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Alcoholic vs Aqueous KOH

The Battle: Substitution vs Elimination.

By chemca Team • Updated Jan 2026

Potassium Hydroxide (KOH) behaves completely differently depending on the solvent used. In water, it acts as a Nucleophile, while in alcohol, it acts as a Strong Base. This dictates whether an Alkyl Halide undergoes Substitution or Elimination.

1. Aqueous KOH ($KOH_{aq}$)

Nucleophilic Substitution

In water, KOH ionizes completely into $K^+$ and $OH^-$. The $OH^-$ ions are highly hydrated (solvated by water). This reduces their basic character but they remain good nucleophiles.

Reaction: The $OH^-$ attacks the carbon bonded to the halogen, displacing the halide ion ($X^-$).

$$ R-X + KOH(aq) \xrightarrow{\Delta} \underset{\text{Alcohol}}{R-OH} + KX $$

Example: Ethyl Chloride $\rightarrow$ Ethanol.

$$ CH_3CH_2Cl + KOH(aq) \rightarrow CH_3CH_2OH + KCl $$

2. Alcoholic KOH ($KOH_{alc}$)

Elimination (Dehydrohalogenation)

In alcohol (ethanol), KOH reacts to form Potassium Ethoxide ($C_2H_5O^-K^+$). The ethoxide ion ($RO^-$) is a much stronger base than $OH^-$. It is also bulkier.

Reaction: The strong base abstracts a proton ($H^+$) from the $\beta$-carbon, causing the halide to leave and forming a double bond.

$$ R-CH_2-CH_2-X + KOH(alc) \xrightarrow{\Delta} \underset{\text{Alkene}}{R-CH=CH_2} + KX + H_2O $$

Example: Ethyl Chloride $\rightarrow$ Ethene.

$$ CH_3CH_2Cl + KOH(alc) \rightarrow CH_2=CH_2 + KCl + H_2O $$

3. Why the Difference?

Feature	Aqueous KOH	Alcoholic KOH
Active Species	Hydrated $OH^-$ (Nucleophile)	Ethoxide $C_2H_5O^-$ (Strong Base)
Role	Nucleophile (Attacks Carbon)	Base (Attacks Proton/H)
Reaction Type	Substitution ($S_N1$ / $S_N2$)	$\beta$-Elimination ($E2$)
Major Product	Alcohol	Alkene

4. Regioselectivity (Saytzeff Rule)

Formation of Major Alkene

If elimination can occur in two directions (unsymmetrical alkyl halide), the major product is the More Substituted Alkene (Saytzeff Product) because it is more stable.

$$ \underset{\text{2-Bromobutane}}{CH_3-CH(Br)-CH_2-CH_3} \xrightarrow{KOH(alc)} $$ $$ \underset{\text{But-2-ene (80\%)}}{CH_3-CH=CH-CH_3} + \underset{\text{But-1-ene (20\%)}}{CH_2=CH-CH_2-CH_3} $$

Hofmann Product: If the base is very bulky (e.g., t-Butoxide) or the leaving group is poor (e.g., Fluorine), the less substituted alkene becomes the major product.

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