Chemistry Alkyl Halides ETEA 2024
PMDC Verified Question 6 of 13
Dehydrohalogenation of alkyl halide is carried out in presence of:
A
Alcoholic KOH
B
Conc. H\(_2\)SO\(_4\)
C
Aqueous KOH
D
Zn dust
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: Alcoholic KOH
Concept:

Dehydrohalogenation is the elimination of a hydrogen atom and a halogen atom from an alkyl halide to form an alkene. This requires a strong base.

Solution:

  • Potassium hydroxide dissolved in an alcohol (like ethanol) generates alkoxide ions (e.g., ethoxide, \( C_2H_5O^- \)).
  • Alkoxide ions are extremely strong bases. Because they are not strongly hydrated by the alcoholic solvent, their basicity dominates over their nucleophilicity.
  • They successfully abstract a \( \beta \)-hydrogen, triggering the elimination of the halogen to form a double bond.


Why other options are incorrect:

  • Option B: Concentrated sulfuric acid is a dehydrating agent used to remove water from alcohols, not hydrohalic acids from alkyl halides.
  • Option C: Aqueous KOH acts as a nucleophile, leading to substitution (forming an alcohol) instead of elimination.
  • Option D: Zinc dust is used for dehalogenation (removing two halogens from vicinal dihalides), not dehydrohalogenation.

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