LECTURE 2

Dr Ali El-Agamey

CHEM-204:

PHYSICAL ORGANIC CHEMISTRY

DAMIETTA UNIVERSITY

2

Homework: Complete these mechanisms by drawing the structure of the products in each case.

Homework: Complete these mechanisms by drawing the structure of the products in each case.

4

Homework: Complete these mechanisms by drawing the structure of the products in each case.

Homework: Complete these mechanisms by drawing the structure of the products in each case.

p133e

Homework: Put in the arrows on these structures to give the products shown.

p163d

  Remember that individual resonance forms do not exist. The molecule does not "resonate" between these structures. It is a hybrid with some characteristics of both (an analogy is a mule).

  Both C-O bond lengths are the same (136 pm) and in between a normal C=O (123 pm) and C-O (143 pm).

Resonance Forms

The electrons can be moved in one of the following ways:

1. Move π electrons toward a positive charge or toward a π bond

2. Move lone-pair electrons toward a π bond

3. Move a single nonbonding electron toward a π bond

Resonance contributors are obtained by moving π electrons toward a positive charge:

resonance hybrid

resonance hybrid

resonance hybrid

Moving π electrons toward a π bond

Chapter 1 11

Resonance Forms •  In a resonance form, only the electrons are moved.

Connectivity between atoms stay the same.

•  The real structure is a hybrid of the different resonance forms.

•  Arrows connecting resonance forms are double headed. •  Spreading the charges over two or more atoms stabilize

the ion.

•  The more resonance forms, the greater the delocalization (spread out) of electrons, the more stable the compound will be. Example: allyl and heptenyl cations (contains 3 double bonds) .

Chapter 1 12

Resonance Forms

Resonance Forms can be compared using the following criteria, beginning with the most important: – Has as many octets as possible. – Has as many bonds as possible. – Has the negative charge on the most

electronegative atom. – Has as little charge separation as

possible.

Chapter 1 13

Major and Minor Contributors

•  The major contributor is the one in which all the atoms have a complete octet of electrons.

•  Minor contributor will contribute so little, to the hybrid.

The carbon atom does not have a complete octet of electrons.

MAJOR MINOR

Chapter 1 14

Major and Minor Contributors (Continued)

•  When both resonance forms obey the octet rule, the major contributor is the one with the negative charge on the most electronegative atom.

MAJOR MINOR

The oxygen is more electronegative, so it should have the negative charge.

Chapter 1 15

•  Opposite charges should be on adjacent atoms.

Non-Equivalent Resonance

The most stable one is the one with the smallest separation of oppositely charged atoms.

MAJOR MINOR

Summary •  The greater the predicted stability of a resonance contributor, the more it contributes to the resonance hybrid

•  The greater the number of relatively stable resonance contributors, the greater is the resonance energy

•  The more nearly equivalent the resonance contributors, the greater is the resonance energy

(1) Draw the important resonance forms for the following molecules and ions.

(i) CO3 2- (ii) CH2=CH-CH2-

(2) For each of the following compounds, draw the important resonance forms. Indicate which structures are major and minor contributors or whether they have the same energy.

(i) [H2COH]+ (ii) CH2=CH-NO2

Homework

Aromaticity—Hückel’s Rule

Chapter 16 19

Unusual Stability Hydrogenation of just one double bond in benzene is endothermic!

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Chapter 16 20

Resonance Energy •  This difference between the predicted

and the observed value is called the resonance energy.

Reactivity of naphthalene as compared to benzene

21

Stability of the compound increases As R.E. increases

Reactivity decreases

Four structural criteria must be satisfied for a compound to be aromatic.

The Criteria for Aromaticity—Hückel’s Rule

[1] A molecule must be cyclic.

To be aromatic, each p orbital must overlap with p orbitals on adjacent atoms.

[2] A molecule must be planar.

All adjacent p orbitals must be aligned so that the π electron density can be delocalized.

Since cyclooctatetraene is non-planar, it is not aromatic, and it undergoes addition reactions just like those of other alkenes.

[3] A molecule must be completely conjugated.

Aromatic compounds must have a p orbital on every atom.

[4] A molecule must satisfy Hückel’s rule, and contain a particular number of π electrons.

Benzene is aromatic and especially stable because it contains 6 π electrons. Cyclobutadiene is antiaromatic and especially unstable because it contains 4 π electrons.

Hückel's rule:

1.  Aromatic—A cyclic, planar, completely conjugated compound with 4n + 2 π electrons.

2.  Antiaromatic—A cyclic, planar, completely conjugated compound with 4n π electrons.

3.  Not aromatic (nonaromatic)—A compound that lacks one (or more) of the following requirements for aromaticity: being cyclic, planar, and completely conjugated.

Considering aromaticity, a compound can be classified in one of three ways:

Pi bond (π) is formed by sideways overlap of two parallel p orbitals

p175c

Chapter 16 30

Energy Diagram for Benzene

•  The six electrons fill three bonding pi orbitals. •  All bonding orbitals are filled (“closed shell”),

an extremely stable arrangement.

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Chapter 16 31

Energy Diagram for Cyclobutadiene

•  Following Hund’s rule, two electrons are in separate orbitals.

•  This diradical would be very reactive.

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Examples of Aromatic Rings

•  Completely conjugated rings larger than benzene are also aromatic if they are planar and have 4n + 2 π electrons.

•  Hydrocarbons containing a single ring with alternating double and single bonds are called annulenes.

•  To name an annulene, indicate the number of atoms in the ring in brackets and add the word annulene.

•  [10]-Annulene has 10 π electrons, which satisfies Hückel's rule, but a planar molecule would place the two H atoms inside the ring too close to each other. Thus, the ring puckers to relieve this strain.

•  Since [10]-annulene is not planar, the 10 π electrons can’t delocalize over the entire ring and it is not aromatic.

Chapter 16 34

[N]Annulenes

•  [4]Annulene is antiaromatic (4N e-’s) •  [8]Annulene would be antiaromatic, but

it’s not planar, so it’s nonaromatic. •  [10]Annulene is aromatic except for the

isomers that are not planar. •  Larger 4N annulenes are not

antiaromatic because they are flexible enough to become nonplanar. =>

Both negatively and positively charged ions can be aromatic if they possess all the necessary elements.

We can draw five equivalent resonance structures for the cyclopentadienyl anion.

Chapter 16 36

Cyclopentadienyl Ions

•  The cation has an empty p orbital, 4 electrons, so it is antiaromatic.

•  The anion has a nonbonding pair of electrons in a p orbital, 6 electrons, it is aromatic.

•  Two or more six-membered rings with alternating double and single bonds can be fused together to form polycyclic aromatic hydrocarbons (PAHs).

•  There are two different ways to join three rings together, forming anthracene and phenanthrene.

•  As the number of fused rings increases, the number of resonance structures increases. Naphthalene is a hybrid of three resonance structures whereas benzene is a hybrid of two.

Indicate which of the following are aromatic?

C is aromatic 4(3)+2=14

Which of the following is aromatic?

C is aromatic 10 pi electrons, 4(2)+2=10 and completely conjugated b/c lone pair is in a p orbital.

Which are antiaromatic?

Which of these is antiaromatic?

D

Pyridine Is Aromatic

•  Pyrrole is another example of an aromatic heterocycle. It contains a five-membered ring with two π bonds and one nitrogen atom.

•  Pyrrole has a p orbital on every adjacent atom, so it is completely conjugated.

•  Pyrrole has six π electrons—four from the π bonds and two from the lone pair.

•  Pyrrole is cyclic, planar, completely conjugated, and has 4n + 2 π electrons, so it is aromatic.

Pyrrole Is Aromatic

Chapter 16 44

Allotropes of Carbon

•  Amorphous: small particles of graphite; charcoal, soot, coal, carbon black.

•  Diamond: a lattice of tetrahedral C’s. •  Graphite: layers of fused aromatic rings

Chapter 16 45

Some New Allotropes

•  Fullerenes: 5- and 6-membered rings arranged to form a “soccer ball” structure.

•  Nanotubes: half of a C60 sphere fused to a cylinder of fused aromatic rings.