The presence of radicals in some familiar looking addition reactions can completely change the product.
Concept: The general mechanism of Allylic Halogenation.7m
So far we've been learning about radical halogenation mechanisms and how the radical is always going to be attracted to the spot that has the most R groups, so the tertiary is going to be more stable than primary. But what we also know is that according to the radical stability trend, allylic radicals are the most stable. So now I want to throw that extra complexity into the mix. What happens if there are allylic positions available to be halogenated. That brings us to allylic halogenation.
Once again, I just want to show you guys, just in case you forgot, that the allylic and benzylic radicals are more stable than primary, secondary or tertiary. What that means is if I have one of these two possibilities, this is going to be my preferred spot, not those three.
Now, what's interesting about these is that these radicals can resonate. What that means is that resonance is going to have to play a central role in this mechanism. We can't just ignore resonance. We're actually going to have to embrace it for this mechanism.
I just want to show you guys what this could look like. Basically, I've got a diatomic halogen. Let's say this is Br2, so it's selective. And we've got a double bond. Well, usually we would just say secondary or primary, we're going to go for the secondary spot. But in this case, I actually have an allylic position. That allylic position is right here. That's the one next to a double bond. That's actually going to be the position that's the most stable and that's going to be the one that I want to react with.
Now let's go ahead and draw the whole mechanism out. I'm going to show you guys how things are a little bit different now.
First of all, my initiation step is really easy. We're just going to use Br2. And where do those first arrows go? We would just make radicals, there and there. So now what I'm going to get is Br radical plus Br radical. That's it for our initiation. We have our target radical.
Now what we have is the propagation step. In the propagation step, I'm going to want to pull off the hydrogen that's going to give me the most stable radical, so obviously, it should be this one. So let's go ahead and react that. I would use three arrows. I would do one into the middle of nowhere. One into the middle of nowhere and then one right there. What I wind up getting is I wind up getting a radical that looks like this. And also we would get HBr.
Now usually what would happen is that my propagation phase would continue and I would have to regenerate the Br radical by reacting with Br2. That's normal. That's what we're expecting. But, wait. There's a little difference. This radical can now resonate. What that means is that I can't actually just continue from here. I need to draw the resonance structures of this radical before I can continue. What that means is that the resonance structure actually looks like this.
What that means is that now, my Br, that's going to continue in the propagation stage, can attach both there and there. So it's going to be able to attack both of those carbons because the radical is hybridized between those two. Remember that resonance structures, it's not in equilibrium, it's a mathematical statistical average of where that radical is.
What that means is that I can go ahead and draw my mechanism with the Br here. And I can draw it from the first one just to keep things easy I'll just to this there, there, and there. What I wind up getting is I'm going to wind up getting a Br right here plus Br radical. But I'm also going to get something else. I'm going to get a mixture of products because the radical on the end could have also attacked. What that means is that I'm also going to get a product that looks like this.
This is one of the complications of allylic halogenation if your double bond isn't perfectly symmetrical, you're going to wind up getting a combination of products. And really we don't deal with major and minor in this section. We're just going to say we're going to get both of them. It kind of sucks.
Then finally, what would the termination step look like? Well, there are a lot of terminations. We could get a Br and a Br. We could get just any of these radicals, I mean any of these. That plus my Br. Sorry, I'm just going to draw this and this. That would give us our product that looked like this. We would also get the other one. The other radical that would like this. The radical here is Br and that would give us the other looking product.
Now I know that there was another one that was the two R groups. That is almost never seen in here. But if you wanted to include it, you could just add R radical, R. And just say that that's going to give you R, R. But honestly, for these reactions, usually professors just neglect this because it's assumed that you're not really going to get much of this at all.
What is important is that now you can detect when you're going to get just one product or when you're going to get a mixture. When you're going to get a mixture is when you're doing an allylic halogenation. If it was just tertiary or secondary, you would only get one, but since it's allylic that means it can resonate.
In this case, you would get two different structures. The way to know how many structures you're going to get is just to draw the resonance structures yourself.
Now what I want to do is go more into depth on allylic halogenation with the specific reagents that we're going to use.
Remember our friendly addition reaction halogenation? Notice that you achieve a vicinal dihalide in this reaction.
Now we see this reaction. Note that the only difference is the presence of a radical initiator.
This one added condition will lead to the formation of a mixture of allylic alkyl halides. Here’s the full mechanism:
Concept: The products of Allylic Chlorination.1m
So now that we know the mechanism predicting these products is going to be easy the number one thing I'm looking for here that I want you guys to know is these reagents, so it happens to be that when you're doing a little chlorination 400 degrees Celsius is just the most common heat that's used so you should recognize that when you see CO2 400 hundred degrees and a double bond and you have Allylic position available we're going to wind up getting our you know we're going to wind up getting an allylic halogenation with multiple products because I could get the radical here or I could also resonate the radical to here, OK? So I'm going to one of getting both products one looks like this and one looks like this, OK? Easy stuff but you guys should be aware of that.
Concept: Mechanism of Allylic Bromination.1m
For Allylic Bromination usually you want low trace amounts of bromine we don't want too much bromine or we might wind up getting like an actual halogenation on the double bond which would be bad so we use NBS and we usually use NBS with a combination of light or heat doesn't matter but in this case, it would be the same exact thing, I would wind up getting a bromine here and I would get...Opps how do I change all this? There we go and I would get a bromine right here, OK? So really these are the same exact reaction just the reagent are a little tiny bit different that's it. Cool so what I want you guys to do is take this problem try to solve it yourself be sure to think about the whole mechanism before you predict your products I don't need the full mechanism as long as you can give me all the products that's what I'm looking for so go for it and then I'll solve it.
Example: Predict the product(s) of the following reaction.5m
Predict the product(s) of light-initiated reaction with NBS in CCI4 for the following starting materials.
When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunlamp is shone on the mixture, three products result.
(a) Give the structures of these three products.
(b) Propose a mechanism that accounts for the formation of these three products.
Draw the major product of each of the following reactions, disregarding stereoisomers:
Draw the product(s) of each of the following reactions, disregarding stereoisomers:
a. Draw the major product(s) of the reaction of 1-methylcyclohexene with the following reagents, disregarding stereoisomers:
1. NBS/Δ/peroxide 2. Br2/CH2Cl2 3. HBr 4. HBr/peroxide
b. For each reaction, show which stereoisomers are obtained.
How many allylic substituted bromoalkenes are formed from the reaction in Problem 18 if stereoisomers are included?
How many allylic substituted bromoalkenes are formed from the reaction of 2-pentene with NBS? Disregard stereoisomers.
Two products are formed when methylenecyclohexane reacts with NBS. Show how each is formed.
a. How many stereoisomers are formed from the reaction of cyclohexene with NBS?
b. How many stereoisomers are formed from the reaction of 3-methylcyclohexene with NBS?
The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products:
(a) Give a mechanism for this reaction, showing how the two products arise as a consequence of the resonance-stabilized intermediate.
(b) The bromination of cyclohexene using NBS gives only one major product. Explain why there is no second product from an allylic shift.
(a) Propose a mechanism for the following reaction:
(b) Use the bond-dissociation enthalpies given in Table 4-2 to calculate the value of for each step shown in your mechanism. (The BDE for is about 280 kJ/mol, or 67 kcal/mol.) Calculate the overall value of for the reaction. Are these values consistent with a rapid free-radical chain reaction?
Write the products of the reactions of the compound shown in the center with the reagents shown on the arrows. Make sure to include stereochemistry, but do not write mechanisms. DON’T LET THE BICYCLIC STRUCTURE INTIMIDATE YOU: it is just an alcohol, and will react like any other alcohol.
Fill in the box with the product(s) that are missing from the chemical reaction equation. Draw only the predominant regioisomer product or products (i.e. Markovnikov or non-Markovnikov products) and please remember that you must draw the structures of all the product stereoisomers using wedges and dashes to indicate stereochemistry. When a racemic mixture is formed, you must write "racemic" under both structures EVEN THOUGH YOU DREW BOTH STRUCTURES.
Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw all answers in skeletal form.
Provide the mechanism of the propagation steps to explain the stereochemistry of the products (why Br can be added from both sides). Be sure to include all arrows, formal charges, radicals and intermediates if needed.
The allylic bromination of the alkene below with NBS gives four different products. Draw the two initially formed free radical intermediates together with any applicable resonance structures and the four products.
Which one of the following is the major product of the reaction below?
Complete the following reaction with the correct structures of the products. Don’t forget to specify the stereochemistry.
Upon treatment of 1-methylcyclopentene with NBS and irradiation with UV light, exactly nine compounds (including stereoisomers) are formed. Draw any four of the possible products.
Write the structure(s) of the product(s) for the reaction below. Be sure to indicate any relevant stereochemistry.