(Z) -3- bromo -3- hexene when treated with `CH_(3)O^(-)" in "CH_(3)OH` gives

Options

A. 3 - hexyne
B. 2 - hexyne
C. 2, 3 - hexadiene
D. 2, 4 - hexadiene

Answer

To solve the question regarding the reaction of (Z)-3-bromo-3-hexene with CH₃O⁻ in CH₃OH, we can follow these steps:

### Step 1: Identify the Structure of (Z)-3-bromo-3-hexene
- (Z)-3-bromo-3-hexene has a double bond at the 3rd carbon, with a bromine atom attached to the same carbon. The structure can be represented as follows:

```
CH3-CH2-C(Br)=C(CH3)-CH2-CH3
```

- The (Z) configuration indicates that the higher priority groups (the bromine and the ethyl group) are on the same side of the double bond.

### Step 2: Recognize the Role of CH₃O⁻
- CH₃O⁻ (methoxide ion) is a strong base and will act to deprotonate the more acidic hydrogen atom adjacent to the double bond.

### Step 3: Determine the Acidic Proton
- The acidic protons are those on the sp² carbon (the carbon involved in the double bond). In this case, the hydrogen atom on the carbon adjacent to the bromine (C-2) is more acidic than the one on the sp³ carbon (C-4) due to the higher s-character of sp² hybridized carbons.

### Step 4: Deprotonation
- The methoxide ion will abstract the acidic proton from the sp² carbon (C-2), resulting in the formation of a carbanion at the C-2 position:

```
CH3-CH2-C(-)=C(Br)-CH2-CH3
```

### Step 5: Bromine as a Leaving Group
- The carbanion formed is now a good nucleophile. The bromine atom (Br) is a good leaving group and will leave, allowing the double bond to shift and form a triple bond between C-2 and C-3.

### Step 6: Formation of the Final Product
- After bromine leaves, the double bond shifts to form a triple bond, resulting in the final product, which is 3-hexyne:

```
CH3-CH2-C≡C-CH2-CH3
```

### Conclusion
- The final product of the reaction is 3-hexyne.

### Final Answer
The product formed when (Z)-3-bromo-3-hexene is treated with CH₃O⁻ in CH₃OH is **3-hexyne**.

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