Question
How do the products differ when ethyl bromide reacts separately with aq. KOH and alc. KOH?
Solution
This can be explain in the following ways:
When alkyl chloride react with aqueous KOH it forms alcohol. it is subsitution reaction.
C2H5Cl +KOH(aq.) ---> C2H5OH +KCl
When alkyl chloride react with alcoholic KOH it forms alkene. In this case elimination reaction take place.
C2H5Cl +KOH(alc.) ---> H2C=CH2 +KCl +H2O