An mixture of two organic chlorine compounds was treated with sodium metal in ether solution. Isobutane was obtained as a product. The two chlorine compounds are:

Options

A. Methyl chloride and propyl chloride
B. Methyl chloride and ethyl chloride
C. Isopropyl chloride and methyl chloride
D. Isopropyl chloride and ehtyl chloride

Answer

To solve the problem of identifying the two organic chlorine compounds that yield isobutane when treated with sodium metal in ether, we can follow these steps:

### Step 1: Understand the Reaction
The reaction described is a Wurtz reaction, where two alkyl halides react with sodium metal to form a higher alkane. The general reaction can be represented as:
\[ R-X + R'-X + 2Na \rightarrow R-R' + 2NaX \]
where \( R \) and \( R' \) are alkyl groups and \( X \) is a halogen (in this case, chlorine).

### Step 2: Identify the Product
The product of the reaction is isobutane, which has the structure:
\[ \text{C}_4\text{H}_{10} \]
Isobutane can be represented structurally as:
```
CH3
|
CH3-CH-CH3
```
This indicates that isobutane has a branched structure.

### Step 3: Determine Possible Chlorine Compounds
To form isobutane, we need to consider which two chlorine compounds could combine to yield this structure. The possible combinations of alkyl groups that can produce isobutane are:

1. **Methyl chloride (CH3Cl)** - contributes a methyl group (CH3).
2. **Isopropyl chloride (C3H7Cl)** - contributes an isopropyl group (C3H7).

### Step 4: Write the Reaction
Combining these two compounds in a Wurtz reaction would look like:
\[ \text{CH}_3\text{Cl} + \text{C}_3\text{H}_7\text{Cl} + 2\text{Na} \rightarrow \text{C}_4\text{H}_{10} + 2\text{NaCl} \]
This reaction shows that the methyl group from methyl chloride and the isopropyl group from isopropyl chloride combine to form isobutane.

### Conclusion
The two organic chlorine compounds that yield isobutane when treated with sodium metal in ether are:
1. **Methyl chloride (CH3Cl)**
2. **Isopropyl chloride (C3H7Cl)**