Question #1e366

1 Answer
Nov 13, 2017

Alcoholic #KOH# is mostly used in elimination reactions.

Explanation:

Reason:
Alcoholic #KOH#(mostly having ethanol) produces #C_2H_5O^-# ions which act as a stronger base. So these have the ability to abstract the #beta#- hydrogen from the substrate molecule(alkyl halide), forming alkene as a product. That's the reason behind the statement; Alcoholic #KOH# give elimination reactions.

Example:

#CH_3CH_2Br + KOH(alcoholic) to H_2C=CH_2 + KBR +H_2O#

This is an example of elimination reactions of alkyl halides.

Hope it helps...