2-chlorobutane is treated with alcoholic koh
Hi aspirant,
Alcoholic KOH is a dehydrating agent and causes dehydrohalogenation in a chemical reaction.
When 2 chlorobutane is treated with alcoholic KOH, HCL gas is released with formation of but-2-ene as the major product.
Note- If aqueous KOH is used then substitution reaction takes place instead of of the dehydrohalogenation.
Hope it helps!!
Hello,
Here the concept is beta- elimination reaction.When an alkyl halide interacts with alcKOH (or alc NaOH), the halide undergoes beta-elimination, which results in the production of an alkene.
The major product is based on Saytzeff's rule.
According to the Saytzeff or Zaitsev Rule, the more substituted an alkene is, the more stable hence it is the major product. The degree of substitution and hence major and minor products may be assessed by counting the number of alkyl groups linked to the alkene.
CH3CHCl CH2 CH3 + KOH (alc.) → CH2-CH= CH-CH3 + HCl (major)
Hence the major product is 2-Butene and minor product will be 1-Butene
use the link for the concept used
Hope it helped
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