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Which halides are unreactive in a Friedel–Crafts alkylation reaction?

Short Answer

Expert verified

The reaction involving the substitution of an alkyl group on a benzene ring by the replacement of a hydrogen atom is named the Friedel-Crafts reaction.

Alkyl halides in combination with a Lewis acid catalyst are used in Friedel-Crafts alkylation reactions.

Step by step solution

01

Friedel–Crafts alkylation reaction

The reaction involving the substitution of an alkyl group on a benzene ring by the replacement of a hydrogen atom is named the Friedel-Crafts reaction.

Alkyl halides in combination with a Lewis acid catalyst are used in Friedel-Crafts alkylation reactions.

02

Halides that undergo Friedel–Crafts alkylation reaction

Only the alkyl halides in which the halogen atom is bonded to a\({\bf{s}}{{\bf{p}}^{\bf{3}}}\)hybridized carbon undergo Friedel-Crafts alkylation reactions.

Carbon atoms bonded to four other atoms/groups are said to be \(s{p^3}\) hybridized.

03

Identifying the unreactive alkyl halides

Here, the halide group is attached to an \(s{p^2}\) hybridized carbon atom and is therefore unreactive towards Friedel-Crafts alkylation reaction.

Here, the halide group is attached to an \(s{p^3}\) hybridized carbon atom and is therefore reactive towards Friedel-Crafts alkylation reaction.

Here, the halide group is attached to an \(s{p^2}\) hybridized carbon atom and is therefore unreactive towards Friedel-Crafts alkylation reaction.

Here, the halide group is attached to an \(s{p^3}\) hybridized carbon atom and is therefore reactive towards Friedel-Crafts alkylation reaction.

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Benzyl bromide\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br)}}\) reacts rapidly with \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\)to afford benzyl methyl ether\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OC}}{{\bf{H}}_{\bf{3}}}{\bf{)}}\). Draw a stepwise mechanism for the reaction, and explain why this 1° alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an\({{\bf{S}}_{\bf{N}}}{\bf{1}}\) mechanism. Would you expect the para-substituted benzylic halides \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{O}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) and \({{\bf{O}}_{\bf{2}}}{\bf{N}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) to each be more or less reactive than benzyl bromide in this reaction?Explain your reasoning.

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