Given:

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In an ether cleavage reaction, the carbon-oxygen bond in an ether molecule is broken, forming two separate compounds.

HI, a strong acid, is involved in the given ether cleavage reaction.

There are two equivalents of HI in the reaction. When two equivalents of a strong acid are involved in ether cleavage, the resulting products will be two haloalkanes and water.

When HI is added to the ether, it protonates the oxygen atom of the ether, making it more susceptible to nucleophilic attack.

Since the given reactant is a tertiary ether (3° C on each side of the oxygen), the ether cleavage will have an SN1 mechanism.

The given ether is not symmetrical. The carbon in the 3° C—O bond that breaks should form the more stable carbocation. It should be highly substituted. The SN1 mechanism produces carbocation and alcohol.

The carbocation produced is then attacked by the nucleophile I− forming the first haloalkane.

Then, the second equivalent of HI protonates the hydroxyl group of the alcohol.

Since the alcohol is a tertiary alcohol, it will undergo an SN1 mechanism.

The 3° C—O bond breaks forming a carbocation and water.

The carbocation produced is then attacked by the nucleophile I− forming the second haloalkane.

Therefore, the products of the ether cleavage reaction given are:

Ether Cleavage Practice Problems