How many acyclic structural isomeric carbonyl compound having molecular formula `C_(6)H_(12)O` can gives haloform test.

Explanation

To determine how many acyclic structural isomeric carbonyl compounds with the molecular formula C6H12O can give the haloform test, we will follow these steps:

### Step 1: Understand the Haloform Test
The haloform test is specific for methyl ketones, which have the structure RCOCH3, where R can be a hydrogen atom, alkyl, or aryl group. Only those carbonyl compounds that contain a methyl group adjacent to the carbonyl group (C=O) can give a positive haloform test.

### Step 2: Identify the Possible Carbonyl Compounds
The molecular formula C6H12O indicates that we are dealing with acyclic carbonyl compounds (aldehydes and ketones). We need to find structural isomers of C6H12O that can be classified as methyl ketones.

### Step 3: Generate Structural Isomers
1. **Hexan-2-one**:
- Structure: CH3-CO-CH2-CH2-CH2-CH3
- This is a methyl ketone.

2. **Hexan-3-one**:
- Structure: CH3-CH2-CO-CH2-CH2-CH3
- This is also a methyl ketone.

3. **2-Pentanone**:
- Structure: CH3-CO-CH(CH3)-CH2-CH3
- This is a methyl ketone.

4. **3-Pentanone**:
- Structure: CH3-CH2-CO-CH(CH3)-CH3
- This is a methyl ketone.

5. **2-Methyl-2-pentanone**:
- Structure: (CH3)2C(=O)CH2CH3
- This is a methyl ketone.

### Step 4: Count the Methyl Substituted Ketones
From the above structural isomers, we have identified the following methyl ketones:
1. Hexan-2-one
2. Hexan-3-one
3. 2-Pentanone
4. 3-Pentanone
5. 2-Methyl-2-pentanone

### Conclusion
Thus, the total number of acyclic structural isomeric carbonyl compounds having the molecular formula C6H12O that can give the haloform test is **5**.

---