To practice working through the early parts of a multistep syn...

Question

To practice working through the early parts of a multistep synthesis, devise syntheses of

(b) 3-ethylpentan-2-one from compounds containing no more than three carbon atoms.

Accepted Answer

Hey everyone and welcome back into this video, we are going to learn how to solve this problem which reads as an introduction to multi step synthesis, devise the synthesis of two metal B two in from propane C. O. First of all, let's start our reaction scheme. The starting material is propane to also we are going to draw a three carbon chain and then add an alcohol group at position number two. So that's propane C. O. And we have to keep in mind that this is an alcohol having classified it as an alcohol. We understand that we are trying to synthesize to method B two in. So we're going to draw a four member chain with a double bond position number two. So we're going to add a double one here and there's also a method group at that same position. So that's our two Method B two in. And we can classify it as an elk in. Now, when you look at this reaction, you may be thinking, okay if we are forming an al keen out of an alcohol, perhaps we can use an acid catalyzed dehydration and generally, yes, but you have to be extra careful because if you look at that, the two compounds indeed have two method groups, but there's an additional ethyl group over here, right. With a double bond as well. So we cannot really perform dehydration because dehydration doesn't add those carbon atoms. So how do we add those carbon atoms? Let's recall several steps that we can perform here, first of all. What if we perform an oxidation of that alcohol so that we form a key tone. Right? We will get our carbonell compound. Now, when we do dad, having obtained our carbon new compound, this is where we can add our alcohol chain. Right. We can perform the Green Yard reaction. And as we said before, it looks like we have two metal groups in each compound. We're just adding two more carbon atoms. So it makes sense to think about ethel magnesium bromide because we want to add to carbon atoms. And then within this granite reaction. Keep in mind that the second step is the protein ation of the Alcock said. And and so we're going to produce an alcohol. And when we produce a new alcohol, we may think about an asset catalyzed dehydration. That's where we may be successful in getting that alkaline. Right. So let's see how that works. Let's try to use these three steps in our solution. First of all, we are going to start with our alcohol and we want to oxidize that. Right? So let's use that structure. If we want to oxidize it, we can oxidize it into a key tone using an oxidizing agent, but it doesn't necessarily have to be strong. You may use BCC in the presence of period in. Now, essentially Pcc in Purdon oxidizes primary alcohols to al the highs and secondary alcohols to ketones so that's sufficient for us. We don't need to use a strong oxidizing agents such as a jones Regent. So it's pretty much sufficient for us here. We can use PCC to produce our kate stone. So because we're performing an oxidation, nothing really changes except from the fact that the alcohol now turns into a ketone group. Right? We have a carbon group and I was thinking about our next step, we want to make sure that we're adding that those two additional carbon atoms as we previously assad, we're going to use ethel magnesium bromide because we need two carbon atoms. Right? So there will be two steps in this reaction step. Number one is the addition of the Green Earth Regent, two carbon atoms bonded to magnesium bromide. Don't forget that it's important to use ether as our solvent in this reaction. And number two would be acidic. Work up. We are using an asset to protein it at elk oxide, intermediate formed to summarize essentially these two carbon atoms would be added at this position where the carbon he'll group is breaking the pi bon and we may draw our intermediate. We get our elk oxide with two metal groups and then an ethnic group. But during the acidic work up, we're going to protein eight hour Alcock side. So we're producing an alcohol. Now this is very close to what we desire to get. We want to get an elk team with a double bonded disposition. So let's identify beta hydrants because we want to perform an acid catalyzed dehydration. So during an acid catalyzed dehydration, we can remove beta hydrants to produce the most substituted elkin. And if this is our alfa opposition which is directly bonded to the alcohol group, there are two unique beta positions in green. The reason why there are two is because the two adjacent methyl groups are equivalent to each other, there's that free bond rotation. So instead of removing this beta hydrogen and producing a less substituted alkaline, I'm going to take this one because it's producing it would produce a more substituted alc in right now if we removed this beta hydrogen, we would essentially have to protein it a wage first using an asset and then water would essentially attack that hydrogen and these electrons would be transferred into the formation of the pi bond, which would neutralize the car broke Orion. Recall that when a wage gets protein ated, it acts as a leading group. That would be an E one mechanism elimination. We're forming our carbon can iron, then removing the beta hygiene and forming a double bond. So this indeed gives us our elkin. We can say that when this reaction is complete, we're forming our al king and the conditions for this reaction, those are sulfuric acid in the presence of heat right, which promotes elimination. So now we see all of our steps, let's just label them in our synthesis reaction. Number one is addition of pCC a mild oxidizing agent in the presence of paradyne. Number two was the treatment of the result on keystone with magnesium bromide in the presence of ether. Number three was the Acidic work up which essentially allowed us to protein eight hour Alcock side and I and so we can say H 30 plus and number four was the addition of sulfuric acid in the presence of heat to perform that elimination. So once again we started with an alcohol, turn it into a key zone using a mild oxidizing agent. Keeping in mind that we wanted to add two more carbon atoms to the chain and the way that we know to add those two carbon atoms is a greener reaction. So we did that formed our Alcock said and iron protein ated it to form a new alcohol and now this is where an asset catalyst dehydration works. We understood that we had to produce a more substituted L. King. So anyone mechanism perfectly works for us and we are doing that. We're forming our date of product right using sulfuric acid in the presence of heat. So, those are the required conditions. We may label them as our final answer because that's our synthesis plan, right? The two given materials are already given to us the alcohol and the elk in. We just wanted to understand what sort of reactions we could perform to get that transformation. Now to conclude the correct answer to this multiple choice question is B However, if you have any questions, please leave your comment down below in the comment section and thank you for watching.

To practice working through the early parts of a multistep synthesis, devise syntheses of
(b) 3-ethylpentan-2-one from compounds containing no more than three carbon atoms.

Key Concepts

Multistep Synthesis

Carbon Skeleton and Functional Groups

Reagents and Reaction Conditions

Watch next