Question

Why does potassium hydroxide react differently with alkyl halides?

Answer 1

You have not asked a question, but I will make an attempt at an answer....

Explanation:

You gots substitution....

`RCH_2CH_2Cl(aq) + KOH(aq) stackrel(H_2O)rarrRCH_2CH_2OH(aq) + KCl(aq)`

...versus elimination....

`RCH_2CH_2Cl(aq) + KOH(aq) stackrel(EtOH)rarrRCH=CH_2(aq) + KCl(aq)+H_2O(l)`

Any manifestation of reactivity is modified by the identity of the solvent. Hydroxide is a more powerful base in the alcoholic solvent, in that the alcohol is LESS able to stabilize positive and negative ions. The hydroxide anion is effectively MORE basic in the alcoholic solvent, and its basicity IS ENHANCED...

In alcohol, the hydroxide thus acts as a base, and elimination occurs. On the other hand, in aqueous solution, hydroxide anion is better solvated, and substitution occurs. Note that we would anticipate BOTH products in such a reaction. Conditions are tweaked to favour one or the other.....