Question

Suggest a product of the following reaction. HI is a very strong acid. $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}+2 \mathrm{HI} \longrightarrow $$

Short answer

Answer: The products of the reaction between ethyl methyl ether and hydroiodic acid are two molecules of alkyl iodide (CH3CH2I) and a molecule of water (H2O).

Step-by-step solution

Identifying the reactants

The reactants for this reaction are the ether, CH3CH2OCH2CH3 (ethyl methyl ether), and two molecules of HI (hydroiodic acid), which is a very strong acid.

Protonation of the ether

Since HI is a very strong acid, it will protonate the oxygen atom in the ether, leading to the formation of an oxonium ion (a positively charged oxygen with three bonds). The reaction proceeds as follows: $$ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OCH}_{2}\mathrm{CH}_{3} + \mathrm{HI} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}^{+}\mathrm{CH}_{2}\mathrm{CH}_{3} + \mathrm{I}^{-} $$

Cleavage of the C-O bond

The positive charge on the oxygen atom in the oxonium ion makes the adjacent C-O bond more susceptible to nucleophilic attack. The iodide ion (I-) will attack one of the carbons attached to the oxygen, causing the C-O bond to break and form a new bond between carbon and iodine. This step results in the formation of an alkyl iodide. There are two possible products depending on which carbon the iodide attacks: $$ \text{Product 1: }\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} + \mathrm{CH}_{3}\mathrm{OH} $$ $$ \text{Product 2: }\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} + \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} $$

Second HI reacts with the remaining hydroxyl group

The second HI molecule will react with the remaining hydroxyl group (OH) in either Product 1 or Product 2. This substitution reaction replaces the hydroxyl group with an iodine atom, forming another alkyl iodide: $$ \text{Product 1: }\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} + \mathrm{HI} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} + \mathrm{H}_{2}\mathrm{O} $$ or $$ \text{Product 2: }\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}+\mathrm{HI} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} + \mathrm{H}_{2}\mathrm{O} $$

Combining the products

Based on Steps 3 and 4, the final product of the reaction will include a combination of two alkyl iodides and a water molecule: $$ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OCH}_{2}\mathrm{CH}_{3} + 2\mathrm{HI} \longrightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} + \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I} + \mathrm{H}_{2}\mathrm{O} $$