Ph-CH(Br)-CH(Br)-Ph -> Zn/CH3OH
Answer (1)
This question involves a chemical reaction, specifically a dehalogenation reaction using zinc (Zn) in methanol (CH₃OH).
In this reaction, we start with a compound that has two bromine atoms attached to adjacent carbon atoms in the structure Ph-CH(Br)-CH(Br)-Ph. The 'Ph' stands for a phenyl group, which is a benzene ring (C₆H₅). Therefore, the full structure is a dibromoalkane where bromine atoms are attached to each carbon in a chain between two phenyl groups.
When this compound is treated with zinc in methanol, a dehalogenation reaction occurs. Here's a step-by-step breakdown of the process:
Starting Compound: The compound is 1,2-dibromo-1,2-diphenylethane.
Reagent: Zn/CH₃OH (zinc in methanol).
Reaction Process: Zinc acts as a reducing agent. It facilitates the removal of the bromine atoms by converting them into zinc bromide (ZnBr₂).
Result: The removal of the bromine atoms leads to the formation of a double bond between the two carbon atoms where the bromine atoms were originally attached. This results in the formation of trans-stilbene or just stilbene (Ph-CH=CH-Ph).
Why This Occurs:
Zinc is a metal that effectively donates electrons. These electrons reduce the bromine atoms to bromide ions, which are then bound by zinc as ZnBr₂. This enables the formation of a carbon-carbon double bond.
Conclusion:
The reaction of Ph-CH(Br)-CH(Br)-Ph with Zn/CH₃OH results in the formation of stilbene (Ph-CH=CH-Ph) by removing the bromine atoms and forming a double bond in their place.
This type of reaction is commonly discussed in organic chemistry as part of learning about halogenated hydrocarbons and methods of forming alkenes.
