BTEC applied science: unit 14 learning aim B: Understand the reactions and properties of aromatic compounds. Distinction grade
title: Aromatic ring chemistry for designer chemicals
contains:
B.P2 Explain the structure of benzene using sigma and pi bonding, providing evidence for the structur...
b understand the reactions and properties of aromatic compounds aromatic ring chemistry for designer chemicals
p2 explain the structure of benzene using sigma and pi bonding
Geschreven voor
BTEC
PEARSON (PEARSON)
Applied Science 2016 NQF
Unit 14 - Applications of Organic Chemistry
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Voorbeeld van de inhoud
B: Understand the reactions and properties of aromatic compounds.
Aromatic ring chemistry for designer chemicals
P2: Explain the structure of benzene using sigma and pi bonding, providing evidence
for the structure.
Benzene is a simplest organic aromatic hydrocarbon and the source
of many significant aromatic compounds. The chemical formula for
benzene is C6H6, which creates a hexagonal molecule which we
usually view as a hexagon with an interior circle.
Kekule was the first to propose a structure of benzene; the carbon
atoms in the benzene ring are arranged in a trigonal planar geometry.
The hybrid orbitals are arranged at an angle of 120° to each other in
a plane while the p orbitals are at right angles to them.
Hybridisation
Benzene is made up of: Carbon atoms (1s2 2s2 2px1 2py1) and hydrogen (1s1). Each
carbon atom needs to combine with two more carbon atoms and one hydrogen atom to form
a compound. It must promote one of the 2s2 pairs into the open 2pz orbital because it does
not have enough unpaired electrons to do so.
Promotion of electron
During this, the carbon atom will enter into an excited state and the electron configuration
will also change to become 1s2, 2s1, 2px1, 2py1, 2pz1.
Now when the electron is promoted from the 2s to the empty 2p orbital, we will get 4
unpaired electrons. These electrons will be used in the formation of the bonds. For
hybridisation to occur the outer orbitals are used. Three of the carbon orbitals are used
rather than all four. They use the 2s electron and the 3p electrons but not the 2p electron;
this dont change.
A s orbital and two p orbitals are rearranged to form the new orbitals; they are referred as
sp2 hybrids. In a plane of 120°to each other, the three sp2 hybrid orbitals align as far apart
as they can. The remaining p orbital is perpendicular to them.
,In benzene each identical carbon atom is connected to two additional carbon atoms rather
than just one. Each carbon atom forms a sigma bond with two additional carbon atoms and
one hydrogen atom using sp2 hybrids.
Delocalised pi bonding in benzene
Each carbon atom has p orbital electrons which overlap over one another on both sides. The
significant sideways overlap produces an arrangement of pi bonds distributed over the entire
carbon ring. The electrons are considered to be delocalized since they are no longer trapped
between two carbon atoms, but rather are dispersed throughout the entire ring.
The six delocalized electrons enter three molecular orbitals, so two in each.
Problems with kekule’s structure
Considering that benzene possesses three double bonds according to Kekule’s structure,
the chemical properties should be alike to those of an alkane. Although in this case it is not
because benzene is extremely stable and easily forms replacement compounds.
The benzene would be uneven, with alternately shorter and longer sides if it had the Kekulé
structure. The problem is the difference in length between the single and double C-C bonds.
In a carbon single bond (C-C) measures 154 nm and for carbon double bonds (C=C)
measures 134 nm. But all the bonds in actual benzene are the same length which is
0.139nm.
Evidence against the Kekule model
Thermochemical evidence
We measure the stability of benzene by comparing the enthalpy change of hydrogenation in
benzene and cyclohexa-1,3,5-triene, because benzene has the molecular formula C6H6 we
assume its structure is cyclohexa-1,3,5-triene as forms 4 bonds.
, If we hydrogenate cyclohexane it has an enthalpy change of ∆H⦵ = -120 kJ mol-1
In other words, the reaction between 1 mole of cyclohexene releases 120 kJ of heat energy.
Therefore if benzene has 2 double bonds we would expect an enthalpy change of
hydrogenation of -360kJmol-1.
However, the actual enthalpy change of hydrogenation for benzene is far lower at
-208kJmol-1 (this is an experimental value). Which means the energy required to break
bonds and energy released to form bonds suggest more energy is required to break bonds
in benzene than cyclohexa-1,3,5-triene.
X ray diffraction
X-ray is used to confirm the ring type structure, however it did throw up new problems.
Because the structure was shown to be planar and it was found that all the C-C bond lengths
were equal in length and in between C – C and C = C bond lengths
X-ray diffraction shows all carbon-carbon bonds in benzene are the same length (140 pm),
between the lengths of a single (147 pm) and double bond (135 pm). This suggests that
Kekule's alternating single and double bonds model cannot be right as the bonds in the
molecule are all the same.
C – C 0.153nm cyclohexane C = C 0.134nm cyclohexene C – C 0.139nm benzene
Infrared spectroscopy
Within its structure, cyclohexene has a peak for the double C=C bond at about 1650 cm-1.
Three double C=C bonds make up benzene's cyclohexa-1,3,5-triene Kekulé structure.
Therefore, a peak for the double C=C bonds should also be visible at about 1650 cm-1.
Instead of exhibiting the double C=C bond peak at roughly 1650 cm-1 that is typical of
arenes, benzene has peaks at about 1450, 1500, and 1580 cm-1.
P3 Explain the chemical properties of industrially important benzene and
monosubstituted benzene compounds.
Nomenclature of aromatic compounds
Identify and name the parent. If it is not one of the common names, then use benzene.
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