When alkyl halide is treated with aqueous KOH , nucleophillic substitution takes place and when it is treated with alcoholic KOH , alkene is formed.Explain why?​

Dear Student,

When an alkyl halide reacts with aqueous KOH, an alcohol is formed because of nucleophilic substitution. Since aqueous KOH is alkaline in nature, it gives OH- ions which act as nucleophiles and replace the halogen group in the hydrocarbon.

RCl + KOH (aq.) = ROH + KCl

When an alkyl halide reacts with alcoholic KOH, it undergoes elimination to give an alkene. This is because ethoxide ions (C2H5O-) ions are formed in alcoholic KOH, which act as stronger bases than OH- ions. These ions take away the 
β-hydrogen in the alkyl halide to give an alkene.

CH3CH2Cl + KOH(alcoholic) = H2C=CH​2 + KCl + H2O

Hope it helps.

Regards

  • 5
when alkyl halide is treated with alcoholic KOH.....RO- ion is formed which is electron-rich
and since nucleophile is already electron-rich...
so it can't undergo nucleophilic substitution
it undergoes electrophilic subsitution
  • 0
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