The preparation of haloalkanes or halogenalkanes from alcohols essentially involves replacing the hydroxyl functional group (R-OH) on the alcohol with a halogen atom (-X). We can summarise this nucleophilic substitution reaction as:

R-OH → R-X     where X= Cl, Br or I

There are a several reagents that will enable you to successfully carry out this substitution reaction. However the ease in which the hydroxyl group (-ROH) in the alcohol is substituted for a halogen atom (-X) varies with the type of alcohol used. Tertiary alcohols easily undergo these substitution reactions but primary and secondary alcohols are not so reactive and so different reaction conditions are needed to produce halogenalkanes using these two classes of alcohol.

Halogenation of alcohols

The hydroxyl functional group in an alcohol can be replaced by a halogen to form a haloalkane. The simplest method would simply be treatment of an alcohol by the corresponding acid (HX), that is hydrochloric (HCl) and hydrobromic (HBr) acids; this is shown below:

R-OH  +  HX  →  R-X  +  H₂O     where X= Cl, Br or I

However as mentioned above this substitution reaction of the hydroxyl functional group for a halogen only really works well for tertiary alcohols where the reaction is very quick and can be as simple as bubbling the hydrogen halide gas (HX_(g)) through a solution of the tertiary alcohol, although the reaction is easier to control and carry out if you were to simply mix the tertiary alcohol with concentrated hydrochloric acid (HCl) or hydrobromic acid (HBr) at room temperature; for example the image below shows the equations for the chlorination and bromination of two tertiary alcohols using hydrochloric and hydrobromic acids.

Primary, secondary and tertiary carbocations and halogenation reactions

Note: To get the most out of this page, especially for students following the Edexcel specification you should have a clear idea of the differences between S_N1 and S_N2 reactions; if you need a quick review then click here. The summary box below summaries the reactions of primary, secondary and tertiary alcohols as they form haloalkanes. It highlights the type of reaction undergone by primary, secondary and tertiary alcohols and gives an indication of the reaction rates.

Bromination of alcohols

As mentioned above tertiary alcohols can be brominated by simply slowly adding hydrobromic acid (HBr_(aq)) to the tertiary alcohol, however a more common method which is used brominate alcohols is to reflux the alcohol (primary, secondary or tertiary) with a mixture of sulfuric acid and sodium or potassium bromide.

You may recall that halide salts such as sodium bromide and potassium bromide will react with concentrated sulfuric acid to produce hydrogen bromide gas (HBr_(g)), an equation for this reaction is shown below:

sodium bromide_(s) + sulfuric acid_(l) → sodium hydrogensulfate_(s) + hydrogen bromide_(g)

NaBr_(s) + H₂SO_{4 (l)} → NaHSO_{4 (s)} + HBr_(g)

However the concentrated sulfuric acid is able to further oxidise the hydrogen bromide gas produced in this reaction to give bromine, sulfur dioxide gas and water, an equation for the reaction is shown below:

2HBr_(g) + H₂SO_{4 (l)} → Br_2(l) + SO_2(g) + 2H₂O_(l)

This second oxidation reaction is an unwanted reaction, since we need the hydrogen bromide gas (HBr) to dissolve in the reaction mixture to form hydrobromic acid. Now since hydrobromic acid is a strong acid it will dissociate fully to form hydrogen ion (H⁺) and bromide ions (Br^-) both of which are needed in the conversion of the alcohol to the halogenalkane.

Concentrated or 50% sulfuric acid?

So how do we stop this second unwanted oxidation reaction from happening? Well instead of using concentrated sulfuric acid simply use a 50% solution of sulfuric acid; this solution will not oxidise the hydrogen bromide gas to bromine.

The bromination of propan-2-ol

As an example of a bromination reaction consider the addition of hydrogen bromide gas to the secondary alcohol propan-2-ol. A reflux experiment will need to be set-up, here a mixture of sodium bromide and propan-2-ol and a 50% solution of sulfuric acid is placed in a pear shaped flask and the whole mixture is heat for around 40 minutes. The apparatus set-up and equations for this reaction are shown below.

An overall equation for the reaction of propan-2-ol with 50% sulfuric acid and sodium bromide is shown below:

CH₃CH(OH)CH_3(aq) +  NaBr_(s)  +   H₂SO_4(l) →   CH₃CH(Br)CH_3(l)  NaHSO_4(s) + H₂O_(l)

Summary of reactions of alcohols with 50% sulfuric acid and sodium bromide

This bromination reaction of alcohols works well with tertiary alcohols but as summarised in the table below more and more vigorous reaction conditions are needed as we try to brominate secondary and primary alcohols.

Reaction mechanism for the bromination of propan-2-ol

So far we have looked at various examples of how a halogen atom can take the place of the hydroxyl functional group (R-OH) in an alcohol but we have not mentioned in any detail about how this reaction actually proceeds. So let’s correct that by looking at how the bromide ion (Br^-) is able to replace the hydroxyl functional group in propan-2-ol to form the haloalkane 2-bromopropane.

At first glance the reaction seems straight forward; a bromide ion (Br^-) acts as a nucleophile and attacks the delta positive carbon atom attached to the hydroxyl functional group in the alcohol molecule and "kicks out" or replaces the hydroxyl group (R-OH). There are just a few problems we need to consider here:

• The bromide ion (Br^-) is not a good enough nucleophile to “kick-out” the hydroxyl group (-OH) in the alcohol molecule.
• Secondly the hydroxyl group (R-OH) is also a poor leaving group. If the hydroxyl group (–OH) group in the alcohol left it would take both the electrons in the C-OH bond and leave as a hydroxide ion (OH^-); unfortunately hydroxide ions (OH^-) are not particular good at stabilising a negative charge and so it is not particularly stable on its own, this means it is unlikely to form. What we need to help drive this reaction forward is a better leaving group than hydroxide ions (OH^-).
• Another barrier to this nucleophilic substitution reaction is the strength of the C-O bond in the alcohol; if we could somehow weaken this bond then the nucleophilic substitution reaction would proceed much faster.

Solving these problems and putting a mechanism in place

This nucleophilic substitution reaction is normally carried out under reflux conditions; as shown in the image opposite, here the mixture of 50% sulfuric acid will react with the sodium bromide to generate hydrogen bromide gas (HBr_(g)) in situ which will then dissolve in the reaction mixture to form hydrobromic acid (HBr_(aq)), which is a strong acid. This acid along with the 50% sulfuric acid (H₂SO₄) will provide the hydrogen ions (H⁺) needed to start this substitution reaction off.

The first step in this reaction will be:

• The addition of a hydrogen ion (H⁺) or protonation of the hydroxyl group (-OH) on the alcohol molecule. The oxygen atom will use one of its lone pairs of electrons to form a dative covalent bond to the hydrogen ion (H⁺). This is an important step because it transforms the hydroxyl group (R-OH) into a water molecule, which is a good leaving group. Also the presence of a positive charge on the oxygen atom means that this electronegative atom will be withdrawing electron density from the C-O bond and this will weaken it. This is outlined in step 1 in the table below.

Now the alcohol propan-2-ol is a secondary alcohol so we may have a dilemma here; will the reaction proceed via a S_N1 or a S_N2 mechanism, well the reaction mixture is strongly acidic and polar, which favours carbocation formation and so the reaction will probably go through the S_N1 route. So the next step in this reaction mechanism will be:

• The formation of the secondary carbocation as the water molecule leaves. This will be the slow step in the reaction. This is shown as step 2 in the table below.

Once the secondary carbocation ion forms the next step will be:

• The attack of the bromide ion (Br^-); which will act as a nucleophile and attack the carbocation to form the product, 2-bromopropane.This final step will be fast.

🧩 SN1 Mechanism: Bromination of propan-2-ol

Step 1: Protonation: The oxygen atom uses a lone pair of electrons to bond to a H+. The –OH group becomes water, which is a good leaving group.	Step 2: Carbocation formation (slow step) A water molecule leaves, this results in the formation of secondary carbocation. This is the rate-determining step; that is it is a slow step.	Step 3: Nucleophilic attack The bromide ion (Br-) attacks the carbocation to form 2-bromopropane. This step is fast.

Chlorination of alcohols

You might imagine that we could carry out a similar reaction to chlorinate alcohols, all that we would have to do is swap the halide salt from sodium or potassium bromide to a chloride and this would then produce hydrogen chloride gas (HCl_(g)); as shown in the equations below; which could then chlorinate the alcohol in a similar way that the hydrogen bromide gas (HBr_(g)) did.

sodium chloride_(s) + sulfuric acid_(l) → sodium hydrogensulfate_(s) + hydrogen chloride_(g)

NaCl_(s) +  H₂SO_4(l) → NaHSO_4(s) + HCl_(g)

Unfortunately although sodium chloride readily reacts with concentrated sulfuric acid to produce hydrogen chloride gas, the use of HCl(g) this is not an effective method for chlorinating alcohols. Hydrogen chloride and in particular the chloride ion (Cl^-) is a poor nucleophile and will not readily displace the hydroxyl group (R-OH) in primary or secondary alcohols. While tertiary alcohols will react with concentrated hydrochloric acid, primary and secondary alcohols are better converted into chloroalkanes using other chlorinating agents reagents such as phosphorus pentachloride (PCl₅), phosphorus trichloride (PCl₃) or thionyl chloride (SOCl₂) which are much more efficient chlorinating agents.

Chlorination of primary and secondary alcohols

Phosphorus trichloride (PCl₃) and phosphorus pentachloride (PCl₅) are common chlorinating agents, phosphorus pentachloride (PCl₅) is a more effective and aggressive halogenating agent than phosphorus trichloride (PCl₃). Thionyl chloride; also called sulfur dichloride oxide (SOCl₂) is also a highly efficient and useful chlorinating agent, for example consider the equations below which show the halogenation of primary and secondary alcohols using these reagents:

We can also use phosphorus tribromide (PBr₃) to produce bromoalkanes from alcohols:

We can also use phosphorus pentachloride (PCl₅) as a reagent to chlorinate alcohols to form chloroalkanes:

Preparing iodoalkanes

Similar reactions as those shown above can be used to prepare iodoalkanes, however phosphorus (III) iodide (PI₃) is unstable, so it needs to be prepared in situ. That is it is made in the reaction flask from a mixture of red phosphorus and solid iodine, once the phosphorus triiodide forms in the flask it reacts with the alcohol to form the iodoalkane. The equations below show how phosphorus (III) iodide is prepared and how it reacts with the primary alcohols propan-1-ol and methanol:

Practical work- The preparation of chloroalkanes

As we have seen above chloroalkanes can be prepared by reacting an alcohol with phosphorus pentachloride (PCl₅) and phosphorus trichloride (PCl₃:

🧪 Practical Method: Preparation of 2-chloro-2-methylpropane (Click to Expand)

However in the lab it is often easier to use the reaction of concentrated hydrochloric acid with tertiary alcohols to form tertiary halogenalkanes. As an example consider the reaction of the tertiary alcohol 2-methylpropan-2-ol with concentrated hydrochloric acid to form the tertiary halogenalkane 2-chloro-2-methylpropane, this reaction can be summarised as:

Below is a basic outline of a method which can be used to prepare the tertiary haloalkane 2-chloro-2-methylpropane using the tertiary alcohol 2-methylpropan-2-ol.

Preparation of 2-chloro-2-methylpropane

Below is a series of steps which should help you understand the basic method:

• Measure out 10ml of the tertiary alcohol 2-methylpropan-2-ol into a clean separating funnel.
• Add 20ml of concentrated hydrochloric acid a few mls at a time then swirl the funnel after each addition of the acid.
• Leave the mixture of the 2-methylpropan-2-ol and acid to stand in a fume cupboard for about 15 minutes; swirling the mixture occasionally.
• Once the reaction is complete there will be two layers inside the separating funnel. Halogenalkanes are usually insoluble in a water or any aqueous layer, so inside the separating funnel there will be an organic layer containing the halogenalkane and an aqueous layer. Open the tap on the bottom of the separating funnel and discard the aqueous layer.
• Next you have to make sure there is no unreacted acid left in the organic layer, to remove the excess acid simply add a few mls of the basic sodium hydrogencarbonate solution to the organic layer in the separating funnel. This will react with any excess acid and neutralise it. As it reacts it will release carbon dioxide gas. Keep adding a few mls of sodium hydrogencarbonate solution until there is no more fizzing, this will indicate that all the acid has been neutralised.
• There will again be two layers inside the separating funnel. The top organic layer containing our product, the 2-chloro-2-methylpropane and a bottom aqueous layer. Open the tap on the separating funnel and discard the aqueous layer.
• It is likely that the organic layer containing the 2-chloro-2-methylpropane will have some water left in it. To dry it off run the organic layer into a clean dry conical flask and add a few spatulas of anhydrous sodium sulfate (Na₂SO₄) and swirl to dry the organic layer. Sodium sulfate is a drying agent and will remove any water still present or mixed with the haloalkane.
• The next step is to purify the organic layer by distilling off the 2-chloro-2-methylpropane. Pour the organic layer through a small filter funnel into a pear shaped flask which contains a few anti-bumping granules.


• Next set-up a simple distillation (shown below) and collect the fraction that is produced between the temperatures of 47-52⁰C, this will be our product.

Summary table

Key Points

• Making a haloalkane from an alcohol is a (nucleophilic) substitution reaction: the –OH group is replaced by Cl, Br or I.
• Tertiary alcohols react the easiest because they can form a stable carbocations, so they usually react by a S_N1 mechanism and these substitution reactions will usually work at room temperature with concentrated HX_(aq).
• Primary alcohols do not form stable carbocations, so they usually react by a S_N2 mechanism and need more vigorous conditions (often heating / reflux).
• Secondary alcohols sit in the middle — they react more slowly than tertiary alcohols and may show S_N1 or S_N2 reaction routes depending on the reaction conditions.
• In the NaBr + 50% H₂SO₄ method, the acid generates HBr_(g) in situ, which provides the Br^- nucleophile needed to form bromoalkanes when the hydrogen bromide gas dissolves in the reaction mixture to form the strong acid hydrobromic acid (HBr_(aq))
• HCl and Cl^- are both poor nucleophiles, so using HCl(g) made from NaCl + H₂SO₄ is not an effective way to chlorinate primary or secondary alcohols.
• To make chloroalkanes from primary/secondary alcohols, use proper chlorinating agents such as PCl₃, PCl₅ or SOCl₂.
• To make iodoalkanes, PI₃ is made in situ (it’s unstable), then it reacts with the alcohol to replace the –OH group.
• Reactivity order for halogenation of alcohols is: tertiary > secondary > primary (mainly down to carbocation stability and steric hindrance).

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