sp2 hybridized systems: aromatic compounds
Major concepts
 sp2 hybridization allows for extended pi systems that are cyclic or acyclic
 If orbitals do not overlap because they are “perpendicular” to each other, they are considered
“orthogonal” to one another.
 Count the number of electrons in a given pi system by counting all pi bonds and lone pairs that
are overlapped, but not those orthogonal to the system
 If there are 4n+2 electrons in a cyclic, planar pi system, then it is aromatic.
 The properties and reactivity of the nucleobases of nucleic acids is dictated by their aromaticity.
Vocabulary
 Aromatic
 Huckel’s Rule
 orthogonal
 Nucleobases
Students should be able to:
 Recognize aromatic and non-aromatic systems
 Draw orbital overlap pictures of pi systems, including aromatic compounds
Daily Problems
1. Draw an orbital picture of the following molecules, drawing all p orbitals and any orbitals holding
lone pairs. Be as clear as possible with their geometry. Indicate which are orthogonal to one another.
O
H3C
C
C
H
H
H
C
H
H
C
C
C
O
H
H
The 2 sets of p orbitals are
orthogonal to each other. The sp
orbital is orthogonal to both p
orbitals.
The sp2 orbitals are
orthogonal to the p orbitals.
2. When we draw an orbital structure for a compound that has multiple resonance structures, we need
to take into account all the resonance structures. To do this, we need to follow this rule: if an atom is
sp3 hybridized in one resonance structure and sp2 hybridized in another, we should consider the atom to
be sp2 hybridized in drawing an orbital picture.
Draw the orbital pictures of structures A and B as if they were simple Lewis structures (if an
atom is sp3 in the Lewis structure, draw it that way.) Then draw the orbital picture of the hybrid taking
into account the rule described above. How does this structure account for the partial double bonds
between each of the carbons?
-
structure A
H
structure B
hybrid
H
H
H
incorrect assumption
If all are sp2 hybridized, then the p orbital
overlap accounts for partial double bond
character
3. Draw orbital overlap pictures for these molecules. Draw top perspective and side perspective.
Indicate which orbitals are orthogonal.
Example:
p orbital
A.
p orbital p orbital
C.
orthogonal
p orbital
p orbital
p orbital
H
H
H
H
N
H
H
p orbital
p orbital
B.
p orbital
p orbital
p orbital
orthogonal
p orbital p orbital
H
orthogonal
H
H
H
N
H
H
H
H
p orbital
4. Draw a VB orbital picture of the pi system of pyridine and explain why the lone pair is not counted in
the pi system.
N
N
pyridine
Side perspective
The lone pair on nitrogen is in an sp2
hybridized orbital, which is 90o (orthogonal) to
the pi system of the ring. This means the lone
pair and the pi system do not overlap.
5. How many electrons are in each pi system. Label any electrons that are orthogonal to the pi system.
all lone pairs drawn in are orthogonal
NH
NH
O
6 electrons
6 electrons
4 electrons
over 5 p orbitals
isolated- no pi system/ conjugation
6 electrons
6 electrons over
7 p orbitals
6. Which of these systems is aromatic? Explain.
Yes- 14 electrons, cyclic, contiguous
No- not cyclic
No- only 8 electrons
Yes- 6 electrons
Lone pair on the
bottom N is not in
conjugation
Yes- 10 electrons
One of the lone
pairs on O is in
conjugations (not
both)
No- 8 electrons
Yes- 6 electrons,
contiguous
The lone pair on N
is in conjugation
Cumulative problems
7. Which of these aromatic systems is “electron rich” compared to benzene? Which is “electron poor”
compared to benzene. Explain with a resonance structure.
NH2
O
NH2
electron rich ring
O
electron poor ring
8. Compare the relative stabilities of the indicated lone pairs. Which is more reactive?
- localized
-not part of the pi
system
-not as stabilized
-more reactive
-very stable = not very
reactive
-part of the 6 electrons
in the aromatic pi
system
9. For the nucleobases below, draw resonance structures in which the amide functional groups have
been redrawn. How many electrons are in each of these nucleobase systems? Indicate which pi
electrons and lone pair electrons are and are not part of the pi system.
Example:
Problems:
O
O
NH ₂
N
N
O
NH
NH
N
N
H
N
H
N
N
NH ₂
guanine
adenine
N
H
NH
O
thymine
N
O
H
6 electron system
10 electron system
O
O
NH ₂
O
N
N
H
cytosine
N
O
NH ₂
N
N
H
O
6 electron system
N
H
NH
NH
N
H
uracil
N
NH ₂
10 electron system
NH
O
N
O
H
6 electron system
Extension problems
10. Draw the structure of guanine (see above.) Label each nitrogen atom as a Lewis acid or Lewis base.
Explain how you came to these conclusions.