One molecule of dialkyl ether produces how many molecules of alkyl halides with excess of halogen acid ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of how many molecules of alkyl halides are produced from one molecule of dialkyl ether when reacted with an excess of halogen acid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Dialkyl Ether**:
A dialkyl ether has the general formula R-O-R', where R and R' are alkyl groups.

**Hint**: Recall that dialkyl ethers consist of two alkyl groups connected by an oxygen atom.

2. **Reaction with Halogen Acid**:
When a dialkyl ether reacts with a halogen acid (such as HCl, HBr, or HI), it undergoes cleavage. The halogen acid donates a proton (H⁺) to the ether oxygen, leading to the formation of an alcohol and an alkyl halide.

**Hint**: Remember that halogen acids can protonate the ether oxygen, facilitating the cleavage of the R-O bond.

3. **Formation of Alkyl Halides**:
The reaction can be represented as follows:
\[ R-O-R' + H-X \rightarrow R-X + R'-OH \]
Here, R-X is the alkyl halide formed from one of the alkyl groups, and R'-OH is the corresponding alcohol.

**Hint**: Identify the products formed in the reaction: one alkyl halide and one alcohol.

4. **Excess Halogen Acid**:
If there is an excess of halogen acid, the alcohol (R'-OH) can further react with another molecule of halogen acid to produce another alkyl halide:
\[ R'-OH + H-X \rightarrow R'-X + H₂O \]
This means that from the alcohol produced, we can generate another alkyl halide.

**Hint**: Consider the role of excess halogen acid in converting the alcohol to another alkyl halide.

5. **Total Alkyl Halides Produced**:
From one molecule of dialkyl ether, we produce:
- One alkyl halide from the initial cleavage.
- One additional alkyl halide from the subsequent reaction of the alcohol with halogen acid.

Therefore, the total number of alkyl halides produced is 2.

**Hint**: Count the alkyl halides produced from both the initial reaction and the subsequent reaction.

### Final Answer:
One molecule of dialkyl ether produces **2 molecules of alkyl halides** when reacted with an excess of halogen acid.