Chemistry - Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?
Solution 1:
In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the alkyl group). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at E2 eliminations.
Solution 2:
$\ce{OH-}$ acts as a nucleophile. Reactions carried out in alcohol tend to be elimination reactions, and reactions carried out in water (aqueous) tend to be substitution reactions.
If water were used as a solvent in an elimination reaction involving $\ce{KOH}$, the equilibrium would be shifted towards the reactants (water reacting with product), so substitution is favored.
Solution 3:
The reason is quite straightforward-
$\ce{OH^-}$ is a weak base and a stronger nucleophile specially under polar protic conditions. Hence substitution occurs.
$\ce{RO^-}$ is a strong base owing to the inductive effect (+ I effect) of the R group. Hence under alcoholic conditions, $\ce{RO^-}$ extracts the $\beta$ hydrogen of the halides and gives an alkene.
Solution 4:
Aqueous KOH is more solvated by water and not able to abstract H+from substrate but act as nucleophile but alc KOH is less solvated and alkoxide ion can abstract acidic hydrogen