Haloalkanes and Haloarenes

Methods of Preparation

What is the most common haloalkane? It is bromoalkane. There are over 1600 halogenated organic compounds existing today. There are different methods of preparation of haloalkanes and haloarenes. We can trace back the presence of haloalkane to the 15th century. The first haloalkane produced was chloroethane. Later on, in the 19th-century preparation and synthesis of such compounds were done in order to understand organic chemistry and structure of alkanes.

Different methods of preparation include conversion of alcohols to alkyl halides, the addition of halogens to alkenes, and hydrohalogenation of alkenes. The preparation techniques were so reliable and efficient that it became an inevitable part of industrial chemistry. So let’s learn about the methods of preparation of Haloalkanes and Haloarenes.

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Methods of Preparation

There are primarily 4 different types of preparation techniques of Haloalkanes and Haloarenes. They include Preparation of Haloalkanes and Haloarenes from:

  • Alcohols
  • Hydrocarbons
  • Alkenes by addition of hydrogen halides and halogens
  • Halogen exchange reaction.

1) Preparation from Alcohols (Haloalkanes)

The most convenient method of preparation of haloalkane is from alcohols. R-OH when reacts with suitable reagents, the reaction results in the formation of R − X. The suitable reagents that help in the reaction are

  • Concentrated halogen acids (HX)
  • Phosphorus halides (PX5 or PX3)
  • Thionyl chloride (SOCl2)

i) The Reaction of Alcohols with Halogen Acid

An organic compound derivative of alcohol reacts with halogen acid (H-X) to form haloalkanes as the major product.


Example-Preparation of Chloroalkanes

Preparation of chloroalkane is an example of the reaction of an alcohol with halogen acid to form haloalkane. In this case, primary alcohol and secondary alcohol react with HCl acid gas to form haloalkane in the presence of anhydrous ZnCl2, which act as a catalyst in this reaction.



Preparation of Bromoalkanes

Hydrogen bromide (HBr) reacts with alcohols to form bromoalkanes. Hydrogen bromide synthesis of the reaction takes place by the reaction of sodium bromide or potassium bromide and H2SO4 (sulphuric acid). In the reaction below, NaBr and H2SO4 react to form HBr which further reacts with alcohol.


Example of Preparation of Bromoethane from Ethanol


The tertiary carbonation which occurs due to tertiary alcohols is more stable than secondary and primary alcohols. Therefore tertiary carbocations offer more stability in comparison to the primary and secondary form of the compound. Hence, tertiary is more reactive than primary and secondary. The order of reactivity is 30 > 20 > 10.

Additionally, the reactivity of haloacids follows the order HI > HBr > HCl > HF. HI is easily and highly reacting haloacid among all of them because the HI bond is weaker than the other three. Therefore, it can be broken easily in comparison to the other halogen bonds to form H+ and I− ions.

However, this preparation method will not be able to synthesize aryl halides/haloarenes. C-OH bond in the phenol structure contains partial double bond character because of the delocalization of lone pair of electrons present on the oxygen atom of the benzene ring. Therefore, the bond formed in the structure cannot be easily broken down by any sort of reaction with haloacids.

Ar – OH + HX → No reaction

ii) The Reaction of Alcohols with Phosphorus halides (PX5 or PX3)

This reaction helps in the formation of Chloroalkanes, bromoalkanes, and iodoalkanes. In this reaction phosphorus halides interchange the functional group of alcohols (–OH) with the corresponding halides. The reaction is as follows:

ROH + PCl5 → RCl + POCL3 + HCl

The above reaction is for the formation of alkyl chloride. Similarly, alkyl bromide or alkyl iodide formation is possible by the reaction of an alcohol with phosphorus tribromide and triiodide. To achieve the reaction, red phosphorus reacts with bromine or iodine by in-situ preparation (during the reaction) of phosphorus tribromide and triiodide.


Note: Fresh preparation of the phosphorus tribromide and phosphorus triiodide is made with red phosphorus and bromine or iodine due to the instability of the compounds. Thus, alcohol reacts with phosphorus trihalides (PX3) to obtain three molecules of alkyl halide. The general overall reaction is


Example: Reaction of ethanol with PCl3 for the formation of chloroethanePreparation

iii) The Reaction of Alcohols with Thionyl chloride as Suitable Reagent

This reagent is the most preferred and suitable in between the three reactions of alcohols. Alcohol reacts with Thionyl chloride (SOCl2) to form alkyl chlorides. However, the by-products formed in this reaction are gaseous in nature. Therefore, the by-products can easily escape into the atmosphere, leaving the pure alkyl halide. This method helps in the generation of pure alkyl halide.


2) Preparation of Haloalkanes & Haloarenes from Hydrocarbons

Preparation of Haloalkanes and haloarenes from hydrocarbons is possible by 3 different methods. They are

  • Free radical halogenation of haloalkanes
  • Electrophilic Substitution Reactions
  • Sandmeyer reaction

i) Free Radical Halogenation

Alkyl bromides and alkyl chloride formation are possible by the free radical halogenation reaction. However, radicals are very non-selective in nature. Moreover, radicals are non-specific and highly reactive intermediates that result in the formation of the mixture of products.

For instance bromination or chlorination of free radical results in the formation of a number 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 product formation takes place.


ii) Electrophilic Substitution Reaction

This method helps in the preparation of haloarenes such as aryl bromides and aryl chlorides. Electrophilic substitution forms the aryl bromides and aryl chlorides by using halogens such as chlorine and bromine in the presence of Lewis acid. However, the reaction requires the following of certain specific condition for the generation of proper electrophile.

For example, the reaction should be conducted in presence of Lewis acid. Additionally, the reaction must be carried out in the dark. The reactions to obtain the electrophiles are


The electrophiles in the above reactions are Cl+ and Br+ and HCl and HBr are the by-products of the reaction. Therefore, the electrophilic substitution reaction for the preparation of aryl bromide and aryl chloride is


Mechanism of Electrophilic Substitution Reaction

In the above reaction, two different isomers of the aryl chlorides are formed. They are Ortho and Para isomer. The π-electron in the benzene ring attacks the Cl+ electrophile to produce an intermediate complex. However, the H+ bond from the intermediate complex moves in order to compensate for the positive charge of the carbon atom.

Thus the reaction forms two different isomers of the product-ortho and para. The melting points of both the isomer differ significantly. And para-isomer has the higher boiling point than ortho-isomer. Therefore, they can be easily separated from each other.


Preparation of aryl chloride and bromide is possible from this reaction. Aryl fluoride formation is not possible due to the high reactivity of the halogen fluorine. Additionally, iodine reaction is also not possible as iodine is reversible in nature. Thus, it requires a strong oxidizing agent such as Conc. HNO3 or HIO4 for the oxidation of HI and converting it to I2. Therefore, driving the reaction in forwarding direction is difficult and requires a strong oxidizing agent

iii) Sandmeyer’s Reaction

Sandmeyer’s Reaction is a two-step method which includes:

  • Diazonium salt formation
  • Diazonium salt reaction with a cuprous halide (Cu2X2)

Primary aromatic amine reacts with sodium nitrite in the presence of cold mineral acid to form the diazonium salt. In this case,  HNO2 is prepared within the reaction by reacting sodium nitrite and HX in the temperature of 273-278K.


Mechanism of the Sandmeyer’s Reaction

In the first step-

NaNO2 + HCl → HNO2 + NaCl

The HNO2 formed in the presence of H+ undergo protonation to form NO+ as the electrophile. The lone pair of the atom from the primary amine will react with the electrophile.to form an intermediate compound which further gives diazonium salt after elimination of H2O. In the second step, the diazonium salt reacts with cuprous halide to form the respective aryl halidePreparation

3) Haloalkanes & Haloarenes from Alkenes

Haloalkanes and haloarenes preparation is possible by the addition of halogens (X2) across the double bond of the alkene. It is also possible by the addition of hydrogen halides (HX). In this halogen can be chlorine, bromine or even iodine.

i) Addition of HX

Alkene can be converted to haloalkane by an electrophilic addition reaction. Alkene reacts with HX to form R-X. The order of reactivity of halides with respect to alkenes follows the order HI > HBr > HCl > HF. The general reaction will be


The reaction, in this case, is an example of a regioselective reaction. In this type of reaction, we get products in major and minor quantity. Additionally, the reaction follows Markovnikov′s rule of addition for the determination of the major product by the addition of across the double bond of the alkene.

According to Markovnikov’s rule, in an addition reaction of unsymmetrical alkenes, the negative part of the reagent or halogen will attach itself to the carbon that contains less number of hydrogen atoms. For example, prop-1-ene reacts with hydrogen bromide to form 2-bromopropane as a major product.

Peroxide effect (Kharash effect)

There is another possibility where the reaction contradicts Markovnikov’s rule. This effect is known as Peroxide effect/ Kharash effect/ anti-markovnikov’s rule. In this reaction, alkene reacts with HBr in the presence of peroxide. The Br- or the negative part of the reagent will attach itself to the carbon 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

Similarly, alkenes can also react with halogens (X2). For example, Bromine reacts with an alkene in the presence of carbon tetrachloride (CCl4) to form vic-dibromide.  It is a common test to determine a double bond or an alkene compound. The reaction will release reddish brown colour during the reaction.


4) Haloalkanes and Haloarenes from Halogen exchange reaction

i) Finkelstein Reaction

The last method of preparation of haloalkane and haloarene is halogen exchange reaction. In this reaction, an alkyl chloride or alkyl bromide reacts with sodium iodide in acetone to form alkyl iodides.


The reaction is an equilibrium reaction so there is a possibility of forming other products. The solubility difference of alkyl halides in acetone is used for driving the reaction in the forward direction. We know that sodium iodide is soluble in acetone but NaCl or NaBr are insoluble. Therefore, they precipitate out in the reaction which is easy to remove from the reaction mixture.

ii) Swartz Reaction

In this reaction, alkyl fluorides formation is possible by heating of Alkyl fluorides RBr/RCl. The reaction is carried out in the presence of metallic fluoride such as SbF3, Hg2F2, AgF, CoF2.


A Solved Question for You

Q: Why the reaction given below takes place in the dark?


Solution: Halogens are capable of undergoing free radical reaction in the presence of light. Therefore, the reaction will form benzyl bromide as the product. As we have seen previously, this result of the reaction is different from the product of electrophilic substitution reaction. The reaction must be carried out in dark in order to get the product of electrophilic substitution reaction.


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