CSUMB
ESSP 311 Organic Chemistry I
Ronald W. Rinehart, Ph.D.

Chapter 10 Conjugated Systems

Chapter 10.  Conjugation in Alkadienes and Allylic Systems

I.  The Allyl Group:  CH₂=CHCH₂−
            A.  Conjugated p-electron systems  
                        1.  Conjugated p-electron systems: electrons are delocalized over a p orbital that spans >2 atoms
                        2.  Examples of conjugated systems:   
                                    a)  Allylic cations.
                                    b)  Allylic free radicals.
                                    c)  Allylic anions    
                                    d)  A 1,3-diene (two sets of C=C joined by a single bond).
                        3.  All have ≥3 mutually parallel p orbitals in a row, which results in delocalization of the p electrons.

                        4.  Due to conjugation allylic carbocations are more stable than simple alkyl cations:         
                                    a)  They are more stable than 3^o carbocations! 
                                    b)  Compare the reaction rates for the following two reactions:

                        c)  The rate constant for reaction (ii) is 123 x that of (i)!
                        d)  In reaction (ii) the vinyl group is e⁻-releasing:  p electrons are delocalized, dispersing positive charge.
                                    (i)  This forms a resonance-stabilized species.          

                        5.  Consider the following reaction:

                                    a)  It is important to note that structures I and II are NOT isomeric carbocations in equilibrium. 
                                                They are RESONANCE STRUCTURES. 
                                    b)  A resonance hybrid exists with qualities of both structures. 
                                    c)  Since structure I is 3^o, it is the more important contributor to the resonance hybrid
                                                and therefore yields the most product.    
                                    d)  Reaction with  (CH₃)₂C=CH-CH₂Cl yields the same two products in the same proportion.  Why?  

                                    Answer:  Both reactions go through the same carbocation intermediate.

            B.  Molecular Orbitals of the Allyl Group:

                        1.  The molecular orbitals and p electrons for the allylic cation, free radical, and anion are shown below.               

                        2.  Three parallel p atomic orbitals combine to form three pi molecular orbitals.
                                    a)  Recall that the lesser the number of nodes, the lower the energy of the orbital.
            C.  Comparison of the Stability of Free Radicals: (Think in terms of the Hammond Postulate)
                        1.  Below are listed heats of formation data for the formation of various free radicals:

                        2.  Notice the low energy required to form the allylic free radical, even compared to the 3^o free radical.

II.  Reactions of Conjugated Systems
            A.  Allylic Halogenation.  
                        1.  At HIGH temperatures, alkenes react with Cl₂ and Br₂ by the SUBSTITUTION of allylic hydrogens.

                         2. Mechanism for allylic substitution:

            Step 1.   Initiation

                                     Cl₂  à   2Cl∙

             Step 2.  Propagation

                                    Cl∙ + CH₂=CHCH₃   à   CH₂=CHCH₂∙ +  HCl

                             CH₂=CHCH₂∙ + Cl₂  à   CH₂=CHCH₂Cl +  Cl∙

            Step 3.  Termination

                             CH₂=CHCH₂∙ + Cl∙  à   CH₂=CHCH₂Cl

                                    2Cl∙  à  Cl₂

                             2CH₂=CHCH2∙   à   CH₂=CHCH₂-CH₂CH=CH₂

                         3. For bromination a special reagent is used:  N-Bromosuccinimide  (NBS).

                                     a)  NBS yields a low concentration of Br₂ by reacting with HBr generated from the overall reaction: 

                                       NBS  +  HBr  à  Succinimide  +  Br₂  

                                    b)  Low concentrations of Br₂ favor allylic substitution:  ionic addition doesn't occur, since the                                            bromonium ion reforms the alkene before Br⁻ can attack it to form the dibromide. 
                                                i)  Free radical attack is irreversible due to irreversible formation of the free radical intermediate.

                                    c)  Procedure for NBS reaction of alkenes: 
                                                i)  The allylic compound is dissolved in CCl₄.
                                                ii)  1 Eq of NBS is added. NBS is denser than CCl₄ and relatively insoluble so it sinks to bottom.

                                                iii)  The reaction is initiated with a sunlamp.

                                                iv)  Solid rises gradually rise to top of CCl₄ layer. (It is succinimide, less dense than CCl₄.)

                                                v)  After all the solid succinimide rises to the top the sunlamp is turned off.

                                                vi)  The solid is filtered off, and the CCl₄ evaporated from the product.

                        4.  Explain how/why the following reaction yields two observed products, and predict the major product:

                                     Answer:                The major product arises via the more important resonace contributor

            B.  Addition of HX to Conjugated Dienes (HCl, HBr):  1,2 Addition and 1,4 Addition.
                        1. Example.

                        2.  1,2 and 1,4 addition occurs since the allylic carbocation forms:

                         3.  At −80^oC the less stable alkene is the major product. 
                                    a)  (I) is the more important contributor, so there is a greater positive character on C-3 than on C-1.
                                                i)  Attack on C-3 by X⁻ (forming product A) is faster than on C-1 (resonance form II).
                                                ii)  This is a KINETICALLY  CONTROLLED [or rate-controlled] reaction:
                                                     product A is favored since it forms faster.
                        3.  At room temperature products A and B interconvert rapidly so that A and B can equilibrate.
                                    a)  The reaction mixture no longer reflects the rates at which A and B formed from the reaction, but
                                                from their relative stabilities instead:
                                                i)  Product B is more stable than product A.  Why?

                                                            Answer:  It is a disubstituted alkene whereas product A is only monosubstituted.

                                                ii)  This is referred to as THERMODYNAMIC CONTROL (or equilibrium control): 
                                                            the favored product is the lower-energy product.
                        4.  Example:

 

      C.  Halogen Addition to Conjugated Dienes
                        1.  In the addition of halogens to conjugated dienes, both 1,2 and 1,4- addition products form.
                        2.  1,4- Addition (conjugate addition) is favored.
                        3.  (E)- [trans] products are favored.
                        4.  Example.
 

                        5.         As an exercise, write the mechanism for the formation of both products.

III.       Classes of Dienes
            A.  Isolated diene units.
                        1.  Examples.
                                    a)  CH₂=CH-CH₂-CH=CH₂
                                    b)  CH₃-CH=CH-CH₂-CH₂-CH=CH-CH₃
                        2.  In isolated dienes, the two C=C units are separated by one or more sp³ hybridized carbons. 

                        3.         These are not conjugated systems.  Why not?
                                    Answer:  The four p-orbitals are not on adjacent carbons.

            B.  Cumulated Dienes:  Allenes.   

                        1.  Example.   CH₂=C=CH₂
                        2.  Allenes are not conjugated. 
                                    a)  Why are they not conjugated when the adjacent carbons all have p orbitals?                                        Answer:  The p orbitals are perpendicular to each other.  To be conjugated, the p orbitals must be parallel 

                        3.  Allenes are high energy compounds. 

            C.  Conjugated dienes: 

                        1.  Conjugated dienes have two C=C groups connected by a single bond.
                        2.         Examples.

                         3.  They are prepared by the same methods as alkene synthesis. 
                        4.  Conjugated alkenes more stable than isolated alkenes due to resonance (or delocalization) energy.
                        5.  The stability is due to delocalization of p electrons over 4 nuclei.
                        6.  Two conformations allow for maximum overlap of all 4 p orbitals: s-cis and s-trans
                                    (s- for conformations around a single bond.)

 

                                     a) The s-cis conformation is less stable than the s-trans due to greater van der Waals repulsions.

IV.  Orbital Symmetry and Chemical Reactions
            A.  Electrocyclic Reactions
                        1.  Upon treatment with heat or light a conjugated polyene can undergo a cyclization reaction.
                                    a)  In this type of reaction one of the double bonds break and the other(s)             shift position in the
                                                 formation of the two new sigma bonds.
                                    b)  The reverse process can also occur: a cyclic alkene's single bond can break, forming an
                                                open-chain polyene.

                        2.  Product stereochemistry can be predicted since these reactions are completely stereospecific.
                        3.  The stereochemistry depends upon two factors:
                                    a)  The number of double bonds in the polyene.
                                    b)  Whether the reaction is initiated with heat (thermal) or light (photochemical.)
                        4.  To make our predictions we must know the p molecular orbitals for the reactants and focus our
                                     attention on the HOMO (the highest occupied molecular orbital.)
                        5.  Look at the p orbitals of the HOMO on the end C's of the polyene which cyclizes to the cycloalkene. 
                                    a)  How must the p orbitals rotate in order to form a sigma bond together (match together in phase)?
                                                i)  Rotate the two p orbitals in the same direction:  conrotatory.
                                                ii)  Rotate the two p orbitals in the opposite direction:  disrotatory.
                        6.  Remember that photochemical processes elevate a p electron to a higher energy molecular orbital;
                                     thermal processes do not. 

                        7.  Write the molecular orbitals of ethylene, 1,3-butadiene, and 1,3,5-hexatriene. 

                         8.  Example.

 

                                    Answer:  Draw the above reactant with the HOMO in place in order to determine            whether conrotatory
                                     or disrotatory rotation of the p orbitals will result in the formation of a sigma bond.

 

                        9.  Woodward-Hoffman Rules.  (Note:  Whether considering ring formation or ring opening, the
                                    # of p electrons below refers to the number of p electrons in the acyclic polyene, not the ring system.)

                                    a)  Looking at example 8 above, thanks to the magic table we can now say that with 4 p e⁻ electrons
                                                (4n) under thermal conditions, conrotatory motion would occur, leading to the same conclusions
                                                 without having to draw the molecular orbitals.  [HINT!!!]

                        10.        Predict the product for the following thermal cyclization of trans,cis,trans-2,4,6-octatriene to the
                                    5,6-dimethyl-1,3-cyclohexadiene.

 

                        Answer:  The reactant has 6 pi electrons            (4n + 2), which under thermal conditions will undergo disrotatory rotation.  This yields the cis product.  

                        11.        Explain the following observations:

            Answers:

            B.  The Diels-Alder Reaction:  Cycloaddition
                        1.  Conjugated dienes react with isolated alkenes and alkynes ("dienophiles") to yield cyclohexenes.
                        2.  The reaction is concerted: both ends of the diene form bonds to the dienophile simultaneously. 
                        3.  The dienophile is usually "activated" (but does not have to be) towards the reaction by being bonded                                          to one or more C=O or CN [or other e⁻ –withdrawing] units.
                        4.  It is considered to be the best way to make six-membered rings.
                        5.  Example.

 

                        5.  Important Facts About the Diels-Alder Reaction:
                                    a)  The reaction is stereospecific: 
                                                i)  cis substituents in the dienophile remain cis in the product.
                                                ii)  trans substituents in the dienophile remain trans in the product.

                                    b)  The dienophile should be activated with electron-withdrawing groups: C=O and –CN
                                    c)  Stereoselective:  The "endo rule" or rule of maximum accumulation of unsaturation.
                                                i)  The unsaturated group in the dienophile takes the syn orientation rather than anti to the diene.

                                    d)  The diene must be able to assume the s-cis conformation.

 

                        6.  The most important fact about the Diels-Alder reaction is that four stereocenters can be produced.
                                    a)  Also, the reaction tends to give predominantly one product.

 

                                    b)  In the above reaction both the diene and the dienophile are unsymmetrical which implies several                                                 different products could be produced.
                                                i)  The use of resonance structures for both reactants can help to predict the favored product:                       

                                    c)  Analysis of the diene resonance structures:

 

                                                            i)  (A) seems best due to inductive effects, but (B) is actually better due to resonance effects.

                                    d)  Analysis of the dienophile resonance structures:

 

                                                 i)  Resonance structure (I) is better than (II) due to inductive effects. 
                                                ii)  Notice that oxygen cannot stabilize the positive charge via resonance in this case.
                                    e)  Combining the two preferred resonance structures yields the observed favored product.

 

                           7.  Orbital Symmetry:  Frontier Orbitals.
                                    a)  The frontier orbitals of the diene are the orbitals which contain the electrons most likely to be
                                                transferred to the dienophile (orbitals containing electrons of highest energy.)
                                                i)  HOMO:  Highest Occupied Molecular Orbital.
                                    b)  The dienophile frontier orbitals: the lowest energy vacant orbitals, which receive the electrons
                                                from the diene.
                                                ii)  LUMO:  Lowest Unoccupied Molecular Orbital.
                                    c)  The frontier orbitals are those then most likely to form bonds.
                                    d)  Bond formation occurs between ends of the diene and the two carbons of the dienophile as
                                                 electrons are transferred from the HOMO of the diene to the LUMO of the dienophile. 
                                                i)  In order for this to occur there must be in-phase overlap of the orbitals:  Symmetry-allowed.
                                    e)  Consider the reaction between 1,3-butadiene and ethylene to form cyclohexene.

 

                                                i)  Check if the diene HOMO and dienophile LUMO allow in-phase overlap to form sigma bonds:

 

                                                            ii)  Notice that this reaction is symmetry allowed:

                        7.  Can this reaction occur? 

                                    a)  A few hints:
                                                i)  One molecule acts as the electron donor (HOMO), the other as the electron acceptor (LUMO).
                                                ii)  The reactant does not have to be activated by an electron-withdrawing group.
                                                iii)  Consider the LUMO and HOMO for the reactants. (Draw the molecular orbitals for ethylene,
                                                            and write in their electron populations for both the thermal and photochemical cases.

                                                iv)        Check both thermally and photochemically driven reactions.

            Answer:

            Thermal case.   The molecular orbitals are in their ground states in the thermal case. Notice that no reaction will occur since in-phase overlap cannot occur.  The reaction is symmetry forbidden.

            Photochemical case.  In this case an electron is elevated to the higher molecular orbital, therefore the HOMO of the excited state now has the same symmetry as the LUMO of the ground state.  The reaction is now symmetry allowed and a reaction is observed to occur.

                        9.  The Diels-Alder reaction has been used to synthesize insecticides (chlorinated cyclic dienes.)

 

            C.  Practice Problems

                        1.  Predict the products:

                        2.  What dienes/dienophiles would produce the following Diels-Alder products?

 

3.  Draw the MOs for ethene; show e- populations for thermal and photochemical cases, and identify the HOMO for both
 1,3-butadiene: show e- populations for thermal and photochemical cases, and identify the HOMO and LUMO for both

Many thanks to Rod Oka of MPC for generously sharing his "Lecture Companion" outline,
reproduced here in extensively modified form by permission, with web references and other goodies added by me.
Structures drawn using mainly CS ChemOffice ChemDraw™, and a few with ACDLabs ChemSketch™ .
Reaction profile drawn with MS Excel™, my favorite program.