W 2/1 M 2/6 Due: HW14, spec03 Due: n/a Lecture Organic Chemistry 1e Section: 17.7 What is the major...

W 2/1

• Due: HW14, spec03

• Lecture

– HW14 grading

– Lect17a Conjugated π systems

• Lab

– Lab02 Qual Analysis II (cont)

M 2/6

• Due: n/a

• Lecture

– quiz

– Lect17b

Copyright 2012 John Wiley & Sons, Inc. 7 -1

• There are three categories for dienes:

– Cumulated: pi bonds are adjacent.

– Conjugated: pi bonds are separated by exactly ONE single bond.

– Isolated: pi bonds are separated by any distance greater than ONE single bond.

17.1 Classes of Dienes

Copyright 2012 John Wiley & Sons, Inc. 17-2

• There are three categories for dienes:

– Cumulated: pi bonds are perpendicular.

– Conjugated: pi bond overlap extends over the entire system.

– Isolated: pi bonds are separated by too great a distance to experience extra overlap.

17.1 Classes of Dienes


• This chapter focuses on conjugated systems.

– Heteroatoms may be involved in a CONJUGATED system.


• A sterically hindered base can be used to form dienes while avoiding the competing substitution reaction.

17.2 Conjugated Dienes


• Single bonds that are part of a conjugated pi system are SHORTER than typical single bonds.

• The hybridization of a carbon affects its bond length.


• The more s-character a carbon has, the shorter its bonds will be. WHY?


• HOW does conjugation affect stability?


• Rank the following compounds in order of increasing stability.


• In general, single bonds will freely rotate.

• The two most stable rotational conformations for butadiene are the s-cis and s-trans.

• Which is more stable? Why? – About 98% s-trans form


• The highest energy conformer will not be conjugated.


• molecular orbital (MO) theory review: – An MO forms when atomic

orbitals overlap.

– An MO extends over the entire molecule.

– Recall the pi bonding and antibonding MOs for ethylene

– Which is more stable?

– WHY?

– What is a node?

17.3 Molecular Orbital Theory


• The number of MOs must be equal to the number of combined atomic orbitals (AOs).

– Note how the

shorthand drawing matches the actual MOs.

– The # of MO’s = 2 x #

pi bonds in conjugated pi systems


Orbitals shorthand

• The four pi electrons in butadiene will occupy the lowest energy MOs. HOW will that affect stability?


• MO also explains why the central C–C single bond is shorter and stronger than a typical C–C single bond.


• Draw 1,3,5-hexatriene.

– Is it conjugated?

– How many MO’s?

– How many nodes in the lowest energy MO?

– How many nodes in the highest energy MO?

– Draw the MO energy diagram and fill in the pi electrons

Copyright 2012 John Wiley & Sons, Inc. 17 -16

• 1,3,5-hexatriene should have six pi MOs.

– What are HOMO and LUMO?

– HOMO – Highest Occupied MO

– LUMO – lowest Unoccupied MO

– The HOMO and LUMO are the frontier MOs.


• Reactions that molecules undergo can often be explained by studying their FRONTIER ORBITALS. – Light can be used to excite an electron from the HOMO to the LUMO.

• We can measure the energy gap for the HOMO LUMO excitation. HOW?

• In general, HOW does that give us information about a molecule’s structure?


• Recall the Markovnikov addition of H–X to a C=C double bond from Section 9.3.

• With conjugated dienes as the substrate, two products are observed.

– What is the first step of the mechanism?

17.4 Electrophilic Addition


• The pi electrons attack the acid to give the most stable carbocation.

• Which product is more stable?

• Which would be formed fastest?


• Predict both MAJOR products for the reaction below. Pay close attention to stereochemistry.


• The ratio of 1,2 vs. 1,4 addition is often temperature dependant.

17.5 Thermodynamic Control vs. Kinetic Control


– Why are the products unequal in free energy?

– The 1,2 addition should occur more often regardless of temperature. WHY?

– Explain why high temperatures preferably yield the 1,4-adduct.


• Predict the MAJOR product for the following reactions.


Klein, Organic Chemistry

1e

Section: 17.5 What are the major products of the following

reaction?

A. B. C. D.

Br2, -78oC

Br

BrBr

Br

BrBr

BrBr

• Many polymerization reactions rely on 1,4 addition.

• Give a reasonable mechanism for the cationic polymerization.


• Pericyclic reactions occur without ionic or free radical intermediates.

• There are three main types of pericyclic reactions:

1. CYCLOADDITION reactions:

– How is it an ADDITION reaction?

17.6 Introduction to Pericyclic Reactions


• There are three main types of pericyclic reactions:

2. ELECTROCYCLIC reactions:

3. SIGMATROPIC rearrangements:

– What is the difference between an ADDITION and a REARRANGEMENT?



• Changes in the number of pi and sigma bonds distinguish the pericyclic reactions from one another.

• Pericyclic reactions have four general features

1. The reaction mechanism is concerted. It proceeds without any intermediates.

2. The mechanism involves a ring of electrons moving around a closed loop.

3. The transition state is cyclic.

4. The polarity of the solvent generally has no effect on the reaction rate. WHY is that significant?


• Diels-Alder reactions can be very useful.

– They allow a synthetic chemist to quickly build molecular complexity.

– What is meant by [4+2] cycloaddition?

17.7 Diels-Alder Reactions


• Like all pericyclic reactions, the mechanism is concerted.

– The arrows could be drawn in a clockwise or counterclockwise direction.


• Why do the products generally have lower free energy?


• Most Diels-Alder reactions are thermodynamically favored at low and moderate temperatures.

• At temperatures above 200°C, the retro Diels-Alder can predominate.

• Will the forward reaction probably be favored or disfavored by enthalpy? entropy?


• The pi sigma conversion provides a (–)ΔH.


• ΔS should be (–):

– Two molecules combine to form ONE.

– A ring (with limited rotational freedom) forms.

• What will the sign be (+/–) for the –TΔS term?


• Given the signs for ΔH and –TΔS, how should temperature affect reaction spontaneity (favorability of reactant vs. product)?

• Are there any disadvantages if the temperature is too low? Think kinetics.


• In the Diels-Alder reaction, the reactants are generally classified as either the DIENE or DIENOPHILE.

– If a dienophile is not substituted with an electron withdrawing group, it will not be kinetically favored (a lot of activation energy or high temperature is required).

– However, high temperatures do not favor the products thermodynamically.


• When an electron withdrawing group is attached to the dienophile, the reaction is generally spontaneous.

– Show how the groups highlighted in red are electron withdrawing using resonance and induction where appropriate.


• Diels-Alder reactions are stereospecific depending on whether the (E) or (Z) dienophile is used.

– Which alkene is (E) and which is (Z)?

– We will investigate the stereochemical outcome later in this chapter.


• A C C triple bond can also act as a dienophile.



1e

Section: 17.7 What is the major product of the following

reaction?

A. B. C. D.

HeatH

O

+

H

O

H

O

H

O

H

O

• Diels-Alder reactions can also be affected by DIENE structure.

– Recall that many dienes can exist in an s-cis or an s-trans rotational conformation.

– Which conformation is generally more stable? WHY?

– Diels-Alder reactions can ONLY proceed when the diene adopts the s-cis conformation.


• Dienes that can only exist in an s-trans conformation cannot undergo Diels-Alder reactions because carbons 1 and 4 are too far apart.

– Dienes that are locked into the s-cis conformation undergo Diels-Alder reactions readily.

– Cyclopentadiene is so reactive, that at room temperature, two molecules will react together.

– Show the reaction and products.


1 4

• Draw four potential bicyclic Diels-Alder products for the reaction below.

– Two of the potential stereoisomers are impossible. WHICH ones? WHY?


• The electron withdrawing groups attached to dienophiles tend to occupy the ENDO position.


Major product Minor Product

• When bicyclic systems form, the terms ENDO and EXO are used to describe functional group positioning.


• The Diels-Alder transition state that produces the ENDO product benefits from favorable pi system interactions.

– Is this a kinetic or thermodynamic argument?

– Draw an appropriate energy diagram.



1e

Section: 17.7 What is the major product of the following

reaction?

A. B. C. D.

HeatCO2Et+

CO2Et CO2Et

CO2Et

CO2Et

• In the Diels-Alder, the HOMO of one compound must interact with the LUMO of the other.

17.8 MO Description of Cycloadditions


• With an electron withdrawing group, the dienophile’s LUMO will accept electrons from the diene’s HOMO.


• The phases of the MOs align to allow for orbital overlap.

• If there is conservation of ORBITAL SYMMETRY, the process is SYMMETRY-ALLOWED.

– Note the carbons that change their hybridization from sp2 to sp3.


• Similar to a Diels-Alder ([4+2] cycloaddition), the reaction below is a [2+2] cycloaddition.

– Draw a reasonable concerted mechanism.

– Unless the reaction is symmetry-allowed, the process will not occur, so let’s analyze the MOs.


• The LUMO of one reactant must overlap with the HOMO of the other.

• WHY can’t both of the HOMOs interact together?

• The phases of the HOMO and LUMO cannot line up to give effective overlap, so the reaction is SYMMETRY-FORBIDEN.


• [2+2] cycloadditions are only possible when light is used to excite an electron.


• Determine how the number of σ and π bonds change for the representative electrocyclic reactions below.

– Explain why the equilibriums favor either products or reactants in the examples above.

17.9 Electrocyclic Reactions


• When substituents are present on the terminal carbons, stereoisomers are possible.

– Note that the use of light vs. heat gives different products.


• Often the energy present at room temperature is sufficient to promote thermal electrocyclic reactions.

– Note that with the use of heat, the configuration of the reactant determines the product formed.


• The symmetry of the HOMO determines the outcome.

– The terminal carbons rotate as they become sp3 hybridized and lobes that are in phase overlap.


• The terminal carbons rotate as they become sp3 hybridized and lobes that are in phase overlap.

• DISROTATORY rotation (one rotates clockwise and the other counterclockwise) gives the cis product.



• Use MO theory to explain the products for the reactions below.

– Will DISROTATORY or CONROTATORY rotation be necessary?


• Predict the major product for the reaction below. Pay close attention to stereochemistry.



• Under photochemical conditions, light energy excites an electron from the HOMO to the LUMO.

• What was the LUMO becomes the new HOMO.


• Will the new excited HOMO react via a disrotatory or conrotatory mode?

• Draw the expected product.


• The Woodward-Hoffmann rules for thermal and photochemical electrocyclic reactions are found in Table 17.2.


• SIGMATROPIC REARRANGEMENT is a pericyclic reaction in which one sigma bond is replaced with another.

• Note that the pi bonds switch locations.

17.10 Sigmatropic Rearrangements


• The notation for sigmatropic rearrangements is different from reactions we have seen so far.

– Count the number of atoms on each side of the sigma bonds that are breaking and forming.

– This is a [3,3] sigmatropic rearrangement.


• The reaction below is a [1,5] sigmatropic rearrangement.

• Practice with CONCEPTUAL CHECKPOINT 17.25.


• The Cope rearrangement is a [3,3] sigmatropic reaction in which all six atoms in the cyclic transition state are CARBONS.

– In general, what factors affect the spontaneity of the reaction (product favored vs. reactant favored)?


• The Claisen rearrangement is a [3,3] sigmatropic reaction in which one of the six atoms in the cyclic transition state is an OXYGEN.

– In general, what factors affect the spontaneity of the reaction?


• Two pericyclic reactions occur in the biosynthesis of vitamin D.


• If light with the NECESSARY ENERGY strikes a compound with pi bonds, an electron will be excited from the HOMO to the LUMO.

• Light energy is converted into potential energy.

17.11 UV-Vis Spectroscopy


• In general, the NECESSARY ENERGY to excite an electron from π π* (HOMO to LUMO) is either in the UV or visible region of the spectrum.

• What might happen after the electron is excited?



• UV-Visible (UV-Vis) spectroscopy gives structural information about molecules:

– A beam of light (200-800 nm) is split in two.

– Half of the beam travels through a cuvette with the analyte in solution.

– The other half of the beam travels through a cuvette with just the solvent (used as a negative control).

– The intensities of light that pass through the cuvettes are compared to determine how much light is absorbed by the analyte.



• UV-Visible (UV-Vis) spectroscopy gives structural information about molecules:

– The resulting data is plotted to give a UV-Vis absorption spectrum.

– Compounds require specific wavelengths of energy to excite. WHY?



• More conjugation gives a smaller π π* energy gap.

• The smaller the energy gap, the greater the lambda max (λmax).



• The group of atoms responsible for absorbing UV-Vis light is

known as the chromophore.

• Woodward and Fieser developed rules to predict λmax for chromophore starting with butadiene as the base.



• Woodward and Fieser developed rules to predict λmax for chromophore starting with butadiene as the base.

• The Woodward-Fieser rules are a guide to ESTIMATE λmax.



• Practice with SKILLBUILDER 17.4.



• The visible region of the spectrum (400-700 nm) is lower energy than UV radiation.

• Lycopene is responsible for the red color of tomatoes.

• β-carotene is responsible for the orange color of carrots.

17.12 Color


• The color observed by your eyes will be the opposite of what is required to cause the π π* excitation. WHY?

• Practice with CONCEPTUAL CHECKPOINT 17.31.

17.12 Color


• Bleaching agents generally work by breaking up conjugation through an addition reaction.

• Destroying long range conjugation destroys the ability to absorb colored light. WHY?

• Does bleach actually remove stains?

17.12 Color


• Rods and cones are photosensitive cells:

– Rods are the dominant receptor in dim lighting. Owls have only rods allowing them to see well at night.

– Cones are responsible for detection of color. Pigeons have only cones providing sensitive daytime vision.

• Rhodopsin is the light-sensitive compound in rods.

17.13 Chemistry of Vision


• Sources of 11-cis-retinal include vitamin A and β-carotene.



• When rhodopsin is excited photochemically, a change in shape occurs that causes a release of Ca2+ ions.

• The Ca2+ ions block channels through which billions of Na+ ions generally travel each second.

• The decrease of Na+ ion flow culminates in a nerve impulse to the brain.

• Our eyes are extremely sensitive.

– Just a few photons can cause a nerve impulse.



W 2/1 M 2/6 Due: HW14, spec03 Due: n/a Lecture Organic Chemistry 1e Section: 17.7 What is the major...

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