- 2067 Chaitra · 5 marks
Primary alkyl halide with alcoholic and aqueous alkali
Short answer
With aqueous alkali, primary alkyl halides undergo SN2 substitution: OH- attacks from backside, forming an alcohol (e.g., CH3CH2OH) via a bimolecular mechanism. With alcoholic alkali, ethoxide ion acts as a strong base, leading to E2 elimination: removal of β-H and departure of Br- yields an alkene (e.g., CH2=CH2).
Explain the reaction mechanism involved when primary alkyl halide react with alcoholic alkali and aqueous alkali.
A primary alkyl halide gives different products with the two reagents.
With aqueous alkali: substitution (SN2)
Hydroxide acts as a nucleophile and replaces Br.
HO⁻ + CH₃CH₂—Br → [HO···CH₂(CH₃)···Br]⁻‡ → CH₃CH₂OH + Br⁻
CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr. Rate = k[RX][OH⁻] (bimolecular).
With alcoholic alkali: elimination (E2)
In alcohol, ethoxide forms (OH⁻ + C₂H₅OH ⇌ C₂H₅O⁻ + H₂O). It is a strong base and removes the β-hydrogen as the C–Br bond breaks.
EtO⁻ H—CH₂—CH₂—Br → [EtO···H···CH₂···CH₂···Br]⁻‡ → CH₂=CH₂ + EtOH + Br⁻
CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O. Rate = k[RX][base] (bimolecular).
Thus aqueous alkali favours substitution, whereas alcoholic alkali and heat favour elimination.
✓ Answer checked by Claude Opus 5.5 (high effort)
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