When doing synthesis, you will often find yourself repeating t...

Question

When doing synthesis, you will often find yourself repeating the same series of steps. To see this in action, synthesize the following aldehydes beginning with an organic molecule containing three carbons or fewer.
(b)

(b) Structural formula of an aldehyde with a three-carbon chain and a carbonyl group at the end.

Accepted Answer

Welcome back everyone. To another video, we often repeat the same steps when performing synthetic analysis to produce the desired product. To see this in action, use the alkin synthesis to determine how the following aldehyde is produced, starting with an organic molecule with three or fewer carbons. We're given the structure of Hep now and we want to synthesize that we can only start with a molecule of no more than three carbon atoms. So we can essentially choose a three member chain such as protein. And the reason why we're choosing an alkin as our starting molecule is because alkins are simply more reactive than alkanes are right. So in the first step of our reaction, we can essentially perform a hydro halogen reaction. So we make it more reactive specifically. What we mean by that is that we will obtain a good leaving group. So our resultant alkyl bromide will be able to participate in substitution reactions. So I suppose that our first step is the addition of hydroponic acid. And what we specifically want to add is bromine at the terminal end of the alkane. So we're going to use peroxide to make sure that this would be the anti Markov of Rio chemistry. The reason why we want to do that is because we want to have that long linear chain, right. So we want to add bromine at the terminal end, anti cognitive rio chemistry tells us that we are going to add bromine to the less substituted carbon and hydrogen gets bonded to the more substituted carbon. Now that we have our alkyl hide, this is where we start thinking about lengthening our chain. So we can use a satellite, right? Let's suppose that for step number two, we're using sodium a satellite. And in particular, we're going to draw our satellite anion with a negative charge. And we can choose a counter ion such as sodium. Here, we are performing an essence of reactions. We have our substitution, we have our strong nuclei which would attack the electrophysi group, we would form our al. So we have 123 carbon atoms and now we are bonding our triple bonded carbon. So we are going to essentially illustrate that bond well done. And now we have to think about it. What could be our next step? Well, we can perform the same step as before we can turn our alky into an alkane, then brominated it again and possibly get another alky in the course of the reaction. So our third step would be the reduction that we want to turn our alky into an alkin. And for that purpose, we can use sodium in ammonia or we can use hygiene in the presence of Lindler S catalyst. Because these regions essentially allow us to perform a partial reduction so that we only transform our LK kind into an L king and not into an L Kane. We have our five member chains, we are going to do 12345 carbons. And the result on double bond Lindler catalyst is actually known for producing c isomers. But in this case, cis and trans is not applicable. So it doesn't really matter if we choose sodium and ammonia, which produces trans iser or Lindler catalyst, which produces Cymer, we have two exactly same hydrogen atoms. So cis and trans, those concepts are not applicable. Now, let's remember what we said we could do. We essentially said that if we get another alkin, we can just repeat hydro halogen in the presence of hydrogen peroxide. So our fourth step would be the addition of hydrogen bromide once again or hydrobromide in the presence of hydrogen peroxide. We're going to add that for agent and we're performing another anti Menikoff addition that will be our chain. And then we're adding bromine to the terminal carbon. OK. And now we can use a satellite once again acting as a nuclear file on the resultant LK bromide. We can say that our fifth step would be the addition of a satellite which makes perfect sense because if we have a five member chain and we need a seven member chain. Then this is a perfect choice. Once again, a satellite acts as a strong Nile, it would attack the electro pilic center followed by the loss of the leaving group. So we can draw our 12345 member chain, right? And then we're bonding two more carbon atoms. Now, finally, let's remember that hydration oxidation allows us to form enos out of terminal alkins which then tori into aldehydes specifically, since we have a terminal alky formed, we can use number six bora in the presence of tetra hydro urine and number seven hydroxide peroxide and the water. Now, this essentially allows us to perform a hydration oxidation reaction, right, in which terminal l kinds would be converted into aldehydes. This reaction takes place in several steps. The first step is the formation of an eo because once again, we have anti cogni of rio chemistry so that we're adding our alcohol group to the less substitute as carbon and hydrogen to the more substitute as carbon we're breaking our pyon. So we get an eno which then tos into an aldehyde, which essentially tells us that we can now draw our product 1234567. And now that lost carbon corresponds to our newly formed aldehyde group. And this is how we obtain our final product. We can label this step step, synthesis of our final answer and conclude the video. Thank you for watching.

When doing synthesis, you will often find yourself repeating the same series of steps. To see this in action, synthesize the following aldehydes beginning with an organic molecule containing three carbons or fewer.
(b)

Key Concepts

Aldehyde Structure and Reactivity

Synthetic Pathways

Functional Group Interconversion

Watch next

When doing synthesis, you will often find yourself repeating the same series of steps. To see this in action, synthesize the following aldehydes beginning with an organic molecule containing three carbons or fewer. (b)

Key Concepts

Aldehyde Structure and Reactivity

Synthetic Pathways

Functional Group Interconversion

Watch next

When doing synthesis, you will often find yourself repeating the same series of steps. To see this in action, synthesize the following aldehydes beginning with an organic molecule containing three carbons or fewer.
(b)