Show how each of the following compounds can be prepared using...

Question

Show how each of the following compounds can be prepared using the given starting material, any needed inorganic reagents, and any organic compound that has no more than four carbons:

a. Reaction scheme showing the conversion of an alkyne to a longer carbon chain compound with a carbonyl group.

Accepted Answer

Hey, everyone. And welcome back, provide the synthesis of only using propine any inorganic regions and any organic compounds with fewer than four carbons. OK. So let's think about the synthesis. We are using propine as the starting material. OK. So we're going to draw our two S P hybridized carbon atoms and a metal group bonded to one of them. So we have a terminal alkin and we are producing a ketone because it says that we want to synthesize tan to your own. So let's draw a five member chain and let's add our carbon group to carbon number two. OK. Now, one of the most important things to understand is that the starting material has three carbon atoms and the final product has five. So if we want to enlarge the chain of our carbon atoms, especially if we start with a terminal alkin, we can first of all use sodium amide, right? And amide essentially allows us to remove the the acidic proton from the terminal alkin. And in that case, we are pro producing the aite and ion. Now, if we want to turn it into an internal alkin with a longer chain length, we can essentially use two more carbon atoms. That's what we need. And we want to perform an essence reaction in which we are going to use an alkyl hali. So let's draw two carbon atoms bonded to a good leaving group, suppose chlorine or bromine. And we're using Chloro ethane because we essentially want to add two carbon atoms to our alki to ace right, to turn it into a five member alkin. So we are performing a basic S N two reaction in which we have our nucleic attack at the electro position, followed by the loss of the leaving group. And now we are producing our triple bond, right? So our product will have a triple bond, a methyl group bond to one end and then an eyl group bond to the other. In other words, we have produced pant to iron and now this is quite good for us because if we have tent to iron, we can produce a corresponding ketone out of it. Let's see how it works. And let's remember the acid catalyzed hydration of alcohol. My apologies, the acid catalyzed hydration of Alkins, in particular, we are using the following set of regions which consists of mercury to sulfate sulfuric acid and water, right? So this would correspond to an acid catalyzed hydration of Alkins and we can produce a ketone, right? And let's see why that works. So, first of all, we have to understand that we have our internal alkin and Now we understand that if it's an internal alkin, that means it is equally substituted, both S P carbon atoms have alk substitutions. So that means the marconi of is not applicable. And we have two options. In one of them, we are going to add hydrogen and our alcohol group to carbons number two and three, irrespective. And then the other possibility for us is to swap them. So one of our products will essentially have a five member chain. Another one will also have a five member chain. Now, for the first one, if we add hydrogen to carbon, number two, we're adding alcohol to carbon number three. So we essentially get an and we have a double bond between carbons two and three, right? Because we still have one of our pi bonds remaining. And now for the green pathway, we're adding our alcohol group to carbon number two. And we have our double bond between carbons two and three. And now because enos are not stable, they were totem into ketones. So for the blue one, we can actually draw a partial mechanism to present how that happens. So the lone pair on oxygen would be de localized into a pi bond, the adjacent pi bond would be broken because we can have that proton capture and then we would transfer oxygen hydrogen electrons in the form of a lone pair on oxygen. And that would give us one of the key to which is three Oh, so that's not the product that we want to get, but it will be formed as a byproduct. And the final step for the green pathway, we expect to get exactly what we wanted 10 to own. So if we think about ization, the long pair and oxygen is used to create our Carboni double bond, we have our protein capture and the formation of a lone pair and that gives us our five member chain. And then we are adding our carbon group and that's exactly the product that we wanted to get. So we understood how the final stop takes place. And we understand that this is a feasible reaction scheme for us. So let's label it as our final answer. And let's review the steps. We are starting with propine and using sodium amide, we are producing an satellite anion in an essence a reaction. We're using Chloro ethane so that we can lengthen our chain and make it a five member alk which is to iron. And then during the acid catalyze hydration, we are forming two ketones tan two on exactly the the product that we want to get and we own. So we essentially have to isolate that product in our synthesis from the reaction mixture. However, the overall synthesis is feasible for us. And we can simply summarize the steps on the main arrow step. Number one addition of sodium amide step, number two addition of chloro ethane step number three, acid catalyzed hydration. So mercury two sulfate, so f acid and water. Thank you for watching and I hope to see you in the next video.

Show how each of the following compounds can be prepared using the given starting material, any needed inorganic reagents, and any organic compound that has no more than four carbons:
a.

Key Concepts

Functional Groups

Reaction Mechanisms

Synthesis Strategies

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