Halogenation

One of the most popular addition reactions, this is the primary method for making vicinal dihalides. 

Concept: General properties of halogenation. 

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Video Transcript

Now we're going to talk about one of the most common addition reactions in this entire section and that's called halogenation. So let's just get right into it. Halogenation is the process of taking a double bond and adding a diatomic halogen and at the end what we're going to get is anti-dihalides, anti vicinal dihalides. So let's go ahead and just talk about the general features of this reaction.
First of all, the mechanism is going to have an intermediate and that intermediate is going to be similar to other ones we've learned. It's going to be a bridged ion. So maybe you can already start to visualize what you think that might be. Sorry about that.
The stereochemistry is going to be anti. As I just showed you guys, you're going to get anti-dihalides, so it would be anti. And then my product is going to be, like I said, vicinal dihalides. Now just to remind you guys, the word vicinal is a word to mean that there's two things next to each other, so this relationship here would be vicinal because they are right next to each other. Vicinal is also the same as saying one, two. Basically, you have something on the one position and something at the two position and that would be vicinal.
So will there be rearrangements in this mechanism? No. There won't be because there's no carbocation. And then finally, since I'm adding two of the same thing, I'm not going to worry about Markovnikov because I'm adding two of the same thing, so it doesn't matter. So let's just scratch that out.
As you guys can see, the reaction looks pretty simple. We've got a double bond, so we know this is addition. We've got an electrophile that we're adding. Now notice that I have CCl4 down here. If you guys remember, this is actually going to be an apolar solvent. Apolar solvents are inert. That's not going to do anything, so don't even worry about it. That's just something that it happens to help the reaction, but it's not going to do anything.
Another common solvent that you might see is CH2Cl2. That just means – it's the same thing. It just means instead of having four chlorines, you have two chlorines and two H's. Regardless, these are just solvents that don't do anything.
At the end what we get is those vicinal dihalides. Let's go ahead and look at the reaction. 

Opening of 3-membered intermediates/molecules always results in anti-addition.

General Reaction:

Concept: A worked-example of the halogenation mechanism.     

4m
Video Transcript

So the mechanism works like this, basically you've got a double bond and this double bond is nucleophilic it's looking for something to give its electrons to and I know that halogens seem to have a lot of electrons but they're also very what's called polarizable so what means is that electrons wind up moving a lot and wind up causing imbalances in the electron cloud so at some point there's going to be a part of the X or a part of the halogen that's not to have too many electrons and the double bond is going to pounce on that, so what's going to happen is that the double bond is going to grab one of the X, now the X doesn't like to have that many bonds so it's going to go ahead and break a bond if you make a bond you have to break a bond but similar to other bridged ions that we've looked at before if we make that bond then we have to make another bond back to the double bond what's going to show is it's going to make a bridge instead of just a single bond to one side, OK? So those are our three arrows, we're going to wind up getting is A bridged ion in this case this is called the halonium ion because instead of you know and I basically have instead of my carbocation I have a 3-membered ring and one of those atoms is a halogen, OK? For example if this was a bromine I would call it a Bromonium Ion, alright? Now this positive charge is going to be distributed throughout all three of those atoms so I'm going to have a partial positive here and I'm going to have a partial positive here, partial positive there, OK? But one of these atoms is going to have the most positive density can you guys guess which one? It's going to be the atom that is the best at stabilizing positive charges and that one's going to be the most substituted side so if they both had the same substitution then this X negative could really attack anywhere but since in this case one of them is tertiary and one of them is secondary what that means is that I'm going to go for the more substituted one so my X negative is going to basically do what we call a backside attack an SN2 reaction if you haven't learned that yet that's OK but just you know the backside attack and then if we make that bond we're going to break a bond so then we break the bond to the X, just like any reaction that involves a three membered ring any time that we break it open from the base we're going to have to make anti-products because there's a lot of potential energy in that three membered ring and once you break it going to snap open and they're going to face opposite directions so what that means is that if my X negative attack from the top my X on the ring is going to go towards the bottom so it's going to cause ante products so over here what I would get for my final product is I would get Let's say the X had attacked from the top then that means that this methyl group would face towards the bottom and that means that this methyl I mean that means that this X would face towards the bottom so what I'm going to get is an anti-product, OK? Now I noticed that do have any chiral centers here, I actually do I have 2 chiral centers so what that means is that I'm also going to get the enantiomer and the enantiomer would just be if it had attacked from the bottom then I would get this X at the bottom of this methyl group at the top and then this X at the top...Oppps oh my God this program can be a little dumb sometimes sorry about that guys really, I'm struggling here there we go, cool? So those are two enantiomers these would be produced in even amounts so that would be racemic, makes sense? Cool so this mechanism really wasn't bad as the ones that we've learned for other addition reactions it's pretty simple, OK? I hope that makes sense let's go ahead move on to the next topic.

1. Electrophilic Addition

2. Nucleophilic Substitution (SN2)

Halogenation Additional Practice Problems

Br2 will also react with alkenes to do electrophilic addition reactions. This is a non-radical reaction that has an interesting stereochemical outcome. Treatment of the optically-active methylcyclohexane shown below with bromine gives the product where Br2 has been added across the double bond.
a. Draw all possible stereoisomers that might be obtained. Use the “flat-ring” wedge/dash convention. Be sure that the methyl group is on the topmost carbon in all your structures. Each structure will be in a separate box with a roman numeral designator.
b. For EACH chiral carbon, label its stereochemistry as R or S.
c. Which structures are diasteromers? Use the roman numeral designators for this answer. If there are no diastereomers, write NONE.
d. Which structures are meso? Use the roman numeral designators for this answer. If there are no meso structures, write NONE.

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Draw the product(s) from the following reaction.

a) clearly label each drawing with the correct stereochemistry-(use R & S for chiral centers, if any).

b) clearly label pairs of enantiomers & diastereomers, if any.

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Complete the following reaction by drawing the structure of the principal major product. Indicate relative stereochemistry where necessary. If there is no reaction, write NR.

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Predict the following products of the following reaction (there may be more than one correct answer): 

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Consider the reaction given below.

(a and b) Give the absolute configuration (R or S) for each of the indicated chiral centers.

(c) Using arrows to show the flow of electrons, write a stepwise mechanism for this reaction. Be sure to show the stereochemistry that exists in each elementary step. Also be sure to show clearly how the chiral centers in the final product are formed.

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Propose a mechanism for the following reaction

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Consider the strucutres below and answer the following questions. 

c. Which compounds each form an achiral product by reaction with chlorine? 

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Provide the major product for each of the following alkene reactions. 

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Predict the product(s) or reagents of the following reactions.

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Provide the  products for the following reaction.

a) clearly label each drawing with the correct  regio- and stereochemistry.

b) clearly label pairs of enantiomers & diastereomers, if any.

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Draw the product(s) from the following reaction.

a) Clearly label each drawing with the correct stereochemistry-(use R & S for chiral centers, if any).

b) Clearly label pairs of enantiomers & diastereomers, if any.

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How many atoms and electrons are directly involved in the bond-making and bond-breaking in the first step of the reaction of bromine with an alkene?

 

(A) three atoms, four electrons

(B) three atoms, six electrons

(C) four atoms, four electrons

(D) four atoms, six electrons

(E) five atoms, four electrons 

(F) five atoms, six electrons

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Predict the major product for the following one-step reaction:

 

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Provide the structural formula for the reactant in the following reaction. 

 

 

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The electrophilic addition of Cl 2 to (E)-2-butene gives a product that is different from the electrophilic addition of Cl2 to (Z)-2-butene.

(a) Using arrows to show the flow of electrons, write a stepwise mechanism for each of these two reactions. Be sure to show clearly the stereochemistry associated with each step in each of your two mechanisms and how it connects the stereochemistry of the starting alkene with the stereochemistry of the product. Be sure to label as R or S each chirality center in each of your products. 

 

(b) Is the product of either of these two reactions optically active? ( yes or no; choose one) Provide a detailed, but concise, explanation for your choice. 

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The reaction of Br2 to cyclohexene would produce the compound(s) represented by structure(s): Circle all that apply

 

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Which of the statements is most correct regarding the products expected from the halogenation reaction shown below?

a) Equal amounts of I and IV are produced.

b) Equal amounts of I and III are produced.

c) Equal amounts of III and IV are produced.

d) Equal amounts of I and II are produced.

e) Equal amounts of II and III are produced.

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