chm243h5-studentnotes_slide001-189

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CHEM243H P. T. GUNNING

Organic Chemistry

CHM243H5F

2012

Prof. Patrick T. Gunning

Rm 4046

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ORGANIC CHEMISTRY

•  The chemistry of carbon-based compounds: hydrocarbons and their derivatives

•  How carbon bonds to itself or different atoms such as oxygen, nitrogen, sulfur and/or halogens

to form different molecules

•  How these molecules interact with the environment that they are in

•  Prof Gunning = Do some cool chemistry = (01/2012) (illegally, see above)

O OH

O

O

Acetylsalicylic Acid (ASA or Aspirin)

N

H3CO

O

O

O

Cocaine

N

H3CO

O

O

O

Crack Cocaine

H

Cl

(free base)

Juicy Fruit

O

HO O

3


Extraction Synthetic

O

N

HN

O

N

N

SO

O N

N

Viagra

HN

N N

N

O

OP

O

OH

O

OH

O

H2N

cGMP

Viagra is a synthetic molecule

designed specifically to fit into

an enzyme and complete with

the naturally produced

molecule: cGMP

- Viagra potently inhibits the enzyme PDE-5, which

destroys cyclic guanosine monophosphate (cGMP),

itself a dilator of blood vessels in the body.

- Viagra allows cGMP to persist and led to its use for

oral treatment of male erectile disfunction.

HO

O

H

O

Vanillin

O

HO

OH

OH

O

OHO

O

H

Vanillin β-D-glucoside

Vanilla is the only edible fruit in the

orchid family. Synthetic vanilla

contains only one organic component

– vanillin. Vanilla contains so many

other flavors and fragrance

components, it has a much richer

smell and taste than vanillin does by

itself. 4


Why Study Organic Chemistry?

•  One of the most dynamic and exciting fields of scientific research, involved in ….

- Medicinal Chemistry - drug discovery (that is what I do!)

- Materials and Polymers - plastics

- Petrochemical Industry

- Explosives Industry - weapons, ejector seats….

- It’s better than Physics

- It’s much better than biology!

Billion Dollar Industries $$$

•  Being a chemist is cool!

•  A chemist gets to design and synthesize molecules - we can manipulate atoms!

•  We can alter Nature’s molecules to do new and wonderful things

•  Organic chemistry is difficult but can be learned with hard work and with practice

PRACTICE, PRACTICE, PRACTICE!

O

N

HN

O

N

N

SO

O N

N

5


Synopsis of Course Objectives

•  The chemistry of benzene, alcohols, aldehydes, ketones, carboxylic acid, esters, acid

chlorides, amides and amines will be covered.

•  As well, electrophilic aromatic substitution, protection and deprotection of alcohols,

nucleophilic acyl substitution, nucleophilic addition, carbonyl alpha-substitution reaction, keto-

enol tautomerism, carbonyl ccondensation and proton NMR will be introduced.

•  The emphasis will be on organic mechanisms and application of organic reactions to

multistep synthesis

•  WE WILL COVER A LOT OF MATERIAL!

•  Organic chemistry is a language – you need to practice in order to improve and to keep on top

of the increasing number of material, terminology, reactions…..etc

•  You MUST come to class!

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CHEM243 - EXAMS

Marking Scheme:

Laboratory - 20 %

Mid-term test # 1 - 15 % - Wednesday, February 29th, 2012, 11:10 - 12:00

Mid-term test # 2 - 15 % - Wednesday, March 21st, 2012, 11.10 – 12.00

Final exam - 50 % - TBA

If you miss the test and cannot provide a medical note, you will get zero for that test. For the final exam you will be responsible for all the material covered in the course beginning in September.

Mid-Term Test Dates:

Wednesday, Feb. 29th, 2012, 11:10 - 12:00.

(For that test you will be responsible for all the work covered up to Monday, 27th Feb.)

Wednesday, Mar 21st, 2012, 11:10 – 12.00.

(For that test you will be responsible for all the work covered up to Monday, 19th Mar)

There will be no make-up term tests in this course. If you miss this the class test due to a medical reason accompanied by a doctor’s note, the final exam will be worth 65% of your final mark.

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•  The course work that presented will follow the text book faithfully-

•  McMurry is an extremely easy to read text book - and you should do so regularly!

•  Practicing problems is the best way to improve your knowledge.

“Organic Chemistry”

John McMurry


Course Textbook

• 

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Chapter 15: Benzene and aromaticity

Chapter 16: Chemistry of benzene: electrophilic aromatic substitution

Chapter 17: Alcohols and phenols

Chapter 19: Aldehydes and ketones: nucleophilic addition reactions

Chapter 20: Carboxylic acids and nitriles

Chapter 21: Carboxylic acid derivatives - nucleophilic acyl substitution reactions

Chapter 22: Carbonyl alpha - substitution reactions

Chapter 23: Carbonyl condensation reactions

Chapter 24: Amines and Heterocycles

Start Wed


Course Syllabus - Brief overview

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Lecturing style

•  I will provide lecturing handouts to every student on blackboard the night before lecture or

earlier

•  You will be expected to fill in the blanks provided with chemical structures and reaction mechanisms

•  I want you to spend more time drawing curly arrow mechanisms rather than writing about the reactions!

•  I will be examining you on your ability to draw mechanisms and on your understanding of how reactions work, so it is important that you get as much practice as possible

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OFFICE HOURS: 4046 – William G Davis Building

Monday 2:00 – 3:45

Wednesday 2:00 – 3:45

Friday 2:00 – 3:45

•  If I am not in my office I will be in my laboratory – 3023C.

EMAIL/PHONE CALL POLICY

-  PLEASE DON’T EMAIL ME! - I’m not good at replying (and can’t guarantee I will!)

-  I don’t reply to my girl friend – I probably won’t reply to you

-  If you have a problem/emergency, come and see me in person

•  Don’t feel intimidated, I am friendly and approachable - I will do all I can to help you!

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FINAL EXAM

•  3-hour final exam

•  Held during the Spring examination period

•  Students will be responsible for all work covered in the course – even labs

•  Mid-term test and final exam: questions will be problem solving and testing understanding as

opposed to essay questions

•  YOU ARE NOT IN BIOLOGY ANYMORE!

•  Memorization will not get you an A, B or even C mark in this class!

•  No extra marks will be given to any student after the term is over. The mark you receive

will be the mark you have earned in the course.

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CHEM243H P. T. GUNNING -15

At a minimum what you need to know from the first term, CHM 242!

•  sp3, sp2, sp hybridization for C, N, O

•  Understand concept of Lewis base and acid, nucleophiles and electrophiles

•  Simple nomenclature

•  Stereochemistry – R, S, cis, trans, E, Z – how to recognize them and know how to draw them

•  How to write detailed mechanisms and how to make proper use of arrows for writing

mechanisms. (This is extremely important)

•  How to calculate degrees of unsaturation – helpful with NMR problem(s)

•  Understand why and which double bonds are more stable

•  Understand the differences in stability between 1°, 2°, 3° carbocations and radicals

•  Understand the importance of resonance structures. Know how to recognize resonance

structures and how to draw them

•  Know the difference between inductive and resonance effects

•  Understand Markovnikov and Zaitsev rules

•  Recognize similarities between allylic and benzylic substituted compounds

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Functional Groups - Learn this asap! Amine Aldehyde Ketone Carboxylic

acid Ester Amide Nitrile

RNH2

R2NH

R3N

RCN

CH3NH2 CH3NH2

Methanamine Ethanal Propanone Ethanoic acid Methyl

ethanoate

Ethanamide Ethanenitrile

Methylamine Acetaldehyde Acetone Acetic acid Methyl acetate Acetamide Acetonitrile

NH

O O

OH

O

O

O

N

O

H

C N

RCH

O

RCR'

O

RCOH

O

RCOR'

O

RCNHR'

O

RCNR'R"

O

RCNH2

O

O

CH3CH

O

CH3CCH3

O

CH3COH

O

CH3COCH3

O

CH3CNH2

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ELECTRONEGATIVITY

-  Understanding electronegativity is the first step to understanding organic rxns and bonding

Measure of tendency for an atom in a molecule to attract a -

The greater the no. of protons the greater the nuclear charge

increases

increases

Electronegativity Attraction between oppositely

charged particles increases with

decreasing distance

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ELECTRONEGATIVITIES OF COMMON ELEMENTS

-  derived from gas phase bond dissociation energies – the amount of energy released when a

bond forms (-ve value) or the amount of energy needed to break a bond (+ve value)

-  Fluorine is the most electronegative (electronegativity of - )

-  Lithium is the least electronegative (electronegativity of - )

How does this relate to bonding?

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BOND TYPE

Difference in electronegativity (ΔEN)

COVALENT BONDS

-  Covalent bonds occur between atoms of similar electronegativity (close together in table)

-  Atoms achieve octets by sharing of valence electrons

-  Molecules result from this covalent bonding

-  Valence electrons can be indicated by dots (electron-dot formula or Lewis structures)

Two electrons in a bond are indicated by a line (one line = two electrons)

Cl

Cl Cl

Cl

C O

C O

δ+ δ−

POLAR COVALENTNONPOLAR COVALENT

0 to 0.4 0.5 to 1.6

H2 H HH H+ H H OR

CCH4 + H C

H

H

H

H C

H

HH

H

OR

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POLAR COVALENT BONDS

-  The sharing of electrons unequally due to differences in electronegativity

-  Results in atoms having a partial negative (-δ) or partial positive charge (+δ)

-  Electron density is pulled towards the oxygen atom, why?

-

-

If a molecule has a –ve charge

like a hydroxide anion, where

would it attack chloromethane?

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WRITING LEWIS STRUCTURES:

-  Atoms bond using their valence electrons

-  The number of valence electrons is equal to the group number of the atom

- Carbon is in Group 4 and has 4 valence electrons

- Hydrogen is in Group 1 and has 1 valence electron

- Oxygen is in group 6 and has 6 valence electrons

- Nitrogen is in group 5 and has 5 valence electrons

-  To construct molecules the atoms are assembled with the correct number of valence electrons

-  The structure is written to satisfy the octet rule for each atom and to give the correct charge

If necessary, multiple bonds can be used to satisfy the octet rule for each atom

Draw the Lewis Structure for CO2

Draw the Lewis Structure for CN-

C 4 valence e’s

O 2 x 6 valence e’s

TOTAL = e’s

Each atom must have 8 e’s

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FORMAL CHARGES

-  created by an imbalance in the no. of protons in the nucleus and e’s surrounding the nucleus

Ammonia NH3 electronic configuration of N = 1s22s22p3

Total number of electrons =

Total number of protons =

Charge on N atom =

Ammonium, NH4 – charged? N should be = 1s22s22p3 1s22s22p2

Total number of electrons =

Total number of protons =

Charge on N atom =

Nitrogen should have 5 valence electrons but only has 4. Since it has one more proton

than electron surrounding the nucleus of the nitrogen atom, it is positively charged

5 valence electrons

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PRACTISE: WHAT IS THE FORMAL CHARGE ON THE FOLLOWING MOLECULES? Which atom bears the charge?

 What is the formal charge on Nitrogen?

  How many lone pairs of electrons are around the oxygen atom?

  What is the charge on the carbon atom? There are no lone pairs on the carbon.

 What is the formal charge on the sulfur atom? (all atoms have full octets)

 What is the formal charge on the nitrogen atom? (all atoms have full octets)

  What is the charge on the carbon atom?

 What is the charge on the oxygen atoms?

CH3OH CH3O CH3OH2

N

O

H3C CH3

CH3

S C N

O N O

H3C CH2

O O

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RESONANCE

As seen in different examples, you may be able to draw molecules in which there are different

combinations of bonding. Which is the correct structure of the carbonate ion (CO32-)?

-  Experimentally, in carbonate, all bonds are of equal length and the charge is spread equally over all three oxygen atoms.

The real structure is a resonance hybrid or a mixture of all three resonance structures

- One resonance contributor is converted to another by the use of curly arrows which show the

flow of electron density in a resonance structure

-  In reaction mechanisms, the arrows will show us the movement of electrons between molecules

-  The use of arrows serves as an electron bookkeeping device and to help us think about where

the electrons can possibly flow

O O

O

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RULES FOR RESONANCE

1) 

2) 

3) 

4) 

-  All structures must be proper Lewis structures

- You cannot exceed the octet rule

for elements in rows 1 and 2

- The overall charge of each

resonance structure MUST stay

the SAME!!

C

H

H H

OH

C

H

H H

OH

Carbon has 5 bonds!!

N

H

H

O Nitrogen has 4 bondsand 1 lone pair of

electrons 10 electrons around the nitrogen

N

H

HO

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RESONANCE STABILIZATION

-  The energy of the molecule as a whole is lower than the energy of any single contributing

resonance form – this lowering of energy is called --

-  Equivalent resonance forms make equal contributions to the structure of the real molecule

Structures with equiv. resonance forms tend to be greatly stabilized: the more equiv.

resonance structures a molecule has, the more stable it is

- The most important resonance contributors are structures with -

-  When charges exist, the structure with the least amount of charges is favoured

(i.e. 1 +ve charge vs 2 +ve charges and 1 –ve charge)

-  When structures have the same no. of charges, the structure with least amount of charge

separation of opposite charges are favoured

-  Also, charges are preferably on atoms with compatible electronegativity

A) No charges on it – therefore it is a major contributor

B) Opposite charges have minimal separation and compatible electronegativity

C) Opposite charges have more separation and positive charge less favourable on an oxygen

versus a carbon (based on electronegativity) 24


π BOND OVERLAP and RESONANCE

Why does the bond length between the carbon-carbon single bonds become smaller as the

hybridization changes from sp3 to sp?

Because the sp3 orbitals have more p character and are longer. The more s character the orbital has the smaller and shorter it is. [sp3 (1/4 s, 3/4 p), sp2 (1/3 s, 2/3 p), sp (1/2 s, 1/2 p)]

Conjugation: systems that have a p-orbital on an atom adjacent to a double bond; Conjugated unsaturated systems have “delocalized π bonds”

saturated

unsaturated

conjugated & unsaturated

H3C CH3

1.54 Α

σ, (sp - sp)

o

H2C CH2

1.34 Α

σ, (sp2 - sp2), π(p-p)

o

H2C CH

1.50 Α

σ, (sp2 - sp3)

o

CH3 H2C CH

1.47 Α

σ, (sp2 - sp2)

o

CH

CH2

HC CH

1.20 Α

σ, (sp - sp), π(p-p)

o1.46 Α

σ, (sp - sp3)

o1.43 Α

σ, (sp - sp2)

o

HC C

1.37 Α

σ, (sp - sp)

o

C CHHC C CH3 HC C CH

CH2

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WHAT EFFECT DOES CONJUGATION HAVE ON THE STABILITY OF A MOLECULE?

H2/Ni

Hydrogenation

H2/Ni

ΔH = -30.3 kcal/mol


H2/NiWE WOULD ASSUME THAT ΔH = 2 x -30.3 kcal/mol

= -60.6 kcal/mol

OBSERVED ΔH = -57.1 kcal/mol

Pote

ntia

l E

nerg

y

Reaction Coordinate



MORE STABLE

THAN CALCULATED

For an exothermic reaction the more heat given

off the less stable the molecule is

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π BOND OVERLAP and RESONANCE

Conclusion: Two conjugated bonds are connected.

HOW? There is some overlap of adjacent π electrons which shortens the C2-C3 bond and helps

stabilize the molecule.

What is causing the overlap of adjacent π electrons? Examine the various resonance forms:

The major resonance contributor shows alternating single and double bonds, however, the three

other resonance structures illustrate a double between C2 and C3. Since these three resonance

structures are all minor contributors, the molecule will illustrate “double bond character” between

C2 and C3.

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RESONANCE

Remember: only lone pairs of electrons, double bonds or triplebonds can resonate!

The use of a double headed arrow shows the movement of two electrons (an electron pair)

Resonates to be more stable – the charge is spread to stablize the whole molecule

H

H

H HH

H

H

H

H HH

H

H

H

H HH

H

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RESONANCE

Draw resonance structures for the following anion:

- The resonance arrow starts at the C atom that has the lone pair of electrons and negative

charge. The arrow ends between the two C atoms to show the formation of a new π bond

- The blue arrow is needed because you can’t have 5 bonds on a carbon atom. The arrow

starts between the C atoms (the π bond) and ends on the terminal C.

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NOW LETS LOOK AT THE STABILITY OF BENZENE, A CONJUGATED RING SYSTEM

Benzene is much more stable than would be expected based on calculations for cyclohexatriene

Theoretical: 3 x hydrogenation of cyclohexene + 3 x double bond stabilization

(3 x 120) + (3 x 10) = -330 kJ/mol

Observed:

Difference = WHY? --

•  Benzene is an aromatic compound – a most important structure in organic chemistry

•  Almost all drugs contain aromatic structures.

Kekulé

Correct Incorrect

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Structure and Stability of Benzene: Molecular Orbital Theory

•  Benzene is unsaturated -

•  Benzene doesn’t undergo usual alkene reactions

•  Electron density maps show that electron density in all six C-C bonds is identical

•  Benzene is a planar molecule - shaped like a regular hexagon C-C-C bonds are 120°

•  All C’s are sp2 hybridized - each C has a p-orbital perpendicular to the plane of the 6- ring

•  Each C is equivalent ∴ impossible to define 3 localized π bonds in which a given p orbital

overlaps with only one neighbouring p orbital

•  Each p-orbital overlaps equally well with both neighbouring p orbitals

•  The 6 π electrons are completely delocalized around the ring!

Br2+ Fe

catalyst

Br

Bromobenzene

+ HBr

HBr

HBr

(Addition product not formed)

Br

BrH

HBr2

Fecatalyst

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WHAT MAKES A MONOCYCLIC COMPOUND AROMATIC?

- Aromaticity and the Huckel 4n + 2 Rule

•  The structure must be --

•  Each atom in the ring must have an unhybridized p-orbital (usually sp2 or sp hybridized)

•  The unhybridized p orbitals must overlap to form a continuous ring of parallel orbitals

•  The structure must be planar or nearly planar

•  Delocalization of the π electrons over the ring must lower the electronic energy

Cyclobutadiene

Cyclobutadiene is not aromatic and is very reactive and unstable

•  If π electrons were delocalized it would be a diradical (two unpaired electrons)

TWO NODESπ 4* ANTI-BONDING ORBITAL

π 3 NON-BONDING ORBITALπ 2

π 1 BONDING ORBITAL

Nodal plane

Nodal plane

No nodes

Both degenerate MOs

have 2 bonding and

2 antibonding for a net bonding order of zero

4 electrons in 4 atomic

orbitals, when combined

will form 4 π molecular orbitals

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Is there an easier way of determining the MOs?

Polygon rule: The energy level diagrams for the molecular orbitals resulting from a combination of

any cyclic arrangement of p orbitals can be deduced from the appropriate sided polygon.

“MO energy diagram of a completely conjugated cyclic system has the same shape as the

compound with one vertex (the all bonding) at the bottom”

•  The energies of the MOs will be where the corners touch the circle

•  Lets look at benzene and cyclooctatetrane again – 2 non-bonding orbitals each with 1 electron

- This is an unstable configuration; adopts a non-planar with localized p orbitals to avoid instability

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WHAT DOES IT MEAN: A COMPOUND IS AROMATIC?

Aromatic Compounds

•  Aromatic compounds are more stable than their open-chained counterparts

•  Delocalization of the π electrons makes it more stable, less reactive (than predicted)

Anti-aromatic Compounds

•  Antiaromatic compounds have continuous p orbitals but delocalization of the π electrons

increases the electron energy (less stable). Thus, are destabilized by conjugation

Numeric Patterns and Aromaticity

Huckel’s Rule: To be aromatic, all atoms must have an unhybridized p orbital, conjugated,

planar and have 4n + 2 π electrons, where n is an integer (n = 0, 1, 2, 3,….)

•  Therefore, molecules with --

•  Molecules with 4n πelectrons are said to be --

•  Even if planar and conjugated!! - Delocalization of their π electrons would lead to their

destabilization

4π 6π 8π 10π 12π 14π

electrons electrons electronselectronselectronselectrons

4n 4n + 2 4n + 24n 4n 4n + 2

n = 1 n = 1 n = 2 n = 2 n = 3 n = 4

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SUMMARY OF AROMATICITY

Huckel’s Rule

•  Planar monocyclic rings with a continuous system of p-orbitals and 4n + 2 π electrons are

aromatic (n = 0, 1, 2, 3 etc)

-  Aromatic compounds have substantial resonance stabilization

-  Benzene is aromatic: it is planar, cyclic, has a p-orbital at every carbon, and 6π electrons (n =1)

•  There is a polygon-and-circle method for deriving the relative energies of orbitals of a system

with cyclic continuous array of p orbitals

- A polygon corresponding to the ring is inscribed in a circle with one point of the polygon

pointing directly down

-  A horizontal line is drawn where vertices of the polygon touch the circle

-  Each line corresponds to the energy level of the π MOs at those atoms

-  A dashed horizontal line half way up the circle indicates the separation of bonding and

antibonding orbitals

•  Benzene has 3 bonding and 3 antibonding orbitals

-  All the bonding orbitals are full and there are no electrons in the antibonding orbitals;

benzene has a closed shell of delocalized electrons and is very stable

chm243h5-studentnotes_slide001-189

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