Nucleophilic Aromatic Substitution

Question 1

Which compound would preferentially react in an electrophilic aromatic substitution reaction?

Answer 1

The most electron rich molecule will preferentially react

Question 2

Nucleophilic aromatic substitution is much less common
Why?

Answer 2

Mechanistically it does not occur via familiar subtituion processes

Question 3

A two-step addition-elimination reaction is possible for electron-poor aromatic compounds with….

Answer 3

….Electron withdrawing (-M) substituets

Question 4

The Rate-determining step is the nucleophilic addition to form an anionic intermediate known a ….

Answer 4

….. Meisenheimer complex

Question 5

Nucleophilic aromatic substition is related/similar to what reaction in an aliphatic system?

Answer 5

Mechanistically, nucleophilic aromatic substitution is directly related to conjugate addition-elimation reactions in aliphatic systems

Question 6

What is the rate determining step in nucleophilic aromatic substituion?

Answer 6

Rate determining step is the addition of the nucleophile

Question 7

Out of the following anions:
Fluoride, Chloride, Bromide and Iodide
Which is the better leaving group and why?

Answer 7

• The leaving group polarises the carbon and the greater the polarisation the easier it is for the nucleophile to attack
• Therefore in order of rate of reaction: F»Cl>Br>I
  (note the F slow down the second step however )

Question 8

What are the most likely nucleophiles used in Nucleophilic Aromatic Substituion

Answer 8

An O, N, S or CN nucleophile

Question 9

Predict the major product formed in the reaction show below

Answer 9

Kinetic product is attack at N
(forms a cationic intermediate that in inert to further electrophilic attack)

Question 10

Diazonium salts have an excellent leaving group and can be used in a ipso-substituion
Why?

Answer 10

N₂ is a good leaving group as it is formed as a gas (entropic) resulting in increaded entropy
N₂ also is very stable due to strong N-N triple bond (ethalpic)

Question 11

How is the nitrosonium ion formed

Answer 11

By mixing sodium nitration with acid (e.g. HCl) it will protonate one of the oxygens
This oxygen can then become protonated again (becoming a LG)
Other oxygen uses LP to form triple bond between N-O and water leaves

Question 12

Describe the mechanism of reacting analine to form a diazonium salt

Answer 12

• The Nitrogen LP on analine will attack the nitrogen of the nitrosonium ion, resulting in one of the 3 N-O bonds to break
• A second N-O bond breaks when the oxygen forms a O-H bond. This allows the electrons from the N-H bond to form a N=N double bomnd
• Oxygen forms another bond with hydrogen (become a LG)
• the electrons can move again from another N-H bond to from a N-N triple bond and water leaves

Question 13

Diazonium salts are useful in synthesis as they can be converted into a wide range of products
What happens is you react Diazonium salt with phosphoric acid

Answer 13

Form Benzene

Question 14

Diazonium salts are useful in synthesis as they can be converted into a wide range of products
What happens is you react Diazonium salt with Copper Cyanide

Answer 14

Form Benzonitrile

Question 15

Diazonium salts are useful in synthesis as they can be converted into a wide range of products
What happens when you react a diazonium salt with Copper halide

Answer 15

You form chloro/bromo/iodobenzene
They are known as the ‘Sandmeyer reactions’ (bit more complicated than Sn1

Question 16

Diazonium salts are useful in synthesis as they can be converted into a wide range of products
What happens when you react a diazonium salt with water and heat?

Answer 16

form phenol

Question 17

Predict the major product formed in the reaction shown below

Answer 17

Electrophilic aromatic substitution occurs on the electron rich pyrrole ring, favouring the 3-position (aromaticity of benzene not disturbed)

Question 18

Nucleophilic Aromatic substitution by Aryne Mechanism utilises the fact that…

Answer 18

…Benzyne is highly electrophilic, hence reacts readily with nucleophiles

Question 19

How are Benzyne intermediates formed?

Answer 19

• Using a strong base to deprotonate on the C adjacent to the LG
• The e⁻ from the C-H are then used to make a second pi bond
• Causes the C-LG bond to break

Question 20

In the decarboxylation of diazonium salts this is the first step?
How does the benzyne intermediate form?

Answer 20

• The LP on oxygen forms a C=O double bond, resulting in the Ar-C bond to break, forming a second pi bond with the electrons
• Leaves as CO₂
• The forming of the second pi bond causes N₂ to leave

Question 21

What type of industrial process utilises Benzyne

Answer 21

Diels-Alder