17.1 classes of dienes - wordpress.com · 2014. 6. 26. · – conjugated: pi bond overlap extends...

• 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. Klein, Organic Chemistry 1e 17-1

• 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.



• This chapter focuses on conjugated systems.

• Heteroatoms may be involved in a CONJUGATED system.

• Draw a picture of the molecule shown to the right indicating where the pi bonds are and how they overlap.

• Practice with CONCEPTUAL CHECKPOINT 17.1.



N

• 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. – What is a pm?

• The hybridization of a carbon affects its bond length.



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




• HOW do heats of hydrogenation provide information about stability?

• WHY is energy released upon hydrogenation?



• HOW does conjugation affect stability? • Practice CONCEPTUAL CHECKPOINTs 17.4 and 17.5.



• Rank the following compounds in order of increasing stability.



A B C D

• In general, single bonds will freely rotate. • The two most stable rotational conformations for

butadiene are the s-cis and s-trans. • What does the s of s-cis and s-trans stand for?

• Are there any other rotational conformations that you think might be possible?



• The s-cis and s-trans both allow for full pi system overlap.

• Other possible conformations will be higher in energy. • WHY?



• Why is s-trans lower in energy?



• About 98% of the molecules are in the s-trans form.



• The highest energy conformer will not be conjugated.



• Let’s review molecular orbital (MO) theory: – 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 four pi electrons in butadiene will occupy the lowest energy MOs. HOW will that affect stability?



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



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

HOMO and LUMO?

• 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 a conjugated diene as the substrate, two products are observed.

17.4 Electrophilic Addition


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

• What intermediate would result if the H were added to any of the other carbons in the molecule?



• The halide can attack the

resonance stabilized carbocation at either site that is sharing the (+) charge.

• How is 1,2-addition different from 1,4-addition?

• Practice with SKILLBUILDER 17.1.



• The addition of bromine to a diene also gives both 1,2 and 1,4 addition.

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



H-Cl

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

• The energy diagram must be examined to see WHY temperature affects the product distribution.

17.5 Thermodynamic Control vs. Kinetic Control


• Why are the products unequal in free energy?



• Draw a structure for each of the transition states of the nucleophilic attack.



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



• Explain how high temperatures yield one product while low temperatures yield the other.



• Predict the MAJOR product for the following reactions.



H-Br

0 C

H-Br40

C° °

• Many polymerization reactions rely on 1,4 addition.

• Give a reasonable mechanism for the cationic polymerization.

• Practice with

CONCEPTUAL CHECKPOINT 17.13.



• 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?



• 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?





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

• Diels-Alder reactions can be very useful.

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

• What is meant by [4+2] cycloaddition? • Like all pericyclic reactions, the mechanism is concerted.

17.7 Diels-Alder Reactions


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

• Can you draw a reasonable transition state?



• 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 reaction probably be favored or disfavored by 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.




• Predict products for the following reactions.



O

high temperature

retro Diels-Alder

NC

• 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 molecule 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?



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



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



Major product Minor Product

• 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. • Practice with CONCEPTUAL CHECKPOINT

17.19.



• Note the MO descriptions below.

17.8 MO Description of Cycloadditions


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



• 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.

• Draw a picture that shows how the reaction is now symmetry allowed.




• 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.



Heat

• 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.



• Make the same MO analysis to predict the symmetry-allowed product for the reaction below. Pay close attention to stereochemistry.



Light

• The disrotatory nature of the photochemical [2+2] electrocyclic ring-closing is difficult to observe directly because the thermodynamics favor ring opening.

• The ring-opening reaction gives products that result from disrotatory rotation. Predict products below.



Light

• 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.




• 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? • Practice with CONCEPTUAL CHECKPOINTs

17.27 and 17.28.



• 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).



217 258 290

• 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.



• 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?


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.



17.1 classes of dienes - wordpress.com · 2014. 6. 26. · – conjugated: pi bond overlap extends...

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