In the given reaction $PhOMe\xrightarrow[2){{H}_{2}}O]{1)BB{{r}_{3}}}\_\_\_\_\_+\_\_\_\_\_+\_\_\_\_\_$
State the products

Hint: Think about what will happen when boron tribromide reacts with an ether. After boron tribromide has caused cleavage of the ether, hydrolysis occurs and the products are formed.

Complete step by step solution:
Boron tribromide is a reagent that is used to cause the cleavage of dialkyl ether to form an alcohol, boric acid and alkyl bromide. This is a two-step reaction and requires boron tribromide in the first step and hydrolysis with water in the second step. The reaction mechanism is as follows:
i) One mole of phenyl methyl ether reacts with one mole of $BB{{r}_{3}}$ to give an intermediate complex that has benzene bonded to an oxygen atom which is further bonded to boron and two bromine atoms. It also gives a mole of methyl bromide as a byproduct.

ii) This complex can only be converted to a phenol, which is the aim of the reaction, if it is hydrolyzed by three molecules of water. Two molecules of water first hydrolyze the bonds between boron and bromine, the hydrogen proton is taken by bromine and the hydroxide ion attacks the boron atom. The remaining hydrogen atom hydrolyzes the bond between oxygen and boron where boron gets the hydroxide ion and oxygen the hydrogen proton. The reaction is as follows:
\[PhOBB{{r}_{2}}+3{{H}_{2}}O\to PhOH+{{H}_{3}}B{{O}_{3}}\]

Hence, the three products that are formed after this reaction are phenol, boric acid, and methyl bromide.

Note: The products formed that we are considering here are all the products that we will find in the vessel of the reaction after both the reagents have been added. We will consider the net products obtained and not just the ones formed in only the first or the second step.