Answer:
The correct answer is A) Organic layer, aqueous solution. Any excess bromine that persists after the reaction will mostly be left in the organic layer and quenched with the aqueous solution of sodium bisulfite.
During the bromination of E-stilbene, a reaction occurs between E-stilbene and bromine to form 1,2-dibromo-1,2-diphenylethane. This reaction is commonly carried out in an organic solvent such as dichloromethane or carbon tetrachloride. Bromine is a strong oxidizing agent and reacts readily with the double bond of E-stilbene.
After the reaction is complete, any excess bromine that remains unreacted will be present in the reaction mixture. To remove this excess bromine, it is necessary to quench it with a suitable reagent. Sodium bisulfite (NaHSO3) is commonly used for this purpose. Sodium bisulfite reacts with bromine to form sodium bromide (NaBr) and sulfurous acid (H2SO3). The reaction can be represented as follows:
Br2 + NaHSO3 → NaBr + H2SO3
The sodium bromide formed is soluble in water, while the sulfurous acid remains in the aqueous phase. Therefore, when the reaction mixture is treated with sodium bisulfite, any excess bromine will transfer from the organic layer to the aqueous layer.
The organic layer refers to the layer containing the organic solvent and the product of the reaction (1,2-dibromo-1,2-diphenylethane). The aqueous layer refers to the layer containing water and any dissolved or reacted species such as sodium bisulfite and sodium bromide.
By adding sodium bisulfite to the reaction mixture and allowing it to react with any excess bromine, we ensure that all unreacted bromine is removed from the organic layer. This step is crucial to prevent any unwanted side reactions or further bromination of the product.
In summary, any excess bromine that persists after the bromination of E-stilbene will mostly be left in the organic layer and quenched with the aqueous solution of sodium bisulfite.
Step-by-step explanation: