Halogenation of Unsaturated Hydrocarbons Using Hydrogen Halides in a Hydrohalogenation Reaction Chemistry Tutorial

Key Concepts

• A hydrocarbon is a molecule containing only carbon and hydrogen.
• Unsaturated hydrocarbon molecules contain either a double bond (C=C) or a triple bond (C≡C)
• Alkenes are unsaturated hydrocarbons with a double bond (C=C).
• Alkynes are unsaturated hydrocarbons with a tripe bond (C≡C)
• The active site in alkenes is the double bond.
      The active site in alkynes is the triple bond.
• The elements of Group 17 are known as halogens.
      Fluorine (F), chlorine (Cl), bromine (Br) and iodine (I) are halogens.
      A halogen atom is usualy given the general symbol X
• Hydrogen halides are covalent molecules with the general formula HX
      HBr is hydrogen bromide, HI is hydrogen iodide
• Unsaturated hydrocarbons (alkenes and alkynes) react with hydrogen halides (HX) in addition reactions producing halogenated compounds.

  alkene	+	hydrogen halide	→	haloalkane
  H |   H |     R - C = C - R'	+	H-X	→	H |   H |     R - C - C - R'     | H   | X

  R and R' represent either hydrogen atoms (H) or alkyl groups (alkane chains).

• If the alkene or alkyne molecule is symmetrical, R is the same as R', then only one product is formed.
• If the alkene or alkyne molecule is unsymmetrical, R is NOT the same as R', then more than one product will be produced.
• Markovnikov's Rule is applied in order to determine which product is produced with the the greater yield, the major product.
• Markovnikov's Rule:

  In additions of HX to unsymmetrical alkenes, the H⁺ of HX goes to the double-bonded carbon that already has the greatest number of hydrogens

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Hydrohalogenation of Symmetrical Alkenes

Ethene (ethylene) is a symmetrical alkene molecule. The valence structure of ethene is shown below:

Each carbon atom is covalently bonded to two hydrogen atoms.

The active site on the ethene molecule is the double bond (C=C).

Ethene (ethylene) reacts with hydrogen bromide (HBr) to produce bromoethane (ethyl bromide).

In this reaction, the hydrogen atom of HBr adds to one of the carbon atoms and the bromine atom of HBr adds to the other carbon atom.

The synthesis of bromoethane from the hydrobromination of ethene (ethylene) is shown in the chemical equations below:

Bromoethane is the only product possible for this reaction.

Hydrohalogenation of Unsymmetrical Alkenes

Propene (propylene) is an unsymmetrical alkene molecule. The structure of propene is shown below:

C⁽¹⁾ is covalently bonded to two hydrogen atoms.

C⁽²⁾ is covalently bonded to one hydrogen atom.

When propene reacts with a halogen halide such as hydrogen bromide, two structural isomers are possible products:

• If the bromine atom bonds with C⁽¹⁾, the product is 1-bromopropane.
• If the bromine atom bonds with C⁽²⁾, the product is 2-bromopropane.

1-bromopropane and 2-bromopropane are position isomers, they are derived from the same parent alkane (propane), but, the Br functional group is located in a different position in the carbon chain (on the first C atom for 1-bromopropane, and, on the second C atom for 2-bromopropane).

The synthesis of 1-bromopropane and 2-bromopropane from the hydrobromination of propene is shown below:

Markovnikov's Rule states that the hydrogen adds to the carbon atom of the double bond which is already bonded to more hydrogen atoms, that is, the hydrogen atom from HBr mostly adds to C⁽¹⁾ because this carbon atom already had 2 hydrogen atoms bonded to it, therefore the bromine atom from HBr mostly adds to C⁽²⁾ so the major product of the reaction will be 2-bromopropane.
1- bromopropane is also produced in smaller amounts so it is known as the minor product.

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