Preparing haloalkanes

Four ways in

Read the map from the outside in. We shall go through the four routes, and say which one a chemist uses and why.

From alcohols: the best route

An alcohol $\text{R}\text{-}\text{OH}$ has an OH group that can be replaced by a halogen. Water is the leaving group, once the oxygen is protonated or converted into something better.

With hydrogen halides. The alcohol is heated with a hydrogen halide.

$\text{R}\text{-}\text{OH}\ +\ \text{HX}\ \rightarrow \ \text{R}\text{-}\text{X}\ +\ \text{H}_{2}\text{O}$

• For the chloride, the alcohol is warmed with concentrated hydrochloric acid and anhydrous zinc chloride (Groove's process). Zinc chloride is a Lewis acid, and it makes the OH leave. Tertiary alcohols react with concentrated HCl without it.
• For the bromide, hydrogen bromide is made in the flask from sodium or potassium bromide and concentrated sulphuric acid.
• For the iodide, potassium iodide is used with 95 percent phosphoric acid. Sulphuric acid cannot be used here, because it would oxidise the hydrogen iodide to iodine.

The order of reactivity of the alcohols is tertiary, then secondary, then primary. The order of the acids is HI, then HBr, then HCl.

With phosphorus halides. $\text{R}\text{-}\text{OH}\ +\ \text{PCl}_{5}\ \rightarrow \ \text{R}\text{-}\text{Cl}\ +\ \text{POCl}_{3}\ +\ \text{HCl}$. Phosphorus trichloride and tribromide work in the same way. Red phosphorus with bromine or iodine makes $\text{PBr}_{3}$ or $\text{PI}_{3}$ in the flask. These reagents leave a phosphorus compound that has to be separated from the product.

With thionyl chloride. This is the cleanest method.

$\text{R}\text{-}\text{OH}\ +\ \text{SOCl}_{2}\ \rightarrow \ \text{R}\text{-}\text{Cl}\ +\ \text{SO}_{2}\ +\ \text{HCl}$

Both by-products, sulphur dioxide and hydrogen chloride, are gases. They leave the flask, and the haloalkane that remains is almost pure.

From hydrocarbons

Free radical halogenation of alkanes. An alkane reacts with chlorine or bromine in ultraviolet light, or at about $573\ \text{K}$. A chain reaction runs, and the energy table in the simulation below shows why chlorine and bromine work, but iodine does not.

At fluorine both steps give out energy, and the reaction is violent. At iodine the first step takes in $140\ \text{kJ/mol}$ and the overall change is positive. The reaction does not go. Chlorine and bromine are the useful halogens.

Free radical halogenation has a drawback for making one compound. Any hydrogen can be replaced, and more than one can be. The product is a mixture of isomers, and of di and tri compounds. Different hydrogen atoms react at different speeds. At room temperature, with chlorine, the reactivity of one hydrogen is in the ratio $1 : 3.8 : 5.0$ for primary, secondary and tertiary. Bromine is far more selective, about $1 : 82 : 1640$ at $125\ ^\circ\text{C}$.

Allylic and benzylic halogenation. A hydrogen on a carbon next to a double bond or a benzene ring is replaced easily, because the radical formed is stabilised. N-Bromosuccinimide (NBS) gives allylic bromides from alkenes. Propene with chlorine at about $773\ \text{K}$ gives 3-chloroprop-1-ene by replacement, not addition. Toluene with chlorine in light gives benzyl chloride on the side chain. In the dark with iron, the ring is chlorinated instead.

Addition to alkenes. Hydrogen halides add across a double bond by Markovnikov's rule. The hydrogen goes to the carbon that already has more hydrogen atoms, and the halogen goes to the other carbon.

$\text{CH}_{3}\text{CH}=\text{CH}_{2}\ +\ \text{HBr}\ \rightarrow \ \text{CH}_{3}\text{CHBrCH}_{3}$

In the presence of a peroxide, hydrogen bromide adds the other way, by a free radical chain. This is the peroxide effect, or anti-Markovnikov addition.

$\text{CH}_{3}\text{CH}=\text{CH}_{2}\ +\ \text{HBr}\ \xrightarrow{\text{peroxide}}\ \text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{Br}$

Only HBr shows it. For HCl and HI the radical steps do not release enough energy. Bromine in carbon tetrachloride adds to an alkene and loses its red-brown colour. That is the test for a C=C bond. The product, such as $\text{CH}_{2}\text{BrCH}_{2}\text{Br}$, is a vicinal dibromide.

Halogen exchange

One halogen can be exchanged for another.

• Finkelstein reaction. $\text{R}\text{-}\text{Cl}\ +\ \text{NaI}\ \rightarrow \ \text{R}\text{-}\text{I}\ +\ \text{NaCl}$ in dry acetone. Sodium iodide dissolves in acetone, but sodium chloride and sodium bromide do not. They precipitate and are removed from the equilibrium, so the reaction goes to completion.
• Swarts reaction. A chloro or bromo compound is heated with a metallic fluoride, such as $\text{AgF}$, $\text{Hg}_{2}\text{F}_{2}$, $\text{CoF}_{2}$ or $\text{SbF}_{3}$. This is the route to fluoroalkanes, which cannot be made by direct fluorination or by HF.

Try it in Python

The program calculates the energy change of the two propagation steps for each halogen. It uses bond enthalpies from a table. It should show the pattern of the simulation.

Common mistakes

• Using sulphuric acid for hydrogen iodide. It oxidises the iodide. Use phosphoric acid.
• Calling free radical halogenation a clean route. It gives a mixture, so it is a poor way to make one haloalkane.
• Writing the wrong product for HBr with a peroxide. The bromine goes to the carbon with more hydrogen atoms.
• Forgetting the dry acetone in the Finkelstein reaction. The precipitate of NaCl or NaBr is what drives it.

Remember these

• Alcohols give the best route. Thionyl chloride gives the cleanest result because both by-products are gases.
• Alcohol reactivity: tertiary, then secondary, then primary. Acid reactivity: HI, then HBr, then HCl.
• Free radical halogenation gives mixtures. Bromine is more selective than chlorine. Iodine does not react, and fluorine reacts violently.
• Markovnikov addition of HX, and anti-Markovnikov addition of HBr with a peroxide.
• Finkelstein makes iodides (NaI, dry acetone). Swarts makes fluorides (AgF, $\text{Hg}_{2}\text{F}_{2}$, $\text{CoF}_{2}$ or $\text{SbF}_{3}$).

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why is thionyl chloride a better reagent than phosphorus pentachloride for making chloroalkanes?

2. Which product forms when propene reacts with HBr without a peroxide, and which with a peroxide?

3. Why can iodoalkanes not be made by direct iodination of an alkane?

4. How would you make iodoethane from chloroethane?

References

• NCERT, Chemistry Part II for Class XII, the chapter Haloalkanes and Haloarenes, on methods of preparation.
• Morrison and Boyd, Organic Chemistry, the chapters on alkyl halides and on free radical halogenation (the reactivity ratios).
• OpenStax, Organic Chemistry, Chapter 10 (Organohalides).