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Substitution Reactions of Haloalkanes (Alkyl Halides)

Key Concepts

  • Haloalkanes (alkyl halides) are compounds containing carbon, hydrogen and a Group 17 (Group VIIA or halogen) atom1.

  • Haloalkanes (alkyl halides) have the general formula RX in which R is the carbon chain and X is the halogen atom.

  • The halogen atom, X, is the functional group and the C-X bond is the site of chemical reactivity.

  • Primary haloalkanes (alkyl halides)2 are haloalkanes in which the halogen atom is attached to a terminal (end) carbon atom in the carbon chain.

  • Primary haloalkanes (alkyl halides) undergo substitution reactions3 in which a different atom, ion or group is substituted for the halogen atom, X:

haloalkane+aqueous hydroxide solutionalkanol+halide salt solution
R-X+MOH(aq)R-OH+MX(aq)
haloalkane+carboxylate salt solutionester+halide salt solution
R-X+R'-COO-M+R'-COO-R+MX
haloalkane+ammoniaalkanaminium halide
R-X+NH3R-NH3+X-  

Alkanol (alcohol) Production

Primary haloalkanes react with hydroxide ions to produce an alkanol.
Aqueous solutions of strong bases such as sodium hydroxide, NaOH(aq), or potassium hydroxide, KOH(aq), are good sources of hydroxide ions for the reaction.
The halogen atom (X) leaves the haloalkane as the halide ion (X-). The halide ion is known as the "leaving group".
The hydroxide ion (OH-) replaces (substitutes for) the lost halide ion.

General Equationhaloalkane+aqueous hydroxide solutionalkanol (alcohol)+halide salt solution
R-X+MOH(aq)R-OH+MX(aq)

Chloroethane
Example
chloroethane
(ethyl chloride)
+aqueoues sodium
hydroxide solution

ethanol
(ethyl alcohol)
+aqueous sodium
chloride solution
CH3-CH2-Cl+NaOH(aq)CH3-CH2-OH+NaCl(aq)
chloroethane
(ethyl chloride)
+aqueous potassium
hydroxide solution

ethanol
(ethyl alcohol)
+aqueous potassium
chloride solution
CH3-CH2-Cl+KOH(aq)CH3-CH2-OH+KCl(aq)

Bromoethane
Example
bromoethane
(ethyl bromide)
+aqueous sodium
hydroxide solution

ethanol
(ethyl alcohol)
+aqueous sodium
bromide solution
CH3-CH2-Br+NaOH(aq)CH3-CH2-OH+NaBr(aq)
bromoethane
(ethyl bromide)
+aqueous potassium
hydroxide solution

ethanol
(ethyl alcohol)
+aqueous potassium
bromide solution
CH3-CH2-Br+KOH(aq)CH3-CH2-OH+KBr(aq)

Ester Production

Primary haloalkanes react with the salts of carboxylic acids (carboxylates) to produce esters4.
The sodium salt of an alkanoic acid is a good source of alkanoate (carboxylate) ions.
The halogen atom, X, leaves the haloalkane molecule as the halide ion, X-.
It is replaced by the alkanoate ion (carboxylate ion) resulting in a solution that contains the ester and the salt of the halide ion.

general reactionhaloalkane+carboxylate salt solutionester+halide salt
solution
R-X+R'-COO-M+R'-COO-R+MX
Exampleschloromethane+sodium acetate solution
(sodium ethanoate solution)
methyl acetate
(methyl ethanoate)
+sodium chloride
solution
CH3-Cl+CH3-COO-Na+CH3-COO-CH3+NaCl
bromomethane+sodium acetate solution
(sodium ethanoate solution)
methyl acetate
(methyl ethanoate)
+sodium bromide
solution
CH3-Br+CH3-COO-Na+CH3-COO-CH3+NaBr
chloromethane+sodium propanoate solutionmethyl propanoate+sodium chloride
solution
CH3-Cl+CH3-CH2-COO-Na+CH3-CH2-COO-CH3+NaCl
bromomethane+sodium propanoate solutionmethyl propanoate+sodium bromide
solution
CH3-Br+CH3-CH2-COO-Na+CH3-CH2-COO-CH3+NaBr

Alkanaminium Halide (amine salt) Production

Primary haloalkanes will react with ammonia to produce alkanaminium salts (amine salts)5.
The halogen atom, X, leaves the haloalkane molecule as the halide ion, X-.
The nitrogen atom of the ammonia molecule shares its lone pair of electrons with the carbon atom. The ammonia molecule substitutes for the lost halide ion.
The resulting organic molecule has a positive charge.
This positive charge is balanced by the negative charge of the halide ion resulting in an amine salt, an alkanaminium halide.

General reactionhaloalkane
(alkyl halide)
+ammoniaalkanaminium halide
(amine salt)
R-X+NH3R-NH3+X-
Exampleschloroethane+ammoniaethanaminium chloride
(ethyl ammonium chloride)
CH3-CH2-Cl+NH3CH3-CH2-NH3+Cl-
bromoethane+ammoniaethanaminium bromide
(ethyl ammonium bromide)
CH3-CH2-Br+NH3CH3-CH2-NH3+Br-

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1Typically the halogen atom to be replaced is Cl, Br or I

2Haloalkanes are classified as:

  • methyl - 1 hydrogen of methane has been replaced by a halogen
  • primary (1o) - one alkyl group bonded to the head carbon atom
  • secondary (2o) - two alkyl groups bonded to the head carbon atom
  • tertiary (3o) - three alkyl groups bonded to the head carbon atom

For the purposes of this tutorial, we will include the halomethanes in the "primary haloalkane" category.

3Tertiary haloalkanes do not undergo SN2 reactions due to steric hindrance.
In order of increasing rate of reaction for SN2 reactions:
methyl > primary > secondary

4A reactive halide must be used.

5The product amine salt of this reaction can exchange a proton with the starting ammonia.
This means there will be two or more nucleophiles competing in the reaction with the haloalkane.
This results in a mixture of mono-, di-, and trialkyl amines, and the quarternary ammonium salt.
For this reason the reaction between a haloalkane and ammonia is not a useful way to synthesise an amine salt.
In this tutorial we will ignore the proton exchange reaction and continue as if only one reaction will occur.

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