Dehydrohalogenation of Haloalkanes (alkyl halides) Chemistry Tutorial

Key Concepts

• Dehydrohalogenation refers to the removal of H-X in an elimination reaction.

  (where X is a halogen atom)

• Dehydrohalogenation of a haloalkane results in the production of an alkene.

  haloalkane	→	alkene	+	hydrogen halide
  R-CHX-CH2-R'	→	R-CH=CH-R'	+	HX

  (where R and R' are alkyl groups)

• In general, secondary and tertiary haloalkanes undergo dehydrohalogenation reactions using a strong base in a polar solvent (such as ethanol, EtOH) at higher temperature (usually heated under reflux).⁽¹⁾

  2° haloalkane	→	alkene	+	hydrogen halide
  R-CHX-CH2-R'	MOH → EtOH reflux	R-CH=CH-R'	+	HX

  (where MOH is a strong base, eg, NaOH or KOH)

• Dehydrohalogenation of a symmetrical haloalkane (alkyl halide) results in only one product.

  R-CHX-CH₂-R → R-CH=CH-R

• Dehydrohalogenation of an unsymmetrical haloalkane (alkyl halide) results in two products.

  R-CH₂-CHX-CH₂-R' → R-CH=CH-CH₂-R' + R-CH₂-CH=CH-R'

  The Saytzeff Rule predicts that the more highly substituted product is favoured:

  R   R \   /   C=C   /   \ R   R	>	R   R \   /   C=C   /   \ R   H	>	R     \       C=CH2   /     R	>	R     \       C=CH2   /     H
  most favoured		→ → → → → → → →			least favoured

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Dehydrohalogenation of a Symmetrical Secondary Haloalkane

Consider 2-bromopropane as an example of a symmetrical secondary haloalkane (2° alkyl halide):

Line formula: CH₃-CHBr-CH₃

It is a haloalkane (alkyl halide) because a halogen atom, Br, has replaced one of the hydrogen atoms of the parent alkane (propane).

It is a secondary haloalkane (2° alkyl halide) because the head carbon atom (C), that is the carbon atom directly bonded to the halogen atom, is itself bonded to 2 other carbon atoms (as well as a hydrogen atom).

It is a symmetrical molecule because the same alkyl group is attached to both sides of the head carbon: a methyl group (CH₃) on one side and a methyl group (-CH₃) on the other side.

If we heat 2-bromopropane with a solution of potassium hydroxide (KOH) dissolved in ethanol (EtOH) under reflux then hydrogen bromide (HBr) will be eliminated from the molecule and 1 organic product is produced, prop-1-ene (1-propene or propene):

For 2-bromopropane, a symmetrical molecule, it doesn't matter whether the eliminated H atom comes from the first or third carbon atom in the chain because the organic product will still be the same, prop-1-ene.

Prop-1-ene is the organic product (the product containing the carbon chain) and hydrogen bromide is the inorganic product (the product that does not contain the carbon chain).

The organic reagent used in this elimination reaction is 2-bromopropane and the inorganic reagent is KOH dissolved in ethanol.

The conditions under which the reaction occurs are heating under reflux.

Dehydrohalogenation of an Unsymmetrical Secondary Haloalkane

Consider 2-chlorobutane as an example of an unsymmetrical secondary haloalkane (2° alkyl halide):

Line formula: CH₃-CH₂-CHCl-CH₃

It is a haloalkane (alkyl halide) because a halogen atom, Cl, has replaced one of the hydrogen atoms of the parent alkane (butane).

It is a secondary haloalkane (2° alkyl halide) because the head carbon atom (C), that is the carbon atom directly bonded to the halogen atom, is itself bonded to 2 other carbon atoms (as well as a hydrogen atom).

It is an unsymmetrical molecule because there is a different alkyl group attached to each side of the head carbon: an ethyl group (CH₃-CH₂) on one side and a methyl group (-CH₃) on the other side.

If we heat 2-chlorobutane with a solution of potassium hydroxide dissolved in ethanol under reflux then hydrogen chloride will be eliminated from the molecule and 2 organic products are produced, but-2-ene (2-butene) and but-1-ene (1-butene):

Saytzeff's Rule predicts that the production of but-2-ene is more favoured than production of but-1-ene because but-2-ene is more highly substituted, that is, but-2-ene is of the form R-CH=CH-R (2 alkyl groups, 2 × R) whereas but-1-ene is of the form R-CH=CH₂ (only 1 alkyl group, 1 × R).
Therefore the yield of but-2-ene is greater than the yield of but-1-ene, and but-2-ene is referred to as the major product while but-1-ene is referred to as the minor product.

Note the reaction above is not a balanced chemical equation, it only indicates the formula of the initial organic reactant and the final organic products. The inorganic product is not shown. The ratios of organic products produced is not indicated.

Dehydrohalogenation of Tertiary Haloalkanes

Consider 2-chloro-2-methylpropane as an example of a symmetrical tertiary haloalkane (3° alkyl halide):

It is a haloalkane (alkyl halide) because a halogen atom, Cl, has replaced one of the hydrogen atoms of the parent alkane (propane).

It is a tertiary haloalkane (3° alkyl halide) because the head carbon atom (C), that is the carbon atom directly bonded to the halogen atom, is itself bonded to 3 other carbon atoms (and no hydrogen atoms).

It is a symmetrical molecule because the same alkyl group is attached to both sides of the head carbon: a methyl group (CH₃) on one side and a methyl group (-CH₃) on the other side.

If we heat 2-chloro-2-methylpropane with a solution of potassium hydroxide (KOH) dissolved in ethanol (EtOH) under reflux then hydrogen bromide (HBr) will be eliminated from the molecule and 1 organic product is produced, 2-methylprop-1-ene (2-methyl-1-propene or 2-methylpropene):

When this elimination reaction occurs it doesn't matter whether the H atom being eliminated is from the first or third carbon atom in the chain because the product in either case will be the same, 2-methylprop-1-ene.

Consider the tertiary haloalkane (3° alkyl halide) 2-chloro-2-methylbutane:

When this molecule undergoes an elemination reaction to remove HCl then the H atom could be removed from the CH₂ of an ethyl group (CH₃CH₂) or from a methyl group (CH₃), so there are 2 possible organic products:

Saytzeff's Rule predicts that the production of 2-methylbut-2-ene is more favoured than production of 2-methylbut-1-ene because 2-methylbut-2-ene is more highly substituted, that is, 2-methylbut-2-ene is of the form R-CR=CH-R (3 alkyl groups, 3 × R) whereas 2-methylbut-1-ene is of the form R-CR=CH₂ (only 2 alkyl groups, 2 × R).
Therefore the yield of 2-methylbut-2-ene is greater than the yield of 2-methylbut-1-ene, and 2-methylbut-2-ene is referred to as the major product while 2-methylbut-1-ene is referred to as the minor product.

Note the reaction above is not a balanced chemical equation, it only indicates the formula of the initial organic reactant and the final organic products. The inorganic product is not shown. The ratios of organic products produced is not indicated.