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Substitution Reactions of Haloalkanes (alkyl halides) Chemistry Tutorial

Key Concepts

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Synthesis of Alcohol Using Haloalkane Substitution Reaction

Primary haloalkanes (alkyl halides) 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.

Examples of the synthesis of ethanol (ethyl alcohol) using this substitution reaction are shown below:

General Equation haloalkane + aqueous hydroxide solution alkanol (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)

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Synthesis of Ester Using Haloalkane Substitution Reaction

Primary haloalkanes (alkyl halides) react with the salts of carboxylic acids (carboxylates) to produce esters(5).

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.

Examples of this substitution reaction to produce various methyl alkanoate esters are shown below:

general reaction haloalkane + carboxylate salt solution ester + halide salt
solution
R-X + R'-COO-M+ R'-COO-R + MX
Examples chloromethane + 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 solution methyl propanoate + sodium chloride
solution
CH3-Cl + CH3-CH2-COO-Na+ CH3-CH2-COO-CH3 + NaCl
bromomethane + sodium propanoate solution methyl propanoate + sodium bromide
solution
CH3-Br + CH3-CH2-COO-Na+ CH3-CH2-COO-CH3 + NaBr

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Synthesis of Alkanaminium Halide (amine salt) Using Haloalkane Substitution Reaction

Primary haloalkanes will react with ammonia to produce alkanaminium salts (amine salts)(6).

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.

Examples of the production of ethanaminium salts are shown below:

General reaction haloalkane
(alkyl halide)
+ ammonia alkanaminium halide
(amine salt)
R-X + NH3 R-NH3+X-
Exampleschloroethane+ammoniaethanaminium chloride
(ethyl ammonium chloride)
CH3-CH2-Cl + NH3 CH3-CH2-NH3+Cl-
bromoethane + ammonia ethanaminium bromide
(ethyl ammonium bromide)
CH3-CH2-Br + NH3 CH3-CH2-NH3+Br-

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Footnotes:

(1) Using IUPAC substitutive nomenclature these compounds are haloalkanes, using the alternative functional class nomenclature they are alkyl halides.

(2) Typically the halogen atom to be replaced is Cl, Br or I

(3) Haloalkanes 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.

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

(5) A reactive halide must be used.

(6) The 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.