No. of acyclic isomers of the compound having the molecular formula `C_(4)H_(10)O` is :

Options

A. 4
B. 6
C. 5
D. 7

Explanation

To determine the number of acyclic isomers of the compound with the molecular formula C4H10O, we can analyze the possible structures based on the functional groups present, namely alcohols (due to the -OH group) and ethers.

### Step 1: Identify the functional groups
The molecular formula C4H10O can represent both alcohols (where -OH is the functional group) and ethers (where the general structure is R-O-R').

### Step 2: Count the isomers for alcohols
For the alcohols, we can derive the following structures:

1. **Butanol (1-butanol)**: CH3-CH2-CH2-CH2-OH
2. **Isobutanol (2-butanol)**: CH3-CH(OH)-CH2-CH3
3. **Tert-butanol (2-methyl-2-propanol)**: (CH3)3C-OH
4. **Cyclobutanol**: A cyclic structure, but not acyclic.

Thus, we have **3 acyclic isomers** for alcohols.

### Step 3: Count the isomers for ethers
For ethers, we can derive the following structures:

1. **Dimethyl ether**: CH3-O-CH3
2. **Ethyl methyl ether**: CH3-CH2-O-CH3
3. **Diethyl ether**: CH3-CH2-O-CH2-CH3

Thus, we have **3 acyclic isomers** for ethers.

### Step 4: Total the acyclic isomers
Now, we add the number of acyclic isomers from both categories:

- Acyclic isomers from alcohols: 3
- Acyclic isomers from ethers: 3

Total acyclic isomers = 3 (from alcohols) + 3 (from ethers) = **6 acyclic isomers**.

### Conclusion
The total number of acyclic isomers of the compound with the molecular formula C4H10O is **6**.

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