What is the major product of a Friedel–Crafts alkylation using...

Question

What is the major product of a Friedel–Crafts alkylation using the following alkyl chlorides?

e. (CH₃)₂CHCH₂Cl

f. CH₂=CHCH₂Cl

Accepted Answer

Hey, everyone and welcome back through the products formed. If benzene undergoes the Friedel Kraft's calculation with the alkyl chlorides given below. The first one is one Chloro three methyl beat two in and the second one is three Chloro 22 dimethyl beat. Let's go ahead and write down the first reaction that we're analyzing benzene is the starting material. So we're drawing benzene and now we are performing the free little crafts calculation. That means we need to use the loose acid catalyst, which is aluminum chloride. And then we are simply going to read for the starting material or the alky chloride because one of our starting materials is benzene well done. So we understand what kind of reaction we are dealing with. But our goal is to actually understand what sort of electro file we will have in that reaction because the alkyl chloride has a longer carbon chain than usually we have. That's really important for us to understand the exact electro file that will participate in the reaction for that purpose. We are going to think about the generation of an electro file. And what we're going to do here is just draw a short mechanism for the formation of an electro file. We know that chlorine has three lone pairs and one of its lone pairs will be placed into the orbital of aluminum to form an add duct. So we can simply state that we are forming an add duct. And to draw it, we can reach for the structure of our alky chloride and bond aluminum chloride set. So that chlorine will have a formal positive charge and aluminum will have a formal negative charge. Now, we have a good living group because chlorine has a formal positive charge. So we can essentially state that we're going to lose the whole fragment, which will be aluminum with four chlorine atoms, bonnets it and we are going to form a carbo ion. So in this case, we can state that we produce aluminum C L four with a negative charge. And what is more important is that we are forming a carbo ion. So we have 1234 carbon atoms in our parent chain. So that'd be 1234, we're going to add the metal group and the double bond because the carbo ion is primary with, we need to think about the resonance and that's why we are performing this mechanistic analysis. We want to find the most stable carbo for us, which will correspond to the actual electro file, we have resin stabilization. So we're going to shift the pipe and we are going to think about the second resonance form that we get upon shifting the pi bond, we noticed that we are forming a more stable tertiary carbo which will actually participate in the reaction because it's more stable. And from here, we are ready to predict the actual product. OK. So now we know what our electro file is and we can just show that the substitution will take place if we consider one of the pi bonds within the ring, which will sag the electro file. So in this case, we can simply read for the benzene ring break our pie bond bond, the result in fragment. So now if we introduce our carbon carbon bond, we notice that the carbon atom from the benzene ring is going to be bonded to the original Carbo Catron, which has two metal groups. And then we can continue our chain. At this point, we have a positive charge within the benzene ring and we can use A L C L four with a negative charge to perform the deep protonation stop. And actually to make it really clear for us, we have to expand our aluminum chlorine bond. So we can say A L C L three with an additional chlorine atom on it. It and we can just show how we are forming H C L, right. If we are essentially transferring our chlorine aluminum electrons for that purpose, we're going to expand the implicit hydrogen which we are removing and essentially show how we are removing it by forming a pi bond and restoring aromatic. Therefore, we are ready to draw the final product where the purse part, well, then we got our first product for the Friedel calculation and we are ready to think about the second one. For the second one, we are given the following reaction where the starting material once again is benzene. We are dealing with the craft calculation. So let's go ahead and draw our alor and at our catalyst, we already know that we have to think about the stuff where we are generating the electro file. Let's remember that in short, we are just losing chlorine and we are forming the carbo ion. So let's go ahead and control the result on carbo ion. So removing chlorine will essentially give us 1234 membered parent chain. And then if we move up the chain, that'd be 123 carbon. Number three, we will add two metal groups there. So now we have our positive charge and we are forming a secondary carbo ion, but apparently it can be stabilized due to the one comma two metal shift. And that's how we can form a more stable terri. So let's draw our four member chain from left to right on carbon number two, we have a methyl group. Carbon number three, we have another methyl group and that we are forming our positive charge. Now that we know what the electro file is exactly, we are ready to predict the product. So now the from the benzene ring, it attacks the electro file, we are forming our intermediate that would be benzene. We're going to break our pie bond and simply bond the resultant side chain, we will have 1 to 3 carbon atoms, 123. Now, at the wordpress position that has responded to the benzene rank, we are also going to have two metal groups. And then for carbon number two, we have an additional methyl group. Since once again, we have a positive charge on carbon, we simply want to remove the proton for that purpose. We're using our resultant A L four and we can identify the proton to be removed and formation of the pi bond to restore aromatic. Let's go ahead and for the final product. So now we are removing hydrogen, we don't have any positive charge and we have our benzene rank reformed and we got our final product. So for the two problems, we have formed our craft calculation products, we had to think about the formation of the electro file and the formation of a more stable carbo for this problem before we could predict the two products given in two green boxes. Thank you for watching and I will see you in the next video.

What is the major product of a Friedel–Crafts alkylation using the following alkyl chlorides?
e. (CH₃)₂CHCH₂Cl
f. CH₂=CHCH₂Cl

Key Concepts

Friedel-Crafts Alkylation

Carbocation Stability

Regioselectivity

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