Aromatic- Nucleophilic aromatic substitution

Question 1

What are the two different reactions nucleophilic aromatic substitution can occur under

Answer 1

1. Under strongly basic conditions, go through a benzyne intermediate- elimination-addition mechanism
2. Other method is when a nucleophile is added to an electron-deficient aromatic substrate with an electronegative leaving group, forms Meisenheimer intermediate prior to elimination- addition-elimination mechanism

Question 2

Describe basics of the elimination-addition mechanism

Answer 2

1. Under harsh basic conditions
2. Aromatics with good leaving group e.g. halogen undergo nucleophilic substitution
3. Strong bases- NaNH2/KOtBu are required and also act as a nucleophile

Question 3

Describe why it is an elimination-addition mechanism

Answer 3

1. Cannot occur by SN1 as phenyl cation is too unstable
2. Cannot be SN2 as C-X sigma* antibonding orbital is unable to be accessed due to steric hindrance of the aromatic ring
3. So 2 step mechanism of elimination-addition

Question 4

Describe the elimination-addition mechanism

Answer 4

1. First, the strong base deprotonates the proton ortho to the leaving group to give an aryl anion
2. Syn elimination then occurs to give an aromatic intermediate with a triple bond- benzyne
3. The overlap of the orbitals making triple bond is poor so the bond is very weak and readily attacked by nucleophiles to give a substituted product.
4. Most cases the base is also the nucleophile- limitation

Question 5

Why isn’t the benzyne triple bond a true alkyne

Answer 5

1. Carbon atoms within the bond are not sp hybridised and the bond is not linear
2. The carbon atoms are still sp2 hybridised and a pi-bond is still incorporated in the aromatic ring
3. The other ‘pi-bond’ is perpendicular to the pi-cloud of the aromatic system and is formed by overlap of the two adjacent sp2 orbitals
4. The overlap is poor so the bond is very weak and readily attacked by nucleophiles to give a substituted product

Question 6

Where can benzyne be attacked and what are the consequences of this

Answer 6

1. It is symmetrical so can be attacked at either end of the triple bond
2. Inconsequential for mono-substituted halobenzenes
3. Product mixtures and issues of regioselectivity can occur for disubstituted benzenes

Question 7

What is produced from a para-disubstituted halobenzene

Answer 7

1. Only one benzyne intermediate can be formed
2. Since the triple bond is too far away from the substituent, minimal electronic and steric factors an affect the regioselectivity
3. Mixture of para and meta products formed

Question 8

What is produced from a meta-disubstituted halobenzene

Answer 8

1. Two benzyne intermediates are possible
2. If substituent is electronegative, or can stabilise charge through inductive effects, the triple bond is formed closest to the substituent
3. If substituent is electropositive or is electron-donating through inductive effects, then the other is preferentially formed

Question 9

What is produced from an ortho-disubstituted halobenzene

Answer 9

1. Only one benzyne intermediate can be formed
2. With regards to regioselectivity of the nucleophilic substitution, electronic effects take precedent over steric effects
3. If substituent is electronegative, substitution will be favoured at the meta position as the resulting negative charge at the ortho position will be stabilised
4. If the substituent is electropositive, substitution will be favoured at the ortho position as the negative charge at the meta position will be less destabilised

Question 10

What are problems with the elimination-addition mechanism

Answer 10

1. Harsh basic conditions
2. Minimal control of regioselectivity for meta-disubstituted halobenzenes
3. Nucleophile is also the base

Question 11

What are two more modern methods of the elimination–addition reaction that overcome its disadvanatages

Answer 11

1. One uses a 2-diazonium benzoate zwitterion which decomposes to benzyne upon heating
2. Other uses 2-(trimethylsilyl)phenyl triflate which collapses to benzyne upon addition of a fluoride source such as a tetrabutylammonium fluoride (TBAF)

Question 12

What do the more modern methods of the elimination-addition reactions allow

Answer 12

1. Allow for different and more useful nucleophiles to be used to trap the benzyne intermediate
2. Amines, alcohols, carboxylic acids and Grignard reagents

Question 13

What can undergo Addition-elimination reactions

Answer 13

1. Halobenzenes which possess electron-withdrawing groups ortho or para to the halogen

Question 14

What is the intermediate in an addition-elimination reaction

Answer 14

1. Meisenheimer intermediate

2.

Question 15

Describe the mechanism of an addition-elimination mechanism

Answer 15

1. Conjugated EWG provides stability, causing the nucleophile to first attack the carbon which is directly bonded to the halogen to give an negatively charged Meisenheimer intermediate
2. As aromaticity is broken during the nucleophilic addition, the first step is the slower, rate determining step
3. Secondly, elimination of the halogen occurs with the restoration of aromaticity- not rate-determining so leaving group ability of the halogen is irrelevant
4. Gives substituted product where the halogen has been substituted with the nucleophile

Question 16

What is needed for the addition-elimination mechanism to occur

Answer 16

1. Stabilisation of the Meisenheimer intermediate by EWGs

Question 17

Which groups work well to stabilise the Meisenheimer intermediate

Answer 17

1. Nitro- NO2
2. Sulfone- SO2R
3. Nitrile- CN
4. Carbonyl- C=O
5. CF3 can but its stabilisation effect is inductive whihc is less effective

Question 18

What is the reactivity series for the halogens in addition-elimination SNAr reaction

Answer 18

1. F»Cl=Br»I

Question 19

Why does the halogen used effect the rate of reaction

Answer 19

1. Steric hindrance and electronegativity
2. Meisenheimer intermediate is negatively charged, so the more electronegative fluorine atom can better stabilise this intermediate leading to a faster rate of reaction
3. F is also smaller, allowing the nucleophile unhindered access to attack the pi* antibonding orbital
4. Iodine is much less electronegative and can’t stabilise the Meisenheimer intermediate to the same extent
5. Iodine is also a much larger atom which prevents the nucleophile from accessing the pi* antibonding orbital on steric grounds
6. Not to do with C-X bond breaking as not rate determining step