Mechanism of the Reaction:
When 2,3-dibromobutene is treated with metallic zinc in CH3OH, it undergoes a reaction known as the reductive dehalogenation. This reaction involves the reduction of the bromine atoms in the dibromobutene molecule to form new carbon-carbon bonds.

Step 1: Formation of an Alkyl Radical
The reaction begins with the formation of an alkyl radical from the dibromobutene molecule. This is achieved by the reaction of zinc metal with the bromine atoms. The zinc metal donates a pair of electrons to one of the bromine atoms, causing it to dissociate from the molecule as a bromide ion. This generates a carbon-centered radical.

Step 2: Formation of a Carbanion
In the presence of the solvent CH3OH, the carbon-centered radical reacts with a molecule of methanol. The lone pair of electrons on the oxygen atom in methanol attacks the carbon atom of the radical, forming a new bond and resulting in the formation of a carbanion.

Step 3: Protonation of the Carbanion
The carbanion formed in the previous step is highly reactive and unstable. In order to stabilize it, a proton from the solvent CH3OH is transferred to the carbanion. This protonation reaction leads to the formation of an alkene intermediate.

Step 4: Reduction of the Alkene Intermediate
Finally, the alkene intermediate undergoes a reduction reaction in the presence of metallic zinc. The zinc metal donates a pair of electrons to the alkene, resulting in the formation of a new carbon-carbon bond. This leads to the formation of the product, which is 2-butene.

Overall Reaction:
The overall reaction can be summarized as follows:
2,3-dibromobutene + Zn + CH3OH → 2-butene

Key Points:
- The reaction involves the reductive dehalogenation of 2,3-dibromobutene.
- Metallic zinc serves as a reducing agent.
- The reaction occurs in the presence of CH3OH as a solvent.
- The bromine atoms in 2,3-dibromobutene are reduced to form new carbon-carbon bonds.
- The product of the reaction is 2-butene.