Alkane Reactions: Videos & Practice Problems

Alkanes, the least reactive hydrocarbons, primarily undergo combustion and halogenation reactions. Combustion involves alkanes reacting with oxygen to yield carbon dioxide and water. Halogenation is a substitution reaction where halogens like bromine or chlorine replace hydrogen atoms in alkanes, requiring heat or UV light to initiate. This process can result in mono or poly substitution, producing alkyl halides. Understanding these reactions is crucial for grasping organic chemistry fundamentals and the behavior of hydrocarbons in various chemical processes.

Common Types of Alkane Reactions

Alkanes, the least reactive hydrocarbons, primarily undergo two significant types of reactions: combustion and halogenation. In combustion reactions, a hydrocarbon reacts with oxygen, yielding carbon dioxide and water. This fundamental reaction is a key concept in general chemistry.

Halogenation, on the other hand, is a substitution reaction where a halogen, such as bromine (Br₂) or chlorine (Cl₂), replaces one of the hydrogen atoms in an alkane. This process requires energy, typically in the form of heat or ultraviolet (UV) light, denoted as HV. The UV light facilitates the breaking of the halogen bond, allowing the halogen to attach to the alkane.

For example, in a halogenation reaction involving methane (CH₄) and a halogen (X₂), the halogen molecule is split into two halogen atoms. One of these atoms substitutes a hydrogen atom in methane, resulting in the formation of an alkyl halide and hydrogen halide (HX). This reaction can lead to mono-substitution, where only one hydrogen is replaced, or poly-substitution, where multiple hydrogens are replaced. It is crucial to understand the context of the question being asked, as the desired product can vary based on whether mono or poly substitution is specified.

In summary, alkanes primarily engage in combustion and halogenation reactions, with halogenation being a key process for producing alkyl halides through the substitution of hydrogen atoms by halogens.

Alkane Reactions Example

In the halogenation reaction of pentane, specifically focusing on monosubstitution, we can identify the major products formed when chlorine replaces hydrogen atoms. Pentane has a symmetrical structure, which is crucial for determining the unique products of this reaction.

When substituting a chlorine atom for a hydrogen atom, we can analyze the carbon chain. The structure consists of two terminal CH₃ groups, two CH₂ groups, and one central CH₂ group. The symmetry of the molecule means that substituting chlorine at either end (the terminal carbons) will yield the same product, which is named 1-chloropentane. This is because both terminal carbons are equivalent, leading to redundancy in product representation.

Next, if we consider the two CH₂ groups adjacent to the terminal carbons, substituting chlorine at either of these positions will result in the same product, 2-chloropentane. Again, this reflects the symmetry of the molecule, allowing us to represent only one of these substitutions.

Finally, substituting chlorine at the central CH₂ group produces a unique product, 3-chloropentane, as this position does not have an equivalent counterpart in the structure.

In summary, the major products from the monochlorination of pentane are 1-chloropentane, 2-chloropentane, and 3-chloropentane. The key takeaway is that unique names correspond to unique products, and due to the symmetry in the structure, we only need to represent one of each equivalent product.

Determine the major product(s) of the following alkane reaction. (Assume monosubstitution.)

• A.
• B.
• C.
• D.

Answer: (D)

Here’s what students ask on this topic:

What are the two major types of reactions that alkanes undergo?

Alkanes primarily undergo two major types of reactions: combustion and halogenation. Combustion involves the reaction of an alkane with oxygen to produce carbon dioxide and water. This is a common reaction that releases energy in the form of heat and light. Halogenation, on the other hand, is a substitution reaction where a halogen (such as bromine or chlorine) replaces a hydrogen atom in the alkane. This reaction requires heat or UV light (represented as HV) to break the bond between the halogen atoms, allowing one of them to attach to the alkane, forming an alkyl halide and releasing HX.

What is the general equation for the combustion of alkanes?

The general equation for the combustion of alkanes is:

$C_n{H}_{2n+2}+3(n+1)O=n_{CO}+2n_{H}O$

In this equation, $C_n$ represents the alkane, and $H_{2n+2}$ represents the hydrogen atoms in the alkane. The products of the reaction are carbon dioxide (${\mathrm{CO}}_2$) and water ($H_{2}O$).

What is halogenation of alkanes and what conditions are required for it?

Halogenation of alkanes is a substitution reaction where a halogen (bromine or chlorine) replaces a hydrogen atom in the alkane. This reaction requires heat or UV light (represented as HV) to break the bond between the halogen atoms in the diatomic halogen molecule (X₂). Once the bond is broken, one of the halogen atoms attaches to the alkane, forming an alkyl halide and releasing HX. For example, in the halogenation of methane (CH₄) with chlorine (Cl₂), the reaction produces chloromethane (CH₃Cl) and hydrogen chloride (HCl).

What is the difference between mono-substitution and poly-substitution in alkane halogenation?

In alkane halogenation, mono-substitution refers to the replacement of only one hydrogen atom in the alkane with a halogen atom. This results in the formation of a single alkyl halide product. For example, in the mono-substitution of methane (CH₄) with chlorine (Cl₂), the product is chloromethane (CH₃Cl). Poly-substitution, on the other hand, involves the replacement of multiple hydrogen atoms in the alkane with halogen atoms. This can result in the formation of multiple alkyl halide products, depending on the number of hydrogen atoms replaced. For example, in the poly-substitution of methane with chlorine, products such as dichloromethane (CH₂Cl₂), trichloromethane (CHCl₃), and tetrachloromethane (CCl₄) can be formed.

What are the products of the halogenation of methane with chlorine?

The halogenation of methane (CH₄) with chlorine (Cl₂) can produce several products depending on the extent of substitution. In mono-substitution, one hydrogen atom in methane is replaced by a chlorine atom, resulting in the formation of chloromethane (CH₃Cl) and hydrogen chloride (HCl). The reaction can be represented as:

${\mathrm{CH}}_4+{\mathrm{Cl}}_2\to {\mathrm{CH}}_3\mathrm{Cl}+\mathrm{HCl}$

In poly-substitution, multiple hydrogen atoms in methane can be replaced by chlorine atoms, leading to the formation of dichloromethane (CH₂Cl₂), trichloromethane (CHCl₃), and tetrachloromethane (CCl₄).