
O
nucleophile = R'
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4. Carbanions as nucleophiles in SN2 reactions with R’X:
a) Corey-House synthesis of alkanes
R2CuLi + R’X R-R’
b) metal acetylide synthesis of alkynes
RCC-M+ + R’X RCCR’
c) Malonate synthesis of carboxylic acids
d) Acetoacetate synthesis of ketones
5. Michael Addition to α,β-unsaturated carbonyl compounds
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Carbanions are the conjugate bases of weak acids and are therefore strong bases and excellent nucleophiles that can react with aldehydes/ketones (nucleophilic addition), esters/acid chlorides (nucleophilic acyl substitution), and alkyl halides (SN2), etc.
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Reactions involving carbanions as nucleophiles:
1. Alpha-halogenation of ketones
2. Nucleophilic addition to aldehydes/ketones
a) aldol and crossed aldol
b) aldol related reactions
c) Grignard synthesis of alcohols
d) Wittig synthesis of alkenes
3. Nucleophilic acyl substitution with esters and acid chlorides
a) Claisen and crossed Claisen
b) R2CuLi + RCOCl
(next slide)
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4. SN2 with alkyl halides
a) Corey-House
b) metal acetylide
c) Malonate synthesis
d) Acetoacetate synthesis
5. Michael Addition to α,β-unsaturated carbonyl compounds
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A-heeti
Ref..Morrison and Boyd
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Carbanions
|— C: –
|
The conjugate bases of weak acids,strong bases, excellent nucleophiles.
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1. Alpha-halogenation of ketones
CCH
O+ X2
OH- or H+
CCX
O+ HX
X2 = Cl2, Br2, I2
-haloketone
H3CC
CH3
O+ Br2, NaOH H3C
CCH2Br
O+ NaBr
acetone -bromoacetone
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O
+ Cl2, H+
OCl
+ HCl
2-chlorocyclohexanone
C CH3
O+ Br2, NaOH C CH2Br + NaBr
O
-bromoacetophenone
cyclohexanone
acetophenone
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Alpha-hydrogens: 1o > 2o > 3o
CH3CH2CH2CCH3
O
2-pentanone
+ Br2, NaOH CH3CH2CH2CCH2Br + NaBr
O
1-bromo-2-pentanone
Hydrogens that are alpha to a carbonyl group are weakly acidic:
H3CC
CH3
O
H3CC
CH2
O+ OH + H2O
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RC
CH2
O
RC
CH2
O
"enolate" anion
Hydrogens that are alpha to a carbonyl are weaklyacidic due to resonance stabilization of the carbanion.
The enolate anion is a strong base and a good nucleophile
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Mechanism for base promoted alpha-bromination of acetone:
H3CC
CH3
O
H3CC
CH2
O+ OH + H2O
RDS
H3CC
CH2
O+ Br Br
H3CC
CH2Br
O+ Br
1)
2)
Rate = k [acetone] [base]
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Mechanism for acid catalyzed halogenation of ketones. Enolization.
H3CC
CH3
O
H3CC
CH3
OH+ H+
H3CC
CH3
OH+ :B
H3CC
CH2
OH+ H:B
H3CC
CH2
OH+ Br Br
H3CC
CH2Br
OH+ :Br
H3CC
CH2Br
OH
H3CC
CH2Br
O+ H
“enol”
1)
2)
3)
4)
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RC
CH3
O
Oxidation of "methyl" ketones. Iodoform test.
+ (xs) OI R CO
O+ CHI3
NaOH + I2
RC
CH2I
O
RC
CHI2
O
RC
CI3
O+ OH
R C CI3
O
OH
goodleavinggroup
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Carbanions. The conjugate bases of weak acids; strong bases, good nucleophiles.
1. enolate anions
2. organometallic compounds
3. ylides
4. cyanide
5. acetylides
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Aldehydes and ketones: nucleophilic addition
Esters and acid chlorides: nucleophilic acyl substitution
Alkyl halides: SN2
CO
+ YZ COY
Z
CW
O+ Z C
Z
O+ W
R X + Z R Z + X
Carbanions as the nucleophiles in the above reactions.
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2. Carbanions as the nucleophiles in nucleophilic addition to aldehydes and ketones:
a) aldol condensation
“crossed” aldol condensation
b) aldol related reactions (see problem 21.18 on page 811)
c) addition of Grignard reagents
d) Wittig reaction
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Carbanions as the nucleophiles in nucleophilic addition to aldehydes and ketones:
c) addition of Grignard reagents
Grignard reagents are examples of organo metallic carbanions.
CO
+ RMgX COMgX
R
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a) Aldol condensation. The reaction of an aldehyde or ketone with dilute base or acid to form a beta-hydroxycarbonyl product.
CH3CH=Odil. NaOH
CH3CHCH2CH OOH
acetaldehyde 3-hydroxybutanal
CH3CCH3
O dil. NaOHCH3CCH2CCH3
OOH
CH3acetone4-hydroxy-4-methyl-2-pentanone
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CH3CH=Odil. NaOH
CH3CHCH2CH OOH
acetaldehyde 3-hydroxybutanal
OH
CH2CH=O CH3CH+ O CH3CHCH2CH OO
+ H2O
+ H2O
nucleophilic addition by enolate ion.
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H3CC
CH3
O
OH
H3CC
CH2
O
H3CC
CH3
O
H3CCO
CH2
CO
CH3
CH3
+ H2O
+ H2O
H3CCO
CH2
COH
CH3
CH3dil. NaOH
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CH3CH2CH=O + dil. NaOH CH3CH2CHCH2CH2CHOH
O
CH3CHCH Oalpha!
CH3CH2CH CH3CH2CHCHCHOH
CH3
OO
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O
dil. OH-
O
OH
OH
O
O
O
O + HOH
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dil. H+
O
+ H2O
O
With dilute acid the final product is the α,β-unsaturated carbonyl compound!
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CH2CH O
phenylacetaldehyde
dil NaOHCH2 C
HCH
OHCH=O
dilute H+
CH2 CH
C CH=O
note: double bond is conjugatedwith the carbonyl group!
+ H2O
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NB: An aldehyde without alpha-hydrogens undergoes the Cannizzaro reaction with conc. base.
CHO
benzaldehyde
conc. NaOH COO- CH2OH+
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Crossed aldol condensation:
If you react two aldehydes or ketones together in an aldol condensation, you will get four products. However, if one of the reactants doesn’t have any alpha hydrogens it can be condensed with another compound that does have alpha hydrogens to give only one organic product in a “crossed” aldol.
CH3CH2CH + H2C OO CH3CHCH2 OHCH ONaOH
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N.B. If the product of the aldol condensation under basic conditions is a “benzyl” alcohol, then it will spontaneouslydehydrate to the α,β-unsaturated carbonyl.
CH=O + CH3CH2CH2CH=Odil OH-
CH=CCH=OCH2CH3
CHCHCH=OOH
CH2CH3
-H2O
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A crossed aldol can also be done between an aldehyde and a ketone to yield one product. The enolate carbanion from the ketone adds to the more reactive aldehyde.
C CH3
O
acetophenone
+ CH3CH=O
acetaldehyde
dil OH-CCH2
OCH
OHCH3
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b) Aldol related reactions: (see problem 21.18 page 811 of your textbook).
CH=O + CH3NO2KOH
CH=CHNO2 + H2O
CH2NO2
CH=O + CH2C NNaOEt
CH=C CN
CHC N
+ H2O
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Perkin condensation
CH=O + (CH3CO)2OCH3COONa
CH=CHCOOH
H2C CO
OCCH3
O
CHOH
CH2 CO
OCCH3
O
+ H2OHC C
HC
O
OCCH3
O
hydrolysis ofanhydride
+ CH3COOH
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d) Wittig reaction (synthesis of alkenes)
1975 Nobel Prize in Chemistry to Georg Wittig
C O + Ph3P=C R'
R
ylide
CO
C R'R
PPh3
C CR
R' + Ph3PO
CH2CH=O + Ph3P=CH2 CH2CH=CH2 + Ph3PO
Ph = phenyl
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CO
C R'R
PPh3
C CR
R' + Ph3PO
PPh
PhPh
CR
R' CO
ylide
nuclephilic addition by ylide carbanion, followed by loss of Ph3PO (triphenylphosphine oxide)
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O + Ph3PCHCH=CH2 CHCH CH2 + Ph3PO
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3. Carbanions as the nucleophiles in nucleophilic acyl substitution of esters and acid chlorides.
a) Claisen condensation
a reaction of esters that have alpha-hydrogens in basic solution to condense into beta-keto esters
CH3COOEtethyl acetate
NaOEtCH3CCH2COOEt
O+ EtOH
ethyl acetoacetate
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CH3COOEtNaOEt
CH3CCH2COOEtO
+ EtOH
CH3 COEt
OCH3 C OEt
O
CH2COOEt
nucleophilic acyl substitution by enolate anion
OEt
CH2COOEt
Mechanism for the Claisen condensation:
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ethyl propionate
CH3CH2CCHCOOEt
CH3
O
ethyl 2-methyl-3-oxopentanoate
OEtCH3CH2COOEt
OEt
CH3CHCOOEt CH3CH2CO
OEt CH3CH2CO
OEtCHCOOEtCH3
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CH2COOEt
NaOEtCH2C
OCHCOOEt
ethyl phenylacetate
CHCOOEt CH2CO
OEtCH2C
O
CHCOOEtOEt
OEt
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Crossed Claisen condensation:
COOEt + CH3COOEtNaOEt
CO
CH2COOEt
ethyl benzoate
HCOOEt + CH3CH2COOEt
ethyl formate
H CO
CHCOOEtCH3
OEt
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COOEt
COOEtCH3CH2COOEt
OC2H5+ C
O
CO
OEtCHCOOEtCH3
COOEt
COOEt2 CH3CH2COOEt
NaOC2H5+ C
O
CO
CHCOOEtCH3
CHCOOEtCH3
ethyl oxalate
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EtOCOEtethyl carbonate
+
COOEt
CH2COOEt
ethyl malonate
NaOEtC CHO COOEt
COOEtEtO
CH3CH2COOEt
ethyl propionate
+O
cyclohexanone
NaOEtCH3CH2C
OO
enolate from ketone in nucleophilic acyl substitution on ester
O
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b) Coupling of lithium dialkyl cuprate with acid chloride
R CCl
O+ R'2CuLi R C
R'
O
nucleophile = R'
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4. Carbanions as nucleophiles in SN2 reactions with R’X:
a) Corey-House synthesis of alkanes
R2CuLi + R’X R-R’
b) metal acetylide synthesis of alkynes
RCC-M+ + R’X RCCR’
c) Malonate synthesis of carboxylic acids
d) Acetoacetate synthesis of ketones
5. Michael Addition to α,β-unsaturated carbonyl compounds
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Carbanions are the conjugate bases of weak acids and are therefore strong bases and excellent nucleophiles that can react with aldehydes/ketones (nucleophilic addition), esters/acid chlorides (nucleophilic acyl substitution), and alkyl halides (SN2), etc.
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Reactions involving carbanions as nucleophiles:
1. Alpha-halogenation of ketones
2. Nucleophilic addition to aldehydes/ketones
a) aldol and crossed aldol
b) aldol related reactions
c) Grignard synthesis of alcohols
d) Wittig synthesis of alkenes
3. Nucleophilic acyl substitution with esters and acid chlorides
a) Claisen and crossed Claisen
b) R2CuLi + RCOCl
(next slide)
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4. SN2 with alkyl halides
a) Corey-House
b) metal acetylide
c) Malonate synthesis
d) Acetoacetate synthesis
5. Michael Addition to α,β-unsaturated carbonyl compounds
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Organic chemistry : third stage/College of science /Baghdad university
By: Ass.prof.Dr.Muna A‐heetiRef..Morrison and Boyd
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Chapter 11: Amines and Related Nitrogen Compounds
The painkiller morphine is obtained from opium, the dried sap of unripe seeps of the poppy Papaver somniferum
Amines are relatives of ammonia and are usually classified as primary, secondary or tertiary
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Classify the following amines as primary, secondary or tertiary
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Structure of Amines
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Nomenclature of Amines
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When other functional groups are present, the amino group is named as a substituent
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Aromatic amines are named as derivatives of aniline. In the CA system, aniline is called benzeneamine
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Physical Properties and Intermolecular Interactions of Amines
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Preparation of Amines
This reactions proceed by the formation of an ammonium salt followed by the treatment of a strong base.
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Similarly secondary amines and tertiary amines can be prepared by the same procedure
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Reaction with tertiary amines leads to the formation of quaternary ammonium salts.
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Alkylation of aromatic amines may be selective
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Alkylation may also be intramolecular an example is shown in the synthesis of nicotine
Tobacco plant, a natural source of nicotine
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Reduction of the nitrobenzene always yields aniline.
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Amides can be reduced to amines with lithium aluminum hydride
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Reduction of nitriles (cyanides) gives primary amines
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Device a synthesis of the compound on the left from the one on the right
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Aldehydes and ketones undergo reductive amination when treated with ammonia, primary amine or secondary amines, to give primary, secondary, or tetiary amines respectively. The commonly used reducing agent for this purpose is metal hydride sodium cyanoborohydride, NaBH3CN.
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Basicity of Amines
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Aromatic amines are weaker bases than aliphatic amines or ammonia. This is mainly due to the resonance delocalization of the unshared electron pair in the aromatic amines.
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Quaternary Ammonium Compounds
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Aromatic Diazonium Compounds
Primary aromatic amines react with nitrous acid at 0 0C to yield aryldiazonium ions. The process is called diazotization
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Examples of conversion of diazonio group to other groups
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How can one prepare m‐dibromobenzene?
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Prepare o‐methylbenzoic acid from o‐toluidine (o‐methylaniline)
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Prepare 1,3,5‐tribromobenzene from aniline?
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Diazo Coupling; Azo Dyes
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Aryldiazonium ions react with strongly activated aromatic rings like phenols and aryl amine to give azo compounds
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A‐heeti
Ref..Morrison and Boyd
Dr. Dina Bakhotmah ١
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Heterocyclic compound
٢
Introduction:The IUPAC Gold Book describes hemical substances that carry the genetic information controlling inheritance, consist of long chains of heterocyclic units
are antibioticsand , vitamins, pigmentsMany naturally occurring held together. heterocyclic compounds.
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Heterocyclic derivatives and classification Heterocyclic derivatives as a group, can be divided into two broad areas:aromatic and non-aromatic. (note : aromatic system must have 4n+2 π-electron)
1) the 1, five-membered rings are shown below, the derivative furan 1 is aromatic, while tetrahydrofuran (2), dihydrofuran-2-one (3) are not aromatic, Why?.
2) The six-membered rings below, pyridine is aromatic (5), while piperidine (6), piperidin-2-one (7) are not aromatic, why? Most heterocycles have the same chemistry as their open-chain counterparts
particularly when the ring is unsaturated.
Dr. Dina Bakhotmah
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In general :heterocyclic is the largest and most varied family of organic compounds, heterocyclic system can be 3, 4, 5, 6, 7 membered rings
Dr. Dina Bakhotmah ٤
in addition to a fused rings (two rings joined at two adjacent atoms)
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Heterocyclic analogues of cyclopentadiene with one heteroatomor
Five membered monoheterocyclic* *Organic chem. Solomons, p 655-
Cyclic compounds that include an element other than carbon are called heterocyclic compound
Pyrrole, furan and thiophene are a five-membered heterocyclic compound, We might expect each of these compounds to have properties of conjugated diene of an amine , an ether or sulphide respectively.
On this basis pyrrole , furan and thiophene must be considered to be aromatic, this is proved by NMR spectrum
N O SCyclopentadiene
PyrroleFuran ThiopheneH
Cyclopentadiene anion
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٦
these heterocyclic compounds in general undergo electrophilic substitution reaction for example; nitration ,sulphonation, halogenation ,Friedel-craft Acylation and coupling with diazonium salts
The Heat of combustion indicate resonance stabilization to the extent 22-28 kcal/mole less than the resonance energy of benzene (36kcal/mol)but much greater than of most conjugated diene (about 3 kcal/mol).
Dr. Dina Bakhotmah
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Dr. Dina Bakhotmah ٧
Pyrrole
The delocalization of the lone pair of Pyrrole pushes electrons from the nitrogen atom into the ring and we expect the ring to be electron-rich and become more nucleophile.Thus, decreased basicity of the nitrogen atom and increased acidity of the NH group as a whole
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Pyrrole is assumed to be an aromatic molecule. While Its protonated form is not aromatic
Using orbital pictures, explain why?
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Class exercise 1) In each of the following, encircle the stronger base :
Dr. Dina Bakhotmah
2) If the proton was removed from Pyrrole to give the structure shown below, would it still be considered aromatic? Why or why not? Use pictures in your explanation.
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Dr. Dina Bakhotmah ١٠
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Dr. Dina Bakhotmah ١٢
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Synthesis of heterocyclic
١٣Dr. Dina Bakhotmah
1. from 1,4-diketones: the Paal-Knorr synthesis
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Paal‐Knorr synthesis: pyrroles and thiophenes
• Class exercise • Write the Paal‐Knorr synthesis of 2‐ethyl‐4‐methyl Furan?• Write the Paal‐Knorr synthesis of 2‐phenyl‐5‐methyl pyrrole starting from 1‐phenyl‐2,5‐
dipentanon? ١٤Dr. Dina Bakhotmah
O2H-
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Dr. Dina Bakhotmah ١٥
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Dr. Dina Bakhotmah ١٦
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2. Knorr pyrrole synthesis
١٧Dr. Dina Bakhotmah
α-
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Dr. Dina Bakhotmah ١٨
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١٩Dr. Dina Bakhotmah
Write the reaction mechanism for each reaction
Examples of pyrroles syntheses by Knorr method:
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٢٠
Electrophilic substitution
Dr. Dina Bakhotmah
The aromatic five-membered heterocycles pyrrole , furan and thiophene all undergo electrophilic substitution, with a general reactivity order:
pyrrole >furan > thiophene > benzene. This Due to the higher electron density at the carbons as shown by the resonance hybrids of pyrrole as an example .
Write the resonance hybrids of furan and thiophene
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Electrophilic attack on aromatic five-membered heterocycles pyrrole , furan and thiophene
Dr. Dina Bakhotmah ٢١
Major isomer
Write the general electrophilic attack on furan and thiophene
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٢٢Dr. Dina Bakhotmah
All these aromatic heterocycles react vigorously with chlorine and bromine, often forming polyhalogenated products reaction 3. while reagent N-bromosuxcinamide (NBS) give monosubstituted bromine
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٢٣Dr. Dina Bakhotmah
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Dr. Dina Bakhotmah ٢٤
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Dr. Dina Bakhotmah ٢٥
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Dr. Dina Bakhotmah ٢٦
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Dr. Dina Bakhotmah ٢٧
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Dr. Dina Bakhotmah ٢٨
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Dr. Dina Bakhotmah ٢٩
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Activity 1in the following slid a top 200 pharmaceutical products by US prescription in 2012 http://cbc.arizona.edu/njardarson/group/top‐pharmaceuticals‐posteropen the link, choose one product, find the kinds of hetero rings system and write a short note on its used (2‐5 lines only) see the example bellow
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Organic chemistry : third stage/College of science /Baghdad
university By: Ass.prof.Dr.Muna A-heeti
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Phenols Ar-OH
Phenols are compounds with an –OH group attached to an aromatic carbon. Although they share the same functional group with alcohols, where the –OH group is attached to an aliphatic carbon, the chemistry of phenols is very different from that of alcohols.
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Nomenclature.
Phenols are usually named as substituted phenols. The methylphenols are given the special name, cresols. Some other phenols are named as hydroxy compounds.
OH
phenol
OH
Br
m-bromophenol
CH3OH
o-cresol
OHCOOH
salicylic acid
OHOH
OH
OH
OH
OH
catechol resorcinol hydroquinone
COOH
OH
p-hydroxybenzoic acid
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physical properties
phenols are polar and can hydrogen bond
phenols are water insoluble
phenols are stronger acids than water and
will dissolve in 5% NaOH
phenols are weaker acids than carbonic acid and
do not dissolve in 5% NaHCO3
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Intramolecular hydrogen bonding is possible in some ortho-substituted phenols. This intramolecular hydrogen bonding reduces water solubility and increases volatility. Thus, o-nitrophenol is steam distillable while the isomeric p-nitrophenol is not.
N
OH
O
O
o-nitrophenolbp 100oC at 100 mm0.2 g / 100 mL watervolatile with steam
OH
NO2
p-nitrophenolbp decomposes1.69 g / 100 mL waternon-volatile with steam
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phenols, syntheses:
1. From diazonium salts
2. Alkali fusion of sulfonates
N2H2O,H+
OH
SO3 Na NaOH,H2O
300o
ONa H+ OH
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Reactions:
alcohols phenols
1. HX NR
2. PX3 NR
3. dehydration NR
4. as acids phenols are more acidic
5. ester formation similar
6. oxidation NR
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Phenols, reactions:
1. as acids
2. ester formation
3. ether formation
4. EAS
a) nitration f) nitrosation
b) sulfonation g) coupling with diaz. salts
c) halogenation h) Kolbe
d) Friedel-Crafts alkylation i) Reimer-Tiemann
e) Friedel-Crafts acylation
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as acids:
with active metals:
with bases:CH4 < NH3 < HCCH < ROH < H2O < phenols < H2CO3 < RCOOH < HF
OHNa
ONa
sodium phenoxide
+ H2(g)
OH
+ NaOH
ONa
+ H2O
SA SB WB WA
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CH4 < NH3 < HCCH < ROH < H2O < phenols < H2CO3 < RCOOH < HF
OH
+ NaOH
ONa
+ H2O
SA SB WB WA
water insoluble water soluble
OH
+ NaHCO3 NR phenol < H2CO3
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insoluble soluble insoluble
insoluble soluble soluble
water 5% NaOH 5% NaHCO3
phenols
carboxylic acids
CH4 < NH3 < HCCH < ROH < H2O < phenols < H2CO3 < RCOOH < HF
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We use the ionization of acids in water to measure acid strength (Ka):
HBase + H2O H3O+ + Base-
Ka = [H3O+ ][ Base ] / [ HBase]
ROH Ka ~ 10-16 - 10-18
ArOH Ka ~ 10-10
Why are phenols more acidic than alcohols?
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ROH + H2O H3O+ + RO-
ArOH + H2O H3O+ + ArO-
OH OH O O O O O
Resonance stabilization of the phenoxide ion, lowers the PE of the products of the ionization, decreases the ΔH, shifts the equil farther to the right, makes phenol more acidic than an alcohol
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effect of substituent groups on acid strength?
OH
G + H2O
O
G + H3O
Electron withdrawing groups will decrease the negative charge in the phenoxide, lowering the PE, decreasing the ΔH, shifting the equil farther to the right, stronger acid.
Electron donating groups will increase the negative charge in the phenoxide, increasing the PE, increasing the ΔH, shifting the equilibrium to the left, weaker acid.
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Number the following acids in decreasing order of acid strength (let # 1 = most acidic, etc.)
OH OH OH OH OH
NO2 CH3 Br
3 5 1 4 2
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SO3H COOH OH CH2OH
1 2 3 4
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2. ester formation (similar to alcohols)
OHCH3
+ CH3CH2CO
OH
H+
CH3CH2CO
O
H3C
+ H2O
OHCOOH
salicyclic acid
+ (CH3CO)2O
OCOOH
CH3CO
aspirin
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OCOOH
CH3CO
aspirin
analgesicanti-inflamatoryantipyrreticanticoagulant
Reye's syndromenot to be used by childrenwith high fevers!
OH
NHCH3CO
acetaminophen
aspirin substituteTylenol
Kidney damage!
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3. ether formation (Williamson Synthesis)
Ar-O-Na+ + R-X Ar-O-R + NaX
note: R-X must be 1o or CH3
Because phenols are more acidic than water, it is possible to generate the phenoxide in situ using NaOH.
OH
CH3
+ CH3CH2Br, NaOH
OCH2CH3
CH3
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4. Electrophilic Aromatic Substitution
The –OH group is a powerful activating group in EAS and an ortho/para director.
a) nitration
OH OHNO2
NO2
O2Npolynitration!
OHdilute HNO3
OH OHNO2
NO2
+
HNO3
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OHBr2 (aq.)
OHBr
Br
Br no catalyst required
use polar solvent
polyhalogenation!
OHBr2, CCl4
OH OHBr
Br
+non-polar solvent
b) halogenation
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c) sulfonation
OH
H2SO4, 15-20oC
OHSO3H
H2SO4, 100oC
OH
SO3H
At low temperature the reaction is non-reversible and the lower Eact ortho-product is formed (rate control).
At high temperature the reaction is reversible and the more stable para-product is formed (kinetic control).
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d) Friedel-Crafts alkylation.
OH
+ H3C C CH3
CH3
Cl
AlCl3
OH
C CH3CH3
H3C
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e) Friedel-Crafts acylation
OH
CH3CH2CH2CO
Cl+
AlCl3
OH
O
Do not confuse FC acylation with esterification:
OH
CH3CH2CH2CO
Cl+ O
O
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OH
O
OH
CH3CH2CH2CO
Cl+ O
O
AlCl3
Fries rearrangement of phenolic esters.
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f) nitrosation
OHHONO
OH
NO
EAS with very weak electrophile NO+
OHCH3 NaNO2, HCl
OHCH3
NO
p-nitrosophenol
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g) coupling with diazonium salts
(EAS with the weak electrophile diazonium)
OHCH3
+
N2 Cl
benzenediazoniumchloride
CH3
OH
NN
an azo dye
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h) Kolbe reaction (carbonation)
ONa
+ CO2125oC, 4-7 atm.
OHCOONa
sodium salicylate
H+
OHCOOH
salicylic acid
EAS by the weaklyelectrophilic CO2
O C O
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i) Reimer-Tiemann reaction
OHCHCl3, aq. NaOH
70oCH+
OHCHO
salicylaldehyde
The salicylaldehyde can be easily oxidized to salicylic acid
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Spectroscopy of phenols:
Infrared:
O—H stretching, strong, broad 3200-3600 cm-1
C—O stretch, strong, broad ~1230 cm-1
(alcohols ~ 1050 – 1200)
nmr: O—H 4-7 ppm (6-12 ppm if intramolecular hydrogen bonding)
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o-cresol
C--O
O--H
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o-cresol
OHCH3 a
b
c
c b a
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ethyl salicylate (intramolecular hydrogen bonding)
OHC
O
O
CH2CH3
abc
d
d c b a
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Organic chemistry : third stage/
College of science /Baghdad university
By: Ass.prof.Dr.Muna A‐heeti
Ref..Morrison and Boyd
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I. Isolated Ring Polynuclear
Hydrocarbons
Biphenyl (diphenyl):
I. Isolated Ring Polynuclear
Hydrocarbons
Biphenyl (diphenyl):
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a) Fittig reaction
b) From benzene diazonium sulphate
c) From benzidine
2Ph-Br + 2Na ethersoln Ph-Ph + 2NaBr
N=N HSO4CuEtOH
Biphenyl
+ N2 + CuHSO4
Preparation of Biphenyl
H2N NH2NaNO2
HClN=N N=N ClCl
Benzidine
H3PO4
Biphenyl
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2 Ph-I2 Cu
Ph-Ph + 2CuI
2 CuI2 sealed tube
4,4'- Dimethyl-biphenyl
H3C CH3H3C
2 CuI2
sealed tube
2,2'- Dinitro-biphenyl
NO2 NO2
O2N
d) Ulmann diaryl synthesis
e) By using Arylmagnesium halide
PhMgBr + PhBr Ph-Ph + MgBr2C OC l2
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Reactions of biphenyl
Biphenyl undergoes substitution reactions,
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• In biphenyl one ring act as electron donating group and the other act as
electron withdrawing group
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Resonance shows that O- and P- are the most reactive positions towards electrophilic substitution.
The electrophilic substitution occurs in 4- position (major) and 2- position (minor) due to steric effect of other benzene ring.
The 2nd substitution occurs in the empty ring in 2 or 4- position.e.g. NO2
+ NO2
NO2
O2N
NO2
NO2NO2O2N + +
conc HNO3conc H2SO4
conc HNO3conc H2SO4
2-Nitro-biphenyl 4-Nitro-biphenyl
4,4'-Dinitro-biphenyl 2,4'-Dinitro-biphenyl 2,2'-Dinitro-biphenyl
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Problem: Explain the products formed when biphenyl is mononitrated and when it is dinitrated.
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Biphenyl derivatives
NH2H2N
COOH COOH
(1) Benzidine (4, 4`-diaminobiphenyl)
(2) Diphenic acid
(3) Diphenyl methaneCH2
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(1) Benzidine (4, 4' diaminobiphenyl)
Methods of preparation
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From dihydrazo benzene
NH-NH
Hydrazano-benzene
H2N NH2
Benzidine
HCl
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Q. Show how could you prepare benzidine from benzene?
Answer
I2
HNO3
I
conc. HNO3
conc. H2SO4
I
NO2I
NO2
2 Cu O2N NO2Zn/HCl
H2N NH2
Benzidine
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Uses of Benzidine
preparation of Congo red
NH2H2N NaNO2/HCl NNNCl N Cl
SO3H
NH2
2
NNN N
SO3H
NH2NH2
SO3HCongo Red
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Organic chemistry : third stage/
College of science /Baghdad university
By: Ass.prof.Dr.Muna A‐heeti
Ref..Morrison and Boyd
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Polynuclear Hydrocarbons
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(2) Diphenic acid
Methods of preparation
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From anthranilic acid
NH2
COOH
NaNO2HCl N
COOH
N Cu
COOH COOH
biphenyl- 2,2'- dicarboxylic acid(Diphenic acid)
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Ulmann diaryl synthesis
I
COOEt
2 Cu
COOEt COOEt
HCl
COOH COOH
biphenyl- 2,2'- dicarboxylic acid(Diphenic acid)
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oxidation of Phenanthrene or phenanthraquinone
50% H2O2AcOH,
COOH
COOH
Diphenic acid
K2Cr2O7H2SO4
O
O
phenanthrene phenanthraquinone
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Chemical Reactions
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1. with acetic anhydride
COOH
COOH
Diphenic acid
Ac2O O
O
O
Diphenic anhydride
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2. Conversion of diphenic acid to dflourenone
COOH
COOH
Diphenic acid
Ca(OH)2 C
C
O
O
O
OCa
80-200C-CaCO3
O
Ca diphenate 9- fluorenone
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3. with conc. H2SO4
COOH
COOH
Diphenic acid
FreeO
9- fluorenone- 4- carboxylic acid
rotation
O
OHOH
O conc. H2SO4-H2O
HOOC
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4. Oxidation of KMnO4
COOH
COOH
Diphenic acid
KMnO4 COOH
COOH
pthalic acid
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5. with sodalime
COOH
COOH
Diphenic acid
Sodalime
biphenyl
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Atropisomers of biphenyl
• Optical isomers produced due to restricted rotation is called atropisomers
• Restricted rotation produce when 0‐ position contains two different bulky groups and hence molecule is optically active.. for example
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CO2H
NO2
NO2
CO2H
A
NO2
CO2HNO2
CO2H
Mirror
Optically activeB
MirrorCl H
H Cl
H Cl
HH2N
Optically active
NO2
CO2H
CO2H
CO2H
Cl
Optically activeno plane of symmetry
NO2
CO2H NO2
Optically activeno plane of symmetry
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• When o‐ position contains two similar groups, themolecule is optically inactive due to presence of plane ofsymmetry .. for example
HO2C
O2N
CO2H
CO2H NO2
CO2HH3OC
H3OC
Optically inactive due to presence of plane of symmetry
Plane of symmetry
Optically activefree rotation is possible
O2N
HO2C NO2
COOH
F
Optically activeF is a small atom so permit
by free rotation
NO2
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(3) Diphenyl methane
Methods of preparation
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1. Friedel‐ Crafte
CH2Cl
+AlCl3 CH2
Diphenyl methaneBenzyl chloride
AlCl3 CH2
Diphenyl methane
2 + CH2Cl2
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2. From benzophenone
CH2
Diphenyl methane
O
HI/ P
or Zn- Hg/ HClor NH2NH2/ NaOEt
Benzophenone
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Chemical Reactions
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1. Nitration
CH2
Diphenyl methane
conc. HNO2conc. H2SO4
CH2 NO2
conc. HNO2conc. H2SO4
CH2 NO2O2N
bis(4- nitrophenyl)methane
1-benzyl-4-nitrobenzene
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2. Halogenation
CH2
Diphenyl methane
hvCH
Diphenylmethylbromide
Br2
Br
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3. Oxidation
CH2
Diphenyl methane
K2Cr2O7H2SO4
C
benzophenone
[O]O
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Organicchemistry:thirdstage/Collegeofscience/BaghdaduniversityBy:Ass.prof.Dr.MunaA‐heetiRef..MorrisonandBoyd
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Preparationofnaphthaquinones
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1,4‐ Naphthaquinone:
O
O1,4- Naphthaquinone
CrO3
AcOH
Naphthalene
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1,2‐ Naphthaquinone:
O
O
1,2- Naphthaquinone
K2Cr2O7
H2SO4 or PbO2
1- Amino- 2-naphthol
NH2
OH
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2,6‐ Naphthaquinone:
O
O
2,6- Naphthaquinone
PbO2
benzene
2,6- dihydroxynaphthalenel
OH
HO
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7.Naphthoicacid
1- Naphthoic acidor
- Naphthoic acid
COOHCOOH
2- Naphthoic acidor
- Naphthoic acid
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Preparationof1‐naphthoicacid
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Frombromonaphthalene
1- Naphthoic acid
COOHBr
Mgdry ether
MgBr
1) CO22) H
1- bromonaphthalene 1- naphthyl magnesium bromide
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From1‐ naphthylamine
NaNO2
HClCuCN
1- Naphthylamine
NH2 N NCl
CN
H2O
H
1- Naphthoic acid
COOH 1- naphthonitrile
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From1‐ acetylacetophenone
COCH3
I2/NaOH
1- Naphthoic acid
COOH
1- Acetylacetophenone
2- Naphthoic acid can be prepared by the same above methods
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Anthracene
1
2
3
4 5
6
7
89
10
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• Anthracene has 4 isomers:
III
Resonance I, II are more stable, contain 2 benzene rings.
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Synthesisofanthracene
• 1. Friedl Crafts
CH2Cl
+ClH2C
AlCl3
-2HBenzyl chloride
Anthracene
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• 2. Elbe reaction
Anthracene
CH3
O
Pyrolysis
o- Methyl- benzenophenone or o- Benzoyl- toluene
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• 3. From 1,4‐ NaphthoquinoneO
O
+
1,3- Butadiene1,4- Naphthaquinone
O
O
O
O
CrO3AcOH
9,10- anthraquinone
Zn
Anthracene
The above method shows presence of naphthalene in anthracene
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• 4. From benzene and phthalic anhydride
HOOC
AlCl3+ O
O
O
O
Zn
Anthracene
O
O anthraquinone
Phthalic anhydride
H2SO4
-H2O
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Chemicalreactions
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• 1) Diels Alder
+ O
O
OAnthracene Maleic anhydride
O
O
OEndo- anthracene- maleic anhydride adduct
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• 2) Addition of one molecule of O2
Anthracene
+ O2
Anthracene epoxide
OO
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• 3) Halogenation of anthracene
Anthracene
X2
X= Cl or Br
X
X
-HX
X
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• 4) Oxidation of anthracene
Anthracene 9, 10- Anthraquinone
Dil HNO3
O
O
In using dil. HNO3 only to obtain 9,10- anthraquinone
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• 4) Reduction of anthracene
Anthracene 9, 10- Dihydroanthracene
Naisopropanol
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Anthraquinone
O
O
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Preparation
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Anthracene 9, 10- Anthraquinone
Dil HNO3
O
O
K2Cr2O7
HOOC
AlCl3+ O
O
O
O
O
O anthraquinone
Phthalic anhydride
H2SO4
-H2O
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ChemicalReactions
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O
O
Zn/ Distillation
or Zn/H/150°C
Anthracene
O
9- Anthrone
Sn/HCl/AcOH
Zn/ NaOH
OH
OHAnthraquinol
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• Nitration
O
O
H2SO4
HNO3
O
O
NO2
H2SO4
HNO3
O
O
NO2
NO2
+
O
O
NO2NO2
1,5- dinitroanthraquinone
1,8- dinitroanthraquinone
Major
1- nitro-9,10- anthraquinone
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• SulphonationO
O
160°C
Oleum
O
O
Major
Anthraquinone-2-sulfonic acid
SO3H
+
O
O
SO3H
small
O
O
Anthraquinone-2-sulfonic acid
SO3HNH3
O
O
NH2
2- amino-anthraquinone
Anthraquinone does not undergo Friedl Craft reaction•Preparation of 2-amino-anthraquinone
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Alizarin
O
O
1,2-dihydroxyanthraquinone Alizarine
OH
OH
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Preparation
O
O
160°C
Oleum, H2SO4
O
OAlizarine
OH
O
O
OH
SO3H
1)NaOH,
2) [O]
9,10- anthraquinone-2- sulfonic acidAnthraquinone
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PreparationofAlizarineBlue
O
OAlizarine
OH
OH
1) conc. HNO3 conc. H2SO4
2) [H]
O
O
OH
OH
NH2
1) Glycerol/ H2SO4
2) PhNO2
O
O
OH
OH
N
Alizarine Blue
Alizarine blue is used for dyeing wool by blue color
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Phenanthrene
1
2
3
4
5
6
7
8
9
10
11
12
13
14
1
2
3 4
5
6
7
8
910
11 12
13
14
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Positionofdoublebond
The most stable 3 benzenoid rings
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Preparationofphenanthrene
• 1) Howrth method
+ O
O
O
AlCl3
COOH
O
Zn-Hg/HCl
COOHconc.H2SO4 O Zn-Hg/HCl
Naphthalene Succinic anhydride
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• Preparation of 2‐ alkyl phenanthrene:
O 1) RMgX2) HOH
OH
R
Se
R
+ O
O
O
AlCl3
COOH
O
Zn-Hg/HCl
COOHconc.H2SO4 O Zn-Hg/HCl
Naphthalene
Se
R
R
R R R
R
Preparation of 1- alkyl phenanthrene:
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• 2) Posher synthesis
NO2
CHO
+
CH2COONa
Ac2O
NO2
COOH
Zn-Hg/HCl
NH2
COOH
NaNO2/H2SO4
N2HSO4
COOH
Cu
Phenanthrene
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ChemicalReactions
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Na/EtOH
9,10-dihydro-phenanthrene
1) O2 2) H2O
CHO
CHO
Biphenyl-2,2'-dicarbaldehyde
Br2 Br
Br
9,10-dibromo-9,10-dihydro-phenanthrene
Br2
FeBr3
Br
9-bromo-9,10-dihydro-phenanthrene
H2O2 AcOH
COOH
COOH
Diphenic acid
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Benzil‐Benzilicrearrangement
CPh O
C OPh
NaOH CPh
Ph
COONa
OH
Benzil Benzilic acid salt
CPh O
C OPh
HO CPh O
C OPh
OHC O
C OPh
OH
Ph
C O
C OHPh
O
Ph
Proton transfer
Mechanism
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O
O
NaOH
H OH
COOH
Phenanthraquinone 9-hydroxy-9H-flourene-9-carboxylic acid
O
O
HOOH
O
OO
COOH
Proton transfer OH
COO
Mechanism
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Phenanthraquinone
• Preparation
K2Cr2O7 H2SO4
O
O
PhenanthraquinonePhenanthrene
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Conditionnecessaryforaromaticity
Any compound to be aromatic, it must be;• 1. Cyclic• 2. Planner• 3. All atoms must be SP2• 4. All double bonds must be conjugated• 5. Obey Huckle rule which state that any aromatic compound must contain 4n+2 pi electrons where n 0,1,2,3,…
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Examples
n=1 electron
n=2 electron
n=3 electron
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Examplesofnon‐ benzenoidaromaticcmpound
All are aromatic( cyclic, planner,1, and agree with Huckle rule: 4n+2= 6 (n=1)
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Examplesofnon‐ aromatic
Not aromatic; both contain Sp3
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Not aromatic
Does not obey Huckel rule
4n+2=8; n=1.5
Not aromatic
Does not obey Huckel rule
4n+2=4; n=0.5
Aromatic; cyclic, planner,
obey Huckel rule
4n+2=2; n=0
Aromatic
4n+2=10; n=2 Aromatic
4n+2=14; n=3
Not Aromatic
8 pi electrons
Aromatic
4n+2=10; n=2
Aromatic
4n+2=10; n=2
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Organicchemistry:thirdstage/Collegeofscience/BaghdaduniversityBy:Ass.prof.Dr.MunaA‐heetiRef..MorrisonandBoyd
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Preparationofnaphthaquinones
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1,4‐ Naphthaquinone:
O
O1,4- Naphthaquinone
CrO3
AcOH
Naphthalene
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1,2‐ Naphthaquinone:
O
O
1,2- Naphthaquinone
K2Cr2O7
H2SO4 or PbO2
1- Amino- 2-naphthol
NH2
OH
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2,6‐ Naphthaquinone:
O
O
2,6- Naphthaquinone
PbO2
benzene
2,6- dihydroxynaphthalenel
OH
HO
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7.Naphthoicacid
1- Naphthoic acidor
- Naphthoic acid
COOHCOOH
2- Naphthoic acidor
- Naphthoic acid
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Preparationof1‐naphthoicacid
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Frombromonaphthalene
1- Naphthoic acid
COOHBr
Mgdry ether
MgBr
1) CO22) H
1- bromonaphthalene 1- naphthyl magnesium bromide
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From1‐ naphthylamine
NaNO2
HClCuCN
1- Naphthylamine
NH2 N NCl
CN
H2O
H
1- Naphthoic acid
COOH 1- naphthonitrile
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From1‐ acetylacetophenone
COCH3
I2/NaOH
1- Naphthoic acid
COOH
1- Acetylacetophenone
2- Naphthoic acid can be prepared by the same above methods
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Anthracene
1
2
3
4 5
6
7
89
10
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• Anthracene has 4 isomers:
III
Resonance I, II are more stable, contain 2 benzene rings.
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Synthesisofanthracene
• 1. Friedl Crafts
CH2Cl
+ClH2C
AlCl3
-2HBenzyl chloride
Anthracene
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• 2. Elbe reaction
Anthracene
CH3
O
Pyrolysis
o- Methyl- benzenophenone or o- Benzoyl- toluene
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• 3. From 1,4‐ NaphthoquinoneO
O
+
1,3- Butadiene1,4- Naphthaquinone
O
O
O
O
CrO3AcOH
9,10- anthraquinone
Zn
Anthracene
The above method shows presence of naphthalene in anthracene
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• 4. From benzene and phthalic anhydride
HOOC
AlCl3+ O
O
O
O
Zn
Anthracene
O
O anthraquinone
Phthalic anhydride
H2SO4
-H2O
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Chemicalreactions
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• 1) Diels Alder
+ O
O
OAnthracene Maleic anhydride
O
O
OEndo- anthracene- maleic anhydride adduct
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• 2) Addition of one molecule of O2
Anthracene
+ O2
Anthracene epoxide
OO
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• 3) Halogenation of anthracene
Anthracene
X2
X= Cl or Br
X
X
-HX
X
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• 4) Oxidation of anthracene
Anthracene 9, 10- Anthraquinone
Dil HNO3
O
O
In using dil. HNO3 only to obtain 9,10- anthraquinone
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• 4) Reduction of anthracene
Anthracene 9, 10- Dihydroanthracene
Naisopropanol
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Anthraquinone
O
O
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Preparation
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Anthracene 9, 10- Anthraquinone
Dil HNO3
O
O
K2Cr2O7
HOOC
AlCl3+ O
O
O
O
O
O anthraquinone
Phthalic anhydride
H2SO4
-H2O
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ChemicalReactions
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O
O
Zn/ Distillation
or Zn/H/150°C
Anthracene
O
9- Anthrone
Sn/HCl/AcOH
Zn/ NaOH
OH
OHAnthraquinol
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• Nitration
O
O
H2SO4
HNO3
O
O
NO2
H2SO4
HNO3
O
O
NO2
NO2
+
O
O
NO2NO2
1,5- dinitroanthraquinone
1,8- dinitroanthraquinone
Major
1- nitro-9,10- anthraquinone
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• SulphonationO
O
160°C
Oleum
O
O
Major
Anthraquinone-2-sulfonic acid
SO3H
+
O
O
SO3H
small
O
O
Anthraquinone-2-sulfonic acid
SO3HNH3
O
O
NH2
2- amino-anthraquinone
Anthraquinone does not undergo Friedl Craft reaction•Preparation of 2-amino-anthraquinone
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Alizarin
O
O
1,2-dihydroxyanthraquinone Alizarine
OH
OH
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Preparation
O
O
160°C
Oleum, H2SO4
O
OAlizarine
OH
O
O
OH
SO3H
1)NaOH,
2) [O]
9,10- anthraquinone-2- sulfonic acidAnthraquinone
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PreparationofAlizarineBlue
O
OAlizarine
OH
OH
1) conc. HNO3 conc. H2SO4
2) [H]
O
O
OH
OH
NH2
1) Glycerol/ H2SO4
2) PhNO2
O
O
OH
OH
N
Alizarine Blue
Alizarine blue is used for dyeing wool by blue color
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Phenanthrene
1
2
3
4
5
6
7
8
9
10
11
12
13
14
1
2
3 4
5
6
7
8
910
11 12
13
14
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Positionofdoublebond
The most stable 3 benzenoid rings
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Preparationofphenanthrene
• 1) Howrth method
+ O
O
O
AlCl3
COOH
O
Zn-Hg/HCl
COOHconc.H2SO4 O Zn-Hg/HCl
Naphthalene Succinic anhydride
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• Preparation of 2‐ alkyl phenanthrene:
O 1) RMgX2) HOH
OH
R
Se
R
+ O
O
O
AlCl3
COOH
O
Zn-Hg/HCl
COOHconc.H2SO4 O Zn-Hg/HCl
Naphthalene
Se
R
R
R R R
R
Preparation of 1- alkyl phenanthrene:
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• 2) Posher synthesis
NO2
CHO
+
CH2COONa
Ac2O
NO2
COOH
Zn-Hg/HCl
NH2
COOH
NaNO2/H2SO4
N2HSO4
COOH
Cu
Phenanthrene
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ChemicalReactions
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Na/EtOH
9,10-dihydro-phenanthrene
1) O2 2) H2O
CHO
CHO
Biphenyl-2,2'-dicarbaldehyde
Br2 Br
Br
9,10-dibromo-9,10-dihydro-phenanthrene
Br2
FeBr3
Br
9-bromo-9,10-dihydro-phenanthrene
H2O2 AcOH
COOH
COOH
Diphenic acid
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Benzil‐Benzilicrearrangement
CPh O
C OPh
NaOH CPh
Ph
COONa
OH
Benzil Benzilic acid salt
CPh O
C OPh
HO CPh O
C OPh
OHC O
C OPh
OH
Ph
C O
C OHPh
O
Ph
Proton transfer
Mechanism
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O
O
NaOH
H OH
COOH
Phenanthraquinone 9-hydroxy-9H-flourene-9-carboxylic acid
O
O
HOOH
O
OO
COOH
Proton transfer OH
COO
Mechanism
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Phenanthraquinone
• Preparation
K2Cr2O7 H2SO4
O
O
PhenanthraquinonePhenanthrene
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Conditionnecessaryforaromaticity
Any compound to be aromatic, it must be;• 1. Cyclic• 2. Planner• 3. All atoms must be SP2• 4. All double bonds must be conjugated• 5. Obey Huckle rule which state that any aromatic compound must contain 4n+2 pi electrons where n 0,1,2,3,…
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Examples
n=1 electron
n=2 electron
n=3 electron
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Examplesofnon‐ benzenoidaromaticcmpound
All are aromatic( cyclic, planner,1, and agree with Huckle rule: 4n+2= 6 (n=1)
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Examplesofnon‐ aromatic
Not aromatic; both contain Sp3
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Not aromatic
Does not obey Huckel rule
4n+2=8; n=1.5
Not aromatic
Does not obey Huckel rule
4n+2=4; n=0.5
Aromatic; cyclic, planner,
obey Huckel rule
4n+2=2; n=0
Aromatic
4n+2=10; n=2 Aromatic
4n+2=14; n=3
Not Aromatic
8 pi electrons
Aromatic
4n+2=10; n=2
Aromatic
4n+2=10; n=2