Topic 18 · page 8 of 10

Friedel-Crafts reactions

A halogenoalkane or an acyl chloride, with an AlCl₃ catalyst, joins a carbon group to the benzene ring. The catalyst makes a carbocation or an acylium ion, which is the electrophile.

15 min read18.418.5

Read first: Electrophilic substitution and bromination

Two Friedel-Crafts reactions 18.4

A Friedel-Crafts reaction forms a new C–C bond between the benzene ring and a carbon group. There are two types, both using anhydrous aluminium chloride, AlCl₃, as the catalyst:

Alkylation Acylation
reagent a halogenoalkane, RCl an acyl chloride, RCOCl
example chloroethane, CH₃CH₂Cl ethanoyl chloride, CH₃COCl
organic product an alkylbenzene an aromatic ketone
example product ethylbenzene, C₆H₅CH₂CH₃ phenylethanone, C₆H₅COCH₃
by-product HCl HCl

Conditions for both: heat under reflux, with dry apparatus and reagents.

Alkylation:

CX6HX6+CHX3CHX2Cl→CX6HX5CHX2CHX3+HCl \ce{C6H6 + CH3CH2Cl -> C6H5CH2CH3 + HCl}

Acylation:

CX6HX6+CHX3COCl→CX6HX5COCHX3+HCl \ce{C6H6 + CH3COCl -> C6H5COCH3 + HCl}

The hydrogen chloride is the by-product, not chlorine.

Generating the electrophile 18.5

AlCl₃ is electron-deficient: the aluminium atom has only six electrons in its outer shell. It accepts a lone pair from a chlorine atom in the halogenoalkane or acyl chloride, and pulls the chloride off. This leaves a positive ion, the electrophile:

CHX3CHX2Cl+AlClX3→CHX3CHX2X++AlClX4X− \ce{CH3CH2Cl + AlCl3 -> CH3CH2^+ + AlCl4^-}
CHX3COCl+AlClX3→CHX3COX++AlClX4X− \ce{CH3COCl + AlCl3 -> CH3CO^+ + AlCl4^-}

The ion CH₃CH₂⁺ is a carbocation. The ion CH₃CO⁺ is an acylium ion. Once formed, either attacks the ring in the same two steps as before.

The mechanism 18.5

The first curly arrow goes from the ring to the positively charged carbon of the electrophile. For the acylium ion that is the carbon of C=O. Here is acylation with E = CH₃CO.

Mechanism of Friedel-Crafts acylation. A curly arrow from the ring to the carbon of CH3C+=O. An intermediate with a positive horseshoe over five carbons and H and the acyl group on the sixth. A curly arrow from the C-H bond back into the ring, giving phenylethanone and H+.
Friedel-Crafts acylation: the electrophile is CH3CO+. Alkylation is the same with CH3CH2+ in place of CH3CO+.

For alkylation, write the same mechanism with CH₃CH₂⁺ as the electrophile. The ring is joined through the CH₂ carbon, giving ethylbenzene.

Regenerating the catalyst

HX++AlClX4X−→HCl+AlClX3 \ce{H^+ + AlCl4^- -> HCl + AlCl3}

The AlCl₃ is regenerated, and the by-product HCl is formed.

Worked exampleAcylation with propanoyl chloride

Write the equation, the formation of the electrophile and the product for benzene reacting with propanoyl chloride, CH₃CH₂COCl, and AlCl₃.

  1. Electrophile: CHX3CHX2COCl+AlClX3→CHX3CHX2COX++AlClX4X−\ce{CH3CH2COCl + AlCl3 -> CH3CH2CO^+ + AlCl4^-}
  2. Curly arrow from the ring to the carbonyl carbon of CH₃CH₂CO⁺. The intermediate has a horseshoe over five carbons, + inside, and H and CH₃CH₂CO on the sixth.
  3. Curly arrow from the C–H bond back into the ring, losing H⁺. Catalyst regenerated: HX++AlClX4X−→HCl+AlClX3\ce{H^+ + AlCl4^- -> HCl + AlCl3}
  4. Overall: CX6HX6+CHX3CHX2COCl→CX6HX5COCHX2CHX3+HCl\ce{C6H6 + CH3CH2COCl -> C6H5COCH2CH3 + HCl}. The product is 1-phenylpropan-1-one, a ketone.

Worked exampleAtom economy of making phenylethanone

Calculate the atom economy of the acylation of benzene with ethanoyl chloride, C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl.

  1. M(C₆H₅COCH₃) = 8 × 12.0 + 8 × 1.0 + 16.0 = 120.0 g mol⁻¹
  2. M(HCl) = 1.0 + 35.5 = 36.5 g mol⁻¹
  3. Atom economy = 120.0 ÷ (120.0 + 36.5) × 100 = 120.0 ÷ 156.5 × 100 = 76.7%

Check yourself

  1. Name the organic product and write the equation for the reaction of benzene with chloroethane and AlCl₃.
  2. Give the reagents and conditions needed to make phenylethanone from benzene.
  3. Explain how AlCl₃ acts as a catalyst in a Friedel-Crafts reaction.
  4. Why must the apparatus be dry?
  5. Calculate the atom economy of making ethylbenzene from benzene and chloroethane, C₆H₆ + CH₃CH₂Cl → C₆H₅CH₂CH₃ + HCl.
  6. Outline the mechanism for the reaction of benzene with ethanoyl chloride and AlCl₃, including the formation of the electrophile.

Answers cover these until you have tried the questions

  1. Ethylbenzene. CX6HX6+CHX3CHX2Cl→CX6HX5CHX2CHX3+HCl\ce{C6H6 + CH3CH2Cl -> C6H5CH2CH3 + HCl}
  2. Ethanoyl chloride, CH₃COCl, and anhydrous aluminium chloride catalyst, heated under reflux in dry conditions.
  3. The aluminium atom in AlCl₃ is electron-deficient, so it accepts a lone pair from chlorine and pulls Cl⁻ off the reagent, forming the electrophile and AlCl₄⁻. After the ring loses H⁺, HX++AlClX4X−→HCl+AlClX3\ce{H^+ + AlCl4^- -> HCl + AlCl3} regenerates AlCl₃.
  4. AlCl₃ reacts with water, so the catalyst would be destroyed. Acyl chlorides also react with water.
  5. M(C₆H₅CH₂CH₃) = 8 × 12.0 + 10 × 1.0 = 106.0. M(HCl) = 36.5. Atom economy = 106.0 ÷ (106.0 + 36.5) × 100 = 106.0 ÷ 142.5 × 100 = 74.4%.
  6. Electrophile: CHX3COCl+AlClX3→CHX3COX++AlClX4X−\ce{CH3COCl + AlCl3 -> CH3CO^+ + AlCl4^-}. Curly arrow from the ring to the carbon of CH₃CO⁺. Intermediate: horseshoe over five carbons, + inside, H and CH₃CO on the sixth carbon. Curly arrow from the C–H bond back into the ring, losing H⁺. Product: phenylethanone. Catalyst regenerated: HX++AlClX4X−→HCl+AlClX3\ce{H^+ + AlCl4^- -> HCl + AlCl3}