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Chapter 14 Lecture Conjugated Compounds and Ultraviolet Spectroscopy

Organic Chemistry Vol II 9th ed John McMurry

E Bryant PhD Houston Community College Houston, TX

1 Introduction: Alkenes

. Hydrocarbons with carbon– carbon double bonds . Carbon–carbon double bond is the functional group of alkenes . Carbon–carbon double bond gives this group its reactivity . Called unsaturated compounds

In a ring, the double bond is assumed to be between carbon 1 and carbon 2.

CH3 3 1 CH3 2

2 1

1-methylcyclopentene 3-methylcyclopentene

. Give the double bonds the lowest numbers possible . Use di-, tri-, tetra- before the ending -ene to specify how many double bonds are present

. Also called geometric isomerism . Similar groups on same side of double bond, alkene is cis . Similar groups on opposite sides of double bond, alkene is trans . Not all alkenes show cis-trans isomerism . All cycloalkenes are assumed to be cis unless otherwise specifically named trans © 2013 Pearson Education, Inc. 6 Introduction: Alkene E-Z Nomenclature . Use the Cahn–Ingold–Prelog rules to assign priorities to groups attached to each carbon in the double bond 1. Look at the four atoms directly attached to the stereogenic center (X) 2. Assign priorities based on atomic number to all four atoms. Priority 1 is assigned to the atom or group of highest atomic number, priority 4 to the lowest 3. If two or more atoms are identical look at all the atoms directly attached to the identical atoms in questions Compare the highest priority atoms, if the highest priority atoms have the same priority,, then compare the second highest priority atoms and so on 4. If a difference still can not be found, move out to the next highest priority group (A- 1 and B-1 in the diagram) and repeat the process 5. Multiple bonds are considered as an equivalent number of single bonded atoms 7 Introduction: Alkene E-Z Nomenclature 1. Assign priorities to groups by the Cahn–Ingold–Prelog rules 2. If high-priority groups are on the same side, the name is Z (for zusammen) 3. If high-priority groups are on opposite sides, the name is E (for entgegen)

8 Introduction: Alkene E-Z Nomenclature Example . Assign priority to the substituents according 1 2 to their atomic number (1 is highest priority). . If the highest priority groups are on opposite 1 2 sides, the isomer is E. . If the highest priority E-1-bromo-1-chloropropene groups are on the same side, the isomer is Z.

. Double bonds outside the ring can show stereoisomerism

. If there is more than one double bond in the molecule, the stereochemistry of all the double bonds should be specified.

. Molecules contains more than one double bond

• Double bonds separated by two or more single bonds are isolated • Double bonds separated by one single bond are conjugated • Conjugated double bonds are more stable than isolated ones

Is there a difference between isolated double bonds and conjugated double bonds?

Yes, conjugated polyenes are more stable than isolated double bonds

Determine by Hhydrog

14 What is Heat of Hydrogenation?

. Compare heat given off on hydrogenation: Ho . Evaluate heat given off when C=C is converted to C-C . Less stable isomer is higher in energy . More stable alkene gives off less heat

Hhydrog . trans-butene generates 5 kJ Hhydrog less heat than cis-butene

15 Stability of Conjugated Compounds

. Conjugated dienes are more stable than nonconjugated based on heats of hydrogenation

. Hydrogenating 1,3-butadiene produces 17 kJ/mol less heat than 1,4-pentadiene

16 Stability of Conjugated Compounds . What accounts for the stability of conjugated dienes? . Valence bond theory states it is due to orbital hybridization

17 Stability of Conjugated Compounds

. Typical C-C bonds result from overlap of sp3 orbitals on both carbons . Conjugated C-C bonds result from overlap of sp2 orbitals on both carbons . sp2 orbitals have more s character (33% s) than sp3 orbitals (25% s) . Greater stability of a conjugated diene results from the greater amount of s character in the orbitals forming the C-C single bond

18 Stability of Conjugated Compounds . What accounts for the stability of conjugated dienes? . Valence bond theory states it is due to orbital hybridization . Molecular orbital theory states it is due to interaction between the  orbitals of the two double bonds

19 Stability of Conjugated Compounds

1,3-Butadiene

. Small amount of overlap across the central C2—C3 bond, giving it a partial double bond character . Electrons are delocalized over the molecule . The C2—C3 single bond is shorter than 1.54 Å

21 Allylic Carbon

22 The Allylic Position

. The allylic carbon is the one directly attached to an sp2 carbon . Allylic cations are stabilized by resonance

The positive charge is delocalized over two carbons by resonance, giving the allylic cation more stability than nonconjugated cations

. Stability of 1 allylic  2 carbocation . Stability of 2 allylic  3 carbocation

26 1,2- and 1,4-Addition to Conjugated Dienes

. Electrophilic addition to the double bond produces the most stable intermediate . For conjugated dienes, the intermediate is a resonance-stabilized allylic cation . Nucleophile adds to either carbon 2 or 4, both of which have the delocalized positive charge

27 Conjugate Addition: 1,2- Addition and 1,4-Addition

Addition by Markovnikov Rule

Addition by allylic cation stabilization

. Conjugated dienes undergo electrophilic addition . Products a mixture of 1,2 and 1,4 addition . Numbers refer to which carbon the addition occurs on 28 Conjugate Addition: 1,2- Addition and 1,4-Addition

. Electrophilic addition . 1,2-addition is by Markovnikov Rule . 1,4 addition is by allylic cation stabilization

29 Conjugate Addition: 1,2-Addition and 1,4-Addition

30 Example 14-1 Give the structures of the likely products from reaction of 1 equivalent of HCL with 2-methyl- 1,3,-cyclohexadiene. Show both 1,2 and 1,4 adducts. SOLUTION 1. Protonate the two ends of the diene and draw resonance forms of the 2 allylic carbocations that result 2. React each resonance form with Cl-

Most likely product More stable form of protonation Allylic cation more stable 31 Electrophilic Addition to Conjugated Dienes

Kinetic or Thermodynamic Reaction

32 Conjugate Addition: Kinetic versus Thermodynamic Control

. The percent of adduct available differs at different temperatures . At low temperatures, Markovnikov Rule product is in abundance . At higher temperature, the product ratio changes and the 1,4 adduct predominates 33 Conjugate Addition: Kinetic versus Thermodynamic Control

1,3 Butadiene

Why would increasing the temperature after the product is formed at lower temperatures change the ratio of the products formed?

34 Conjugate Addition: Kinetic versus Thermodynamic Control

. Kinetic control is an irreversible reaction that depends on relative rates, not stability . Thermodynamic control is reversible reaction that depends on stability not on relative rates 35 Conjugate Addition: Kinetic versus Thermodynamic Control

. Transition state for the 1,2-addition

has a lower Ea because it is a more stable secondary carbocation

37 Kinetic Control at – 80oC

. Transition state for the 1,2-addition

has a lower Ea because it is a more stable secondary carbocation . The 1,2-addition will be the faster addition at any temperature . The nucleophilic attack of the bromide on the C2 allylic carbocation is irreversible at this low temperature . The product that forms faster predominates (kinetic product) . Because the kinetics of the reaction determines the product, the reaction is said to be under kinetic control

38 Thermodynamic Control at 40oC

. The 1,2-addition is still the faster addition, but at 40oC, the bromide attack is reversible . The 1,2-product ionizes back to the allylic cation . At 40oC an equilibrium is established, which favors the most stable product . The 1,4-addition is the most stable product (thermodynamic product) because it has a more substituted double bond . Because the thermodynamics of the reaction determines the product, the reaction is said to be under thermodynamic control

Diels-Alder Reaction

40 Diels–Alder Reaction

. The reaction is between a diene and an electron-deficient alkene or alkyne (dienophile)

dienophile diene

41 Diels–Alder Reaction

. Conjugate dienes combine with alkenes to form six- membered cyclic compounds . Reaction called a cycloaddition . The formation of the ring involves no intermediate (concerted formation of two bonds)

42 Diels–Alder Reaction

. Also called a [4 + 2] cycloaddition because a ring is formed by the interaction of four pi electrons of the alkene with two pi electrons of the alkene or alkyne . Two more sigma bonds . Two fewer pi bonds

43 Diels–Alder Reaction: Examples

44 s-Cis Conformation of the Diene

. Positions of the two double bonds around the single bond in the diene are “cis” or “trans” to each other . Conformations are called s-cis and s-trans (“s” stands for “single bond”) . s-trans conformation is 12 kJ/mol more stable than the s-cis

45 Diels–Alder Reaction: Stereochemical Requirements

. Diene’s C1 and C4 p orbitals must overlap with dienophile’s p orbitals to form new sigma bonds . Dienes react in the s-cis conformation in the Diels-Alder reaction

46 Diels–Alder Reaction: Orbital Overlap

47 Diels–Alder Reaction: Mechanism

. One-step, concerted mechanism . A diene reacts with an electron-poor alkene (dienophile) to give cyclohexene or cyclohexadiene rings

48 Diels–Alder Reaction: Mechanism

Substituents on diene or dienophile affect mechanism due to resonance structures

49 Diels–Alder Reaction: Mechanism

50 Diels–Alder Reaction: Mechanism

51 A Diels–Alder product always contains one more ring than the reactants. The two ends of the diene form new bonds to the ends of the dienophile. The center (formerly single) bond of the diene becomes a double bond. The dienophile’s double bond becomes a single bond (or its triple bond becomes a double bond).

52 Example 14-2 Predict the product of the following Diels-Alder reaction:

SOLUTION 1. Draw diene with two double bonds near dienophile double bonds 2. Draw new bonds between diene and dienophile and convert diene bond types

s-cis diene

Attached hydrogens from dienophile are cis because dienophile double bond was cis 53

Diels-Alder Rate of Reaction

54 Diels–Alder Reaction: Effect of Substituents on Rate

. Reaction rate increased by electron releasing substituents on the diene . Increase the electron rich character of the diene . Substituents are on the carbon atoms of the diene. . Electron donating groups increase electron density . Electron withdrawing groups decrease electron density

http://www.organicchem.org/oc2web/lab/exp/da/dades.html 55 Diels–Alder Reaction: Effect of Substituents on Rate

. Cyclopentadiene undergoes the Diels–Alder reaction readily because of its fixed s-cis conformation. . When the diene is sterically hindered, the reaction slows down even though the conformation can be s-cis. . s-trans dienes cannot undergo the Diels–Alder reaction.

56 Diels–Alder Reaction: Effect of Substituents on Rate

. Reaction rate increased by electron withdrawing substituents on the dienophile . Increase electron deficient character of dienophile . Substituents are on the carbon atom of the alkene . Electron withdrawing groups decrease electron density . Electron donating groups increase electron density

http://www.organicchem.org/oc2web/lab/exp/da/dades.html 57 Diels–Alder Reaction: Effect of Substituents on Rate

very slow

. Electron-withdrawing substituents on dienophile . Reaction occurs rapidly if dienophile has an electron withdrawing group on allylic carbon

Diels-Alder Stereochemistry

59 Diels–Alder Reaction: Stereochemistry

. Reaction is stereospecific, product is a single stereoisomer . Stereochemistry of dienophile is retained

60 Diels–Alder Reaction: Bicyclic Stereochemistry

1 carbon bridge anti to larger bridge exo

2 carbon bridge endo syn to larger bridge

Stereochemistry for bicyclic products described by the terms exo and endo

61 Diels–Alder Reaction: Bicyclic Stereochemistry

. Diene and dienophile orient to form the endo product . Orbital overlap is greater when reactants lie directly on top of each other . Electron withdrawing substituent on dienophile is underneath the diene double bond

62 Diels–Alder Reaction: Bicyclic Stereochemistry

p orbitals of the electron-withdrawing groups on the dienophile have a secondary overlap with the p orbitals of C2 and C3 in the diene, which make the endo state more stable than the exo 63 Diels–Alder Reaction: Bicyclic Stereochemistry Examples

64 Pericyclic Reactions

. Diels–Alder reaction is an example of a pericyclic reaction . MOs must overlap constructively to stabilize the transition state

65 DIENE POLYMERS

66 Diene Polymers: Natural and Synthetic Rubbers

. Conjugated dienes can be polymerized . Product cis or trans configuration . Polymerization: 1,4 addition of growing chain to conjugated diene monomer

67 Diene Polymers: Natural and Synthetic Rubbers Rubber . Naturally occurring diene polymer . Produced by more than 400 different plants . Latex, a crude form of rubber

68 Diene Polymers: Natural and Synthetic Rubbers

Synthetic rubber . Similar to natural material . Cis and trans material can be made . Synthetic alternatives include neoprene, polymer of 2-chloro-1,3-butadiene

69 Diene Polymers: Natural and Synthetic Rubbers

Vulcanization of rubber . Crosslinking rubber chains by forming carbon sulfur bonds . Hardens natural and synthetic rubbers . Degree of hardening can be varied

70 ULTRAVIOLET SPECTROSCOPY

71 How to determine identity of unknown solution . Mass spectrometry – molecular size and formula . Infrared spectroscopy – functional groups . NMR spectroscopy – carbon-hydrogen framework . Ultraviolet (UV) spectroscopy – nature of conjugated  electron system

72 Ultraviolet Spectroscopy: Electromagnetic Range

73 Ultraviolet Spectroscopy: Electromagnetic Range

Energy promotes an electron from a low energy orbital to a high energy molecule

74 Ultraviolet Spectroscopy: Electron Excitation

Energy promotes an electron from a low energy orbital to a high energy molecule

75 Ultraviolet Spectroscopy: Electron Excitation

Absorption wavelength that excites electron is recorded

76 Ultraviolet Spectroscopy: Operation

. The spectrometer measures the intensity of a reference beam through solvent only (Io) and the intensity of a beam through a solution of the sample (I)

. Absorbance is the log of the ratio Io/I 77 Ultraviolet Spectroscopy: Data

Formula used to calculate absorbance

78 Ultraviolet Spectroscopy: Spectrum

Ultraviolet Spectrum

79 Ultraviolet Spectroscopy: Data

.Amount of UV light absorbed . = molar absorptivity (intensive property) .A = Absorbance .C = concentration of sample in mol/L . l = sample pathlength in cm

80 Ultraviolet Spectroscopy: Data

Solve for c to determine the concentration of your sample!

81 Ultraviolet Spectroscopy: Effect of Conjugation . Conjugated dienes have MOs that are closer in energy . A compound that has a longer chain of conjugated double bonds absorbs light at a longer wavelength . Remember: longer wavelength means lower energy

82 Ultraviolet Spectroscopy: Data

UV absorption maxima of some representative conjugated molecules

83 Some good rules of thumb: An additional conjugated C═C increases lmax about 30 to 40 nm; an additional alkyl group increases it about 5 nm. Useful base values:

84 Example The molar absorptivity of 2,4-dimethyl-2,4-hexadiene in methanol is 13,100 M-1cm-1. What concentration of this diene in methanol is required to give an absorbance of 1.6? Assume a light path of 1.00 cm. Calculate concentration is units of: (a) moles per liter (b) milligrams per milliliter

SOLUTION (a) Solve the Beer-Lambert equation for concentration and substitute appropriate values for length, absorbance and molar absorptivity

A 1.6 c    1.22 104 mol / L l  1.00 cm 13,100 Lmol1 cm 1

(b) The molecular weight of 2,5-dimethyl-2,4-hexadiene is 110 g/mol. The concentration of the sample in milligrams per milliliter is mol 110g 1 L 1000 mg 1.22 104     1.34 102 mg / mL L mol 1000 mL g

85 86 Identify the structure with the isolated double bonds.

a) Penta-1,2-diene b) trans-Penta-1,3-diene c) Penta-1,4-diene d) Pent-1-yne

87 Identify the structure with the isolated double bonds.

a) Penta-1,2-diene b) trans-Penta-1,3-diene c) Penta-1,4-diene d) Pent-1-yne

Explanation: Isolated double bonds are separated by two or more single bonds.

88 Identify the alkene with the smallest heat of hydrogenation.

a) Penta-1,2-diene b) trans-penta-1,3-diene c) Penta-1,4-diene d) Pent-1-yne

89 Identify the alkene with the smallest heat of hydrogenation.

a) 1,2-Pentadiene b) trans-1,3-Pentadiene c) 1,4-Pentadiene d) 1-Pentyne

Explanation: Conjugated dienes have the lowest heat of hydrogenation.

90 Identify the alkene with the highest heat of hydrogenation.

a) Penta-1,2-diene b) trans-Penta-1,3-diene c) Penta-1,4-diene d) Pent-1-yne

91 Identify the alkene with the highest heat of hydrogenation.

a) Penta-1,2-diene b) trans-Penta-1,3-diene c) Penta-1,4-diene d) Pent-1-yne

Explanation: Allenes, also known as cumulenes, have the highest heat of hydrogenation.

92 Identify the highest energy orbital of buta-1,3-diene.

a) 1 bonding MO b) 2 bonding MO * c) 3 antibonding MO * d) 4 antibonding MO

93 Identify the highest energy orbital of buta-1,3-diene.

a) 1 bonding MO b) 2 bonding MO * c) 3 antibonding MO * d) 4 antibonding MO

Explanation: * The 4 antibonding MO has the highest energy.

94 + Name CH2CH-CH2 .

a) Allylic anion d) Benzylic cation b) Allylic radical e) Benzylic anion c) Allylic cation

95 + Name CH2CH-CH2 .

a) Allylic anion d) Benzylic cation b) Allylic radical e) Benzylic anion c) Allylic cation

Explanation: The allylic cation has a double bond next to a positive carbon.

96 How does HX add to a conjugated diene?

a) 1,2-Addition b) 1,3-Addition c) 2,3-Addition d) 1,4-Addition e) 1,2- and 1,4-addition

97 How does HX add to a conjugated diene?

a) 1,2-Addition b) 1,3-Addition c) 2,3-Addition d) 1,4-Addition e) 1,2- and 1,4-addition

Explanation: Both products are formed by addition of HX to a conjugated diene.

98 Explain the reaction of HBr with

CH2CHCHCH2. a) 1,2-Product is the kinetic product; 1,4-product is the thermodynamic product. b) 1,4-Product is the kinetic product; 1,2-product is the thermodynamic product. c) 1,2- and 1,4-products are thermodynamic products. d) 1,2- and 1,4-products are kinetic products.

99 Explain the reaction of HBr with

100 Identify the best nucleophile.

a) An alkene b) An alkyne c) A conjugated diene with an electron-withdrawing group d) A conjugated diene with an electron-donating group e) An isolated diene

101 Identify the best nucleophile.

a) An alkene b) An alkyne c) A conjugated diene with an electron-withdrawing group d) A conjugated diene with an electron-donating group e) An isolated diene

102 List criteria for the Diels–Alder reaction. a) Four electrons in the conjugated diene and two electrons in the dienophile b) Two electrons in the conjugated diene and four electrons in the dienophile c) Eight electrons in the conjugated diene and four electrons in the dienophile d) Four electrons in the conjugated diene and eight electrons in the dienophile

103 List criteria for the Diels–Alder reaction. a) Four electrons in the conjugated diene and two electrons in the dienophile b) Two electrons in the conjugated diene and four electrons in the dienophile c) Eight electrons in the conjugated diene and four electrons in the dienophile d) Four electrons in the conjugated diene and eight electrons in the dienophile

104 +

a) 4-Ethylcyclohexene b) Ethylbenzene c) 5-Ethylcyclohexa-1,3-diene d) 1-Ethylcyclohexene

105 +

a) 4-Ethylcyclohexene b) Ethylbenzene c) 5-Ethylcyclohexa-1,3-diene d) 1-Ethylcyclohexene

Explanation: A cyclohexene is formed from a diene and alkene in a Diels–Alder reaction.

106 Identify the groups that increase the wavelength in a UV spectrum. a) Alkyl groups, but not conjugated double bonds b) Conjugated double bonds, but not alkyl groups c) Both alkyl groups and conjugated double bonds d) Neither alkyl groups nor conjugated double bonds

107 Identify the groups that increase the wavelength in a UV spectrum. a) Alkyl groups, but not conjugated double bonds b) Conjugated double bonds, but not alkyl groups c) Both alkyl groups and conjugated double bonds d) Neither alkyl groups nor conjugated double bonds

108 Identify the alkene with the highest UV wavelength.

a) Hex-1-ene b) Hexa-1,4-diene c) Hexa-2,4-diene d) Hexa-1,3,5-triene

109 Identify the alkene with the highest UV wavelength.

a) 1-Hexene b) Hexa-1,4-diene c) Hexa-2,4-diene d) Hexa-1,3,5-triene

Explanation: The wavelength increases with the addition of conjugated double bonds.

110 Let’s Work a Problem

Explain why 2,3-di-tert-butyl-1,3-butadiene does not undergo Diels-Alder reactions.

111 Answer This compound has the tertiary butyl groups in a cis relationship with each other. This brings about a steric strain from the proximity of the bulky substituents. As a result, the compounds undergoes a change to a trans conformation to relieve strain, but Diels-Alder reaction cannot take place.

112