Boron tribromide$\left({\mathrm{BBr}}_3\right)$cleaves ethers to give alkyl halides and alcohols.

The reaction is thought to involve attack by a bromide ion on the Lewis acid-base adduct of the ether with$\mathbf{\left(}{\mathbf{BBr}}_{\mathbf{3}}\mathbf{\right)}$(a strong Lewis acid). Propose a mechanism for the reaction of butyl methyl ether with$\mathbf{\left(}{\mathbf{BBr}}_{\mathbf{3}}\mathbf{\right)}$to give (after hydrolysis) butan-1-ol and bromomethane.

Butyl methyl ether on reaction with boron tribromide forms an intermediate in which oxygen of ether is involved in back bonding with boron of boron tribromide as boron has vacant-p orbital and oxygen has lone pairs, hence, back bonding occurs. Bromide ion acts as a leaving group and further, oxygen neutralizes the positive charge that is formed due to back bonding and bromide ion which left as leaving group, now acts as a nucleophile and attacks at methyl group and forms methyl bromide or bromomethane as side product.

Reaction of butyl methyl ether with boron tribromide

In next step, again back bonding occurs between oxygen of ether and boron due to which oxygen acquires positive charge and after charge neutralization, butanol leaves from the molecule as leaving group and forms as the product along with ${\mathbf{HOBBr}}_{\mathbf{2}}$.

${\mathrm{HOBBr}}_2$on reaction with water forms boric acid and hydrogen bromide as the side products.

Formation of butan-1-ol