Preparation of Haloalkanes: Meaning, Methods, Chemical Equation

Preparation of Haloalkenes: Did you know Bromoalkane is the most common haloalkane? The first haloalkane to be produced was chloroethane. Haloalkanes are halogen derivatives of hydrocarbons. Based on the bonding of the halogen atom to the carbon chain, haloalkanes can be classified into primary \(\left( {{1^ \circ }} \right)\), secondary \(\left( {{2^ \circ }} \right)\), and tertiary \(\left( {{3^ \circ }} \right)\). There are different methods of preparing these haloalkanes. Let’s learn the different methods employed in preparing haloalkanes.

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Preparation Methods of Haloalkanes

Haloalkanes are prepared by the following methods:

Preparation of Haloalkanes From Alcohols

The most convenient method of preparation of haloalkane is from alcohol. \({\rm{R}} – {\rm{OH}}\), when reacts with suitable reagents, the reaction results in the formation of \({\rm{R}} – {\rm{X}}\). The suitable reagents that help in the reaction are:

1. Concentrated halogen acids \({\rm{(HX)}}\)
2. Phosphorus halides \({\rm{P}}{{\rm{X}}_5}\) or \({\rm{P}}{{\rm{X}}_3}\)
3. Thionyl chloride \(\left( {{\rm{SOC}}{{\rm{l}}_2}} \right)\)

Out of the above three reagents, thionyl chloride is the most suitable reagent for the preparation of haloalkanes.

i) Reaction of Alcohols with Halogen Acid

Alcohol reacts with halogen acid \(({\rm{H}} – {\rm{X}})\) to form haloalkanes as the major product.

Example: Preparation of Chloroalkanes

The preparation of chloroalkane is an example of the formation of haloalkane by the reaction of an alcohol with halogen acid.

1. Primary and secondary alcohols react with \({\rm{HCl}}\) acid in the presence of anhydrous \({\rm{ZnC}}{{\rm{l}}_2}\) (Lucas Reagent), which act as a catalyst in this reaction.

\({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{OH}} + {\rm{HX}}\mathop \to \limits^{{\rm{ZnC}}{{\rm{l}}_2}} {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{X}} + {{\rm{H}}_2}{\rm{O}}\)

\({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}}({\rm{OH}}){\rm{C}}{{\rm{H}}_3} + {\rm{HX}}\mathop \to \limits^{{\rm{ZnC}}{{\rm{l}}_2}} {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}}({\rm{X}}){\rm{C}}{{\rm{H}}_3} + {{\rm{H}}_2}{\rm{O}}\)

2. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated \({\rm{HCl}}\) at room temperature.

\({\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} – {\rm{OH}} + {\rm{HCl}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} – {\rm{Cl}} + {{\rm{H}}_2}{\rm{O}}\)

Preparation of Bromoalkanes

Bromoalkanes are prepared by the reaction of alcohols with Hydrogen bromide (HBr). Hydrogen bromide required for the reaction is prepared by the reaction of sodium bromide or potassium bromide with \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\) (sulphuric acid). In the reaction below, \({\rm{NaBr}}\) and \({{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\) react to form HBr, which further reacts with alcohol.

\({\rm{R}} – {\rm{OH}} + {\rm{NaBr}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{R}} – {\rm{Br}} + {\rm{NaHS}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{O}}\)

Example of Preparation of Bromoethane from Ethanol

\({\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2} – {\rm{OH}} + {\rm{NaBr}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2} – {\rm{Br}} + {\rm{NaHS}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{O}}\)

3. The order of reactivity of alcohols with halo acid \(3^\circ > 2^\circ > 1^\circ \).

\({3^ \circ }\) or tertiary carbocations are the most stable carbocations among secondary and primary carbocations. Hence, tertiary alcohols are more reactive than primary and secondary alcohols.

4. The order of reactivity of halo acids with alcohol is in the order:

\({\rm{HI}} > {\rm{HBr}} > {\rm{HCl}}\)

\({\rm{HI}}\) is easily and highly reacting haloacid because of the weak HI bond. It can be broken easily compared to the other halogen bonds to form \({{\rm{H}}^ + }\) and \({{\rm{I}}^ – }\) ions.

ii) Reaction of Alcohols with Phosphorus Halides (\({\bf{P}}{{\bf{X}}_{\bf{5}}}\) or \({\bf{P}}{{\bf{X}}_3}\))

In this reaction, the halide functional group of phosphorus halides interchange the functional group of alcohols \(( – {\rm{OH}})\) with the corresponding halides. The reaction is as follows:

\({\rm{ROH}} + {\rm{PC}}{{\rm{l}}_5} \to {\rm{RCl}} + {\rm{POC}}{{\rm{l}}_3} + {\rm{HCl}}\)

The above reaction is only for the formation of alkyl chloride. However, alkyl bromide, alkyl iodide and alkyl chlorides are formed by the reaction of an alcohol with phosphorus tribromide and triiodide.

\(3{\rm{ROH}} + {\rm{P}}{{\rm{X}}_3} \to 3{\rm{RX}} + {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_3}\quad ({\rm{X}} = {\rm{Br}},{\rm{I}},{\rm{Cl}})\)

\({\rm{PB}}{{\rm{r}}_3}\) and \({\rm{P}}{{\rm{I}}_3}\) is generated in situ (produced in the reaction mixture) by the reaction of red phosphorus with bromine and iodine, respectively.

iii) Reaction of Alcohols with Thionyl Chloride or Sulphonyl Chloride

It is the most preferred reagent. This is because the by-products formed in this reaction are gaseous in nature and can easily escape into the atmosphere, leaving behind the pure alkyl halide. This reaction is known as Darzen’s halogenation.

However, these methods are not used to synthesise aryl halides/haloarenes. \({\rm{C}} – {\rm{OH}}\) bond in the phenol structure contains a partial double bond character, which involves the delocalisation of lone pair of electrons present on the oxygen atom of the benzene ring. It is stronger than a single bond and cannot be broken down by any reaction with haloacids.

\({\rm{Ar}} – {\rm{OH}} + {\rm{HX}} \to \) No reaction

Preparation of Haloalkanes From Alkenes (Markovnikov’s rule)

Alkene can be converted to haloalkane by an electrophilic addition reaction. Alkene reacts with \({\rm{HX}}\) to form \({\rm{R – X}}\). The order of reactivity of halides with respect to alkenes follows the order:

\({\rm{HI}} > {\rm{HBr}} > {\rm{HCl}} > {\rm{HF}}\)

Alkyl halides can be prepared from alkenes in the following ways:

I) Addition of Hydrogen Halides

The addition of halides to alkenes takes place through Markovnikov’s rule. It states that, in hydrohalogenation of an unsymmetrical alkene, the hydrogen atom of the hydrogen halide molecule forms a bond with the doubly bonded carbon atom in the alkene, bearing the greater number of hydrogen atoms.

Peroxide effect (Kharash effect): The peroxide effect or Kharash effect is also known as the anti-markovnikov’s rule. In this reaction, alkene reacts with HBr in the presence of peroxide. The \({\rm{B}}{{\rm{r}}^ – }\) or the negative part of the addendum will attach itself to the doubly bonded carbon atom of the alkene having more number of hydrogen atoms. For example, Prop-\(1\)-ene reacts with hydrogen bromide to form \(1\)-bromopropane as a major product in the presence of peroxide.

II) Addition of Halogens

The haloalkanes are also prepared from alkene in the laboratory by the addition of bromine in \({\rm{CC}}{{\rm{l}}_4}\) to the alkene. This reaction is used in organic chemistry to test for unsaturation. In this reaction, decolourisation of the reddish-brown colour of bromine takes place. Colourless vic-dibromides are formed as a product of this reaction.

Preparation of Haloalkanes By Radical Halogenation of Alkanes

Alkanes react with halogens (\({\rm{C}}{{\rm{l}}_2}\) or \({\rm{B}}{{\rm{r}}_2}\)) in the presence of ultraviolet light to form haloalkane. This is a radical substitution reaction and gives a mixture of mono, di or polysubstituted haloalkane.

Example: Chlorination of methane gives different products that have different boiling points. Hence, these can be separated by fractional distillation.

The radical halogenation of alkanes results in the formation of a mixture of haloalkanes. This causes difficulty in the isolation of a single product. Therefore it is not the preferred method for the preparation of haloalkanes.

Example: When butane reacts with chlorine in the presence of light as energy, a mixture of products is formed.

Preparation of Haloalkanes by Halogen Exchange Reactions

1. Finkelstein Reaction

Alkyl iodides are prepared by heating Chloro or bromoalkane with a concentrated sodium iodide solution in dry acetone. This reaction is called the Finkelstein reaction.

\({\rm{R}} – {\rm{X}} + {\rm{Nal}} \to {\rm{R}} – {\rm{I}} + {\rm{NaX}}\)
\({{\rm{X}}_1} = {\rm{Cl}},{\rm{Br}}\)

2. Swarts Reaction

Alkyl chlorides or bromides on heating with metallic fluorides like \({\rm{AgF, Sb}}{{\rm{F}}_3}\) or \({\rm{H}}{{\rm{g}}_2}\;{{\rm{F}}_2}\) forms alkyl fluorides. This reaction is called the Swarts reaction.

Example

Preparation of Haloalkanes From Silver Salts of Fatty Acids (Hunsdiccker reaction)

Silver salts of fatty acids when refluxed with bromine in \({\rm{CC}}{{\rm{l}}_4}\) forms an unstable intermediate which undergoes decarboxylation further to form alkyl bromides.

Summary

Haloalkanes consist of halogen atom (s) that are attached to the \({\rm{s}}{{\rm{p}}^3}\) hybridised carbon atom of an alkyl group. These organic compounds find wide applications in industry as well as in everyday life. They are used as starting materials for the synthesis of a wide range of organic compounds and as solvents for relatively non-polar compounds. This article taught us the essential methods through which these halogen derivatives of hydrocarbon could be prepared.

FAQs

Q.1. Which method is preferred for the preparation of chloroalkanes and why?
Ans: Thionyl chloride is preferred in the preparation of chloro alkanes because the by-products of the reaction i.e. \({\rm{S}}{{\rm{O}}_2}\) and \({\rm{HCl}}\) are gases and escape into the atmosphere leaving behind alkyl chlorides in almost pure form.

Q.2. What are the methods of preparation of alkyl halides?
Ans: Some of the methods through which alkyl halides are prepared by:
1. The Reaction of Alcohols with Halogen Acid, Phosphorus halides (\({\rm{P}}{{\rm{X}}_5}\) or \({\rm{P}}{{\rm{X}}_3}\)) and Thionyl Chloride.
2. The addition of Hydrogen halides or halogens to alkenes.
3. radical halogenation of alkanes.
4. Halogen exchange reactions (Frinkleistein and Swarts reaction).

Q.3. Why is the Darzens method the best method to prepare haloalkanes?
Ans: Darzen’s process is the best way to prepare alkyl halide because when alcohol reacts with \({\rm{SOC}}{{\rm{l}}_2}\), it leaves alkyl halide in the purest form and the by-product in the gaseous form escapes.

Q.4. Why is sulphuric acid not used during the reaction of alcohols with KI?
Ans: Sulphuric acid \(\left( {{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}} \right)\) is an oxidising agent. It oxidises HI produced during the reaction to \({{\rm{I}}_2}\).
\(2{\rm{KI}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to 2{\rm{KHS}}{{\rm{O}}_4} + 2{\rm{HI}}\)
\(2{\rm{HI}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {{\rm{H}}_2}{\rm{O}} + 2{\rm{KI}} + {{\rm{I}}_2} + 2{\rm{KI}} + {\rm{S}}{{\rm{O}}_2}\)
To prepare an alkyl iodide, a non-oxidising agent such as \({{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}\) is used instead of sulphuric acid.

Q.5. Why don’t we need anhydrous \({\rm{ZnC}}{{\rm{l}}_2}\) as a catalyst when the alcohol is tertiary?
Ans: Since the carbocation stability follows the order : \({3^ \circ } > {2^ \circ } > {1^ \circ }\), the cleavage of the \({\rm{C – O}}\) bond in tertiary alcohols occurs easily.

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