The total number of isomers for the compounds of the formula `C_4H_10O` are

Options

A. 7
B. 6
C. 3
D. 4

Answer

To determine the total number of isomers for the compound with the formula \( C_4H_{10}O \), we will examine both alcohols and ethers that can be formed from this molecular formula.

### Step 1: Identify the Types of Compounds
The formula \( C_4H_{10}O \) can represent two types of compounds:
1. Alcohols
2. Ethers

### Step 2: Draw Alcohol Structures
We will first identify the possible alcohols that can be formed:

1. **Butanol (1-butanol)**:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \]
(This is a primary alcohol.)

2. **Isobutanol (2-butanol)**:
\[ \text{CH}_3\text{CH}\text{(OH)}\text{CH}_2\text{CH}_3 \]
(This is a secondary alcohol.)

3. **Tert-butanol (2-methyl-2-propanol)**:
\[ \text{(CH}_3)_3\text{COH} \]
(This is a tertiary alcohol.)

4. **Sec-butanol (3-butanol)**:
\[ \text{CH}_3\text{CH}_2\text{CH(OH)}\text{CH}_3 \]
(This is another secondary alcohol.)

### Step 3: Count Alcohol Isomers
From the above structures, we have identified **4 alcohol isomers**:
- 1-butanol
- 2-butanol
- 3-butanol
- tert-butanol

### Step 4: Draw Ether Structures
Next, we will identify the possible ethers:

1. **Diethyl ether**:
\[ \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \]

2. **Ethyl methyl ether**:
\[ \text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3 \]

3. **Methyl tert-butyl ether**:
\[ \text{CH}_3\text{O(CH}_3)_3 \]

### Step 5: Count Ether Isomers
From the above structures, we have identified **3 ether isomers**:
- Diethyl ether
- Ethyl methyl ether
- Methyl tert-butyl ether

### Step 6: Calculate Total Isomers
Now, we add the number of alcohol isomers and ether isomers:
\[ \text{Total Isomers} = \text{Number of Alcohol Isomers} + \text{Number of Ether Isomers} = 4 + 3 = 7 \]

### Conclusion
The total number of isomers for the compounds of the formula \( C_4H_{10}O \) is **7**.

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