(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 problem of what happens when (Z)-3-bromo-3-hexene is treated with CH₃O⁻ in CH₃OH, we can follow these steps:

### Step 1: Understand the Structure of (Z)-3-bromo-3-hexene
- (Z)-3-bromo-3-hexene has a double bond at the 3rd carbon of the hexane chain, with a bromine atom attached to the same carbon. 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: Identify the Reactant
- The reactant CH₃O⁻ is a methoxide ion, which is a strong base. It can abstract protons from the molecule.

### Step 3: Proton Abstraction
- The methoxide ion will abstract a proton from the carbon adjacent to the double bond (C-2 or C-4). Since the double bond is at C-3, we will focus on the proton at C-2, which is sp² hybridized and more acidic than the sp³ hybridized carbons.

### Step 4: Formation of a Carbanion
- When the proton is abstracted from C-2, a carbanion is formed at that position. This carbanion will be stabilized by the adjacent double bond.

### Step 5: Elimination of Bromine
- The presence of the carbanion at C-2 will facilitate the elimination of the bromine atom at C-3. This results in the formation of a triple bond between C-2 and C-3.

### Step 6: Resulting Product
- The final product will be 3-hexyne, which is a linear alkyne. The structure will have the triple bond between C-2 and C-3, with the ethyl groups remaining on C-1 and C-4.

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

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