outline - national university
TRANSCRIPT
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Outline Aromaticity
Huckel’s rule
The Reactions(Electrophilic Substitution)
Halogenation
Friedel-Craft’s Reaction
Alkylation and acylation
Nitration and sulphonation
Oxidation and reduction of
benzene derivates
Disubstitution (Ortho, meta,
para directing groups)
Phenol and aniline
The relative acidity of phenol
The relative basicity of aniline
Diazoniums compounds
L-29,30
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The Main Features The bond length is between C – C and C=C (1.38 A)
Due to delocalised electron (resonance structure)
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The Main Features The structure is planar
Each carbon has p orbital that forms π bonding
Maximum bonding benzene should planar
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p Cloud Formation in Benzene
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Aromaticity (Hückel’s Rules) Huckel’s rules define the classification of aromatic and non-aromatic
molecule.
The criteria of aromatic molecule:
All the atoms are sp2 hybridised and in planar cyclicarrangement.
All atoms are sp2 but not acyclic.
Hence, non-aromatic
There is non-sp2 atom.
Hence, non-aromatic
All atoms are sp2 and a cyclic.
Hence, could be aromatic
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Huckel’s rules
Huckel’s rule
Number of π-electrons is (4n+2),
How to calculate π-electrons?
based on the structure, p-orbitals in sp2 arrangement has 1electron
Has 6 π-electrons (4n+2, n=1)
Hence, aromatic
Has 4 π-electrons (4n, n=1)
Hence, anti-aromatic
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Huckel’s Rule (summary)
Is the molecule
has no non-sp2
atoms?
YES NO
How many π-
electrons in the
molecule?
4n+2 Not 4n+2
aromatic Anti-
aromatic
non-aromatic
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Huckel’s rules
Porphyrin ring in the haem
group
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Huckel’s rule
Which molecules are aromatic?
6 π-electrons
Is this molecule aromatic?
2 π-electrons
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The reactions
Benzene undergoes electrophilic substitution
Doesn’t undergo electrophilic addition
The consequence of aromatic properties
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The reactions - Halogenation Halogenation. E.g. chlorination
Via:
The presence of Lewis acid (e.g. AlCl3) helps benzene to react with Cl2
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The Reactions – Friedel-Crafts Reaction
Friedel-Crafts Reaction (Alkylation)
To substitute with hydrocarbon chain
Via:
Electrophilic
generation
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The reactions Friedel-Crafts Reaction
There is a problem for this reaction when longer alkyl halide is used
Rearrangement of the electrophile (carbocation)
Trying to find the most stable carbocation
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The Reactions Friedel-Crafts Reaction (Acylation)
To substitute with R-CO –
Via:
Electrophilic generation acylium ion
stabilised by resonance. Both structures are valid.
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The reactions
Acylation can be used to get around the ‘messy’ long chain alkylation.
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The Reactions The nitration (concentrated sulphuric acid as catalyst)
Via:
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The Reactions
Sulphonation
Via:
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The reactions Sulphonation
Producing strong sulphonic acid
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The Reactions The Oxidation of toluene
ROH
O
1) KMnO4, OH-, Heat
2) H3O+
Where R is alkyl group
The Reduction of Aniline
NO2 NH2Fe
HCl
aniline
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The Reactions• H2O
• 0 - 15 oC
Diazonium salts is a good precursor compound for:
Halogenation
formation of phenol
deamination
coupling reaction of arenediazonium salts
Formation of Diazonium saltsNH2 N
+ N
Cl-
NaNO2, HCl
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The Reactions
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The Reactions Coupling reaction of arenediazonium salts
Where Q is activating group ( –OH, –NR3).
E.g.:
N+ N
Cl- +
Q
N
N Q
N+ N
Cl- +
OH
N
O+ H
N
H
Cl- OH
NN
N+ N
Cl- +
OH
OH
NN
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Disubstitution of Benzene
ortho (1,2)
CH3O OHO
The benzene ring can be substituted with another FG more than once.
The second position is determined by the first FG
O NH2
CH3 NH2 OH Cl HN CH3
O
Three possible positions:R
R1
R
R1
meta (1,3)
R
R1
para (1,4)
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Disubstitution of Benzene The determining factor
The nature of FG electron withdrawing (EW) or electron
donating (ED) group
EW: the FG generally has partial positive charge
It deactivate the benzene ring, so it is less reactive
ED: the FG generally has partial negative charge
It activate the benzene ring, so it is more reactive
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Disubstitution of Benzene
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Disubstitution of Benzene E.g. Application for synthesis route
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Phenol
The relative acidity
Acidity The easiness to release H+ (proton)
The stability of the acid conjugate determine the relative acidity.
The comparison with water and alcohol (e.g. ethanol)
The structureOH
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Phenol
Let’s put water as the standard and the conjugate.
More stable the conjugate, more acid the substance.
In ethoxide ions the alkyl group push the electrons increasing the charge
In phenoxide ions, it forms a bigger resonance structure due to unbonding
p-orbital
OH
H H3C
O H OH
O-
O-
HH3C
O-
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Phenol
The effect of substituent
The principle: The reduction of the charge
The deactivating benzene substituent will make phenol more acidic
The activating benzene substituent will make phenol less acidic.
Phenol 3-methylphenol 3-nitrophenol 3-chlorophenol
pKa = 9.89 pKa = 10.01 pKa = 8.28 pKa = 8.80
OH
CH3
OH
Cl
OH OH
NO2
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Phenol
Predict the pKa of 2,4 dinitrophenol.
(a) 10.17
(b) 9.31
(c) 8.11
(d) 3.96
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Phenol
Esterification of Phenol
No reaction with carboxylic acid
Only react with acyl chloride or acetic anhydride
OH
+ H3C
OH
O H3O+
No reaction
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Phenol
Suggest the products from the reactions below
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Phenol
How to distinguish with alcohol?
Since the phenol is more acidic
than alcohol, so it can reacts
with weaker base (e.g.
NaHCO3)
Both of them can react with Na
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Aniline
The Basicity of amines
Basicity >< Acidity
Basicity How easy a compound can accept H+
The case: The relative Basicity of ethylamine, amine, and aniline
NH2
aniline
The easiness of compound to accept H+
The availability of lone pair electrons on N atom
NH3
ammonia
H3C NH2
methanamine
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Aniline
The reactions
Phenylamine cannot react in the similar way like amine.
Phenylamine is not a better nucleophile than amine
the availability of the electrons on N atom to do the reaction
HNO3
concd H SO2 4
N -
O
+Fe/Sn
O
HCl
NH2 2NaNO , HCl
0 - 15 oC
N+ N
Cl-
This reaction can produce the
other amines. Could you
draw the other products?
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Aniline
Phenylamine could form an amide with acyl chloride.
Important synthetic pathway for aniline-based compound