why does on treating 2 bromo pentane with alc. Koh major product is 2 pentene.
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Answer: On treating 2-bromo-pentane with alc. KOH, major product is pent-2-ene because when we treat 2-bromo-pentane with alcoholic KOH then Br and beta position of H gets removed and form double bond in that area of the carbon chain. K and OH gets broken and K forms ionic bond with bromine and H₂O is formed.
Ch₃-Ch(Br)-Ch₂-Ch₂-Ch₃ + KOH(alc.) = Ch₃-Ch=Ch-Ch₂-Ch₃ + KBr + H₂O
This is also known as Beta - Elimation Reaction.
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