Show how you would synthesize the following compounds,
starting ...

Question

Show how you would synthesize the following compounds, 
starting with acetylene and any compounds containing 
no more than four carbon atoms.
(g) pentanal, CH3CH2CH2CH2CHO
(h) pentan-2-one, CH3-CO-CH2CH2CH3
(i) (±) 3,4-dibromohexane

Accepted Answer

Hey everyone and welcome back in today's video, we are going to analyze the following problem using a settling as the starting material show how each of the following compounds can be synthesized in your synthesis. You can use any compounds with four or less carbon atoms. Number one, pro penile, number two, assets zone number three, a pair of an and humors of two comma three die bro mobile saying we can start outlining our synthesis and looking at problem number one, we are starting with a settling. So that would be CDC tried to carbon atoms bonded by triple bond, the simplest possible al kind forming propane al prop analysis and al height. So we can show that ch oh group there must be a total of three carbon atoms. So that's 123, right? So that's pro panel the world. First thing we notice is that the starting material has two carbon atoms, the final product has three. And the way that we form a longer chain, al kind essentially is the way to go, right? If we want to directly add an additional carbon atom, we will first form a longer chain al kind. And because we only need one carbon atom. There's only one possibility to do that. So let's recall that first of all, we can use a settling which has that acidic hygiene, a disposition. And we may remove that acidic hydrogen using a strong base such as sodium Amytal sodium, a mite D protein. It's a settling to produce a satellite and iron that is a good nuclear file and it participates in an SN two reaction. So because we had our negative charge, the next step is to think about that addition of one extra carbon atom because we need just one carbon atom. We are going to use methyl group Bonded to some good leaving groups such as bromine, for example. So we can say C3BR and as we said before, that would be an essence a reaction because we have a negatively charged nuclear file, that means it's a strong nuclear file. So there will be an attack followed by the loss of the leaving group. And indeed, we get our three carbon atoms that we needed. However, we still have that triple bond. So we will need to think about several more reactions, but don't worry about it. So we got our pro pine right? We have a triple bond with three carbon atoms total. And now if we want to think about the production of an al to hide out of it, we have to recall the fact that al the hides are produced very easily from terminal al kinds via hydro abrasion oxidation. So what we can do here, we can essentially state that if we manage to properly form an Enel out of it, right. We want to make sure that the alcohol group is attached at that less substituted position. So we're going to add an alcohol group over here and over here, we're going to add hydrogen. This is the Ante Markovic of edition and the reason why we want to make sure that it's ante Markovic of addition is because we're forming an al to hide out of a terminal al kind. That's why we're going with hydro operation oxidation. Okay. So now how does that work? Let's recall that we need to reactions in the stop. The first one is the treatment of the terminal al kind with dice. I am ill boring. That would be side to be H and number two is the addition of hydrogen peroxide in the presence of base. This is very similar to what you're used to see in alkaline reactions except from the fact that we're using a less hindered boring, which is the three, but for all kinds were generally used by Simon boring. Now, in this high preparation oxidation reaction, if we simply draw our product below, as we previous decide, we're going to add our alcohol group at carbon. Number one and hydrogen carbon. Number two, we will have a total of three carbon atoms. We can throw that chain three carbon atoms, there would still be one of the pi bonds remaining. So let's add that the alcohol group goes at carbon. Number one hydrogen is implicit. So don't worry about it. We definitely have added it because we have a double bond now. And from here because this is an email which is not very stable. We have torch memorization into a more stable on which in this case would be. And al the heights we can show equilibrium arrows. I'm showing that the al to hide is more favorable. That would imply our final product would be Hi HN three Carbon atoms. That's 1 2 three. So we got, we got our found product, right? Essentially, if you're wondering there's a proton transfer, it can take place multiple ways. But one of the ways is just that the lumpy and oxygen let's retry that so that be our oxygen, hydrogen bond and then the lone parent oxygen restores the double bond, the double bond within the eternal captures that proton. And there's the transfer of these bonding electrons on the oxygen to form the al to hide. So we got all of our steps, let's Label them. So a reaction number one is the treatment of the alkaline were settling in this case where sodium A might To produce a satellite. # two, treatment of a satellite with metal bromine To give us an essence to reaction that would be CH three BR reaction. # three was addition of the same old boring keeping in mind that our goal is to produce a hydro operation oxidation reaction. And then number four, addition of hydrogen peroxide in the presence of hydroxide and the email formed produces a more stable alla hide. So we got our product, we can label these as our answers for the first reaction. And now we're going to think about the production of acetone instead of penile for reaction. Number two, the starting material is exactly the same. It's a set line. We went to this acetone. So that would be our kids zone with three carbon atoms. And we may immediately notice that once again, there's an additional carbon atom. So we can use the same sequences before You may look at the steps above. We essentially are going to add number one sodium Amytal to produce a satellite number two because we need one carbon atom. We can use methyl bromide once game and this would give us propane. But now this is where we have to be careful. If we draw pro pine, we now want to produce acid, own out of it, right? We want to produce a key tone instead of an al to hide. So rather than using the anti mechanical edition, we are now going to the more cognitive edition and more cosmic of re geochemistry tells us that hygiene must be added to that less substituted position and the alcohol group will be added to a more substituted position to do that. We're not going to use hydro abrasion oxidation which produces the ante Markovic of product. Here, we are simply going to use the acid catalyzed hydration. And you may simply recall that the acid catalyzed hydration of all kinds implies the treatment of that Al Qaeda with water. So Fairy Cast said in the presence of Mark Uric salt in this case, mercury to sulfate, which essentially allows us to speed up that reaction because hydration of al kinds is much slower than the hydration of al kinds. That's why we need it. And this gives us our Elkin. First of all, we are going to draw Elkin. We understand that at position number two, we are adding our alcohol. There's that implicit hygiene at this position, right? And this is where total memorization takes place favoring the formation of the ketones. We have that keto Enel equilibrium that Enel which is not really stable forms a key tone by that proton transfer the double bond and the formation of the carbon yield groups. So we can say that indeed our final product that's asset on here. So a very similar reaction, meaning we can immediately say that the world final stop, which is different is that instead of using hydrogenation oxidation, we are now treating the result. And pro Pine with water still very CASS said, and Marie Curie sold to give us that more comical product, we may now label these as our finally answers for Reaction # two. And let's move on to reaction number three. For reaction number three, our starting material is exactly the same as settling. And now We want to produce a pair of finance remarks. It says 2, 3, dye bromo beating essentially what you can see to easily find out what the structure is. You can just draw that central carbon carbon bond. There will be four carbon atoms and beating Position # 2 & three. These are the positions we're interested in. And if you draw the remaining two carbon atoms, the two metal groups above that central carbon carbon bond, you will be able to assign two bromine atoms. One of them must be pointing up and another one must be pointing down. We're not really worried about the actual configuration, don't worry about it. You essentially want to understand that because we're producing a pair of finance, this structure cannot have an internal plane of symmetry, right? So if I assign two wedges or two dashes that would have an internal line of symmetry, that would be a measles compound would know an anthem or right. But if you make sure that the two bonds have different configurations or different stereo chemistry, right? In that case, there's no internal line of symmetry, meaning there must be an answer to it. So we're saying that our next product would have an inversion of configuration. This blooming would be pointing down And the other one would be pointing up. And indeed, they are non superimposes about mirror images of each other. So that works for us, these are the products that we want to form. And let's think about our sequence for this one. It's better to go in reverse and think about the fact that it's not only that we're going from two carbon atoms to four, but it's also the fact that we're adding blooming, right? So if we're adding roaming, we may think about the electrolytic edition mechanism, the way to add two bromine atoms would be via an Al king. So if we go backwards, we can say if we start with a Cis Al King and in particular data, B says B two in, if we add blooming in the presence of carbon tetrachloride, that's our solvent. Essentially what we're going to do here, we're going to form a bridge, right? We are going to form our bro ammonium bridge followed by the backside that sag at the two positions, right? That's why we form a pair of finance commerce. The backside attack implies that the arrangement of the two bedroom of substitue ints with essential B and C. And that's why we're getting a pair of financial marks where to booming atoms are in N C arrangement. We can either check this carbon or that carbon. And that's indeed why we are going with the sis Al King. Now, having understood that we now narrow down our problem and we understand that the world lost up is the formation of that Al King. The question is how do we form an al king out of an alkaline? And because it says we can use landlords catalyst, recall that that landlords catalyst, we can label it as hydrogen in the presence of landlubbers. Essentially, it allows us to reduce al kinds into sis Calkins only. And if you notice the position of the double bond starts at carbon, number two, meaning the al kind that we want to use must have a triple bond with two metal groups. On the right and on the left, let's add it correctly. That would be our Muslim group over here. So that would be One metal group, another metal group. And we understand that we have beat two Ein. And the question is now, how do we form, beat to ein if our starting material is a settling? And the answer is very simple. Once again, if we're going backwards towards our starting material, we want to have a settling. And if we recall, we are essentially d protein eating that hydrogen forming our satellite that will attack a primary substrate to form an SNC product. So we can add two metal groups performing that reaction twice. So essentially, we can say that number one we are going to D protein ate it form a satellite. When the satellite is formed, we want to add our first metal group at this position. So that would be methyl bromide or any other good leaving group that you know, metal chloride, for example, and you're going to repeat those steps because when you add that metal group, you want to add another metal group, repeating those steps. So that would be D protein ating the remaining is to the high chain. And then using a satellite form to in the reaction with methyl bromide once again, because you want one more carbon atom And this would give you two butin. So everything makes sense. We now understand how it works. We can just label our reactions. Number one reaction. Number one, if we go in reverse order just following the arrows, we're going to use sodium Amytal number two, addition of metal bromine number three, Once again, sodium Amytal number four, once again, metal bromide. So this is how we are forming. Beat two. I'm Now when we form B2, I mean, we are going to perform reaction number five edition of hydrogen in the presence of landlords catalyst. When we form our sister Elkin, we essentially want to add bro mean in the presence of carbon tetrachloride. So these are all regions because when we form that sis Alki in additional grooming would form that ammonium bridge, the backside attack would ensure that the two bromine atoms are antsy and arrangement. So this would give us a pair of finance us because there might be an attack at each electrolytic position. So these would be our final answers for reaction. # three, this was quite a long problem, but I hope you found it enjoyable. And the final answer to this problem is c however, if you have any questions, don't hesitate sleeve your comment down below in the comments section. Thank you for watching.

Show how you would synthesize the following compounds, starting with acetylene and any compounds containing no more than four carbon atoms. (g) pentanal, CH3CH2CH2CH2CHO (h) pentan-2-one, CH3-CO-CH2CH2CH3 (i) (±) 3,4-dibromohexane

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