Week 2 lecture slides - Organic Chemistry PDF

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Week 2/Lecture 1: Aromaticity

Weekly objectives 1. Draw benzene and understand its structure and unique stability 2. Identify aromatic compounds and heterocycles and apply Hückel’s rule 3. Identify the structure of phenols and anilines 4. Draw the resonance structures of phenol 5. Connect reactivity and pKa of phenols and anilines to resonance

Original sample of benzene isolated by Faraday.

Pre-workshop activity recap • Isolated in 1825 by Michael Faraday, from distillation of “illuminating gas” residue, was an unknown material with the empirical formula C6H6

A number of possible structures for benzene were proposed in the following years

all are C6H6 Claus 1867

Dewar 1867

Ladenburg 1869

Benzene – alternating single and double bonds? In 1865, Kekulé proposed a structure with three C−C single bonds and 3 C−C double bonds.

Benzene – trouble with the model. In 1865, Kekulé proposed a structure with three C−C single bonds and 3 C−C double bonds. • BUT Kekulé’s structure does not fit the thermochemical evidence from hydrogenation studies.

H2

ΔH° = −120 kJ mol -1 (experimental) three double bonds!

H2

ΔH° = −360 kJ mol -1 (prediction) ΔH° = −208 kJ mol-1 (experimental)

Benzene is more stable than predicted for a simple triene!

Benzene – more trouble with the model. In 1865, Kekulé proposed a structure with three C−C single bonds and 3 C−C double bonds. • NOR does Kekulé’s structure explain how benzene reacts.

Br Br2 Br

Br Br2

HBr

Benzene – Resonance structures • In 1899, the proposal was refined with a model in which each bond is intermediate between single and double.

HC or

CH

HC

CH HC

CH

This is the resonance hybrid (a mix of the two other structures). The dotted lines mean “half a bond”

Benzene – a modern model This model can be refined further by considering orbital hybridisation to provide an explanation of benzene’s properties. • Not isolated alkenes, but a cloud of electrons in a delocalised p-system.

Determining aromaticity Benzene is not the only molecule to have stability due to such delocalisation. Molecules of this type we call aromatic. Hückel’s rules determine if a compound is aromatic: 1. Is the molecule a planar ring system with conjugated p electrons? 2. Does the number of p electrons = (4n + 2), where n is an integer? If Yes to both it is aromatic!

Note. The rule can be used with all molecules. Neutral, positive or negative.

Note. Analyse each ring separately.

Hetero-aromatic compounds 6 π-electrons

N

pyridine

O

furan

N

sp2 electron pair not involved in aromaticity. Why?

NH

pyrrole 6 π-electrons

6 π-electrons

N H

O sp2 electron pair not involved in aromaticity.

Pairs

ACTIVITY 1: Determining aromaticity Determine whether the molecule is aromatic.

10 mins.

H3C N H

CH3

N

N N

N H

H

ACTIVITY 1: Discussion and feedback

10 π electrons ∴ n=2 (integer) ∴ aromatic

H3C N H

6 π electrons ∴ n=1 (integer) but not conjugated ∴ not aromatic

8 π electrons ∴ n=1.5 (not an integer) ∴ not aromatic

CH3

N

N N

N H

Answers for the following materials are explained on the following slides.

H

Bromination of Benzene

Br Br2 Br

Br Br2

Why does the bromination of benzene proceed differently?

HBr

Bromination of Benzene

Br Br2 Br

Bromination of Benzene

Br Br

Br

HBr

Summary

Today we have: •

discussed the discovery of benzene.

•

Introduced Hückel’s rules for determining if a molecule is aromatic.

Week 2 /Lecture 2: Phenols and anilines

Weekly objectives 1. Draw benzene and understand its structure and unique stability 2. Identify aromatic compounds and heterocycles and apply Hückel’s rule 3. Identify the structure of phenols and anilines 4. Draw the resonance structures of phenol 5. Connect reactivity and pKa of phenols and anilines to resonance

Phenol Phenols are commonly occurring materials with many uses. OH

phenol

OH O H 3CO OH

resorcinol (Antibacterial used in acne treatment)

CH3

N H

CH3

HO

capsaicin

Phenol Phenol’s are strong acids. Phenol is about ~106 times more acidic than ethanol!

OH

O

Ka = H 3O

H 2O AH

H 3C

OH AH

A

H 2O

H 3C

H

O A

H 3O H

[A ][H ] [AH]

= 1.02 X10-10 pKa = 9.99

Ka =

[A ][H ] [AH]

= 1.3 X10-16 pKa = 15.9

Phenol Consider the resonance stabilisation of the phenoxide anion. O

O

O

δ

O

δ

δ δ

Resonance hybrid structure The negative charge has been stabilised by spreading it around.

O

Resonance structures Resonance is a model used commonly to explain the structure and reactivity of molecules where a single Lewis structure is inadequate. Consider the following example H3C

H3C

O18

I

C H3C

H3C

O18 I

C

O16

H3C

O16

C H3C

O18

O18 and

O16

C H3C

CH3 O16

Instead a mixture is observed.

C

C H3C

Notes on resonance

O

O H3C

O 1/2

O or

C H3C

O 1/2

O

1) These are structures related through resonance (resonance structures) 2) They are hypothetical. 3) Neither is real, a mix (the resonance hybrid) can be useful to explain reactivity.

Resonance structures When must we consider resonance structures? Below is a common case that will come across a few times in CHM1022.

A

A B

B C

O

O C

1. Atom A has a lone pair (can be an anion or neutral). 2. Atoms B and C are connected via a double bond.

i.e.

C

C C

C

This is a resonance arrow. • One line with an arrow at each end.

ACTIVITY 2: Acidity of phenol derivatives. In your group you need to: 1. Decide whether phenol 1 or its derivative 2 is more acidic. Explain why. 2. Give an explanation of this answer using resonance structures.

OH

OH

N O 1

O 2

Pairs

5 mins.

Pairs

ACTIVITY 2: Discussion and feedback

5 mins. a)

OH

b)

OH

pKa = 9.99

O

C

C

N

N O

O

O

pKa = 7.15

O

O

O

N O

We can draw a resonance structure, where the phenoxide negative charge is delocalised onto the electron-withdrawing nitro group.

Aniline Anilines are common in the dyeing industry and have properties explainable by resonance NH2

• Badische Anilin und Soda Fabrik (BASF) • Largest chemical manufacturing company in the world. • 120 000 employee’s worldwide. • 4.7 billion euro revenue in 2018

Aniline is a weak base. Aniline is ~106 times less basic than cyclohexylamine!

NH 2

NH 3

Kb = OH

H 2O B

B

NH 2

OH

B

B

OH

[B]

pKb = 9.36

Kb = OH

]

= 4.5 X10-10

NH 3 H 2O

[B ][OH

[B ][OH [B]

= 4.5 X10-4 pKb = 3.34

]

Aniline Consider the resonance structures of aniline. NH2

NH2

NH2

δ

NH2

δ

δ δ

Resonance hybrid structure

NH2

ACTIVITY 3: Basicity of aniline derivatives In your group you need to: 1. Decide whether aniline 1 or the derivative 2 is more basic. Explain why. 2. Give an explanation of this answer using resonance structures.

NH2

NH2

H3C 1

O

O 2

Pairs

5 mins.

Pairs

ACTIVITY 3: Discussion and feedback NH2

a)

NH2

b)

H3C

Most basic.

O

O

O

Least basic. N lone pair is withdrawn by ester.

NH2

H3C

O

NH2

H3C

O

O

NH2

H3C

O

O

NH2

H3C

O

O

Bromination of aniline

The aromatic ring of aniline is electron rich. This increases the rate of bromination compared to benzene. NH2

NH2 Br

Br

Which position(s) of aniline will be brominated?

Br

HBr

Bromination of aniline NH2

NH2 Br Br

Br

HBr

The nitrogen lone-pair electrons direct bromination to the ortho and para positions.

Bromination of aniline NH2

NH2 Br

HBr

Br Br

The nitrogen lone-pair electrons direct bromination to the ortho and para positions.

Summary Today we have: •

Discussed how resonance structures can help predict reactivity.

•

Building upon this concept we have looked at the acidity of phenol.

•

Building upon this concept we have looked at the basicity of aniline....