Show how you would convert hex-1-ene to each of the following ...

Question

Show how you would convert hex-1-ene to each of the following compounds. You may use any additional reagents and solvents you need.

(a) 2-methoxyhexane

(b) 1-methoxyhexane

(c) 1-methoxyhexan-2-ol

Accepted Answer

Hey everyone and welcome back in today's video. We are going to show how each of the following compounds can be synthesized from Butte one in number one to meth Oxy Beauty. Number 2 1/8 oxy butane and number 31, meth oxy beauty. And see, oh, now the starting material in each case is exactly the same. So we control our four member chain because that's butte one in and the position of the double bond starts at carbon. Number one over here, right from 1 to 2. Now we have three reactions and we can simply say that the first reaction will be here and then we have our second reaction and then we have our third reaction. We want to propose the regions for the following products to meth oxy butane. Let's troll beads chain chain position. Number two has meth oxy groups. We're going to add O CH three one with oxy beat saying we are going to draw another beat sane. But now position number one over here contains an ether oxy group zero cc C three. That's our eighth oxy group. And number three beat N C O. So we are going to draw a four member chain with an alcohol group at position number two. And that meth oxy group at position number one. Okay. Well done. We have already done a lot. We got our structures, all that we want to do. Now, we want to understand what kind of regions we can use for each reaction starting with reaction. Number one, we can think of this reaction as the addition across the double bond of two groups. One of them would be hygiene, right. We're essentially adding hydrogen over here and then we are adding meth oxy group at the adjacent position. Notice that this product follows the more cosmic of regio selectivity, right? Because hydrant is added to a less substituted position and meth oxy group is added to a more substituted position because the reaction follows the more cosmic of radio selectivity we are going to use oxen, mercury ation dimmer curation Which allows us to perform such a transformation. So recall that an oxymoron curation curation reaction requires two steps. First of all, in the first step, we're using mercury to acid eight in the presence of the neutral form of our group that we have added to a more substituted position. So if you look at meth oxy group, we want to turn it into a molecule. You are going to get methanol, right? You take o negative siege three, you're going to add hydrant and you will get your nuclear file that participates in that reaction that would be methanol. So C 30 H, that would be our first part of the reaction. And number two, we want to get rid of mercury in our intermediate by performing the reduction reaction or the dimmer curation stop using sodium borrow hydrate. And this allows us to achieve the required transformation, right? So this one was pretty simple. Let's think about the second one. The second one is a bit more difficult because you understand that the addition of the ethyl group, my apologies, the toxic group is not at a more substituted position is it's actually at a less substituted position. So what we're going to do here, we, we want to think of a possible essence a reaction and let's see how that looks like for reaction number two, you will just break it down in part because it's a bit harder to visualize you. First of all want to make sure that you produce a substrate with a good leaving group and then make use of an essence your reaction. So if you want to make a primary substrate, that must be a primary substrate because the product has an toxic group bonded to a primary carbon, you will definitely use the anti more cosmic of addition. In particular, what you're going to do here, you're going to treat your al king with hydrodynamic acid in the presence of peroxide. So if you use hydrogen peroxide, you may recall that this the product form will be the ante Markovic of addition product. If mark optical product here without peroxide would have been hydrogen added at this position and then added at this position. In other words, two bromo beating, now we will have one bromo beating because hydrogen is actually added to a more substituted position and grooming is added to a less substituted position. So in the first step, we are getting our primary substrate and that's really what we wanted because you have your good leaving group now, right? So when you have your good leaving group, you can essentially think of an essence your reaction where you're simply using S oxide. So if you're using S oxide which attacks the primary electra filic position for by the loss of the leaving group, you get the final product, right? And that's it. So we understand the two steps required for the second reaction, we can just list them. So we will need two reactions. Number one addition of hydrogen bromide in the presence of hydrogen peroxide. And number two, because if oxide is negatively charged, we can just add some iron spectator to it such as sodium, right, to make sure that we are listing a neutral compound, actually neutrally looking compound basically because that would be sold. But in a solution that associates and sith oxide. So we can say sodium with oxide and A oh CH two CH three. So that's sodium meth oxide which contains that keith oxide and I am. And we can also use solvent, the corresponding solvency thoughts that would be ethanol. So we can simply add C H three C H 20 H which corresponds to our servant. And we are done as we understand in the first step, we performed that ante Markovic of edition where booming is added to a less substituted position. And that allows us to make use of an essence to reaction very easily to produce the required product. And now listening about the next one, for the next one, we understand that we have both an alcoholic group and an ether group and a very common way to do that is via an IP oxide. So first of all, what we're going to do, we're going to form any Parkside out of our L king. Meaning for reaction number three, our first step is to form an Ipaq side and to form any box side, we are going to use a very well known region meta chloral proxy, Ben's OIC acid em CPB in the presence of the very common solvent that we use for this reaction. That would be dick Laurel, methane, ch two cl T. And what we're doing here, we're just forming our IP oxide. So across the double bond, we are adding our oxygen and that's the box that, that reform. And now because we want to make sure that the alcohol group goes here and then we are adding a substation at this position, which is a myth oxy substitution, we have to evaluate how substituted each electrolytic position is. So the less substituted position is over here, which is indeed the one that we want to attack because look, you want to open your ring and make sure that the alcohol group, let's draw it out. So you can visualize it. If we open it, You will have oxygen over here, right? And then you will have some substitution over here, which gets bonded to it. And that's exactly what we want. We want to have OC three over here and we want to have a wage here. So we are going to attack a less substituted position. This is a more substituted position. So to perform that nuclear attack, we have to appropriately choose our conditions and the conditions that we're going to choose here, they must be basic because in basic conditions, basic conditions imply that attack take, takes place at a less substitute at carbon. And this is the one that is less substituted, meaning we are going to use meth oxide with a negative charge because meth oxide is basic, it's just sufficient to use meth oxide here which attacks that position. We have that ring opening and we are forming our bond. And then because we are releasing elk oxide, right. Auction will have a negative charge. We also need to use a source of protons. So we also want to make sure that we're using an appropriate polar products, solvents such as methanol because it will donate the proton to oxygen to form an alcohol group. So we are done, we can also fill out the final reaction. And we can say that for the third one, we need to carry out two reactions. Number one addition of meta chloral proxy benzoate asset in the presence of di chloral methane. Number two addition of sodium myth oxide in the presence of methanol. Okay. So that will be it. We can just label these as our final answers. We have our analysis done and we see all the required steps. So these would be our required regions for each transformation. Number one, we're using mercury to acid eight with methanol followed by the dimmer creation steps. So the more hydrate, right? So this is our Oxenberg curation demarcation reaction very basic because we understood that we only needed to add a meth oxy group at a more substantive position. So that's why we went with Oxenberg creation creation reaction because it follows the more cosmic of regional chemistry reaction. Number two, we understood that addition takes place at a different carbon atom at a less substituted carbon atoms. So we needed to use peroxide with hydrogen bromide to make sure that we're producing a primary substrate first with a good leaving group. And then that primary substrate can be attacked by a strong nuclear file which is S oxide. And that's how we get our product. And number three, we realized that we have two groups and an alcohol group and an adjacent meth oxy group, which is commonly achieved via an IP oxide. So we said for reaction, number one, let's produce our IP oxide. And when we get it, we can just use basic ring opening of IP oxides to give us the desired product. Right, the basic ring opening of IP oxide takes place at a less substituted carbon so that these are our answers. The correct answer choice to the small pool choice question is c However, if you have any questions, don't hesitate. Sleep there. Comment down below. Thank you for watching.

Show how you would convert hex-1-ene to each of the following compounds. You may use any additional reagents and solvents you need.
(a) 2-methoxyhexane
(b) 1-methoxyhexane
(c) 1-methoxyhexan-2-ol

Key Concepts

Electrophilic Addition Reactions

Nucleophilic Substitution Reactions

Stereochemistry and Regioselectivity

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