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:

c. Reaction scheme showing the conversion of an alkyne to a primary alcohol with a four-carbon chain.

Accepted Answer

Hey, everyone. And welcome back to another video, describe the steps required to make hex N 30 using B one, any inorganic regions and any organic molecule with less than four carbons. Now, the starting material is bet one ion and we have to understand that be one ion essentially corresponds to an al kine which has a triple bond starting at carbon one. And we have a total of four carbon atoms in the parent chain because we want to form hexen 30 specifically, which is an alcohol. We are going to draw a six member chain because we have hexen. So 123456. And we understand that the alcohol group or our O H group is bonded to carbon number three. So now because the final product has a total of six carbon atoms in the parent chain, we first of all want to make sure that we add two more carbons to our alk. And for that purpose, we are first of all going to use sodium amide. Now, the reason why we're using sodium amide is because we have a terminal alk. So we can easily remove the acidic hydrogen to form the ayite and ion. And then we are going to perform an essence of reaction. Since we need two carbon atoms, we're going to use ethyl chloride. OK. So just to give you an idea in the first reaction, sodium amide de proin the alkin and we get our aide and ion, then a set will perform an S N U reaction being a strong nu phy it will attack the electrolytic position of ethyl chloride followed by the loss of the leaving group. And this is how we are going to get and in internal alpine. So now we are going to have our alkin with a total of six carbon atoms. So we have 1234, we're going to add one more and an additional one. So in this step, we are essentially forming hacks three I OK. Well done. So let's eliminate the mechanism so that we can clearly follow the synthesis scheme. And now that we have formed the internal alkin because our goal is to finally get an alcohol, we actually want to form an alk out of it performing the partial reduction. Now, because stereo chemistry is irrelevant, we can actually use hydrogen in the presence of Landler's catalyst. So whenever you want to perform a partial reduction from an alk into an alkin, if stereo chemistry is irrelevant, you can either use hydrogen in the presence of Lin's catalyst or sodium in liquid ammonia. And in this case, we are going to get our six member chain. And now our double bond starts at carbon three and this is really good for us because to form the final alcohol, we can simply think about the acid catalyzed hydration of als. Now, the acid catalyze hydration of Alkins requires us to use water in the presence of sulfuric acid. What's happening is that we're going to add the hydrogen to the less substituted carbon. But actually, in this case, they're equally substituted. So we will get at first glance, a mixture of products. But in reality, if we think about the addition of hygiene and then the alcohol group, in each case, we will get the same alcohol, right? Because the Alcaine is symmetrical specifically, let's think about it. So we're going to draw our team and then when the alkin reacts with sulfuric acid, we're going to add hydrogen to any of the two positions so that we are forming a carbo Canion. Let's suppose that we add hydrogen to carbon number four, then we will have a positive charge that carbon number three and then water will act as a nu phy taking the Carbocaine. And after the deep rotation stopped by another molecule of water, we are going to get our alcohol where the alcohol group is going to be bonded at position three. And even even if we get a positive charge at carbon number four, that would still us that would still give us the same alcohol, right? Because the structure is symmetrical. So let's eliminate the partial mechanism that we have drawn and let's review the steps. So we're going to first of all label our final answer and the final synthesis scheme and conclude that to perform the required census. First of all, we are going to treat the given alkin with sodium amide and ethyl chloride to form an internal alkin with a longer parent chain using hydrogen gas in the presence of landlords catalysts were performing a partial reduction from an internal alkin into an alkin. There's actually one more thing and I'm sorry, but I did a mistake here. Let's remember that Lindor's catalyst essentially forms C Alkins. So instead of drawing a trans alkin that I had previously, even though it doesn't matter, and even if you have a trans alkin, it will still form the same product. But here it will be better to draw the correct product. So we're drawing our double bond and then on each end, we have an L group bonded to it in this arrangement, right? Because Lindler catalyst produces Alkins and that's really important if we want to draw the correct structure. So my apologies is that at the very beginning, I drew the trans king in reality, that would be the and then the, the al undergoes the acid catalyze hydration to produce the final alcohol. That will be our final corrected census scheme. And thank you for watching. 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:
c.

Key Concepts

Functional Groups

Reaction Mechanisms

Synthesis Strategies

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