Which of the following statements is true about alkyl halides? A. They undergo addition reactions B. They undergo hydration reactions C. The undergo elimination reaction when treated with KOH D. They form alcohols when oxidized E. They undergo elimination reaction when refluxed with NaOH 2
The correct answer and explanation is:
Correct Answer: C. They undergo elimination reaction when treated with KOH
Explanation (Approx. 300 Words):
Alkyl halides, also known as haloalkanes, are organic compounds containing one or more halogen atoms (F, Cl, Br, I) attached to an alkyl group. Their reactivity is mainly influenced by the polarity of the carbon-halogen bond, where the carbon bears a partial positive charge and the halogen bears a partial negative charge.
Among the options given, Option C is correct: Alkyl halides undergo elimination reactions when treated with KOH, specifically hot alcoholic KOH. In this reaction, the halide ion is eliminated along with a hydrogen atom from an adjacent carbon, forming a double bond (alkene). This is known as the E2 (bimolecular elimination) mechanism.
The general reaction can be written as:
R–CH₂–CHX–R’ + KOH (alcohol, heat) → R–CH=CH–R’ + KX + H₂O
Here’s a brief analysis of why other options are incorrect:
- A. Addition reactions – These are typical of alkenes and alkynes, not alkyl halides.
- B. Hydration reactions – This involves adding water across a double bond, again a feature of alkenes, not alkyl halides.
- D. Oxidation to alcohols – This is incorrect. Alkyl halides are usually converted to alcohols via nucleophilic substitution, not oxidation.
- E. Elimination with NaOH 2 – While this sounds plausible, the more standard base for elimination is alcoholic KOH, not aqueous NaOH or “NaOH 2,” which seems unclear or possibly a typo.
Thus, when an alkyl halide is heated with alcoholic KOH, it undergoes dehydrohalogenation (removal of HX) to form an alkene — a classic elimination reaction.
✅ Final Answer: C. They undergo elimination reaction when treated with KOH