conjugated systems and aromaticityww2.chemistry.gatech.edu/~cf77/8833/set1.pdf · 1-1 conjugated...

1-1

Conjugated Systems and Aromaticity

1-2

Allylic SubstitutionAt low temperature, propene reacts with Br2 or Cl2 to give the 1,2-addition product:

At high temperature, or under very dilute conditions, a substitution reaction of the allylic hydrogen atom occurs:

10.8

F

Br�–Br Br

BrCCl4

Br�–Br

high temp.

Br + H-Br

CHEM 2312 Spring 2016 Notes: C.J. Fahrni

1-3

Chain Initiating Step:

Chain Propagation Steps:

10.8A

1-4

Allylic Bromination with N-Bromosuccinimide

Mechanism:

10.8B

Flight or ROOR

N

O

O

Br+ Br N

O

O

H+


1-5

energy

0

A -bond is formed through overlap of two adjacent p-orbitals:

MO Description of Ethene

1-6

energy

0

In the allyl radical, 3 p-orbitals overlap to form a set of 3 molecular orbitals with -symmetry

The three electrons of the -bond are delocalized over all three carbon atoms

MO Description of the Allyl Radical13.2


1-7

A single Lewis structure cannot illustrate the delocalized nature of the -electrons. Instead, we use a combination of two structures to describe the allyl radical:

Note: Neither of the two structures is an accurate description of the allyl radical, only the hybrid of both structures

Resonance Description

13.2B

H

H

HH

H

1-8

The neighboring benzene -system stabilizes the benzylic radical:

Resonance Description: Benzylic Radical

10.9

H H H H


1-9

The homolytic bond dissociation energy is a good measure for the relative stability of radicals

Relative Stability of Radicals

10.8C/Fig.10.3

R H R + H H°

H

H H H3C

CH3

HH3C

H3C CH3

H H H

H H H

H

369 kJmol�–1 400 kJmol�–1 413 kJmol�–1 423 kJmol�–1 465 kJmol�–1H°

CH2

CH2CH3> > > >

allylic benzylic tertiary secondary primary

1-10

Problem (Practice Problem13.1): What product(s) would you expect to obtain if cyclohexene labeled at C-3 with 13C would be subjected to allylic bromination?

NBS, ROORCH H

heat

13


1-11

Problem (13.26): What product(s) with formula C5H9Br would you expect for the reaction of 1-pentene with NBS?

1-12

Challenge:

Designing Multistep Syntheses: Retrosynthesis

productstartingmaterial

inexpensive,commercially available

reaction A reaction B reaction C

Solution(s): Retrosynthetic Analysis product

intermediateB-1

intermediateB-2

intermediateA-1

intermediateA-2

intermediateA-3

intermediateA-4

intermediateA-5

startingmaterial

startingmaterial

startingmaterial

S-1 S-2 S-3


1-13

Problem (13.22e): How would you carry out the following (multistep) transformation?

Br

1-14

energy

0

Similar to the allyl radical, the allyl cation is also resonance stabilized:

Allyl Cation

13.3

CH2

CH2CH3~ > >

allylicbenzylic tertiary secondary primary

>


1-15

Carbocations as Reactive Intermediates

13.3, 15.12, 15.15

Mechanism:

Mechanism:

F

F

BrH2O

Br H2O

1-16

Regiochemistry

8.1-2, 15.13A, 15.13B, 15.15

Mechanism:

The relative stability of the carbocation intermediates dictates the product distribution:

FHBr


1-17

1. Resonance structures exist only on paper

2. Use double-headed arrows to connect individual resonance structures

3. Only move non-bonding electrons or -bond electrons. The connectivity (=constitution) of the molecule remains unchanged.

4. All structures must be proper Lewis structures. Strictly follow the octet rule for 2nd period elements.

5. All resonance structures must have the same number of unpaired electrons (and identical net charge)

6. All atoms that are part of the delocalized -system must lie in a plane

7. Equivalent resonance structures make equal contributions to the hybrid and are associated with a large resonance stabilization

8. The more stable a resonance structure is, the greater its contribution to the hybrid

Rules for Writing Resonance Structures13.4A

1-18

Problem (13.3): Write resonance structures for each of the following:

13.4B

O

H

H

Br


1-19

Problem: Which of the following pairs are valid resonance structures?

13.4

O

O

O

O

N

O

and

and

and

andN

OH

H

1-20

1. The more covalent bonds a structure has, the more stable it is.

2. Structures in which all of the atoms have a complete valence shell of electrons (the noble gas structure) are especially stable and make a larger contribution to the hybrid

3. Charge separation decreases stability

Relative Stability of Contributing Resonance Structures

13.4B


1-21

Formal charges should be seen for book-keeping purposes and are not a realistic description of the charge distribution in a molecule:

Formal Charges

13.4B

+0.62 �–0.50�–0.12Calculated Mulliken net-charges:

electrostatic potential map

Example:

N N O N N O

1-22

Problem (13.4): Circle the structure that contributes most to the resonance hybrid:

13.4B

N N

O

OH

O

OH


1-23

Problem (13.31): What products with molecular formula C4H7Cl are formed when 2-buten-1-ol is treated with HCl? Provide a mechanism for this reaction.

1-24

A hydrocarbon which contains two double bonds is called an alkadiene. Depending on the relative positions of the double bonds, three classes of alkadienes exist:

Alkadienes and Polyunsaturated Hydrocarbons

13.5

Isolated double bonds

Conjugated double bonds

Cumulated double bonds (not common)

1,5-hexadiene

1,3-pentadiene

1,2-propadiene (or allene)

1,4-cyclohexadiene

1,3-cyclohexadiene

CH

H H

H


1-25

energy

0

1,3-Butadiene can be described with 4 p orbitals that overlap to form a set of 4 molecular orbitals with -symmetry

MO Description of 1,3-Butadiene13.6

Structural implication:

1.34 Å

1.47 Å

1-26

Conjugated dienes are thermodynamically more stable than the corresponding isomeric isolated dienes:

Stability of Conjugated Dienes13.7

+ 2 H2 H° = �–226 kJmol�–1

+ 2 H2 H° = �–253 kJmol�–1

Difference: 27 kJmol�–1

Enthalpy of hydrogenation:


1-27

Ultraviolet-Visible Spectroscopy

13.8

1-2813.8

UV absorption spectrum of 2,5-dimethyl-2,4-hexadiene in methanol (concentration 59.5 M)


1-29

Photochemistry of Vision

13.8

Three-dimensional model of rhodopsin (chromophore shown in red):

1-30

Mechanism:

Electrophilic Attack on Conjugated Dienes: 1,4-Addition

13.9

F

HBrBr

H

1 molar equiv HBrBr

H H+ Br


1-31

Kinetic vs Thermodynamic Control

13.9A

The product distribution for the addition of HBr to 1,3-Butadiene strongly depends on the reaction temperature

HBrBr

H H+ Br40°C 20% 80%

HBrBr

H H+ Br�–80°C 80% 20%

1-32

Potential Energy Reaction Prole

13.9A

1,2-addition

1,4-addition

BrH

HBr

H


1-3313.9A

BrH H

Br

40°C

20% 80%

BrH H

HH

Br

1-34

Problem (13.32): Provide a mechanism to account for the formation of the following conversion:

Cl2

CH3OHCl

OCH3Cl

OCH3+


1-35

In 1928, Otto Diels and Kurt Alder developed a 1,4-cycloaddition reaction of dienes (Nobel Prize of chemistry in 1950):

Diels-Alder Reaction13.10

Example:

The reaction is favored with dienophiles carrying an electron withdrawing group:

F+ O

O

O

heat

benzeneO

O

Odiene dieneophile adduct

CNNC CHO NO2etc.

1-36

1. The Diels-Alder reaction is stereospecic: the reaction is a syn addition and the conguration of the dienophile (cis or trans isomer) is retained in the product:

Stereochemistry of the Diels-Alder Reaction13.10B

F

F

+heatOCH3

O

O

OCH3

O

OCH3

O

OCH3

dimethyl maleate

+heatOCH3

O O

OCH3

O

OCH3

dimethyl fumarate

O

H3CO+

O

OCH3

O

OCH3


1-37

2. The diene can react only in the s-cis rather than s-trans conformation:

13.10B

3. The reaction occurs primarily in an endo rather than an exo fashion when the reaction is kinetically controlled.

R

s-ciss-trans

RR

+R

1-38

MO Theory offers insights into the endo preference:

13.10B


1-39

Problem (13.16, 13.44b): Which diene and dienophile are required to prepare each of the following?

COOCH3

COOCH3

COOCH3

COOCH3

13.10D

1-40

Problem: Suggest a multistep route for the synthesis of the following molecule starting from materials with 5 carbon atoms.

COOCH3HHCOOCH3


1-41

Problem: Because of its stereospecifcity, the Diels-Alder reaction is often used in the synthesis of acyclic building blocks containing several stereocenters. Suggest a synthesis for the following (racemic) compound from starting materials with four or less carbon atoms.

H

O CN

CN

H

O

1-42

Benzene was isolated 1823 from distillation of whale oil by Michael Faraday (named bicarburet of hydrogen )

The structure was an unsolved puzzle until 1865, when Kekulé dreamed of carbon-chain snakes and nally proposed the correct structure

Some older versions of benzene:

Aromatic Compounds: Kekulé�’s Dream�…

14.1


1-43

Aromatic compounds: Why the name?

14.1

O H O OCH3

OH

OCH3OH OCH3

OHOCH3

O

H

benzaldehyde methyl salicylate vanillin

H

O

eugenol anethole cinnamaldehyde

(almonds) (wintergreen) (vanilla beans)

(cloves) (anis) (cinnamon)

1-44

Many drugs contain benzene as building block:

14.1

O OH

O

tylenol aspirin (±)-fluoxetine (prozac)

OHO

HN

O O

CF3NH


1-45

Aromatic Compounds: Nomenclature

14.2

Benzene as a substituent is called a phenyl group (Ph):

FF Br

NO2

1,2-

1,3-

1,4-

OH

1-46

Important trivial parent names:

14.2

F

toluene

CH3

OHNH2

phenolaniline

OCH3

anisole

benzoic acid

COOH

acetophenone

O

benzaldehyde

HO

benzylalcohol

CH2OH CH3

xylene

CH3


1-47

Enthalpy of hydrogenation may serve as an indicator for the intrinsic stability of benzene:

Stability of Benzene

14.5

1-48

Isodesmic reactions can be used to estimate the resonance stabilization energy:

An isodesmic reaction is an actual or hypothetical reaction in which the types of bonds that are made in forming the products are the same as those which are broken in the starting material

http://www.iupac.org/goldbook/I03272.pdf

3 + 2

H°f (kJ/mol) �–5.0 82.4 �–123.0


1-49

Resonance Description of Benzene

14.6B

In the valence bond model, six sp2-hybridized carbon atoms are joined in a ring, Each carbon atom contributes a p orbital, which form a delocalized -system containing a total of six electrons:

electrostatic potential map of benezene

1-50

NMR Spectroscopy: Evidence for Electron Delocalization in Benzene

In the valence bond model, six sp2-hybridized carbon atoms are joined in a ring, Each carbon atom contributes a p orbital, which form a delocalized -system containing a total of six electrons:

14.7C


1-51

MO Description of Benzene14.6B

1-52

Hückel�’s Rule: 4n+2 -Electrons14.7

Planar monocyclic rings containing 4n+2 -electrons, where n = 0,1,2,3 �…, have closed shells of delocalized electrons and should have substantial resonance stabilization energy.

Planar rings with 4n -electrons are greatly destabilized and very reactive.

cyclobutadiene benzene cyclooctatriene

6 electronsplanar

aromaticanti-aromatic non aromatic


1-5314.7A

Polygon/circle method provides relative energies of the -MO s in cyclic -systems:

1-54

Annulenes14.7B

Monocylic compounds that are represented by alternating single and double bonds are called annulenes:

The following [10]annulenes are not planar and therefore not aromatic:

benzene cyclooctatetraene[6]annulene [8]annulene

HH


1-55

Anti-Aromaticity14.7E

Planar cyclic -systems containing 4n -electrons are called anti-aromatic.

Isodesmic reaction: 2 +

H°f (kJ/mol) 156.9 477.0 28.0

1-56

Problem (14.24a): Cyclooctatetraene undergoes a two electron reduction when treated with potassium metal to give a stable, planar dianion C8H8

2�–. Use a MO diagram to explain this result.


1-57

Other Aromatic Compounds14.7D, 14.8

1. Aromatic Ions:

HH

HpKa = 16

cyclopentadiene

+ base + base-H

HH

cycloheptatriene

pKa = 37+ base + base-HH

1-5814.8A/C

2. Fused Ring Systems:

buckminsterfullerene

F

napthalene anthracene phenanthrene pyrene


1-5914.9

3. Heterocyclic Aromatic Compounds: F

pyridine

N

pyrrole

NH

thiophene

S

1-60

Problem (14.26): Cycloheptatrienone (A) is very stable. Cyclopentadienone (B) is quite unstable and rapidly undergoes a Diels-Alder reaction with itself. Provide an explanation for the different stabilitites, and draw a structure of the Diels-Alder adduct of B.

O O

A B


1-61

NMR Spectroscopy: 1H-NMR: 6.5-9.0 ppm 13C-NMR: 135-175 ppm

Infrared Spectroscopy: C�–H (stretch) 3050 cm�–1 680-860 cm�–1

Spectroscopy of Aromatic Compounds

14.11

Example: 13C NMR of tribromo benzene

1-62

If a pair of atoms (or atom groups) are related through a rotational or mirror symmetry, they exhibit identical chemical shifts.

Symmetry in NMR Spectroscopy

14.11

Cl Cl

Cl

ClCl

Cl

Br

ClBr


1-63

Problem (14.36): Compound N has the molecular formula C9H10O and reacts with osmium tetroxide. Propose a structure for N that is consistent with the following spectra:


conjugated systems and aromaticityww2.chemistry.gatech.edu/~cf77/8833/set1.pdf · 1-1 conjugated...

Documents