Halogenation Reaction - Online Tutor, Practice Problems & Exam Prep

Halogenation reactions involve the addition of halogens (X₂, such as Br₂ or Cl₂) to alkenes and alkynes. In alkenes, each double-bonded carbon gains one halogen, resulting in a dihalide. For alkynes, which contain two pi bonds, two moles of halogen are required, yielding a tetrahalide with four halogens. This process exemplifies an addition reaction, where the structure of the original hydrocarbon is modified by the introduction of functional groups, enhancing its chemical properties.

Halogenation Reactions Concept 1

In this reaction, we're going to talk about the halogenation reactions. Here, under this type of reaction, 2 halogens coming from either Br₂ or Cl₂ are added to 1 pi bond. So here we have our starting alkene, and we're reacting it with X₂. In the process, each double-bonded carbon gains 1 halogen. At the end of this, we have a structure that has 2 halogens on it, which we call a dihalide. Now, if we have an alkyne that means we have 2 pi bonds. So remember, we need 1 mole of reagent for every pi bond. And since there's 2 pi bonds here, we'd need 2 moles of X₂. The first mole would add, to give us 2 halogens, and then the second mole would add again, to give us another 2 halogens. At the end, you have a structure that possesses 4 halogens on it. So we'd call this a tetrahalide. So remember, halogenation is just adding 2 halogens per pi bond within your structure. This could give us a dihalide if you're dealing with an alkene, or a tetrahalide if you're dealing with an alkyne.

Halogenation Reactions Example 1

Complete the following halogenation reaction. So here we have our alkene, and we're reacting it with 1 mole of chlorine. Remember, all that's going to happen here is we're going to sacrifice our pi bond in order to add 2 halogens to my structure. Each double bonded carbon gets a chlorine. Here it doesn't matter the orientation that you place the chlorine if it's up or down or whatever, all that matters is the connection. It's these 2 double bonded carbons that each need to have their chlorines. Here I decide to show them as being on opposite sides of each other, but I could easily show them being on the same side. We don't go into spatial orientation when it comes to a reaction like this. That's left for when we do organic 1 and organic 2. For right now, this would be our dihalide structure completed from the following halogenation reaction.

Write a halogenation reaction of the following alkyne with Br2 and name the product formed.

• A.
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Here’s what students ask on this topic:

What is a halogenation reaction in organic chemistry?

A halogenation reaction in organic chemistry involves the addition of halogen atoms (such as Br₂ or Cl₂) to a hydrocarbon. This reaction typically occurs with alkenes and alkynes. For alkenes, each carbon in the double bond gains one halogen, resulting in a dihalide. For alkynes, which have two pi bonds, two moles of halogen are required, leading to a tetrahalide with four halogens. This process is an example of an addition reaction, where the original hydrocarbon structure is modified by the introduction of halogen atoms, enhancing its chemical properties.

How does halogenation of alkenes differ from halogenation of alkynes?

Halogenation of alkenes and alkynes differs primarily in the number of halogen atoms added. In alkenes, each double-bonded carbon gains one halogen atom, resulting in a dihalide. The general reaction can be represented as:

$R\mathrm{CH}=\mathrm{CH}R+X_2\to R\mathrm{CH}{X}_2\mathrm{CH}R$

For alkynes, which contain two pi bonds, two moles of halogen are required. The first mole adds two halogens, and the second mole adds another two, resulting in a tetrahalide:

$RC\equiv CR+2X_2\to RC{X}_{2}C{X}_{2}R$

What are the products of halogenation of an alkene?

The products of halogenation of an alkene are dihalides. In this reaction, each carbon atom in the double bond of the alkene gains one halogen atom. For example, if we start with ethene (C₂H₄) and react it with bromine (Br₂), the product will be 1,2-dibromoethane (C₂H₄Br₂):

$\mathrm{CH}{2}_{}=\mathrm{CH}{2}_{}+{\mathrm{Br}}_2\to \mathrm{CH}{2}_{}\mathrm{Br}\mathrm{CH}{2}_{}\mathrm{Br}$

This reaction is an example of an addition reaction, where the double bond is broken, and two new single bonds are formed with the halogen atoms.

Why are halogenation reactions important in organic chemistry?

Halogenation reactions are important in organic chemistry because they allow for the functionalization of hydrocarbons, making them more reactive and versatile for further chemical transformations. By adding halogens to alkenes and alkynes, chemists can create dihalides and tetrahalides, which serve as key intermediates in the synthesis of various organic compounds. These reactions are also used in the production of pharmaceuticals, agrochemicals, and polymers. Additionally, halogenation can help in the study of reaction mechanisms and the development of new synthetic methodologies.

What is the mechanism of halogenation of alkenes?

The mechanism of halogenation of alkenes involves a three-step process:

1. **Formation of the Halonium Ion:** The alkene reacts with a halogen molecule (X₂), forming a cyclic halonium ion intermediate. This occurs when the pi electrons of the alkene attack the halogen molecule, leading to the formation of a three-membered ring with a positive charge on the halogen.

2. **Nucleophilic Attack:** The halonium ion is then attacked by a halide ion (X^-), which opens the three-membered ring and forms a vicinal dihalide.

3. **Formation of the Dihalide:** The final product is a dihalide, where each carbon of the original double bond is bonded to a halogen atom.

This mechanism ensures the anti-addition of halogens, meaning the halogens add to opposite sides of the former double bond.