Benzene - Electrophilic Substitution (Edexcel IAL Chemistry)

Spec point — 18.5

Benzene - Electrophilic Substitution

• The main reactions which benzene will undergo include the replacement of one of the 6 hydrogen atoms from the benzene ring

  • This is different to the reactions of unsaturated alkenes, which involve the double bond breaking and the electrophile atoms 'adding on' to the carbon atoms

• These reactions are called electrophilic substitution reactions

  • This is where at least one of the H atoms of benzene are substituted by the electrophile

• You must be able to provide the mechanisms for specific examples of the electrophilic substitution of benzene

General Electrophilic Substitution Mechanism:

• The delocalised π system is extremely stable and is a region of high electron density

• Electrophilic substitution reactions involve an electrophile, which is either a positive ion or the positive end of a polar molecule

• There are numerous electrophiles which can react with benzene

  • However, they usually cannot simply be added to the reaction mixture to then react with benzene
  • The electrophile has to be produced in situ, by adding appropriate reagents to the reaction mixture

• The electrophilic substitution reaction in arenes consists of steps:

  • Generation of an electrophile
  • Electrophilic attack
  • Regenerating aromaticity

Nitration of benzene mechanism

• One hydrogen atom is substituted by a nitro group - NO₂
• The overall reaction is:

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

• The reaction is conducted under reflux at 55 °C or 60 °C

Step 1: Electrophile generation

• The electrophile NO₂⁺ion is generated by reacting concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄)
• The equation for the generation of the electrophile is:

Exam Hint There are 2 accepted equations for the formation of the eletrophile:

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

OR

HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄^- + H₃O⁺

Step 2: Electrophilic attack

• A pair of electrons from the benzene ring is donated to the electrophile to form a covalent bond

  • This disrupts the aromaticity in the ring as there are now only four π electrons and there is a positive charge spread over the five carbon atoms

A pair of electrons from the benzene ring is donated to the nitronium ion electrophile forming a covalent bond and a loss in aromaticity

Exam Hint Examiners are looking for very specific points in an electrophilic substitution mechanism:

1. The initial arrow:

• The curly arrow must start from on or within the circle/hexagon of the benzene ring and point to the NO₂⁺ electrophile.
• Do not forget to draw the + charge on the electrophile.

1. The horseshoe shape:

• The intermediate must be drawn with a 'horseshoe' (an incomplete circle) that faces the tetrahedral carbon, which is the carbon atom attached to the H and NO₂ group.

1. Horseshoe size:

• The horseshoe must be large enough to cover at least 3 carbon atoms.
• Ideally it should cover the 5 carbon atoms that have not been substituted.

1. The charge:

• The positive charge (+) must be drawn inside the horseshoe

Step 3: Regenerating / restoring aromaticity

• Aromaticity is restored by heterolytic cleavage of the C-H bond

  • This means that the bonding pair of electrons goes into the benzene π bonding system

Exam Hint The curly arrow must start from the C–H bond and point to anywhere within the ring to show the reforming of the delocalised structure

Step 4: Regenerating the catalyst

• The H⁺ ion (released in Step 3) reacts to reform the original catalyst:

H⁺ + HSO₄⁻ → H₂SO₄

Halogenation of benzene mechanism

• One hydrogen atom is substituted by a halogen atom
• For bromine, the overall reaction is:

C₆H₆ + Br₂ → C₆H₅Br + HBr

• The reaction is conducted using a halogen carrier catalyst and requires heat / heating under reflux

Step 1: Electrophile generation

• The electrophile X⁺ion is generated by reacting the halogen with a halogen carrier

• The common halogen carriers are:

  • AlBr₃
  • AlCl₃
  • FeCl₃ (or iron and bromine, which react to form FeBr₃)

• The halogen molecules form a dative bond with the halogen carrier by donating a lone pair of electrons from one of its halogen atoms into an empty 3p orbital of the halogen carrier

Mechanism step showing Br₂ reacting with AlBr₃ via a dative covalent bond to form Br⁺ electrophile and [AlBr₄]⁻ in electrophilic substitution

Exam Hint The halogen carrier used must correspond to the halogen that is involved in the substitution reaction:

Br-Br + AlBr₃ → Br⁺ + [AlBr₄]^-

Cl-Cl + FeCl₃ → Cl⁺ + [FeCl₄]^-

Step 2: Electrophilic attack

• A pair of electrons from the benzene ring is donated to the electrophile to form a covalent bond

  • This disrupts the aromaticity in the ring as there are now only four π electrons and there is a positive charge spread over the five carbon atoms

A pair of electrons from the benzene ring is donated to the bromine ion electrophile forming a covalent bond and a loss in aromaticity

Exam Hint Examiners are looking for similar specific points to the nitration mechanism:

1. The initial arrow:

• The curly arrow must start from on or within the circle/hexagon of the benzene ring and point to the Br⁺ or Cl⁺ electrophile.
• Do not forget to draw the + charge on the electrophile.

1. The horseshoe shape:

• The intermediate must be drawn with a 'horseshoe' (an incomplete circle) that faces the tetrahedral carbon, which is the carbon atom attached to the H and halogen atom.

1. Horseshoe size:

• The horseshoe must be large enough to cover at least 3 carbon atoms.
• Ideally it should cover the 5 carbon atoms that have not been substituted.

1. The charge:

• The positive charge (+) must be drawn inside the horseshoe

Step 3: Regenerating / restoring aromaticity

• Aromaticity is restored by heterolytic cleavage of the C-H bond

  • This means that the bonding pair of electrons goes into the benzene π bonding system

Diagram of step 3 electrophilic substitution: [AlBr4]– removes H from bromobenzene sigma complex, restoring benzene aromaticity and forming HBr and AlBr3

Step 4: Regenerating the catalyst

• The H⁺ ion (released in Step 3) reacts to reform the original catalyst:

H⁺ + [AlBr₄]^- → HBr + AlBr₃

Friedel-Crafts acylation of benzene mechanism

• One hydrogen atom on the benzene ring is substituted by an acyl group (e.g., an ethanoyl group, −COCH₃).
• For the reaction with ethanoyl chloride, the overall reaction is:

C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl

• The reaction is conducted using an anhydrous aluminium chloride (AlCl₃) catalyst and requires heating under reflux

Step 1: Electrophile generation

The electrophile is an acylium ion (e.g., CH₃CO⁺).

It is generated by reacting the acyl chloride with the halogen carrier catalyst, which accepts a lone pair of electrons from the chlorine atom.

The equation for the generation of the electrophile is:

CH₃COCl + AlCl₃ → CH₃CO⁺ + [AlCl₄]⁻

• In the Friedel-Crafts acylation reaction, an acyl group is substituted into the benzene ring

  • An acyl group is an alkyl group containing a carbonyl, C=O group

Step 2: Electrophilic attack

• A pair of electrons from the benzene ring is donated to the electrophile to form a covalent bond.

  • This disrupts the aromaticity in the ring as there are now only four π electrons and there is a positive charge spread over the five carbon atoms.

Exam Hint Examiners are looking for similar specific points to the nitration mechanism:

1. The initial arrow:

• The curly arrow must start from on or within the circle/hexagon of the benzene ring and point to the positively charged carbon atom of the CH₃CO⁺ electrophile.
• Do not forget to draw the + charge on the electrophile.

1. The horseshoe shape:

• The intermediate must be drawn with a 'horseshoe' (an incomplete circle) that faces the tetrahedral carbon, which is the carbon atom attached to the H and acyl group.

1. Horseshoe size:

• The horseshoe must be large enough to cover at least 3 carbon atoms.
• Ideally it should cover the 5 carbon atoms that have not been substituted.

1. The charge:

• The positive charge (+) must be drawn inside the horseshoe

Step 3: Regenerating / restoring aromaticity

• Aromaticity is restored by heterolytic cleavage of the C–H bond.

  • This means that the bonding pair of electrons goes into the benzene π bonding system.

Step 4: Regenerating the catalyst

• The H+ ion (released in Step 3) reacts with the complex ion to reform the original catalyst and produce hydrogen chloride gas:

H⁺ + [AlCl₄]⁻ → HCl + AlCl₃