The number of isomer (including stereoisomers) of `C_(5)H_(12)O` which can give positive haloform test is

Explanation

To solve the question regarding the number of isomers (including stereoisomers) of C5H12O that can give a positive haloform test, we will follow these steps:

### Step 1: Understand the Haloform Test
The haloform test is positive for compounds that contain either a methyl ketone (CH3C=O) or a secondary alcohol that can be oxidized to a methyl ketone (CH3CHOH).

### Step 2: Identify the General Formula
The molecular formula given is C5H12O. This corresponds to the general formula for alcohols and ethers, which can be represented as CnH2n+2O. For n = 5, we have C5H12O.

### Step 3: Determine Possible Structures
Since ethers do not give a haloform test, we will focus on monohydric alcohols. We need to find alcohols that can either be a methyl ketone or have the structure CH3CHOH.

### Step 4: Identify Relevant Alcohols
We need to find monohydric alcohols that contain the CH3CHOH group. The only alcohol that fits this criterion is 2-pentanol (CH3-CHOH-CH2-CH2-CH3).

### Step 5: Check for Stereoisomers
The carbon atom bearing the -OH group in 2-pentanol is a chiral center because it has four different substituents (CH3, OH, CH2, and CH2CH3). This means that 2-pentanol can exist as two enantiomers (stereoisomers).

### Step 6: Count the Isomers
Thus, we have:
1. 2-pentanol (first isomer)
2. The enantiomer of 2-pentanol (second isomer)

### Conclusion
The total number of isomers (including stereoisomers) of C5H12O that can give a positive haloform test is **2**.

### Final Answer
The answer is **2**.

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