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Q7.31 (iii):
Write equations of the following reactions: Bromination of anisole in ethanoic acid medium.

Solution :

Step 1: Chemical Equation Representation

The bromination of anisole ($C_6H_5OCH_3$) in an ethanoic acid medium is an electrophilic aromatic substitution reaction. Anisole contains a methoxy ($-OCH_3$) group, which is a strongly activating group and an ortho/para directing group due to the resonance effect ($+R$ effect), which increases electron density at the ortho and para positions.

The chemical equation for the reaction is as follows:

$C_6H_5OCH_3 + Br_2 \xrightarrow{CH_3COOH} C_6H_4(Br)OCH_3 + HBr$

Step 2: Analysis of Regioselectivity

Due to the steric hindrance provided by the $-OCH_3$ group at the ortho position, the para-isomer is the major product. The reaction proceeds as follows:

Product Name Yield Characterization
$o-Bromoanisole$ $1-bromo-2-methoxybenzene$ Minor product
$p-Bromoanisole$ $1-bromo-4-methoxybenzene$ Major product

Step 3: Visual Representation of the Substitution

Anisole OCH₃ → Br₂, CH₃COOH OCH₃ Br

Step 4: Theoretical Justification

  • Activating Nature: The lone pair of electrons on the oxygen atom of the methoxy group participates in delocalization into the benzene ring [Resonance Effect].
  • Directing Influence: The ortho and para positions experience an increase in electron density, making them highly susceptible to electrophilic attack by the bromonium ion ($Br^+$) generated in the medium [Electrophilic Aromatic Substitution mechanism].
  • Solvent Effect: Ethanoic acid ($CH_3COOH$) acts as a polar solvent that facilitates the heterolytic cleavage of $Br_2$ to initiate the reaction.

Final Solution: The bromination of anisole in ethanoic acid yields a mixture of $o-bromoanisole$ (minor) and $p-bromoanisole$ (major), with $p-bromoanisole$ being the predominant product due to decreased steric hindrance at the para position.


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