Transcript Organic Chemistry - City University of New York

Chapter 21, Benzene
and and the Concept of
Aromaticity
21-1
Benzene - Kekulé
 In
1872, August Kekulé proposed the following
structure for benzene.
H
H
C
C
H
C
C
H
C
C
H
H
H
H
C
C
H
C
C
H
C
C
H
H
 This
structure, however, did not account for the
unusual chemical reactivity of benzene.
21-2
Benzene - Resonance
 We
often represent benzene as a hybrid of two
equivalent Kekulé structures.
• Each makes an equal contribution to the hybrid and
thus the C-C bonds are neither double nor single, but
something in between.
Ben zene as a hybrid of two equ ivalen t
contributin g s tru ctures
21-3
Benzene - Resonance Model
 The
concepts of hybridization of atomic orbitals
and the theory of resonance, developed in the
1930s, provided the first adequate description of
benzene’s structure.
• The carbon skeleton is a planar regular hexagon.
• All C-C-C and H-C-C bond angles 120°.
H 120°
H
sp2 -sp2
120° C
C

120°
H
C
C H
C
C
109 pm
sp2 -1 s
H
139 pm H
21-4
The Pi System of Benzene
• (a) The carbon framework with the six 2p orbitals.
• (b) Overlap of the parallel 2p orbitals forms one torus
above the plane of the ring and another below it
• this orbital represents the lowest-lying pi-bonding
molecular orbital.
21-5
Benzene-Molecular Orbital Model
 The
molecular orbital representation of the pi
bonding in benzene.
21-6
Orbitals of the pi System of Benzene
Number of
nodal surfaces
3
2
1
0
21-7
Benzene - Resonance
 Resonance
energy: The difference in energy
between a resonance hybrid in which the
electrons are delocalized
and
 the most stable one of its hypothetical
contributing structures in which electrons are
localized on particular atoms and in particular
bonds.
• One way to estimate the resonance energy of benzene
is to compare the heats of hydrogenation of benzene
and cyclohexene.
21-8
Benzene- Resonance Energy
Experimental
data
Model
21-9
Concept of Aromaticity
 The
underlying criteria for aromaticity were
recognized in the early 1930s by Erich Hückel,
based on molecular orbital (MO) calculations.
 To be aromatic, a compound must
• Be cyclic.
• Have one p orbital on each atom of the ring.
• Be planar or nearly planar so that there is continuous
or nearly continuous overlap of all p orbitals of the
ring.
• Have a closed loop of (4n + 2) pi electrons in the cyclic
arrangement of p orbitals.
21-10
Frost Circles
 Frost
circle: A graphic method for determining
the relative order of pi MOs in planar, fully
conjugated monocyclic compounds.
• Inscribe in a circle a polygon of the same number of
sides as the ring to be examined such that one of the
vertices of the polygon is at the bottom of the circle.
• The relative energies of the MOs in the ring are given
by where the vertices of the polygon touch the circle.
 Those
MOs
• Below the horizontal line through the center of the ring
are bonding MOs.
• on the horizontal line are nonbonding MOs.
• above the horizontal line are antibonding MOs.
21-11
Frost Circles
• Frost circles describing the MOs for monocyclic,
planar, fully conjugated four-, five-, and six-membered
rings.
21-12
Relationship of hexa-1,3,5-triene to benzene
How does the linear triene relate
to benzene?
21-13
Relationship of hexa-1,3,5-triene to benzene
?
Relationship of hexa-1,3,5-triene to benzene
Look at orbitals 2 and 3.
p3
?
p2
Bonding,
stabilizing
Curve
around
Antibonding,
destabilizing
Aromatic Hydrocarbons
 Annulene:
A cyclic hydrocarbon with a
continuous alternation of single and double
bonds.
• [14]Annulene is aromatic according to Hückel’s
criteria.
n=3
H
H
H
H
HH
H
H
HH
H
H
H
H
[14]Ann ulene
(aromatic)
21-16
Aromatic Hydrocarbons
• [18]Annulene is also aromatic.
H
H
H
H
H
H
H
H
H
H
H
H
H
H
n=4
H
H
H
H
[18]A nnu lene
(aromatic)
21-17
Aromatic Hydrocarbons
• According to Hückel’s criteria, [10]annulene should be
aromatic; it has been found, however, that it is not.
• Nonbonded interactions between the two hydrogens
that point inward toward the center of the ring force
the ring into a nonplanar conformation in which
overlap of the ten 2p orbitals is no longer continuous.
[10]A nnu lene
21-18
Aromatic Hydrocarbons
• What is remarkable relative to [10]annulene is that if
the two hydrogens facing inward toward the center of
the ring are replaced by a methylene (CH2) group, the
ring is able to assume a conformation close enough to
planar that it becomes aromatic.
CH2
Bridged [10]ann ulen e
21-19
Antiaromatic Hydrocarbons
 Antiaromatic
hydrocarbon: A monocyclic, planar,
fully conjugated hydrocarbon with 4n pi
electrons (4, 8, 12, 16, 20...).
• An antiaromatic hydrocarbon is especially unstable
relative to an open-chain fully conjugated hydrocarbon
of the same number of carbon atoms.
 Cyclobutadiene
is antiaromatic.
• In the ground-state electron configuration of this
molecule, two electrons fill the p1 bonding MO.
• The remaining two electrons lie in the p2 and p3
nonbonding MOs.
21-20
Cyclobutadiene
• The ground state of planar cyclobutadiene has two
unpaired electrons, which make it highly unstable and
reactive.
21-21
Cyclooctatetraene
• Cyclooctatetraene, with 8 pi electrons is not aromatic;
it shows reactions typical of alkenes.
• X-ray studies show that the most stable conformation
is a nonplanar “tub” conformation.
• Although overlap of 2p orbitals occurs to form pi
bonds, there is only minimal overlap between sets of
2p orbitals because they are not parallel.
view ed from above
view ed throu gh an edge
21-22
Cyclooctatetraene
 MO
energy diagram for a planar conformation of
cyclooctatetraene.
21-23
Heterocyclic Aromatics
 Heterocyclic
compound: A compound that
contains more than one kind of atom in a ring.
• In organic chemistry, the term refers to a ring with one
or more atoms that differ from carbon.
 Pyridine
and pyrimidine are heterocyclic analogs
of benzene; each is aromatic.
4
3
5
2
6
1
N
••
Pyridine
4
3
N
2
5
N
6
1 ••
Pyrimidine
21-24
Pyridine
• The nitrogen atom of
pyridine is sp2
hybridized.
• The unshared pair of
electrons lies in an sp2
hybrid orbital and is not
a part of the six pi
electrons of the
aromatic system (the
aromatic sextet).
• Resonance energy of
pyridine is134 kJ (32
kcal)/mol.
21-25
Furan and Pyrrole
• The oxygen atom of furan is sp2 hybridized.
• one unshared pairs of electrons on oxygen lies in an
unhybridized 2p orbital and is a part of the aromatic
sextet.
• The other unshared pair lies in an sp2 hybrid orbital
and is not a part of the aromatic system.
• The resonance energy of furan is 67 kJ (16 kcal)/mol.
21-26
Other Heterocyclics
CH2 CH2 NH 2
HO
N
N
H
H
Serotonin
(a neurotransmitter)
Indole
N
N
N
H3 C
N
O
H
Purine
O
CH3
N
N
CH3
N
N
Caffeine
21-27
Aromatic Hydrocarbon Ions
 Any
neutral, monocyclic, unsaturated
hydrocarbon with an odd number of carbons
must have at least one CH2 group and, therefore,
cannot be aromatic.
CH2
Cycloprop ene
CH2
Cyclop entadien e
CH2
Cycloh eptatriene
• Cyclopropene, for example, has the correct number of
pi electrons to be aromatic, 4(0) + 2 = 2, but does not
have a closed loop of 2p orbitals.
21-28
Cyclopropenyl Cation
• If, however, the CH2 group of cyclopropene is
transformed into a CH+ group in which carbon is sp2
hybridized and has a vacant 2p orbital, the overlap of
orbitals is continuous and the cation is aromatic.
H
H
+
H
H
+
H
H
H
+
H
H
Cycloprop enyl cation represented as a h yb rid
of three equ ivalen t contributin g s tru ctures
21-29
Cyclopropenyl Cation
• When 3-chlorocyclopropene is treated with SbCl5, it
forms a stable salt.
H
+
Cl
3-Chlorocyclopropene
Sb Cl 5
Antimony(V)
chloride
(a Lewis acid)
+
H Sb Cl 6
-
Cyclopropenyl
hexachloroantimonate
• This chemical behavior is to be contrasted with that of
5-chloro-1,3-cyclopentadiene, which cannot be made
to form a stable salt.
21-30
Cyclopentadienyl Cation
H
+ AgBF4
Cl
5-C h loro-1,3cycl ope n tadi e n e
+
H BF4
-
+ Ag Cl
Cycl open tadie n yl
tetraflu oroborate
• If planar cyclopentadienyl cation were to exist, it would
have 4 pi electrons and be antiaromatic.
• Note that we can draw five equivalent contributing
structures for the cyclopentadienyl cation. Yet this
cation is not aromatic because it has only 4 pi
electrons.
21-31
Cyclopentadienyl Anion, C5H5 To
convert cyclopentadiene to an aromatic ion, it
is necessary to convert the CH2 group to a CH
group in which carbon becomes sp2 hybridized
and has 2 electrons in its unhybridized 2p orbital.
H •
•
•
H
H
•
H
H
th e origin of th e 6 pi electrons
in the cyclopen tadienyl anion
H
H
H
:
H
H
H
H
H
H
H
Cyclopentad ienyl anion
(aromatic)
n=1
21-32
Cyclopentadienyl Anion, C5H5• As seen in the Frost circle, the six pi electrons of
cyclopentadienyl anion occupy the p1, p2, and p3
molecular orbitals, all of which are bonding.
21-33
Cyclopentadienyl Anion, C5H5 The
pKa of cyclopentadiene is 16.
• In aqueous NaOH, it is in equilibrium with its sodium
salt.
H
H
CH2 + NaOH
:
H
p K a 16.0
H
H Na + + H2 O
pK a 15.7
• It is converted completely to its anion by very strong
bases such as NaNH2 , NaH, and LDA.
21-34
Cycloheptatrienyl Cation, C7H7+
 Cycloheptatriene
forms an aromatic cation by
conversion of its CH2 group to a CH+ group with
its sp2 carbon having a vacant 2p orbital.
H
H
H
H
H
H
H
+
H
H
+
H
H
H
H
H
Cyclohep tatrien yl cation
(Tropylium ion )
(aromatic)
21-35
Nomenclature
 Monosubstituted
alkylbenzenes are named as
derivatives of benzene.
• Many common names are retained.
Tolu e n e
OH
Ph e n ol
Ethyl ben z e n e
N H2
An i li n e
CHO
Cu me n e
Styre n e
COOH
Be n z al de h yde Be n z oi c acid
OCH3
An i sol e
21-36
Nomenclature
 Benzyl
and phenyl groups
CH3
Benzene
Phenyl group, PhO
Toluene
CH2
Benzyl group, Bn-
O
H 3 CO
Ph
1-Phe nyl-1-pe ntanone 4-(3-Me thoxyphe nyl )2-butanone
(Z)-2-Phenyl 2-butene
21-37
Disubstituted Benzenes
 Locate
two groups by numbers or by the
locators ortho (1,2-), meta (1,3-), and para (1,4-).
• Where one group imparts a special name, name the
compound as a derivative of that molecule.
CH3
NH2
COOH
NO2
CH3
Cl
CH3
Br
4-Bromotolu ene 3-Chloroan iline
2-N itrobenzoic acid m-Xylene
(p-Bromotoluen e) (m-Chloroan iline) (o-N itrob enzoic acid )
21-38
Disubstituted Benzenes
• Where neither group imparts a special name, locate
the groups and list them in alphabetical order.
CH2 CH3
4
3
2
NO2
Br
1
2
1
Cl
1-Chloro-4-ethylben zene
(p-Ch loroethylbenzen e)
1-Bromo-2-nitrob enzene
(o-Bromon itroben zene)
21-39
Polysubstituted Derivatives
• If one group imparts a special name, name the
molecule as a derivative of that compound.
• If no group imparts a special name, list them in
alphabetical order, giving them the lowest set of
numbers.
1 2
N O2
OH
CH3
N O2
Br
6
1
2
4
Br
2
4
4
Cl
Br
4-C hl oro-2-n i trotol u e n e
2,4,6-Tri bromoph e nol
1
Br
CH2 CH3
2-Bromo-1-e th yl -4n itrobe n z e n e
21-40
Phenols
 The
functional group of a phenol is an -OH group
bonded to a benzene ring.
OH
OH
OH
OH
OH
CH3
OH
Ph e n ol 3-Me th ylph e n ol 1,2-Ben z e n e diol 1,4-Ben z e n e diol
(m-C re sol )
(C ate ch ol )
(Hydroqu in on e )
21-41
Acidity of Phenols
 Phenols
are significantly more acidic than
alcohols.
OH + H2 O
O- + H3 O+
CH3 CH2 OH + H2 O
CH3 CH2 O + H3 O
-
pK a = 9.95
+
pK a = 15.9
21-43
Acidity of Phenols
 Separation
of waterinsoluble phenols
from water-insoluble
alcohols.
21-44
Acidity of Phenols (Resonance)
• The greater acidity of phenols compared with alcohols
is due to the greater stability of the phenoxide ion
relative to an alkoxide ion.
O
O
O
O
O
H
H
These 2 Kekulé
s tru ctures are
equivalent
H
These th ree con trib utin g s tru ctures
delocalize th e negative charge
on to carb on atoms of th e rin g
21-45
Phenol Subsitituents (Inductive Effect)
 Alkyl
and halogen substituents effect acidities by
inductive effects:
• Alkyl groups are electron-releasing.
• Halogens are electron-withdrawing.
OH
OH
OH
CH3
Phen ol
pK a 9.95
m-Cres ol
p Ka 10.01
OH
OH
Cl
CH3
p-Cres ol
pK a 10.17
Cl
m-Chlorop henol p-Chororophen ol
pK a 8.85
p Ka 9.18
21-46
Phenol Subsitituents(Resonance, Inductiion)
• Nitro groups increase the acidity of phenols by both an
electron-withdrawing inductive effect and a resonance
effect.
OH
OH
OH
NO2
Ph e no l
p K a 9.95
NO2
m - N itrop h e n ol p- N itrop h e n ol
p K a 8.28
p K a 7.15
21-47
Acidity of Phenols
• Part of the acid-strengthening effect of -NO2 is due to
its electron-withdrawing inductive effect.
• In addition, -NO2 substituents in the ortho and para
positions help to delocalize the negative charge.
O
O
O
N+
N+
O
O
delocalization of negative
charge onto oxygen further
increases the resonance
stabilization of phenoxide ion
O
21-48
Synthesis: Alkyl-Aryl Ethers
 Alkyl-aryl
ethers can be prepared by the
Williamson ether synthesis:
• but only using phenoxide salts and haloalkanes.
• haloarenes cannot be used because they are
unreactive to SN2 reactions.
X + RO - N a+
n o re action
21-51
Synthesis: Alkyl-Aryl Ethers
OH + CH2 = CHCH2 Cl
Ph e n ol
N aOH, H2 O, CH2 Cl 2
3-C hl oroprope n e
(All yl ch l ori de )
OCH2 CH= CH2
Ph e n yl 2-prope n yl e th er
(All yl ph e n yl eth e r)
O
OH + CH3 OSOCH3
Ph e n ol
NaOH, H2 O, CH2 Cl 2
O
Di me th yl s u lfate
OCH3 + Na2 SO 4
Me th yl ph e n yl eth e r
(An is ole )
21-52
Synthesis: Kolbe Carboxylation
 Phenoxide
ions react with carbon dioxide to give
a carboxylate salt.
-
OH
O Na
+
CO2
NaOH
H2 O
Phenol
OH O
+
CO Na
H2 O
Sodiu m
phen oxid e
Sodium salicylate
HCl
H2 O
OH O
COH
S alicylic acid
21-53
Mechanism: Kolbe Carboxylation
• The mechanism begins by nucleophilic addition of the
phenoxide ion to a carbonyl group of CO2.
O
O
O
+
C
O
Sodium
phenoxide
O
C
(1)
O
keto-enol
tautomerism
H
OH
O
C
O
(2)
A cyclohexadienone
intermediate
Salicylate anion
Go back to aromatic
structure
21-54
Synthesis: Quinones
 Because
of the presence of the electron-donating
-OH group, phenols are susceptible to oxidation
by a variety of strong oxidizing agents.
OH
O
H2 Cr O 4
Ph e n ol
O
1,4-B en z oqui n on e
(p-Q u i n on e )
21-55
Quinones
O
OH
OH
O
K2 Cr2 O7
H2 SO4
1,2-Benzen ediol
(Catechol)
1,2-Benzoquin on e
(o-Qu inone)
OH
O
K2 Cr2 O7
H2 SO4
OH
1,4-Ben zenediol
(Hydroquin on e)
O
1,4-Benzoquin on e
(p-Qu inone)
21-56
Quinones
 Readily
reduced to hydroquinones.
O
OH
Na2 S2 O4 , H2 O
(reduction )
O
1,4-Benzoqu inone
(p-Qu inone)
OH
1,4-Benzen ediol
(Hydroq uinone)
21-57
Coenzyme Q
 Coenzyme
Q is a carrier of electrons in the
respiratory chain.
O
OH
MeO
MeO
reduction
MeO
O
Coenzyme Q
(oxid ized form)
n
H
oxidation
MeO
OH
Coenzyme Q
(redu ced form)
n
H
21-58
Benzylic Oxidation
 Benzene
is unaffected by strong oxidizing agents
such as H2CrO4 and KMnO4
• Halogen and nitro substituents are also unaffected by
these reagents.
• An alkyl group with at least one hydrogen on its
benzylic carbon is oxidized to a carboxyl group.
CH3
O2 N
COOH
H2 Cr O4
Cl
2-C h loro-4-n i trotol u en e
O2 N
Cl
2-C h loro-4-n itrobe n z oi c aci d
21-60
Benzylic Oxidation
• If there is more than one alkyl group on the benzene
ring, each is oxidized to a -COOH group.
H3 C
CH3
1,4-Dime th ylbe n ze n e
(p -xyle n e )
K2 Cr 2 O 7
H2 SO 4
O
HOC
O
COH
1,4-Ben z e n e di carboxyl ic acid
(tere ph th ali c aci d)
21-61
Benzylic Chlorination
 Chlorination
and bromination occur by a radical
chain mechanism.
CH3
+
Cl2
h eat
or ligh t
Toluen e
CH2 Cl
+
HCl
Benzyl ch loride
Br
NBS
( PhCO2 ) 2 , CCl4
Ethylbenzen e
1-Bromo-1-p henylethan e
(racemic)
21-62
Mechanism: Benzylic Reactions
 Benzylic
radicals (and cations also) are easily
formed because of the resonance stabilization of
these intermediates.
• The benzyl radical is a hybrid of five contributing
structures.
C
C
C
C
C
21-63
Benzylic Halogenation
• Benzylic bromination is highly regioselective.
Br
NBS
(PhCO2 ) 2 , CCl4
Eth ylb enzene
1-Bromo-1-phen yleth ane
(the only product formed )
• Benzylic chlorination is less regioselective.
Cl
+ Cl2
Eth ylb enzene
heat
or ligh t
Cl
+
1-Chloro-1p henylethan e
(90%)
1-Ch loro-2ph enylethan e
(10%)
21-64
Hydrogenolysis
 Hydrogenolysis:
Cleavage of a single bond by H2
• Benzylic ethers are unique in that they are cleaved
under conditions of catalytic hydrogenation.
this bond
is cleaved
O
Benzyl butyl ether
+ H2
Pd/ C
Me
OH +
1-Butanol
Toluene
21-65
Synthesis, Protecting Group: Benzyl Ethers
 The
value of benzyl ethers is as protecting
groups for the OH groups of alcohols and
phenols.
• To carry out hydroboration/oxidation of this alkene,
the phenolic -OH must first be protected; it is acidic
enough to react with BH3 and destroy the reagent.
2 . BH3 • THF
1 . ClCH2 Ph
OH
2-(2-Propen yl)p henol
(2-A llylp henol)
Et 3 N
O
OH
O
Ph
H2
Pd/ C
Ph
3 . H2 O2 / NaOH
OH
OH
2-(3-Hyd roxyprop yl)p henol
21-66