The reaction of 1,2-dibromoethane with alcoholic KOH yields

Join Our Telegram Channel

ethene

ethyne

1-bromo-2-hydroxyethane

1-bromoethene

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC Geoscientist – 2023

Download PDF Attempt Online

1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene).

Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.

Step 1: CH₂Br-CH₂Br + KOH (alc.) → CH₂=CHBr + KBr + H₂O
Step 2: CH₂=CHBr + KOH (alc.) → HC≡CH + KBr + H₂O
The final product is ethyne. Ethene would result from single dehydrohalogenation of a mono-halogenated alkane or dehalogenation of a vicinal dihalide using a different reagent (e.g., Zn dust). 1-bromo-2-hydroxyethane would be a substitution product, which is less favored with alcoholic KOH, which promotes elimination. 1-bromoethene is an intermediate in the reaction, not the final product under excess alcoholic KOH.

More similar MCQ questions for Exam preparation:-