27.7: Reactions of alkanes (2023)

  1. last change
  2. save as PDF
  • Side-id
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!- \!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{ span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart }{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\ norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm {span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\ mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{ \ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{ \unicode[.8,0]{x212B}}\)

    Alkanes (the most basic of all organic compounds) undergo very few reactions. The two more important reactions are combustion and halogenation (ie replacing a single halogen with a single hydrogen on the alkane) to form a haloalkane. The halogen reaction is very important in organic chemistry as it opens the door to further chemical reactions.


    When a hydrocarbon burns completely (if there is enough oxygen), carbon dioxide and water are formed. Being able to write correctly balanced equations for these reactions is very important as they are often part of thermochemical calculations. Some are easier than others. For example, in alkanes, those with an even number of carbon atoms are slightly harder than those with an odd number!

    Example \(\PageIndex{1}\): Propane combustion

    For example with propane (C3H8), you can balance the carbon and hydrogen while writing the equation. Your first draft would be:

    \[ C_3H_8 + O_2 \højrepil 3CO_2 + 4H_2O\]

    Counting the oxygens leads directly to the final version:

    \[ C_3H_8 + 5O_2 \højrepil 3CO_2 + 4H_2O\]

    Example \(\PageIndex{2}\): Butane combustion

    Mit Butan (C4H10), you can rebalance the carbon and hydrogen atoms while writing the equation.

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

    Counting the oxygen atoms leads to a small problem - with 13 on the right. The simple trick is to indulge in "six and a half" O's2Molecules 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 larger the molecules become, the more difficult it is to ignite the hydrocarbons. This is because the larger molecules do not evaporate so easily - the reaction goes much better when the oxygen and hydrocarbon gases are well mixed. Unless the liquid is very volatile, only the surface molecules can react with the oxygen. Larger molecules have greater van der Waals attraction, making it harder for them to break away from their neighbors and become a gas.

    In complete combustion, all hydrocarbons burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecule increases. That is, the larger the hydrocarbon, the more likely it is to produce a yellow, smoky flame. Incomplete combustion (where there is not enough oxygen) can result in the formation of carbon or carbon monoxide. Simply put, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets what's left! The presence of glowing carbon particles in a flame makes 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 through the blood by hemoglobin, which unfortunately binds to exactly the same place on hemoglobin as oxygen. The difference is that carbon monoxide binds irreversibly (or very tightly) - making the hemoglobin molecule in question useless for oxygen transport. If you inhale enough carbon monoxide, you will die from a form of internal suffocation.

    Halogenation of alkanes

    Halogenation is the replacement of one or more hydrogen atoms in an organic compound with a halogen (fluorine, chlorine, bromine or iodine). In contrast to the complex combustion transformations, the halogenation of an alkane occurs straightforwardlysubstitution reactionA CH bond is broken and a new C-X bond is formed. The chlorination of methane shown below is a simple example of this reaction.

    CH4+ Cl2+ Energy → CH3Cl + HCl

    Since only two covalent bonds are broken (C-H and Cl-Cl) and two covalent bonds are formed (C-Cl and H-Cl), this reaction seems to be an ideal case for mechanistic studies and speculation. A complication, however, is that all the hydrogen atoms in an alkane can be substituted, resulting in a mixture of products as shown belowunbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of hydrocarbon favors the formation of methyl chloride as the main product; whereas an excess of chlorine promotes the formation of chloroform and carbon tetrachloride.

    CH4+ Cl2+ Energy → CH3Cl + CH2Cl2+ I WANT3+ CCl4+ HCI

    In the presence of a flame, the reactions are more like those for fluorine - a mixture of carbon and hydrogen halide is formed. The intensity of the reaction decreases significantly when going from fluorine to chlorine to bromine. The interesting reactions take place in the presence of ultraviolet light (sunlight will do it). These are photochemical reactions that take place at room temperature. We will look at the reactions with chlorine, although the reactions with bromine are similar but slower.

    In substitution reactions, hydrogen atoms in methane are individually replaced by chlorine atoms. Finally, a mixture of chloromethane, dichloromethane, trichloromethane and carbon tetrachloride is formed.

    27.7: Reactions of alkanes (1)

    The original mixture of a colorless and a green gas would produce vapors of hydrogen chloride and a mist of organic liquids. All organic products are liquid at room temperature with the exception of the gaseous chloromethane.

    When using bromine, you can either mix methane with bromine vapor or bubble the methane through liquid bromine - in both cases exposed to UV light. The original gas mixture would naturally be reddish-brown rather than green. One would not use these reactions as a means of producing these organic compounds in the laboratory because it would be too difficult to separate the product mixture. The mechanisms of the reactions are explained on separate pages.

    Larger alkanes and chlorine

    You would again get a mixture of substitution products, but it is worth looking briefly at what happens when only one of the hydrogen atoms is substituted (monosubstitution) - just to show that things are not always as simple as they seem ! For example, with propane you can get one of two isomers:

    27.7: Reactions of alkanes (2)

    If chance were the only factor, you would expect to end up with three times as much chlorine isomers at the end. There are 6 hydrogen atoms that could be replaced on the carbons at the end compared to only 2 in the middle. In fact, you get about the same amount of each of the two isomers. If you use bromine instead of chlorine, most of the product is where the bromine is attached to the middle carbon atom.


    The reactions of the cycloalkanes are generally the same as for the alkanes, except for the very small ones - especially cyclopropane. As a non-cyclic alkane, cyclopropane undergoes substitution reactions with chlorine or bromine in the presence of UV light. However, it also has the ability to react in the dark. In the absence of UV light, cyclopropane can undergo addition reactions that break the ring. For example, cyclopropane with bromine gives 1,3-dibromopropane.

    27.7: Reactions of alkanes (3)

    This can also happen in the presence of light, but substitution reactions also occur. The ring is broken because cyclopropane suffers severely from ring strain. The bond angles in the ring are 60° instead of the normal value of about 109.5° when the carbon forms four single bonds. The overlap between the atomic orbitals in the formation of the carbon-carbon bonds is less than usual, and there is considerable repulsion between the bond pairs. If the ring breaks, the system becomes more stable.

    Contributors and attributions


    Top Articles
    Latest Posts
    Article information

    Author: Sen. Emmett Berge

    Last Updated: 06/02/2023

    Views: 6144

    Rating: 5 / 5 (80 voted)

    Reviews: 95% of readers found this page helpful

    Author information

    Name: Sen. Emmett Berge

    Birthday: 1993-06-17

    Address: 787 Elvis Divide, Port Brice, OH 24507-6802

    Phone: +9779049645255

    Job: Senior Healthcare Specialist

    Hobby: Cycling, Model building, Kitesurfing, Origami, Lapidary, Dance, Basketball

    Introduction: My name is Sen. Emmett Berge, I am a funny, vast, charming, courageous, enthusiastic, jolly, famous person who loves writing and wants to share my knowledge and understanding with you.