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.
(d) trans-hex-2-ene
(e) 1,1-dibromohexane
(f) 2,2-dibromohexane

Accepted Answer

everyone. And welcome back in today's video, we are looking at the problem which reads using a settling as a 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. And we have three different products formed. Each of them is formed from a settling. So we may 1st of all indicate these three reactions and we will draw a settling as our starting material. A settling is the simplest alkaline, which consists of two carbon atoms. And we'll be forming our products from a settling and some additional regions that we have to determine. So for each reaction we're drawing a settling and we're drawing a reaction arrow. We will complete these regions a bit later. The very first thing we want to do is just to ride the products and we will do that shortly. So these are settling molecules. These are reaction arrows. We will have to fill the regions over here. But now let's draw our products. First of all we have trans VT in so we are drawing A. B. Team chain with double bond at position number two, that's trans arrangement. The tooele keel groups are pointing in opposite directions. Then we have one common one libro Moby chain. We are drawing a bead chain chain in position number one has two bromo substitutions. That's what it says. So that's our second product and our third product is to to dye bromo butane. So we are drawing a bead chain chain, position number two has two bromo substitute prints. These are our reactions. Now we have three questions. What sort of regions are we using in these steps. And now essentially we will analyze each reaction one by one. We will start with reaction number one let's go down and and lives reaction one step by step and let's think what's happening over here. First of all we saw what that sets lean and we want to notice that we are producing a longer chain alki and therefore carbon atoms here there too. So to increase the chain land we use that simple method where we will be the protein ating the initial AL kind and we will be performing an essence reaction on some electro Felix center notice that the position of the double bond tells us everything in order to start in order to get a double bond out of a triple bond. We will essentially be reducing it. So notice that because the double bond starts at position two, that means that position to you. We will be adding a methyl group and the position we will be adding another metal group. So across that triple bond we will be adding two methyl groups. Recall that there's an acidic hydrogen vanity and S. P carbon atom. So we may remove that acidic hydrogen atom by using sodium. I mean this is a classical base whenever we wish to D. Protein eight and AL kind using sodium Hamid will essentially D protein ated because we have a strong base with a negative charge. We will have the deep rotation step to produce our good nuclear file and the nuclear file that we're getting here is a negatively charged nuclear file, meaning it will undergo an essence reaction very well. That's sad. If we want to add a method group, we need to use a substrate which has a muscle group on it. We also need a good leaving group. So let's choose bombing although you may also use chlorine or iodine here. So notice that the negative charge would attack the electra Felix center. And simultaneously we would have the loss of the leaving group to make our chain longer. That would give us our metal group that we desire to get. There will be a method group over here and now we want to add another method group. So why don't we just repeat that process? Why don't we just add another molecule? My apologies. Another formal unit of sodium amid. So we would D protein ate this remaining acidic hydrogen and we would produce our intermediate which would act as a nuclear file. In the next step. This would be our intermediate with a negative charge. And now because we need another metal group, we are adding another molecule of methyl bromide. That said, let's indicate that the next step is the addition of methyl bromide. Once again, we already know that this is an essence to reaction. So the negative charge would attack the electra Felix center. We would have the loss of the leaving group and we would form the desired al kind. Okay, so we have our Al Qaeda. Now, the question is how do we get a trans AL keen out of it? We call that we need to use a selective reduction. We may either use landlords catalyst or sodium in ammonia. Depending on the outcome that we wish to get. We choose one of the two. And because we want to get a trans AL king, we use sodium in ammonia. That was essential, reduce that triple bond and say double bond in trans arrangement. And that will be our final product. We are done with our steps, we have all the regions needed and we are ready to look at the second reaction. Now, if we look at the second reaction reaction number two, we want to get one comma one, one comma one. Di bromo beating. My apologies. The actual compound is over here. That's one comma one dye bromo beating. So to get one comma one dye bromo beating, we first of all need to realize that it's important to make sure that we only remove one of the two acidic hydrogen because addition of two bromine atoms implies electra filic edition. So we want to form a double bond at this position. If we want to form a double bond a disposition, that means we don't want to add any AL kill substations here. But because we have a four carbon chain, that means we would need to add an ethyl group here instead, we would not be adding two method groups as before. We would be adding a single ethyl group at this position. So going back to our reaction. If we draw, I said to lean, we are going to dip protein eight that using sodium Amytal, we will form our Nicola file that could participate in an essential reaction. And as we previously discussed, we need an ethyl group. So why don't we use an ethyl group bonded to another roaming? Because it's a good leaving group. We would have an essential reaction at this step and we would make a longer chain Al kind. So that's exactly what we wanted to see. And now our goal is to add two bromine atoms in this position and two hydrogen atoms in this position. We call that this is the ansi more cognitive addition. According to the more Kafelnikov's rule, we're actually adding hydrogen is to a less substituted position. So there will be two hydrogen atoms added over here according to the more cognitive school. But in this case we want to have the anti more cognitive product. We want to have this what we discussed in reverse in other words, instead of adding hydrogen atoms. So we're here, we will be adding them over here and to do that. We are using the electra Filic edition mechanism in which we are adding HPR But to make sure that the product is anti Markov. Markov, we are using peroxide in general. We may label them as R O. R. It stands for any kind of peroxide. And because we want to make sure that we are forming single bonds and adding to bro means in this position we are going to use two equivalents or two moles of hydrochloric acid that would allow us to add two bromine atoms, making this a single bond. And we are nearly done. That means our final product would essentially be a beating chain with two bromo substitue ints added Well done. And now we're ready to look at the third reaction and the information covered in reaction to see will be very helpful for us. If we start with a settling. The main difference is that if you look at the structure that we want to form this is too to die bro mo beat saying we want to add these two blooming atoms according to the more cosmic of school. If we had an elk in here, that would be great because we would be able to use the more cosmic of addition, meaning similarly to what we had before. We will not be affecting this acidic hydrogen in this position. We will only be removing one of them and because we need a four member chain, we'll be adding an ethyl group. So let's do that. We will d protein ate it using sodium Amytal to form our nuclear file. We will perform an sN two reaction in which we will be using ethyl bromide. Or you may use ethyl chloride or fluoride. That's perfectly fine. You will have that sn two reaction at this stop just as before you will form the required al kind. And now here is where we have our difference. Everything was exactly the same. But now our goal is to actually add bruning Adams at these positions and hydrogen atoms at position number one. So essentially that means more comic of sedition because this is a less substituted position in order to add two hydrogen atoms here and two bromine atoms here, we're simply adding hydroponic acid itself. We don't need peroxide anymore because we want to make sure that we follow the more cosmic approval. And again we need to equivalence because two hydrants and two bromine atoms were added. That's how we get the desired products. There will be two Roaming Adams bonded to position # two. Great job. That was really long, wasn't it? So now let's consider all of this information. Let's summarize our findings to make it easier for us to understand that required steps for this problem. If we fill out our regions for reaction number one, we had five steps. Step # one was using sodium. I mean adding methyl bromide afterwards, repeating steps one and two. So we may just right step three s again. So do I mean Step four Methyl bromide and finally we use sodium in ammonia to produce that trans elkin Great job five steps Now for the second reaction. We first will dip protein eight. That using sodium amid we increase the chain land of the al Qaeda and using ethyl bromide. And then the final step we use the ante Markovic of addition. We use two equivalents of hydro bro Marcus said in peroxide to promote the ante Markovic of addition. And for number three we performed three similar steps. We used sodium Hamid, two. D. protein ate it. We added ethyl bromide to increase the chain land and finally we did not need peroxide because we wanted to get the more cognitive product. So we used the equivalence of hydrochloric acid itself. We may now state that these are the required regions for the formation of these products that we were given and you may now notice that the correct answer to this multiple choice question is B however, if you still have any questions, don't hesitate to leave them down below in the comment section, Thank you for watching. And I'll see you in the next video

Show how you would synthesize the following compounds, starting with acetylene and any compounds containing no more than four carbon atoms. (d) trans-hex-2-ene (e) 1,1-dibromohexane (f) 2,2-dibromohexane

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