The organic reactions involving benzene emphasising on the electrophilic substitution on benzene ring, phenol and aniline.

1. Organic Chemistry IV
Benzene and Its Derivates
Indra Yudhipratama

2. 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

3. The Main Features
 The bond length is between C – C and C=C (1.38 A)
 Due to delocalised electron (resonance structure)

4. The Main Features
 The structure is planar
 Each carbon has p orbital that forms π bonding
 Maximum bonding benzene should planar

5. p Cloud Formation in Benzene

6. 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 cyclic arrangement.
All atoms are sp2 but not a cyclic.
Hence, non-aromatic
There is non-sp2 atom.
Hence, non-aromatic
All atoms are sp2 and a cyclic.
Hence, could be aromatic

7. 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 1 electron
Has 6 π-electrons (4n+2, n=1)
Hence, aromatic
Has 4 π-electrons (4n, n=1)
Hence, anti-aromatic

8. 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

9. Huckel’s rules
Porphyrin ring in the haem
group

10. Huckel’s rule
 Which molecules are aromatic?
 Is this molecule aromatic?
6 π-electrons 2 π-electrons

11. The reactions
 Benzene undergoes electrophilic substitution
 Doesn’t undergo electrophilic addition
 The consequence of aromatic properties

12. The reactions - Halogenation
 Halogenation. E.g. chlorination
 Via:
 The presence of Lewis acid (e.g. AlCl3) helps benzene to react with Cl2

13. The Reactions – Friedel-Crafts
Reaction
 Friedel-Crafts Reaction (Alkylation)
 To substitute with hydrocarbon chain
 Via:
Electrophilic
generation

14. 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

15. The Reactions
 Friedel-Crafts Reaction (Acylation)
 To substitute with R-CO –
 Via:
 Electrophilic generation  acylium ion
 stabilised by resonance. Both structures are valid.

16. The reactions
 Acylation can be used to get around the ‘messy’ long chain alkylation.

17. The Reactions
 The nitration (concentrated sulphuric acid as catalyst)
 Via:

18. The Reactions
 Sulphonation
 Via:

19. The reactions
 Sulphonation
 Producing strong sulphonic acid

20. The Reactions
 The Oxidation of toluene
 Where R is alkyl group
 The Reduction of Aniline
R
OH
O
1) KMnO4, OH-
, Heat
2) H3O+
NO2 NH2
Fe
HCl
aniline

21. The Reactions
 Formation of Diazonium salts
 Diazonium salts is a good precursor compound for:
 Halogenation
 formation of phenol
 deamination
 coupling reaction of arenediazonium salts
NH2 N
+ N
Cl
-
NaNO2, HCl
H2O
0 - 15 o
C

22. The Reactions

23. 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
O
+
N
N
H
H
Cl
- OH
N
N
N
+ N
Cl
-
+
OH
OH
N
N

24. Disubstitution of Benzene
 The benzene ring can be substituted with another FG more than once.
 The second position is determined by the first FG
 Three possible positions:
ortho (1,2) meta (1,3) para (1,4)
CH3 NH2 OH Cl
CH3O OHONH2O
NH CH3
O
R
R1
R
R1
R
R1

25. 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

26. Disubstitution of Benzene

27. Disubstitution of Benzene
 E.g. Application for synthesis route

28. Phenol
 The structure
 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)
OH

29. 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
O H
H CH3
OH
O
-
O
-
H
CH3
O
-

30. 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 OH
CH3
OH
Cl
OH
NO2

31. Phenol
 Predict the pKa of 2,4 dinitrophenol.
(a) 10.17
(c) 8.11
(d) 3.96
(b) 9.31

32. Phenol
 Esterification of Phenol
 No reaction with carboxylic acid
 Only react with acyl chloride or acetic anhydride
OH
+ CH3
OH
O H3O+
No reaction

33. Phenol
 Suggest the products from the reactions below

34. 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

35. 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
 The easiness of compound to accept H+
 The availability of lone pair electrons on N atom
CH3 NH2 NH3
NH2
aniline
ammoniamethanamine

36. 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 H2SO4
N
+
O
-
O
Fe/Sn
HCl
NH2 NaNO2, HCl
0 - 15 o
C
N
+ N
Cl
-
This reaction can produce the
other amines. Could you
draw the other products?

37. Aniline
 Phenylamine could form an amide with acyl chloride.
 Important synthetic pathway for aniline-based compound