Cleavage of Phenyl Ethers - Online Tutor, Practice Problems & Exam Prep

Cleavage of Phenyl Ethers Concept 1

Hey, everyone. So in this video, let's talk about the acidic cleavage of phenyl ethers. Now recall when an ether is treated with HBr or HI, it will undergo acidic cleavage. Here, the difference now is we're dealing with a Phenyl ether. We're going to say the products of acidic cleavage of phenyl ether will always produce phenol as a product plus an alkyl halide.

Up to this point, we knew that excess acidic cleavage could change an ether into 2 alkyl halides at the end. Because of the special properties of phenol, once we get to phenol, it would not undergo further reaction to make an alkyl halide. It would stay in its phenyl state. If we take a look here, we have our phenyl ether. We're using HX, so that could be HBr or HI.

We know that this oxygen would grab an H⁺ from our acid becoming protonated. Because it's protonated now, it's susceptible to nucleophilic attack. This alkyl halide would come in and it would have to hit the carbon, the ether carbon tha

Cleavage of Phenyl Ethers Example 1

Hey, everyone. So here it says, provide a mechanism for the acidic cleavage of propoxybenzene. Alright. So benzene is our conjugated ring. Propoxy, so that's oxygen connected to a propyl group, so 3 carbons.

Here we're doing acidic cleavage, let's say we're using HBr as our strong acid of choice. Now, remember what's going to happen here is oxygen is going to grab an H⁺ from our acid, and what we're going to get here is our protonated oxygen. It's making 3 bonds, so it's positively charged. At this point, normally, we would look and see what kind of ether, oxygens, and carbons I'm dealing with to decide which one gets attacked by the incoming halogen.

But because this is a phenyl ether, we don't have to worry about that. The incoming nucleophile will always, always attack the ether carbon that is not part of the benzene ring. So, here, this bromine is going to come in, it's going to hit the ether carbon, this carbon because it's not part of the benzene ring. Because remember, you can either hit this carbon or this carbon. They're the ether carbons.

So we hit that carbon there which causes this bond to break and go to oxygen. This ensures that we're going to make phenol as a product, plus we're going to make our alkyl halide. Okay. So then here goes our alkyl halide. These will be our 2 final products for this particular acid or acidic cleavage of propoxybenzene.

Alright. So just remember, when it's phenyl ethers, it's pretty straightforward. Acidic cleavage will always make a phenol product plus an alkyl halide product.

Predict the product formed from the following acidic cleavage reaction. 

Alright, everyone. So here in this practice question, it says, predict the product formed from the following acidic cleavage reaction. So here we're dealing with a Phenyl ether and we're reacting it with excess strong acid in the form of hydroiodic acid. Now remember, when it comes to a Phenyl ether, our products will always be a Phenol plus an Alkyl halide. So here, this is our ether carbon that's part of Benzene. It has to be what's produced at the end in the form of phenol, so it's an OH group attached to it, plus, and then this ether carbon, it's going to gain the halogen. The halogen in this case would be the iodine. So here, the iodine would attach to it. And here's the thing. The iodine when it attaches, it could attach in one of two ways. Right? It could have come from the top, or it could come from the bottom. And because of that, there's a possibility of two answers. So maybe iodine attaches this way or iodine attaches this way where this bond is wedged and this becomes dashed. Both are possibilities. You have to show both. So you'd have a like a racemic mixture of these two alkyl halides involved because you're attaching a halogen to a chiral center. Right? It becomes chiral when it attaches. Right? So these will be our products. We have our phenol that has to be created plus our two alkyl halides. Alright. So these would be our final answers.

Determine which of the following choices represents the major product from the following reaction. 

• A.
• B.
• C.
• D.

Choose the correct chemical sequence necessary for the following synthesis. 

• A. 1. HBr (xs)/ heat2. KOH3. Benzyl bromide
• B. 1. HI (xs)/ heat2. KOH3. Bromobenzene
• C. 1. H₃O⁺2. Br₂ / FeBr₃3. NaOEt
• D. 1. HI (xs)/ heat2. Benzyl bromide3. KOH