Hydrohalogenation Reaction - Video Tutorials & Practice Problems
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concept
Hydrohalogenation Reaction Concept 1
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In this video, we're gonna take a look at hydro halogen reactions. So under this type of reaction, one hydrogen and one halogen, either bromine or chlorine are added to one pi bond. So here in our general type of reaction, we have an keen 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 alkin 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 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 haid product. So when it comes to hydro halogen nation, we're trying to create alkyl haid products as our final answer.
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example
Hydrohalogenation Reaction Example 1
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Complete the following hydro halogen reaction. So here we have cyclops reacting with hydro Bromma acid. Here, we're going to say we're gonna 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 Broin would go here in the process. We've just created an alky haid. Here, we've created BROMO cyclo pentane as our final product.
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concept
Markovnikov's Rule Concept 2
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Now, the addition of hydrogen and a halogen to a non symmetrical alkene or alky follows Markov's rule. Under Markov's rule, the H atom is added to the carbon, whether it be double bonded or triple bonded with more hydrogens. And then the eight 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 Markov, Niko's rule, H would go to the double bonded carbon with more hydrogen. So go here and then the halogen would go to the one with fewer hydrogens. So it would go here in this process. We made an Al Kal Haid. Now, when we have an al kind, we have two py boss. So we'd need two moles of HX. Still following Mako Niko'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 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 die haid with both halogens adding to the same formally triple bonded carbon. So just remember, we use Markov'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.
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example
Markovnikov's Rule Example 2
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Complete the following hydro halogen 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 two double bodied carbons have a difference, different number of hydrogens. We're gonna use Markov Niko's rule to determine where the H and the CL will go. So following Marko Niko's rule, it says that the hydrogen will go to the double bonded carbon with more hydrogens. So we'd go here, of course, the that was originally there is still there and then the chlorine would go to the double bonded carbon with fewer hydrogens. So we go here. This will represent the alko haid product formed from the hydro halogen nation 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.
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Problem
Problem
Write a hydrohalogenation reaction with excess HCl and name the organic product formed.
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