Hydrohalogenation Reaction - Online Tutor, Practice Problems & Exam Prep

Hydrohalogenation reactions involve the addition of hydrogen and a halogen (bromine or chlorine) to a pi bond in alkenes or alkynes, resulting in alkyl halide products. For non-symmetrical alkenes, Markovnikov's rule dictates that the hydrogen atom attaches to the carbon with more hydrogens, while the halogen attaches to the carbon with fewer hydrogens. In alkynes, two moles of HX are required, leading to the formation of dihalides. Understanding these principles is crucial for predicting product formation in organic reactions.

Hydrohalogenation Reaction Concept 1

In this video, we're going to take a look at hydrohalogenation reactions. So under this type of reaction, one hydrogen and one halogen, either bromine or chlorine, are added to one pi bond. Here in our general type of reaction, we have an alkene to start with, and we're reacting it with HX. Here, one hydrogen would go to one of these double-bonded carbons and the halogen to the other. Both of them are the same; it's a symmetrical molecule, both alkene carbons have the same number of hydrogens. So here you could either add the H here or here, and then you'd add the halogen to the other side. Here I choose to add them in these positions, and at the end of it, what do we just make? Well, we transitioned from an alkene reactant to an alkyl halide product. So when it comes to hydrohalogenation, we're trying to create alkyl halide products as our final answer.

Hydrohalogenation Reaction Example 1

Complete the following hydrohalogenation reaction. So here we have cyclopentene reacting with hydrobromic acid. Here we're going to say, we're going to break the double bond, and both of these double bonded carbons are symmetrical, so the H could go to either one, and the Br could go to the other. Here I decide to add the hydrogen here, and then the bromine will go here. In the process, we've just created an alkyl halide. Here we've created bromocyclopentane as our final product.

Markovnikov's Rule Concept 2

Now the addition of hydrogen and a halogen to a non-symmetrical alkene or alkyne follows Markovnikov's rule. Under Markovnikov's rule, the hydrogen atom is added to the carbon, whether it be double bonded or triple bonded, with more hydrogens. And then the addition, and then the halogen atom or X atom is added to the triple bonded or double bonded carbon with fewer hydrogens. If we take a look here, we have an alkene to start, the alkene on the left only has 1 hydrogen, and the one on the right has 2. Following Markovnikov's rule, hydrogen would go to the double bonded carbon with more hydrogens, so it'd go here, and then the halogen would go to the one with fewer hydrogens, so it'd go here. In this process, we made an Alkyl Halide. Now, when we have an Alkyne, we have 2 π bonds, so we'd need 2 moles of HX. Still following Markovnikov's Rule, hydrogen would go to the now triple bonded carbon with more hydrogens, and there'd be 2 of them adding, and then we'd have 2 halogens adding to the triple bonded carbon with fewer hydrogens. So this one here. As a result of using 2 moles of HX, we wind up with a dihalide with both halogens adding to the same formally triple bonded carbon. So just remember, we use Markovnikov's rule if our double or triple bonded carbons have a different number of hydrogens each. Use this rule to determine what your final answer will be.

Markovnikov's Rule Example 2

Complete the following hydrohalogenation reaction. Here we have our alkene reacting with HCl. What we do first is we look to see what kind of double bonded carbons we have. The one on the left is making 4 bonds, so it has no hydrogens. The one on the right is making 3 bonds that we see, so it has 1 hydrogen. Since the two double-bonded carbons have different numbers of hydrogens, we are going to use Markovnikov's rule to determine where the H and the Cl will go. Following Markovnikov's rule, it states that the hydrogen will go to the double-bonded carbon with more hydrogens, so it would go here. Of course, the hydrogen that was originally there is still there, and then the chlorine would go to the double bonded carbon with fewer hydrogens. So it would go here.

This will represent the alkyl halide product formed from the hydrohalogenation between our reactants. Now remember, your hydrogens when they are connected to carbon are invisible, so you could erase them, and this would be our full skeletal formula for our product.

Write a hydrohalogenation reaction with excess HCl and name the organic product formed. 

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Here’s what students ask on this topic:

What is hydrohalogenation in organic chemistry?

Hydrohalogenation is a type of addition reaction in organic chemistry where a hydrogen atom (H) and a halogen atom (X, such as bromine or chlorine) are added to a carbon-carbon double bond (alkene) or triple bond (alkyne). The general reaction involves an alkene or alkyne reacting with a hydrogen halide (HX) to form an alkyl halide. For example, when ethene (C₂H₄) reacts with hydrogen bromide (HBr), the product is bromoethane (C₂H₅Br). This reaction is important for synthesizing various organic compounds.

How does Markovnikov's rule apply to hydrohalogenation reactions?

Markovnikov's rule states that in the addition of HX to an unsymmetrical alkene, the hydrogen atom (H) will attach to the carbon with more hydrogen atoms, while the halogen (X) will attach to the carbon with fewer hydrogen atoms. This rule helps predict the major product of the reaction. For example, in the hydrohalogenation of propene (CH₃-CH=CH₂) with HBr, the hydrogen will add to the carbon with two hydrogen atoms (CH₃-CH₂-CH₂Br), and the bromine will add to the carbon with fewer hydrogen atoms, resulting in 2-bromopropane as the major product.

What is the product of hydrohalogenation of an alkyne?

In the hydrohalogenation of an alkyne, two moles of HX are required to fully saturate the triple bond. The reaction follows Markovnikov's rule, where the hydrogen atoms add to the carbon with more hydrogen atoms, and the halogen atoms add to the carbon with fewer hydrogen atoms. For example, when acetylene (C₂H₂) reacts with two moles of HBr, the product is 1,1-dibromoethane (CH₂Br-CH₂Br). This results in a dihalide with both halogens attached to the same carbon atom.

What are the key differences between hydrohalogenation of alkenes and alkynes?

The key differences between hydrohalogenation of alkenes and alkynes are the number of moles of HX required and the resulting products. For alkenes, only one mole of HX is needed, and the product is an alkyl halide. For alkynes, two moles of HX are required, leading to the formation of a dihalide. Additionally, both reactions follow Markovnikov's rule, but in alkynes, the addition of two moles of HX results in both halogen atoms attaching to the same carbon atom, whereas in alkenes, only one halogen atom is added.

Why is Markovnikov's rule important in predicting the outcome of hydrohalogenation reactions?

Markovnikov's rule is important because it helps predict the major product of hydrohalogenation reactions involving unsymmetrical alkenes or alkynes. By determining which carbon atom the hydrogen and halogen will attach to, chemists can accurately predict the structure of the final product. This rule is based on the stability of the carbocation intermediate formed during the reaction, where the more stable carbocation leads to the major product. Understanding this rule is crucial for designing and synthesizing specific organic compounds in a controlled manner.