Halogenation Reactions - Video Tutorials & Practice Problems
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concept
Halogenation Reactions
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In this reaction, we're gonna talk about the halogenation reactions. Here, under this type of reaction, 2 halogens coming from either b r 2 or c l 2 are added to 1 pi bond. So here we have our starting alkene, and we're reacting it with X 2. 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 X 2. So 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 tetrahalad. So remember, halogenation is just adding 2 halogens per pi bond within your structure. This could give us a dihalid if you're dealing with an alkene, or a tetrahalid if you're dealing with an alky.
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example
Halogenation Reactions Example
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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.
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Problem
Problem
Write a halogenation reaction of the following alkyne with Br2 and name the product formed.
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