How many alkynes isomers are formed with molecular formula `C_(4)H_(6)` ?

Answer

To determine how many alkyne isomers can be formed with the molecular formula \( C_4H_6 \), we will follow these steps:

### Step 1: Calculate the Double Bond Equivalent (DBE)
The formula for calculating the double bond equivalent (DBE) is:
\[ \text{DBE} = \frac{(2C + 2 - H)}{2} \]
Where \( C \) is the number of carbon atoms and \( H \) is the number of hydrogen atoms.

For \( C_4H_6 \):
- \( C = 4 \)
- \( H = 6 \)

Substituting the values:
\[ \text{DBE} = \frac{(2 \times 4 + 2 - 6)}{2} = \frac{(8 + 2 - 6)}{2} = \frac{4}{2} = 2 \]

### Step 2: Identify Possible Structures
A DBE of 2 indicates that there can be:
- 2 double bonds, or
- 1 triple bond, or
- 1 double bond and 1 ring.

However, since we are specifically looking for alkynes (which contain a triple bond), we will focus on structures that contain a triple bond.

### Step 3: Draw the Possible Alkynes
1. **1-Butyne**:
- Structure: \( CH_3C \equiv CH \)
- This structure has the triple bond between the first and second carbon atoms.

2. **2-Butyne**:
- Structure: \( CH_3C \equiv CCH_3 \)
- This structure has the triple bond between the second and third carbon atoms.

### Step 4: Count the Isomers
From the structures drawn, we can see that there are two distinct alkynes:
- 1-Butyne
- 2-Butyne

### Conclusion
Thus, the total number of alkyne isomers that can be formed with the molecular formula \( C_4H_6 \) is **2**.

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