# 16.2 Reactivity of Alkanes Alkanes are non-polar hydrocarbons because the electronegativity difference between carbon and hydrogen is not significant enough to create polar bonds. They are also known as **paraffins**, a term derived from the Latin words "parum" (little) and "affinis" (reactivity), which aptly describes their low reactivity. The low reactivity of alkanes is attributed to two main factors: ### 1. Non-polar Bonds The electronegativity values of carbon and hydrogen are 2.5 and 2.1, respectively. The small difference in electronegativity means the bonding electrons in C-H bonds are shared almost equally, making the bonds **non-polar**. Because alkanes are composed of non-polar C-C and C-H bonds, they do not react with polar reagents such as acids, bases, oxidizing agents, or reducing agents under normal conditions. ### 2. Strength of Sigma (σ) Bond Alkanes contain only single covalent bonds, which are **sigma (σ) bonds**. They do not have any pi (π) bonds. A sigma (σ) bond is very strong because the shared electrons are located directly between the two bonded nuclei, resulting in a strong electrostatic attraction. These strongly held electrons are not easily available for chemical reactions, which contributes to the chemical inertness of alkanes. For more on orbital overlaps, see . Despite their general lack of reactivity under normal conditions, alkanes undergo reactions at high temperatures or in the presence of ultraviolet (UV) light. Two important reactions of alkanes are: 1. **Free Radical Substitution** 2. **Cracking** Free radical substitution reactions are initiated by UV light, which generates highly reactive free radicals. Cracking involves heating large alkane molecules to high temperatures to break them down into smaller, more useful molecules. ### 16.2.1 Free Radical Substitution of Ethane This type of reaction involves highly reactive **halogen free radicals** that attack ethane molecules, replacing hydrogen atoms with halogen atoms. The mechanism occurs in three distinct steps. #### 1. Initiation The reaction is initiated by ultraviolet (UV) light, which provides the energy to break the chlorine-chlorine bond through **homolytic fission**. This process creates two highly reactive chlorine free radicals. $$\mathrm{Cl-Cl} \xrightarrow{h\nu} \mathrm{Cl^\cdot} + \mathrm{Cl^\cdot}$$ (Here, $h\nu$ represents a photon of UV light) #### 2. Propagation This step is a self-sustaining chain reaction. Once a chlorine radical is formed, it attacks an ethane molecule, abstracting a hydrogen atom to form hydrogen chloride ($\mathrm{HCl}$) and an ethyl free radical. $$\mathrm{CH_3-CH_3 + Cl^\cdot \rightarrow CH_3-CH_2^\cdot + HCl}$$ The newly formed ethyl free radical is also highly reactive. It attacks a neutral chlorine molecule, abstracting a chlorine atom to form chloroethane and a new chlorine free radical, which continues the chain. $$\mathrm{CH_3-CH_2^\cdot + Cl_2 \rightarrow CH_3-CH_2Cl + Cl^\cdot}$$ **Further Substitution:** If excess chlorine is used, the propagation steps can continue, substituting more hydrogen atoms. This leads to a mixture of chloro-substituted products like 1,1-dichloroethane, 1,1,1-trichloroethane, and eventually, 1,1,1,2,2,2-hexachloroethane ($\mathrm{C_2Cl_6}$). #### 3. Termination The chain reaction eventually stops when two free radicals collide and combine to form a stable molecule. This can happen in several ways, removing the reactive radicals from the system. - Two chlorine radicals combine: $$\mathrm{Cl^\cdot + Cl^\cdot \rightarrow Cl_2}$$ - An ethyl radical and a chlorine radical combine: $$\mathrm{CH_3-CH_2^\cdot + Cl^\cdot \rightarrow CH_3-CH_2Cl}$$ - Two ethyl radicals combine to form butane: $$\mathrm{CH_3-CH_2^\cdot + CH_3-CH_2^\cdot \rightarrow CH_3-CH_2-CH_2-CH_3}$$