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Chapter 15Conjugated Systems,

Orbital Symmetry, andUltraviolet Spectroscopy

Chapter 15: Conjugated Systems Slide 15-2

Definitions

• Conjugated double bonds are separated by one single bond.Example: 1,3-pentadiene.

• Isolated double bonds are separated by two or more singlebonds. Example: 1,4-pentadiene.

• Cumulated double bonds are on adjacent carbons. Example:1,2-pentadiene. =>

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Stabilities of Dienes

• Heat of hydrogenation for 1,4-pentadiene is –252 kJ/mol, abouttwice that of 1-pentene.

• ΔH for 1-pentene is -126 kJ/mol and for trans-2-pentene is -116kJ/mol, so expect -242 kJ/mol for trans-1,3-pentadiene.

• Actual ΔH is -225 kJ/mol, so the conjugated diene is morestable.

• Difference, -242 kJ (-225 kJ) = 17 kJ/mol, is the resonanceenergy. =>


Relative Stabilities

twice 1-pentene

more substituted

=>

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Structure of 1,3-Butadiene

• Most stable conformation is planar.• Single bond is shorter than 1.54 Å.• Electrons are delocalized over molecule.

=>


ConstructingMolecular Orbitals

• Pi molecular orbitals are the sideways overlap of p orbitals.• p orbitals have 2 lobes. Plus (+) and minus (-) indicate the

opposite phases of the wave function, not electrical charge.• When lobes overlap constructively, (+ and +, or - and -) a

bonding MO is formed.• When + and - lobes overlap, waves cancel out and a node

forms; antibonding MO. =>

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Ethylene Pi MO’s

• The combination of two porbitals must give twomolecular orbitals.

• Constructive overlap is abonding MO.

• Destructive overlap is an anti-bonding MO. =>


π1 MO for 1,3-Butadiene

• Lowest energy.• All bonding

interactions.• Electrons are

delocalized overfour nuclei.

=>

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π2 MO for 1,3-Butadiene

• 2 bondinginteractions.

• 1 antibondinginteraction.

• A bonding MO.

=>


π3* MO for 1,3-Butadiene

• Antibonding MO.• Empty at ground state.• Two nodes. =>

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π4* MO for 1,3-Butadiene

• All antibondinginteractions.

• Highest energy.• Vacant at ground state.

=>


MO Energy Diagram

The average energy of electrons is lower in the conjugated compound.

=>

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Conformations of1,3-Butadiene

• s-trans conformer is more stable than the s-cis by 12 kJ/mol(2.8 kcal/mol).

• Easily interconvert at room temperature.

HH

H

H

H

H

s-trans s-cis

H

H

H

H

HH

=>


Allylic Cations

• Carbon adjacent to C=C is allylic.• Allylic cation is stabilized by resonance.• Stability of 1° allylic ≈ 2° carbocation.• Stability of 2° allylic ≈ 3° carbocation.

H2C C

H

CH2

+

H2C C

H

CH2

+

=>

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1,2- and 1,4-Additionto Conjugated Dienes

• Electrophilic addition to the double bond produces the moststable intermediate.

• For conjugated dienes, the intermediate is a resonancestabilized allylic cation.

• Nucleophile adds to either carbon 2 or 4, both of which havethe delocalized positive charge. =>


Addition of HBr

Br_

Br_

H3C C

H

C

H

CH2

Br

H3C C

H

C

H

CH2

Br

1,2-addition product 1,4-addition product

=>

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Kinetic vs.Thermodynamic Control

Major product at 40°C

Major product at -80°C

=>


Allylic Radicals

• Stabilized by resonance.• Radical stabilities: 1° < 2° < 3° < 1° allylic.• Substitution at the allylic position competes with addition to double

bond.• To encourage substitution, use a low concentration of reagent with

light, heat, or peroxides to initiate free radical formation. =>

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Allylic Bromination

Br

H

H

HH

H

H

H

H

H

H

+ HBr

Br Br Br Br

H

H

BrH

H

H

H

Br + Br •

=>

Br2h!

Br2


Bromination Using NBS

• N-Bromosuccinimide (NBS) provides a low, constantconcentration of Br2.

• NBS reacts with the HBr by-product to produce Br2 andprevent HBr addition.

=>

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MO’s for the Allylic System

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SN2 Reactions of Allylic Halides andTosylates

=>

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Diels-Alder Reaction

• Otto Diels, Kurt Alder; Nobel prize, 1950• Produces cyclohexene ring• Diene + alkene or alkyne with electron-withdrawing group

(dienophile)

=>


Examples ofDiels-Alder Reactions

=>

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Stereochemical Requirements• Diene must be in s-cis conformation.• Diene’s C1 and C4 p orbitals must overlap with

dienophile’s p orbitals to form new sigma bonds.• Both sigma bonds are on same face of the diene: syn

stereochemistry.

=>


Concerted Mechanism

=>

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Endo RuleThe p orbitals of the electron-withdrawing groups on the dienophile

have a secondary overlap with the p orbitals of C2 and C3 in thediene.

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RegiospecificityThe 6-membered ring product of the Diels-Alder reaction will

have electron-donating and electron-withdrawing groups1,2 or 1,4 but not 1,3.

=>

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Pericyclic Reactions

• Diels-Alder reaction is example.• Woodward and Hoffmann predicted reaction products using

their theory of conservation of orbital symmetry.• MO’s must overlap constructively to stabilize the transition

state. =>


Symmetry-Allowed Reaction• Diene contributes electrons from

its highest energy occupiedorbital (HOMO).

• Dienophile receives electrons inits lowest energy unoccupiedorbital (LUMO).

=>

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“Forbidden” Cycloaddition

[2 + 2] cycloaddition of two ethylenes to form cyclobutene has anti-bonding overlap of

HOMO and LUMO.

=>


Photochemical Induction

Absorption of correct energy photon will promote an electronto an energy level that was previously unoccupied.

=>

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[2 + 2] Cycloaddition

Photochemically allowed, butthermally forbidden.

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Ultraviolet Spectroscopy

• 200-400 nm photons excite electrons from a π bondingorbital to a π* antibonding orbital.

• Conjugated dienes have MO’s that are closer in energy.• A compound that has a longer chain of conjugated double

bonds absorbs light at a longer wavelength. =>

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=>

π → π* forethylene

andbutadiene


Obtaining a UV Spectrum

• The spectrometer measures the intensity of a reference beamthrough solvent only (Ir) and the intensity of a beam through asolution of the sample (Is).

• Absorbance is the log of the ratio

I

I

s

r =>

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The UV Spectrum

• Usually shows broad peaks.• Read λmax from the graph.• Absorbance, A, follows Beer’s Law:

A = εclwhere ε is the molar absorptivity, c is the sampleconcentration in moles per liter, and l is the length of thelight path in centimeters.


UV Spectrum of Isoprene

=>

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Sample UV Absorptions

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Woodward-Fieser Rules

=>

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End of Chapter 15

Homework: 25, 26, 29, 30, 31, 33, 36, 38

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