Hydrohalogenation Reactions - Online Tutor, Practice Problems & Exam Prep

Hydrohalogenation is a chemical reaction where hydrogen (H) and a halogen (X), typically bromine or chlorine, are added across a pi bond of an alkene or alkyne to form an alkyl halide product. When dealing with symmetrical alkenes, the addition of hydrogen and halogen can occur on either carbon with no preference. However, with unsymmetrical alkenes or alkynes, Markovnikov's rule applies: the hydrogen atom attaches to the carbon with more hydrogens, and the halogen attaches to the carbon with fewer hydrogens. For alkynes, two moles of HX are required, resulting in a dihalide with halogens adding to the same carbon. This rule is essential for predicting the outcome of hydrohalogenation and understanding the formation of alkyl halides in organic chemistry.

Hydrohalogenation Reaction

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 O. Here in our general type of reaction we have an alkene to start with and we're reacting it with Ajax 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, you can either add the H here or here and then you'd have 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 halid product. So when it comes to hydrohalogenation, we're trying to create alkyl Halid products as our final answer.

Hydrohalogenation Reactions Example

Complete the following hydrohalogenation reaction. So here we have a 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 would go here. In the process we've just created an alkyl halid.

Here we've created Bromo Cyclo Pentane as our final product.

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Now the addition of hydrogen and a halogen to a non-symmetrical alkene or alkyne follows Markovnikov's rule. Under Markovnikov's rule, the H atom is added to the carbon, whether it be double-bonded or triple-bonded, with more hydrogens, 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 one hydrogen, and the one on the right has two. Following Markovnikov's rule, H would go to the double-bonded carbon with more hydrogens, so it would go here, and then the halogen will go to the one with fewer hydrogens, so it would go here. In this process, we made an alkyl halide.

Now, when we have an alkyne, we have two π bonds, so we need 2 moles of HX. Still following Markovnikov's rule, H would go to the now triple-bonded carbon with more hydrogens, and there'd be two of them adding, and then we'd have two halogens adding to the triple-bonded carbon with fewer hydrogens. So this one here, as a result of using two moles of HX, we wind up with a dihalide with both halogens adding to the same formerly 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.

Hydrohalogenation Reactions Example

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. This one on the left is making four bonds, so it has no hydrogens. This one is making three bonds that we see, so it has one.

Since the 2 double bodied carbons have a difference different number of hydrogens, we're going to use Markovnikov's rule to determine where the H and the CL will go. So following Markovnikovs rule it says that the hydrogen will go to the double bonded carbon with more hydrogens, so would go here. Of course the H that was originally there is still there. And then the chlorine would go to the double bonded carbon with fewer hydrogens, so would go here.

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

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:

Hydrohalogenation mechanisms proceed through a carbocation intermediate, which is _________. Therefore, what are generally the products of these reactions?

Hydrohalogenation mechanisms proceed through a carbocation intermediate, which is a positively charged, electron-deficient carbon atom that occurs when an alkene reacts with a hydrogen halide (HX, where X is a halogen). In this process, the double bond of the alkene breaks, and the electrons from the double bond are used to form a bond with a hydrogen atom, leaving a carbocation behind. The positively charged carbocation then quickly reacts with the halide ion (X^-) that was generated when the hydrogen halide dissociated.

The products of these reactions are typically haloalkanes (alkyl halides), where the halogen (X) is added to the carbon atom that had the greatest number of hydrogens (and thus the least amount of alkyl substituents) to begin with. This follows Markovnikov's rule, which states that in the addition of HX to an unsymmetrical alkene, the hydrogen atom bonds to the carbon with the greater number of hydrogen atoms, and the halogen bonds to the carbon with fewer hydrogen atoms. The formation of the most stable carbocation intermediate during the reaction often leads to regioselectivity, favoring the more substituted alkyl halide as the major product.

Which of the following statements characterize hydrohalogenation reactions in general?

Hydrohalogenation reactions are a type of chemical reaction where a hydrogen halide (HX, where X is a halogen such as chlorine, bromine, or iodine) is added to an unsaturated hydrocarbon, such as an alkene or alkyne. These reactions typically follow Markovnikov's rule, which states that in the addition of HX to an unsaturated hydrocarbon, the hydrogen (H) will attach to the carbon with the greater number of hydrogen atoms, and the halogen (X) will attach to the carbon with fewer hydrogen atoms.

Additionally, hydrohalogenation reactions are generally exothermic due to the formation of new sigma bonds, which are stronger than the pi bonds present in the double or triple bonds of the starting materials. The reaction mechanism often involves the formation of a carbocation intermediate, which can lead to rearrangements if more stable carbocations can form. These reactions do not typically occur with significant stereochemistry control, meaning that the addition of the halogen can lead to a mixture of stereoisomers when chiral centers are involved.