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

R-OH → R-X       where X= Cl, Br or I

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

Halogenation of alcohols

The hydroxyl functional group in an alcohol can be replaced by a halogen to form a halogenalkanes. The simplest method would simply be treatment of the alcohol by the corresponding acid (HX), this is shown below:

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

However this only really works well for tertiary alcohols where the reaction is very quick and can be as simple as bubbling the HX gas through a solution of the tertiary alcohol, for example:

Reactions of primary and secondary alcohol with acids (HX) are slow and the conversion of these alcohols to the corresponding halogenalkanes are best done using different reagents. There are many other types of reagents that will enable you to carry out the conversion of primary and secondary alcohols to halogenalkanes, these include for example the phosphorus halides (PX₃ and PX₅).

Primary and secondary alcohols

There are other reagents that are useful in converting primary and secondary alcohols into their corresponding halogenalkanes. These include phosphorus trichloride(PCl₃) and phorphorus pentachloride (PCl₅) as well as thionyl chloride, also called sulfur dichloride oxide (SOCl₂). For example consider the reactions below for the halogenation of primary and secondary alcohols:

Similar reactions as 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 iodine, once it forms in the flask it reacts with the alcohol to form the iodoalkane.

These halogenation reactions all take place in mild conditions, unlike the reactions involving concentrated acids. This is likely to ensure that any "fragile" molecule stay intact. The examples below show the use of other halogenating agents to form halogenalkanes from primary alcohols.

Practical details

Preparation of Chloroalkanes

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

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 tertiary halogenalkanes using tertiary alcohols.

Preparation of 2-chloro-2-methylpropane

Below is a series of steps which should help you understand the basic method needed to carry out the preparation of 2-chloro-2-methylpropane from the tertiary alcohol 2-methylpropan-2-ol:

• 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 acid.
• leave the mixture of the 2-methylpropan-2-ol and acid to stand in a fume cupboard for about 15 minutes. Swirl 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 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. Add a few mls of sodium hydrogencarbonate solution to the organic layer in the separating funnel. This will react will 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 or aqueous layer. Open the tap 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.
• 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.

Bromoalkanes and iodoalkane

As mentioned above you can react an alcohol with concentrated acid to form the halogenalkane e.g.

R-OH + HBr → R-Br + H₂O

You may recall from the module of work you did on the halogens that concentrated sulfuric acid, as well as acting as an acid in its reactions can also act a a reasonably strong oxidising agent. It will for example react with sodium and potassium bromide salts to form hydrogen bromide gas:

H₂SO_4(aq) + HBr_(aq) → NaHSO_4(s) + HBr_(g)

We can make use this reactions to form a bromoalkane. If a mixture of 50% sulfuric acid and sodium bromide in place in a pear shaped flask and then an alcohol such as butan-1-ol is added and the mixture heated under reflux conditions for approximately 45 minutes then the halogenalkane 1-bromobutane will be produced in the flask.

CH₃CH₂CH₂CH₂OH_(aq) + HBr_(aq) → CH₃CH₂CH₂CH₂Br _(l) + H₂O_(l)

You might imagine that we could use a similar method to produce iodoalkanes, however you may recall that if we add concentrated sulfuric acid to sodium iodide the reaction is completely chaotic and you end up with a complete mess!! The sulfuric acid will be reduced all the way down to sulfur and even hydrogen sulphide gas, this means that there will be little hydrogen iodide available to react with any alcohol added to this mixture. The solution to this problem is simple; just change the acid to concentrated phosphoric acid. This mixture of phosphoric acid and sodium iodide will form hydrogen iodide (HI) which will react with an alcohol to form the iodoalkane e.g.

CH₃CH₂CH₂CH₂OH_(aq) + HI_(aq) → CH₃CH₂CH₂CH₂I _(l) + H₂O_(l)

We can of course use phosphorus triiodide (PI₃) and simply react it with the alcohol as we did above with phosphorus trichloride, however phosphorus(III)iodide is unstable. So a mixture of the alcohol, red phosphorus is mixed with iodine (we can do this with bromine as well to produce PBr₃) and the heated under reflux to form the iodoalkane (or the bromoalkane if bromine is used instead of iodine).

Key Points

• Tertiary alcohols under halogenation reactions easily using concentrated acids (HX).
• Primary and secondary alcohols react slowly with concentrated acid and it is better to use alternative halogenating agents.

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