revision notes about benzene and its derivates

1. Chapter 15

2. • In early 19th century, the term aromatic was used
to describe some fragrant compounds
– Not correct: later they are grouped by chemical behavior
(unsaturated compounds that undergo substitution rather
than addition).
Introduction
coal distillate cherries, peaches and almonds Tolu balsam

3. • Currently, the term aromatic is used to refer to the
class of compounds related structurally to benzene
– They are distinguished from aliphatic compounds by
electronic configuration
steroidal hormone analgesic tranquilizer

4. 1. Sources of Aromatic Hydrocarbons
• There are two main sources of simple
aromatic hydrocarbons:
i. coal
ii. petroleum

5. i. High temperature distillation of coal tar
– Coal is a mixture of benzene-like rings joined together.
Under high temperature, it produces coal tar which,
upon fractional distillation, yields:

6. ii. Heating petroleum at high temperature under
high pressure over a catalyst
– Petroleum consists mainly of alkanes which, at high
temperature under pressure over a catalyst, convert
into aromatic compounds.

7. 2. Naming Aromatic Compounds
• Aromatic compounds are named according to the
system devised by the International Union of Pure and
Applied Chemistry (IUPAC).

8. • Aromatic compounds have many common names
that have been accepted by IUPAC:
• Toluene = methylbenzene
• Phenol = hydroxybenzene
• Aniline = aminobenzene

10. • Monosubstituted benzenes, like hydrocarbons, are
systematically named with –benzene as the parent
name
C6H5Br C6H5NO2 C6H5CH2CH2CH3
Monosubstituted benzenes

11. • Arenes are alkyl-substituted benzenes
– If # Csubstituent < or = 6, then the arene is named as
an alkyl-substituted benzene
– If # Csubstituent > 6, then the arene is named as a
phenyl-substituted alkane
Arenes

12. Aryl groups
• “Phenyl” refers to C6H5
– It is used when a benzene ring is a substituent
– “Ph” or “f” can also be in place of “C6H5 ”
• “Benzyl” refers to “C6H5CH2
”

13. • Relative positions on a disubstituted benzene ring:
– ortho- (o) on adjacent carbons (1,2 disubstituted)
– meta- (m) separated by one carbon (1,3 disubstituted)
– para- (p) separated by two carbons (1,4 disubstituted)
Disubstituted benzenes

14. • The ortho- (o), meta- (m), and para- (p) nomenclature
is useful to describe reaction patterns
Example: “Reaction of toluene with Br2 occurs at the para
position”

15. • Multisubstituted benzenes (more than two substituents)
are named as follows:
– Choose the sequence when the substituents have the lowest
possible number
– List substituents alphabetically with hyphenated numbers
– Use common names, such as “toluene”, as parent name (as in
TNT)
Multisubstituted benzenes

16. – Use common names, such as “toluene”, as parent name
 The principal substituent is assumed to be on C1
 See Table 15.1

17. Practice Problem: Tell whether the following compounds are
ortho-, meta-, or para-disubstituted
(a) Meta (b) Para (c) Ortho

18. Practice Problem: Give IUPAC names for the following
compounds
(a) m-Bromochlorobenzene (b) (3-Methylbutyl)benzene
(c) p-Bromoaniline (d) 2,5-Dichlorotoluene
(e) 1-Ethyl-2,4-dinitrobenzene (f) 1,2,3,5-Tetramethylbenzene

19. Practice Problem: Draw structures corresponding to the
following IUPAC names:
a) p-Bromochlorobenzene
b) p-Bromotoluene
c) m-Chloroaniline
d) 1-Chloro-3,5-dimethylbenzene

20. 3. Structure and Stability of Benzene
• Benzene is very stable
– It undergoes substitution rather than the rapid addition reaction
common to compounds with C=C, suggesting that in benzene
there is a higher barrier
– Example: Benzene reacts slowly with Br2 to give bromobenzene
(where Br replaces H)

21. Heats of Hydrogenation as Indicators of Stability
• The addition of H2 to C=C normally gives off about 118
kJ/mol – 3 double bonds would give off 356 kJ/mol
– Two conjugated double bonds in cyclohexadiene add 2 H2

22. • Benzene has 3 unsaturations but gives off only 206
kJ/mol on reacting with 3 H2 molecules
– Therefore it has about 150 kJ/mol more “stability” than an
isolated set of three double bonds

23. • Benzene has 150 kJ/mol more “stability” than expected
for “cyclohexatriene”

24. Benzene’s Unusual Structure
• All its C-C bonds are the same length: 139 pm —
between single (154 pm) and double (134 pm) bonds
• Electron density in all six C-C bonds is identical

25. • Structure is planar, hexagonal
• All C–C–C bond angles are 120°
• Each C is sp2-hybridized and has a p orbital
perpendicular to the plane of the six-membered ring

26. Drawing Benzene and Its Derivatives
• The two benzene resonance forms can be represented by
a single structure with a circle in the center to indicate the
equivalence of the carbon–carbon bonds
– This does not indicate the number of  electrons in the ring
but shows the delocalized structure
– One of the resonance structures will be used to represent
benzene for ease in keeping track of bonding changes in
reactions

27. 4. Molecular Orbital Description of
Benzene
• In benzene, 6 p orbitals combine to form 6 
molecular orbitals (MO):
– 3 bonding orbitals with 6  electrons
– 3 antibonding orbitals

28. – 3 bonding, low-energy MO: y1, y2, and y3
– 3 antibonding high-energy MO: y4*, y5*, and y6*

29. • Orbitals with the same energy are degenerate
– 2 bonding orbitals, y2 and y3
– 2 antibonding orbitals, y4* and y5*

30. – y3 and y4* have no  electron density on 2 carbons
because of a node passing through these atoms

31. Practice Problem: Pyridine is flat, hexagonal with bond angles
of 120°. It undergoes electrophilic substitution
rather than addition and generally behaves
like benzene. Draw the  orbitals of pyridine.

32. Recall: Key Ideas on Benzene
• Benzene is a cyclic conjugated molecule
• Benzene is unusually stable - DHhydrogenation = 150 kJ/mol
less negative than a cyclic triene
• Benzene is planar hexagon: bond angles are 120°;
carbon–carbon bond lengths, 139 pm
• Benzene undergoes substitution rather than electrophilic
addition
• Benzene is a resonance hybrid with structure between two
line-bond structures
• Benzene has 6  electrons, delocalized over the ring

33. 5. Aromaticity and the Hückel 4n + 2 Rule
The Hückel 4n + 2 rule:
– was devised by Eric Hückel in 1931
– states that planar, monocyclic conjugated
systems with a total of 4n + 2  electrons where
n is an integer (n = 0, 1, 2, 3,…) are aromatic

34. Aromatic compounds with 4n + 2  electrons
• Benzene
– It has 6  electrons: 4n + 2 = 6, thus n = 1
– It is aromatic: it is stable and the electrons are
delocalized

35. Compounds with 4n  electrons are NOT aromatic
(May be Antiaromatic)
• Planar, cyclic conjugated molecules with 4n  electrons
are antiaromatic
– They are much less stable than expected
– They will distort out of plane and behave like ordinary
alkenes
cyclobutadiene
cyclooctatetraene
Which of the above is antiaromatic?

36. • Cyclobutadiene
– It has 4  electrons: 4n + 2 = 4, thus n = ½ (not an integer)
– It is antiaromatic: The  electrons are localized into two
double bonds
localized 
electrons

37. • Cyclobutadiene
– It has 4  electrons: 4n + 2 = 4, thus n = ½ (not an integer)
– It is antiaromatic: The  electrons are localized into two
double bonds
• It is so unstable that it dimerizes by a self-Diels-Alder
reaction at low temperature
dienophile diene

38. • Cyclooctatetraene
– It has 8  electrons: 4n + 2 = 8, thus n = 3/2 (not an integer)
– It is nonaromatic:
• the  electrons are localized into four double bonds
• it is tub-shaped not planar
• it has four double bonds, reacting with Br2, KMnO4, and HCl as if it
were four alkenes
p orbitals
not parallel
for overlap

39. Practice Problem: To be aromatic, a molecule must have 4n + 2
 electrons and must have cyclic conjugation.
Is cyclodecapentaene aromatic?
• It has 10  electrons: 4n + 2 = 10, thus n = 2 (an integer)
• It is not planar due to steric strain, thus the neighboring p
orbitals are not properly aligned for overlap. It is not
conjugated. Thus it is not aromatic.

40. 6. Aromatic Ions
• The Hückel 4n + 2 rule applies to ions as well
as to neutral species:
• To be aromatic, a molecule must be planar,
cyclic conjugated system with 4n + 2 
electrons
• Example: Both the cyclopentadienyl anion and
the cycloheptatrienyl cation are aromatic.

41. • Example: Both the cyclopentadienyl anion and the
cycloheptatrienyl cation are aromatic.
The key feature of both is that they contain 6 
electrons in a ring of continuous p orbitals

42. Aromaticity of the Cyclopentadienyl Anion
Unstable and nonaromatic Stable and aromatic
Not fully conjugated and not aromatic

43. • Cyclopentadiene is relatively acidic (pKa = 16) because its
conjugate base, the aromatic cyclopentadienyl anion, is so
stable.
– Other hydrocarbons have pKa > 45

44. Aromaticity of the Cyclopentadienyl Anion
• 1,3-Cyclopentadiene contains
conjugated double bonds joined
by a CH2 that blocks
delocalization
• Removal of H+ at the CH2
produces a cyclic 6-electron
system, which is stable
• Removal of H- or H• generate
nonaromatic 4 and 5 electron
systems
• Relatively acidic (pKa = 16)
because the anion is stable

45. Aromaticity of the Cycloheptatrienyl Cation
Not fully conjugated and not aromatic
Unstable and nonaromatic
Stable and aromatic

46. • The cycloheptatrienyl cation (six  electrons) is aromatic
and very stable
– Reaction of cycloheptatriene with Br2 yields cycloheptatrienylium
bromide, an ionic substance containing the cycloheptatrienyl
cation

47. Aromaticity of the Cycloheptatrienyl Cation
• Cycloheptatriene has 3
conjugated double bonds joined
by a CH2
• Removal of H- at the CH2
produces the cycloheptatrienyl
cation
• The cation is a cyclic 6-electron
system, which is stable and is
aromatic
• Removal of H+ or H• generate
nonaromatic 7 and 8 electron
systems

48. Practice Problem: Draw the five resonance structures of the
cyclopentadienyl anion. Are all carbon-carbon
bonds equivalent? How many absorption lines
would be in the 1H and 13C NMR spectra of
the anion?

49. Practice Problem: Cyclooctatetraene readily reacts with potassium
metal to form the stable the cyclooctatetraene
dianion, C2H8
2-. Why does the reaction occur so
easily? What is the geometry of the dianion?

50. 7. Aromatic Heterocycles: Pyridine and
Pyrrole
• A heterocycle is a cyclic compound that contains
an atom or atoms other than carbon in its ring,
such as N, O, S, P
• There are many heterocyclic aromatic compounds
and many are very common
• Cyclic compounds that contain only carbon are
called carbocycles (not homocycles)
• Nomenclature is specialized
• Example: Pyridine and Pyrrole

51. • Pyridine is a six-membered heterocycle with a nitrogen
atom in its ring
•  electron structure resembles benzene (6 electrons)
• The nitrogen lone pair electrons are in sp2 orbital, not part
of the  aromatic system (perpendicular orbital)
• Pyridine is a relatively weak base compared to normal
amines but protonation does not affect aromaticity
Pyridine

52. • Pyrrole is a five-membered heterocycle with a nitrogen
atom in its ring
•  electron system is similar to that of cyclopentadienyl
anion
• Four sp2-hybridized carbons with 4 p orbitals
perpendicular to the ring and 4 p electrons
Pyrrole

53. • Nitrogen atom is sp2-hybridized, and lone pair of electrons
occupies a p orbital (6  electrons)
• Since lone pair electrons are in the aromatic ring,
protonation destroys aromaticity, making pyrrole a very
weak base
Pyrrole

54. Practice Problem: Thiophene, a sulfur-containing heterocycle,
undergoes typical aromatic substitution
reactions rather than addition reactions.
Explain why thiophene is aromatic.
It has 6  electrons: 4n + 2 = 6, thus n = 1 (an integer)
It has a lone pair of electrons in a p orbital perpendicular to the plane

55. Practice Problem: Draw an orbital picture of furan to show how
the molecule is aromatic
It has 6  electrons: 4n + 2 = 6, thus n = 1 (an integer)
It has a lone pair of electrons in a p orbital perpendicular to the plane

56. Practice Problem: Draw an orbital picture of imidazole, and account
for its aromaticity. Which nitrogen atom is
pyridine-like, and which is pyrrole-like? Which
nitrogen atom is more electron-rich, and why?

57. 8. Why 4n + 2?
• According to the molecular orbital (MO) theory:
– Cyclic conjugated molecules always have a single
lowest-lying MO, above which the MOs come in
degenerate pairs
– When electrons fill the various molecular orbitals,
it takes two electrons (one pair) to fill the lowest-
lying orbital and four electrons (two pairs) to fill
each of n succeeding energy level
• Only a total of 4n + 2 electrons fill bonding MOs

58. • Benzene has its bonding orbitals filled (6  electrons)
 y1, the lowest-energy MO, is single and has two electrons
 y2 and y3, the next two lowest-energy MOs, are degenerate
and have four electrons
Benzene

59. • Cyclopentadienyl anion has its bonding orbitals filled (6
 electrons)
 y1 is single and has two electrons; y2 and y3 are degenerate
and have four electrons
- Cyclopentadienyl cation and radical do not have their bonding
orbitals filled
Cyclopentadienyl anion

60. Practice Problem: Show the relative energy levels of the seven 
MOs of the cycloheptatrienyl system. Tell
which of the seven orbitals are filled in the
cation, radical, and anion, and account for the
aromaticity of the cycloheptatrienyl cation

61. 9. Polycyclic Aromatic Compounds:
Naphthalene
• Polycyclic aromatic compounds are
– aromatic compounds with rings that share a set
of carbon atoms (fused rings)
– compounds from fused benzene or aromatic
heterocycle rings
carcinogenic
(tobacco)

62. • They are cyclic, planar and conjugated molecules
• They are unusually stable
• They react with electrophiles to give substitution
products, in which cyclic conjugation is retained,
rather than electrophilic addition products
• They can be represented by different resonance
forms
• They have 4n + 2  electrons, delocalized over the
ring
Characteristics of Polycyclic Aromatic Compounds

63. • Naphthalene has three resonance forms
Naphthalene
• Naphthalene reacts slowly with electrophiles to give
substitution products

64. • Naphthalene is a cyclic, conjugated  electron system,
with p orbital overlap both around the ten-carbon
periphery of the molecule and across the central bond
– It has ten delocalized  electrons (Hückel number)
Naphthalene

65. Practice Problem: Azulene is an isomer of naphthalene. Is
azulene aromatic? Draw a second resonance
form of azulene.

66. Practice Problem: Naphthalene is sometimes represented with
circles in each ring to represent aromaticity.
How many  electrons are in each circle?

67. 10. Spectroscopy of Aromatic Compounds
Aromatic compounds can be identified by:
– Infrared (IR) Spectroscopy
– Ultraviolet (UV) Spectroscopy
– Nuclear Magnetic Resonance (NMR) Spectroscopy

68. • Aromatic rings have C–H stretching at 3030 cm1 and
peaks in the range of 1450 to 1600 cm1
• Substitution pattern of the aromatic ring:
Monosubstituted: 690-710 cm-1
730-770 cm-1
o-Disubstituted: 735-770 cm-1
m-Disubstituted: 690-710 cm-1
810-850 cm-1
p-Disubstituted: 810-840 cm-1
Infrared Spectroscopy

69. • Example: Toluene (IR)
Monosubstituted: 690-710 cm-1
730-770 cm-1
3030 cm1

70. • Aromatic rings have peaks near 205 nm and a less
intense peak in 255-275 nm range
– Aromatic compounds are detectable by UV spectroscopy
since they have a conjugated  electron system
Ultraviolet Spectroscopy

71. 1H NMR:
– Aromatic H’s are strongly deshielded by ring and
absorb between 6.5 and 8.0 
– Peak pattern is characteristic positions of substituents
Nuclear Magnetic Resonance Spectroscopy

72. • Ring Current is a property unique to aromatic rings
– When aromatic ring is oriented perpendicular to a
strong magnetic field, delocalized  electrons
circulate producing a small local magnetic field
– This opposes applied field in middle of ring but
reinforces applied field outside of ring

73. • Ring Current accounts for the downfield shift of
aromatic ring protons in the 1H NMR spectrum
– Aromatic 1H’s experience an effective magnetic field
greater than applied
– It results in outside H’s resonance at lower field

74. • Ring Current produces different effects inside and
outside the ring
– Outside 1H are deshielded and absorb at a lower field
– Inside 1H are shielded and absorb at a higher field

75. • Ring Current is characteristic of all Hückel aromatic
compounds
– Aryl 1H absorb between 6.5-8.0 
– Benzylic 1H absorb between 2.3-3  downfield from other
alkane 1H

76. • Example: p-bromotoluene (1H NMR)
The 4 aryl protons:
Two doublets at
7.02 and 7.45 
The benzylic CH3
protons: a singlet at
2.29 
Integration 2:2:3

77. 13C NMR
– Carbons in aromatic ring absorb between 110 to 140 
– Shift is distinct from alkane carbons but is in same
range as alkene carbons

79. Chapter 15