Primer on Pericyclic Reactions

C343 Professor Kennan
<strong>Primer</strong> on Pericyclic Reactions
How do we determine whether a particular electrocyclization should be conrotatory or disrotatory?
How do we determine which cycloadditions are “allowed” and which are “forbidden”?
I. Electrocyclizations
Bottom Line: cyclization and ring-opening of the same molecule proceeds via the same rotation
molecules with symmetric HOMOs give disrotatory cyclization/opening
molecules with antisymmetric HOMOs give conrotatory cyclization/opening
orbital symmetries (sym/anti) alternate as you increase energy level
symmetric
need disrotatory rotation to
ensure bonding interaction
antisymmetric
need conrotatory rotation to
ensure bonding interaction
• Determine the number of electrons or π bonds involved – consider the ring-opened structure
2 π bonds
4 electrons
(only bonds in
circle are involved)
4 π bonds
8 electrons
3 π bonds
6 electrons
2 π bonds
4 electrons
• A) Memorize that butadiene electrocyclization is conrotatory
∆
Conrotatory
looks like a 'C'
• B) Preferably, work out that butadiene is antisymmetric/conrotatory
4 π electrons => Ψ 2 is HOMO => 1 node
antisymmetric
need conrotatory rotation to
ensure bonding interaction

C343 Professor Kennan
• Since adding one π bond (and thus two electrons) makes a molecule’s HOMO one level higher, and
since the symmetries alternate, the HOMO of hexatriene (below) must be symmetric. Thus the reaction
is disrotatory. If you add two more π bonds (to get 5) then the HOMO will flip from symmetric to
antisymmetric (for 4 π bonds) and back to symmetric. Since this alternation continues indefinitely, all
conjugated π systems with an odd number of double bonds will have symmetric HOMOs. Put more
succinctly:
Molecules with an even number of π bonds (4n electrons) in the ring opened form have antisymmetric
HOMOs, while those with an odd number of π bonds (4n+2 electrons) have symmetric HOMOs
6π electrons => Ψ 3 is now the HOMO
orbitals alternate symmetry => if Ψ 2 is antisymmetric Ψ 3 must be symmetric
symmetric => disrotatory
• Similarly, since photochemical conditions involve promotion of one electron the effect is the same as
adding a new π bond – in other words the next highest orbital now becomes the HOMO. Since the
symmetry alternates, this has the effect of switching between conrotatory and disrotatory. Thus:
Reactions which are conrotatory thermally will be disrotatory photochemically and vise versa
new LUMO
opposite
symmetries
by definition
LUMO
HOMO
hν new HOMO => change in symmetry
• So now we can determine if any reaction is conrotatory or disrotatory. Start with the number of π bonds,
and determine if it is like butadiene (even number π bonds, 4n electrons) or not. Remember or figure
out that butadiene is antisymmetric/conrotatory (thermally). Then ask two questions:
(1) Is it like butadiene? antisymmetric/conrotatory; if not symmetric/disrotatory.
(2) Is it thermal (heat)? leave above decision alone; if photochemical (light) switch to opposite
II. Cycloadditions
• Count the π bonds/electrons in each molecule
• Determine if the HOMOs for each molecule are symmetric or antisymmetric as above
• The LUMO for each molecule will have the opposite symmetry to the HOMO (since it alternates)
• Compare HOMO of molecule A with LUMO of molecule B (doesn’t matter how you assign A/B)
If the HOMO of A and the LUMO of B have the same symmetry, the reaction is “allowed”
If the HOMO of A and the LUMO of B have the opposite symmetry, the reaction is “forbidden”
• Since photochemical (light) conditions involve altering the symmetry of molecule A or B (not both):
Thermally allowed (heat) cycloadditions are photochemically (light) forbidden and vise versa

C343 Professor Kennan
Examples of Pericyclic Reactions
H 3 C
CH 3
∆
?
H 3 C
CH 3
H CH 3 H CH 3
∆
H 3 C
H
H
CH 3
CH 3
CH 3
4 π bonds / 8 electrons => same as => antisymmetric/conrotatory
thermal conditions => still antisymmetric/conrotatory
CH 3
CH 3
∆
?
CH 3
CH 3
H
H 3 C
H
CH 3
∆
H
H CH 3
CH 3
CH 3
CH 3
H
H
3 π bonds / 6 electrons => opposite of => symmetric/disrotatory
thermal conditions => still symmetric/disrotatory
CH 3
CH 3
CH 3
?
CH 3
CH 3
CH3
CH 3
CH 3
CH 3
H 3 C
H
H
CH 3
CH 3
hν
H
CH 3
H 3 C CH 3
H
CH 3
CH 3
CH 3
3 π bonds / 6 electrons => opposite of => symmetric/disrotatory
observed product results from conrotatory cyclization => must be photochemical conditions