17.1 Classes of Dienes

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1 W 2/1 Due: HW14, spec03 Due: n/a M 2/6 Lecture HW14 grading Lect17a Conjugated π systems Lecture quiz Lect17b Lab Lab02 Qual Analysis II (cont) 7-1

2 17.1 Classes of Dienes 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. 17-2

3 17.1 Classes of Dienes 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. 17-3

4 This chapter focuses on conjugated systems. Heteroatoms may be involved in a CONJUGATED system. 17-4

5 17.2 Conjugated Dienes A sterically hindered base can be used to form dienes while avoiding the competing substitution reaction. 17-5

6 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. 17-6

7 The more s-character a carbon has, the shorter its bonds will be. WHY? 17-7

8 HOW does conjugation affect stability? 17-8

9 Rank the following compounds in order of increasing stability. 17-9

10 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 17-10

11 The highest energy conformer will not be conjugated

12 17.3 Molecular Orbital Theory 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? 17-12

13 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 17-13

14 The four pi electrons in butadiene will occupy the lowest energy MOs. HOW will that affect stability? 17-14

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

16 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 17-16

17 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

18 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? 17-18

19 17.4 Electrophilic Addition 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? 17-19

20 The pi electrons attack the acid to give the most stable carbocation. Which product is more stable? Which would be formed fastest? 17-20

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

22 17.5 Thermodynamic Control vs. Kinetic Control The ratio of 1,2 vs. 1,4 addition is often temperature dependant

23 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

24 Predict the MAJOR product for the following reactions

25 Section: 17.5 What are the major products of the following reaction? Br 2, -78 o C A. B. Br C. D. Br Br Br Br Br Br Br Klein, Organic Chemistry

26 Many polymerization reactions rely on 1,4 addition. Give a reasonable mechanism for the cationic polymerization

27 17.6 Introduction to Pericyclic Reactions 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? 17-27

28 There are three main types of pericyclic reactions: 2. ELECTROCYCLIC reactions: 3. SIGMATROPIC rearrangements: What is the difference between an ADDITION and a REARRANGEMENT? 17-28

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

30 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? 17-30

31 17.7 Diels-Alder Reactions Diels-Alder reactions can be very useful. They allow a synthetic chemist to quickly build molecular complexity. What is meant by [4+2] cycloaddition? 17-31

32 Like all pericyclic reactions, the mechanism is concerted. The arrows could be drawn in a clockwise or counterclockwise direction

33 Why do the products generally have lower free energy? 17-33

34 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? 17-34

35 The pi sigma conversion provides a ( )ΔH

36 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? 17-36

37 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

38 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

39 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

40 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

41 A C C triple bond can also act as a dienophile

42 Section: 17.7 What is the major product of the following reaction? + O H Heat A. O B. O C. O D. O H H H H Klein, Organic Chemistry

43 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

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

45 Draw four potential bicyclic Diels-Alder products for the reaction below. Two of the potential stereoisomers are impossible. WHICH ones? WHY? 17-45

46 The electron withdrawing groups attached to dienophiles tend to occupy the ENDO position. Major product Minor Product 17-46

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

48 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

49 Section: 17.7 What is the major product of the following reaction? + CO 2 Et Heat A. B. C. D. CO 2 Et CO 2 Et CO 2 Et CO 2 Et Klein, Organic Chemistry

50 17.8 MO Description of Cycloadditions In the Diels-Alder, the HOMO of one compound must interact with the LUMO of the other

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

52 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 sp 2 to sp

53 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

54 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

55 [2+2] cycloadditions are only possible when light is used to excite an electron

56 17.9 Electrocyclic Reactions 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

57 When substituents are present on the terminal carbons, stereoisomers are possible. Note that the use of light vs. heat gives different products

58 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

59 The symmetry of the HOMO determines the outcome. The terminal carbons rotate as they become sp 3 hybridized and lobes that are in phase overlap

60 17.9 Electrocyclic Reactions The terminal carbons rotate as they become sp 3 hybridized and lobes that are in phase overlap. DISROTATORY rotation (one rotates clockwise and the other counterclockwise) gives the cis product

61 Use MO theory to explain the products for the reactions below. Will DISROTATORY or CONROTATORY rotation be necessary? 17-61

62 17.9 Electrocyclic Reactions Predict the major product for the reaction below. Pay close attention to stereochemistry

63 Under photochemical conditions, light energy excites an electron from the HOMO to the LUMO. What was the LUMO becomes the new HOMO

64 Will the new excited HOMO react via a disrotatory or conrotatory mode? Draw the expected product

65 The Woodward-Hoffmann rules for thermal and photochemical electrocyclic reactions are found in Table

66 17.10 Sigmatropic Rearrangements SIGMATROPIC REARRANGEMENT is a pericyclic reaction in which one sigma bond is replaced with another. Note that the pi bonds switch locations

67 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

68 The reaction below is a [1,5] sigmatropic rearrangement. Practice with CONCEPTUAL CHECKPOINT

69 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)? 17-71

70 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? 17-72

71 Two pericyclic reactions occur in the biosynthesis of vitamin D

72 17.11 UV-Vis Spectroscopy 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

73 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? 17-75

74 17.11 UV-Vis Spectroscopy UV-Visible (UV-Vis) spectroscopy gives structural information about molecules: A beam of light ( 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

75 17.11 UV-Vis Spectroscopy 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? 17-77

76 17.11 UV-Vis Spectroscopy More conjugation gives a smaller π π* energy gap. The smaller the energy gap, the greater the lambda max (λ max )

77 17.11 UV-Vis Spectroscopy The group of atoms responsible for absorbing UV-Vis light is known as the chromophore. Woodward and Fieser developed rules to predict λ max for chromophore starting with butadiene as the base

78 17.11 UV-Vis Spectroscopy Woodward and Fieser developed rules to predict λ max for chromophore starting with butadiene as the base. The Woodward-Fieser rules are a guide to ESTIMATE λ max

79 17.11 UV-Vis Spectroscopy Practice with SKILLBUILDER

80 17.12 Color The visible region of the spectrum ( 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

81 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

82 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? 17-84

83 17.13 Chemistry of Vision 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

84 17.13 Chemistry of Vision Sources of 11-cis-retinal include vitamin A and β- carotene

85 17.13 Chemistry of Vision When rhodopsin is excited photochemically, a change in shape occurs that causes a release of Ca 2+ ions. The Ca 2+ 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

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