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Q6.22(iv):
What happens when (iv) ethyl chloride is treated with aqueous KOH,

Solution :

Step 1: Chemical Formulae and Reactants Identification

The reaction involves ethyl chloride, which is a primary alkyl halide, and an aqueous solution of potassium hydroxide ($KOH$).

  • Reactant 1: Ethyl chloride (Chloroethane) $\rightarrow$ $CH_3CH_2Cl$
  • Reactant 2: Aqueous Potassium Hydroxide $\rightarrow$ $KOH_{(aq)}$

Step 2: Reaction Mechanism Analysis

When an alkyl halide is treated with an aqueous solution of a strong base (like $KOH$ or $NaOH$), the hydroxide ion ($OH^-$) acts as a strong nucleophile. Because the medium is aqueous, the hydroxide ion is fully dissociated and available to attack the electrophilic carbon atom bonded to the halogen. This results in a nucleophilic substitution reaction, specifically $S_N2$ for primary halides. [Per the Nucleophilic Substitution mechanism: $Nu^- + R-X \rightarrow R-Nu + X^-$].

Step 3: Chemical Equation Construction

The hydroxide ion replaces the chloride ion ($Cl^-$) in the ethyl group to form ethanol, with potassium chloride ($KCl$) as the byproduct.

Balanced chemical equation:

$CH_3CH_2Cl + KOH_{(aq)} \xrightarrow{\Delta} CH_3CH_2OH + KCl$

Step 4: Visualizing the Molecular Transformation

CH₃CH₂Cl + OH⁻(aq) CH₃CH₂OH + Cl⁻ Nucleophilic Substitution

Step 5: Theoretical Justification

In aqueous solution, $KOH$ provides $OH^-$ ions which are excellent nucleophiles. In alcoholic $KOH$, the hydroxide ion acts as a strong base, favoring an elimination reaction ($E2$) to form an alkene (ethene). However, in the aqueous medium specified, the solvent polarity favors the ionization and nucleophilic attack, leading exclusively to the formation of an alcohol. [Per the Solvation Effect Theory: Aqueous media favors substitution while alcoholic media favors elimination].

Final Solution: Ethyl chloride reacts with aqueous $KOH$ to form ethanol ($CH_3CH_2OH$) and potassium chloride ($KCl$).


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