w 2/1 m 2/6 due: hw14, spec03 due: n/a lecture organic chemistry 1e section: 17.7 what is the major...
TRANSCRIPT
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
Copyright 2012 John Wiley & Sons, Inc. 17-3
• This chapter focuses on conjugated systems.
– Heteroatoms may be involved in a CONJUGATED system.
Copyright 2012 John Wiley & Sons, Inc. 17-4
• A sterically hindered base can be used to form dienes while avoiding the competing substitution reaction.
17.2 Conjugated Dienes
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• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-6
• The more s-character a carbon has, the shorter its bonds will be. WHY?
Copyright 2012 John Wiley & Sons, Inc. 17-7
• Rank the following compounds in order of increasing stability.
Copyright 2012 John Wiley & Sons, Inc. 17-9
• 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
Copyright 2012 John Wiley & Sons, Inc. 17-10
• 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
Copyright 2012 John Wiley & Sons, Inc. 17-12
• 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
Copyright 2012 John Wiley & Sons, Inc. 17-13
Orbitals shorthand
• The four pi electrons in butadiene will occupy the lowest energy MOs. HOW will that affect stability?
Copyright 2012 John Wiley & Sons, Inc. 17-14
• MO also explains why the central C–C single bond is shorter and stronger than a typical C–C single bond.
Copyright 2012 John Wiley & Sons, Inc. 17-15
• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-17
• 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?
Copyright 2012 John Wiley & Sons, Inc. 17-18
• 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
Copyright 2012 John Wiley & Sons, Inc. 17-19
• The pi electrons attack the acid to give the most stable carbocation.
• Which product is more stable?
• Which would be formed fastest?
Copyright 2012 John Wiley & Sons, Inc. 17-20
• Predict both MAJOR products for the reaction below. Pay close attention to stereochemistry.
Copyright 2012 John Wiley & Sons, Inc. 17-21
• The ratio of 1,2 vs. 1,4 addition is often temperature dependant.
17.5 Thermodynamic Control vs. Kinetic Control
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– 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.
Copyright 2012 John Wiley & Sons, Inc. 17-23
• Predict the MAJOR product for the following reactions.
Copyright 2012 John Wiley & Sons, Inc. 17-24
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.
Copyright 2012 John Wiley & Sons, Inc. 17-26
• 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
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• There are three main types of pericyclic reactions:
2. ELECTROCYCLIC reactions:
3. SIGMATROPIC rearrangements:
– What is the difference between an ADDITION and a REARRANGEMENT?
Copyright 2012 John Wiley & Sons, Inc. 17-28
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• 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?
Copyright 2012 John Wiley & Sons, Inc. 17-30
• 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
Copyright 2012 John Wiley & Sons, Inc. 17-31
• Like all pericyclic reactions, the mechanism is concerted.
– The arrows could be drawn in a clockwise or counterclockwise direction.
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• Why do the products generally have lower free energy?
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• 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?
Copyright 2012 John Wiley & Sons, Inc. 17-34
• 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?
Copyright 2012 John Wiley & Sons, Inc. 17-36
• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-37
• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-38
• 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.
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• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-40
Klein, Organic Chemistry
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.
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• 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.
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1 4
• Draw four potential bicyclic Diels-Alder products for the reaction below.
– Two of the potential stereoisomers are impossible. WHICH ones? WHY?
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• The electron withdrawing groups attached to dienophiles tend to occupy the ENDO position.
Copyright 2012 John Wiley & Sons, Inc. 17-46
Major product Minor Product
• When bicyclic systems form, the terms ENDO and EXO are used to describe functional group positioning.
Copyright 2012 John Wiley & Sons, Inc. 17-47
• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-48
Klein, Organic Chemistry
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
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• With an electron withdrawing group, the dienophile’s LUMO will accept electrons from the diene’s HOMO.
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• 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.
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• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-53
• 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.
Copyright 2012 John Wiley & Sons, Inc. 17-54
• [2+2] cycloadditions are only possible when light is used to excite an electron.
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• 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
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• When substituents are present on the terminal carbons, stereoisomers are possible.
– Note that the use of light vs. heat gives different products.
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• 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.
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• The symmetry of the HOMO determines the outcome.
– The terminal carbons rotate as they become sp3 hybridized and lobes that are in phase overlap.
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• 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.
17.9 Electrocyclic Reactions
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• Use MO theory to explain the products for the reactions below.
– Will DISROTATORY or CONROTATORY rotation be necessary?
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• Predict the major product for the reaction below. Pay close attention to stereochemistry.
17.9 Electrocyclic Reactions
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• Under photochemical conditions, light energy excites an electron from the HOMO to the LUMO.
• What was the LUMO becomes the new HOMO.
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• Will the new excited HOMO react via a disrotatory or conrotatory mode?
• Draw the expected product.
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• The Woodward-Hoffmann rules for thermal and photochemical electrocyclic reactions are found in Table 17.2.
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• 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
Copyright 2012 John Wiley & Sons, Inc. 17-68
• 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.
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• The reaction below is a [1,5] sigmatropic rearrangement.
• Practice with CONCEPTUAL CHECKPOINT 17.25.
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• 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)?
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• 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?
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• Two pericyclic reactions occur in the biosynthesis of vitamin D.
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• 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
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• 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?
17.11 UV-Vis Spectroscopy
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• 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.
17.11 UV-Vis Spectroscopy
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• 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?
17.11 UV-Vis Spectroscopy
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• More conjugation gives a smaller π π* energy gap.
• The smaller the energy gap, the greater the lambda max (λmax).
17.11 UV-Vis Spectroscopy
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• 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.
17.11 UV-Vis Spectroscopy
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• 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.
17.11 UV-Vis Spectroscopy
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• Practice with SKILLBUILDER 17.4.
17.11 UV-Vis Spectroscopy
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• 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
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• 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
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• 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
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• 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
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• Sources of 11-cis-retinal include vitamin A and β-carotene.
17.13 Chemistry of Vision
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• 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.
17.13 Chemistry of Vision
Copyright 2012 John Wiley & Sons, Inc. 17-87