What is the basic difference between aqueous and alcoholic $\ce{KOH}$? Why does alcoholic $\ce{KOH}$ prefer elimination whereas aqueous $\ce{KOH}$ prefers substitution?


5 Answers 5


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.

  • $\begingroup$ I don't get how the alcoholate salts are formed. Where is the 'R' group coming from? $\endgroup$
    – User
    Jul 30, 2017 at 4:21
  • 2
    $\begingroup$ @InternetGuy $\ce{ROH}$ is the notation for a generic alcohol, where $\ce{R}$ is usually an alkyl chain. Thus, when $\ce{KOH}$ deprotonates an alcohol, you get an alcoholate salt $\ce{ROK}$, i.e. $\ce{KOH + ROH <<=> ROK + H2O}$. $\endgroup$
    – Philipp
    Jul 30, 2017 at 11:15
  • $\begingroup$ Hmm, isn't $\ce{RO^-}$ often considered a stronger nucleophile that $\ce{OH^-}$? (or did you factor in solvent interactions as well?) :-) $\endgroup$ Dec 31, 2017 at 7:01
  • 1
    $\begingroup$ Does that mean that in ethanolic KOH, it is the ethoxide that is the primary basic species involved in proton abstraction from the substrate. Does that also mean that the concentration of the alkoxide ion would be factored into the rate law? $\endgroup$ Jun 12, 2019 at 3:45
  • $\begingroup$ In carbylamine reaction, why $\ce{OH-}$ is used in mechanism instead of $\ce{EtO^-}$? $\endgroup$
    – Apurvium
    Nov 12, 2021 at 12:34

$\ce{OH-}$ 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.


The reason is quite straightforward-

  1. $\ce{OH^-}$ is a weak base and a stronger nucleophile specially under polar protic conditions. Hence substitution occurs.

  2. $\ce{RO^-}$ 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.


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


See, we know that nucleophiles which are in polar protic solvents are solvated and therefore nucleophilicity is directly proportional to the size of nucleophile 'cause they will be less solvated. And in polar aprotic solvent the anions are not solvated so they will show their usual nucleophilicity behaviour and is inversely proportional to the stability of anion. Hence, in alcoholic medium the $\ce{OH-}$ is less solvated and acts as a strong base and eliminates acidic $\ce{H+}$ whereas in aqueos medium $\ce{OH-}$ shows nucleophilicity and prefers substitution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.