Question

C2h4br2+koh????????????

Answer 1

Ответ:

CH2Br-CH2Br + 2KOH --> CH(тройная связь) СН + 2KBr + 2H2O  

Объяснение:

CH(тройная связь) СН - ацетилен

Answer 2

When ethylene dibromide (C2H4Br2) reacts with potassium hydroxide (KOH), it undergoes an elimination reaction known as dehydrohalogenation. The reaction can be represented by the following equation:

C2H4Br2 + 2 KOH -> C2H4 + 2 KBr + 2 H2O

In this reaction, the potassium hydroxide (KOH) acts as a base, abstracting a hydrogen atom from one of the ethylene dibromide molecules. This generates a bromide ion (Br-) and an alkoxide ion (C2H4O-) as intermediate species. The alkoxide ion then reacts with another ethylene dibromide molecule, leading to the elimination of a bromine atom and the formation of ethylene gas (C2H4).

The reaction also produces potassium bromide (KBr) and water (H2O) as byproducts.