Substitution Reactions of Haloalkanes (alkyl halides) Chemistry Tutorial

Key Concepts

Haloalkanes are also known as alkyl halides.⁽¹⁾

Haloalkanes (alkyl halides) are compounds containing carbon, hydrogen and a Group 17 (Group VIIA or halogen) atom⁽²⁾.
Halogens are fluorine, chlorine, bromine, iodine and astatine.

Haloalkanes (alkyl halides) have the general formula RX in which R is the carbon chain and X is the halogen atom.
Typically X is Cl, Br or I

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

Primary haloalkanes (alkyl halides)⁽³⁾ 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 reactions⁽⁴⁾ in which a different atom, ion or group is substituted for the halogen atom, X:

haloalkane	+	aqueous hydroxide solution	→	alkanol	+	halide salt solution
R-X	+	MOH_(aq)	→	R-OH	+	MX_(aq)
haloalkane	+	carboxylate salt solution	→	ester	+	halide salt solution
R-X	+	R'-COO^-M⁺	→	R'-COO-R	+	MX
haloalkane	+	ammonia	→	alkanaminium halide
R-X	+	NH₃	→	R-NH₃⁺X^-
haloalkane	+	cyanide ion	→	alkanenitrile	+	halide ion
R-X	+	CN^-	→	R-CN	+	X^-

Alcohols, esters, amine salts and nitriles, can be synthesized using substitution reactions of haloalkanes (alkyl halides).

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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
general reaction	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)
General reaction	R-X	+	NH₃	→	R-NH₃⁺X^-
Examples	chloroethane	+	ammonia	→	ethanaminium 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^-

Note that you can then react the alkanaminium halide with hydroxide to produce an amine:

R-NH₃⁺X^-

OH^-
→

R-NH₂

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Synthesis of Alkanenitriles Using Haloalkane Substitution Reaction

The cyanide ion, CN^-, can displace the halide atom in the haloalkane which results in the production of an alkanenitrile.⁽⁷⁾

General:	haloalkane	+	cyanide ion	→	alkanenitrile	+	halide ion
General:	R-X	+	CN^-	→	R-CN	+	X^-
Example:	1-bromobutane	+	cyanide ion	→	pentanenitrile	+	bromide ion
Example:	CH₃-CH₂-CH₂-CH₂Br	+	CN^-	→	CH₃-CH₂-CH₂-CH₂-CN	+	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 (1^o) - one alkyl group bonded to the head carbon atom
secondary (2^o) - two alkyl groups bonded to the head carbon atom
tertiary (3^o) - 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 S_N2 reactions due to steric hindrance.
In order of increasing rate of reaction for S_N2 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.

(7) High yeilds are only obtained for primary haloalkanes, and, to a lesser extent, secondary haloalkanes.