Question

Alkyl chloride reacts with aqueous KOH to form alcohol while in the presence of alcoholic KOH, alkene is obtained as major product. Explain. (2)

Hint:

Aqueous \( OH^- \) acts as nucleophile (substitution to alcohol); alcoholic KOH gives alkoxide, a strong base that removes a beta-hydrogen (elimination to alkene).

Answer 1

Step 1: With aqueous KOH (substitution). In water, KOH is almost fully ionised to give hydroxide ions \( OH^- \). The \( OH^- \) ion is a good nucleophile. It attacks the carbon bearing the chlorine and replaces \( Cl^- \) by nucleophilic substitution, giving an alcohol:
\[ R\text{-}CH_2\text{-}CH_2Cl + KOH\ (aq) \rightarrow R\text{-}CH_2\text{-}CH_2OH + KCl \]
Step 2: With alcoholic KOH (elimination). When KOH is dissolved in alcohol, it produces alkoxide ions (for example ethoxide, \( C_2H_5O^- \)). Alkoxide ion is a bulky, strong base (a poor nucleophile because of its size). Instead of attacking carbon, it abstracts a hydrogen from the \( \beta \)-carbon. Loss of this \( \beta \)-hydrogen together with the chlorine (dehydrohalogenation) forms a carbon to carbon double bond, giving an alkene as the major product:
\[ R\text{-}CH_2\text{-}CH_2Cl \xrightarrow{alc.\ KOH} R\text{-}CH=CH_2 + KCl + H_2O \]
Step 3: Reason for the difference. The medium decides whether the reagent acts mainly as a nucleophile or as a base. In water \( OH^- \) behaves as a nucleophile favouring substitution (alcohol). In alcohol the alkoxide behaves as a strong base favouring elimination (alkene). Hence the products differ.
\[\boxed{\text{aq. KOH } \rightarrow \text{ substitution (alcohol);\quad alc. KOH } \rightarrow \text{ elimination (alkene)}}\]