17-1 Klein, Organic Chemistry 1e 17-2 Klein, Organic ... · 17.5 Thermodynamic Control vs. ... 40...

• 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 1e17-1

• 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.



• 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.



• 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.



• 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.




• 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 features1. 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.



• 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.



14

• 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 EXOare 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.




• 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.



• 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).



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

known as the chromophore.

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



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

• The Woodward‐Fieser rules are a guide to ESTIMATEλmax.






• 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 Klein, Organic Chemistry 1e 17-2 Klein, Organic ... · 17.5 Thermodynamic Control vs. ... 40...

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