# 20.06 Reactions of Alkanes


As a class, alkanes generally are unreactive. The names saturated hydrocarbon, or "paraffin," which literally means "not enough affinity" [L. par(um), not enough, affins, affinity], arise because their chemical "affinity" for most common reagents may be regarded as "saturated" or satisfied. Thus none of the or bonds in a typical saturated hydrocarbon, for example ethane, are attacked at ordinary temperatures by a strong acid, such as sulfuric acid (), or by an oxidizing agent, such as bromine (in the dark), oxygen, or potassium permanganate (KMnO4). Under ordinary conditions, ethane is similarly stable to reducing agents such as hydrogen, even in the presence of catalysts such as platinum, palladium, or nickel.

However, all saturated hydrocarbons are attacked by oxygen at elevated temperatures and, if oxygen is in excess, complete combustion to carbon dioxide and water occurs. Vast quantities of hydrocarbons from petroleum are utilized as fuels for the production of heat and power by combustion, although it is becoming quite clear that few of the nations of the world are going to continue to satisfy their needs (or desires) for energy through the use of petroleum the way it has been possible in the past.

## Combustion

Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water. It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!

Example $$\PageIndex{1}$$: Propane Combustion

For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:

$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

Counting the oxygens leads directly to the final version:

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Example $$\PageIndex{2}$$: Butane Combustion

With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down.

$C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O$

Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" O2 molecules on the left.

$C_4H_{10} + 6\dfrac{1}{2}\, O_2 \rightarrow 4CO_2 + 5H_2O$

If that offends you, double everything:

$2C_4H_{10} + 13 O_2 \rightarrow 8CO_2 + 10 H_2O$

The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules don't vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.

Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas.

Note: Why carbon monoxide is poisonous

Oxygen is carried around the blood by hemoglobin, which unfortunately binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.

## Substitution Reactions

A substitution reaction, which is the same as a single replacement reaction in inorganic reactions, is a reaction in which one or more atoms in a molecule are replaced with another atom or group of atoms. Alkyl halides are formed by the substitution of a halogen atom for a hydrogen atom. When methane reacts with chlorine gas, ultraviolet light can act as a catalyst for the reaction.

$\ce{CH_4} \left( g \right) + \ce{Cl_2} \left( g \right) \overset{\text{UV light}}{\rightarrow} \ce{CH_3Cl} \left( g \right) + \ce{HCl} \left( g \right)$

The reaction produces chloromethane and hydrogen chloride. When the mixture is allowed to react for longer periods of time, further substitution reactions may occur.

$\ce{CH_3Cl} \left( g \right) + \ce{Cl_2} \left( g \right) \overset{\text{UV light}}{\rightarrow} \ce{CH_2Cl_2} \left( g \right) + \ce{HCl} \left( g \right)$

The product above is dichloromethane. Further substitution products trichloromethane and tetrachloromethane, commonly called carbon tetrachloride. A mixture of products occurs in the reaction, with the relative amounts dependent upon the time that the reaction is allowed to proceed. Chlorofluorocarbons are produced by reacting chloroalkanes with $$\ce{HF}$$, because the fluorine atom makes a stronger bond to the carbon atom than chlorine does.

$\ce{CCl_4} \left( g \right) + \ce{HF} \left( g \right) \overset{\ce{SbF_5}}{\rightarrow} \ce{CCl_3F} \left( g \right) + \ce{HCl} \left( g \right)$

The fluorine atom substitutes for a chlorine atom in the reaction.

## Elimination Reactions

Some reactions involve the removal, or “elimination,” of adjacent atoms from a molecule. This results in the formation of a multiple bond and the release of a small molecule, so they are called elimination reactions. They have the general form

and are similar to cleavage reactions in inorganic compounds. A typical example is the conversion of ethyl chloride to ethylene:

$CH_3CH_2Cl \rightarrow CH_2=CH_2 + HCl$

Much of the approximately 26 million tons of ethylene produced per year in the United States is used to synthesize plastics, such as polyethylene. In Equation 24.2, the A–B molecule eliminated is HCl, whose components are eliminated as H+ from the carbon atom on the left and Cl from the carbon on the right. When an acid is produced, as occurs here, the reaction is generally carried out in the presence of a base (such as NaOH) to neutralize the acid.

## Contributors

• Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)

• John D. Robert and Marjorie C. Caserio (1977) Basic Principles of Organic Chemistry, second edition. W. A. Benjamin, Inc. , Menlo Park, CA. ISBN 0-8053-8329-8. This content is copyrighted under the following conditions, "You are granted permission for individual, educational, research and non-commercial reproduction, distribution, display and performance of this work in any format."