Diethyl ether , when heated with excess HI produces __________.

Options

A. ethanol
B. iodoform
C. ethyl iodide
D. methyl iodide

Answer

To determine the product formed when diethyl ether is heated with excess HI, we can follow these steps:

### Step 1: Understand the Structure of Diethyl Ether
Diethyl ether has the molecular formula C2H5OC2H5. It consists of two ethyl groups (C2H5) connected by an oxygen atom (O).

### Step 2: Reaction with HI
When diethyl ether is treated with hydroiodic acid (HI), the hydrogen ion (H⁺) will protonate the oxygen atom in the ether. The lone pairs of electrons on the oxygen will attract the H⁺ ion, leading to the formation of a protonated ether.

### Step 3: Formation of Carbocations
The protonation of the ether makes the oxygen positively charged, which destabilizes the ether bond. This can lead to the cleavage of the C-O bond, resulting in the formation of ethyl carbocations (C2H5⁺) and an alcohol (C2H5OH).

### Step 4: Nucleophilic Attack by Iodide Ion
In the presence of excess HI, the iodide ion (I⁻) will act as a nucleophile. It will attack the ethyl carbocation, leading to the formation of ethyl iodide (C2H5I).

### Step 5: Final Products
Since there are two ethyl groups in diethyl ether, the reaction will produce two molecules of ethyl iodide (C2H5I) and one molecule of water (H2O) as a byproduct.

### Conclusion
The final products when diethyl ether is heated with excess HI are:
**2 Ethyl iodide (C2H5I) and 1 Water (H2O)**.

---