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Question

1-Chlorobutane on reaction with alcoholic potash gives?


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Solution

Dehydrohalogenation reaction

  • Dehydrohalogenation is an elimination reaction that removes a hydrogen halide from a substrate.
  • The reaction is usually associated with the synthesis of alkenes

The reaction of 1-Chlorobutane with alcoholic potash

  • Alcoholic potash (KOH) is a dehydrohalogenation agent. It reacts with an alkyl halide to form an alkene.
  • Alcoholic potash (KOH) reacts with 1-Chlorobutane to form butene.
  • The reaction of the process:

Step 1: KOH removes a slightly acidic hydrogen proton from the Chlorobutane via an acid‐base reaction.

OH-hydroxideion(base)+CH3-CH2-CH2(acidicH)-CH2ClH2Owater+CH3-CH2-C+H-CH2Clcarbocation

Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free, leading to the formation of the double bond.

CH3-CH2-C+H-CH2ClcarbocationCH3-CH2-CH=CH2butene+Cl-chlorideion

Step 3: The overall reaction is given as:

K+potassiumion+OH-hydroxideion(base)+CH3-CH2-CH2-CH2Cl1-chlorobutaneCH3-CH2-CH=CH2butene+KClpotassiumchloride+H2Owater

Therefore, 1-Chlorobutane on reaction with alcoholic potash gives butene.


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