Pure (S)-2-bromo-2-fluorobutane reacts with methoxide ion in meth...

Question

Pure (S)-2-bromo-2-fluorobutane reacts with methoxide ion in methanol to give a mixture of (S)-2-fluoro-2-methoxybutane and three fluoroalkenes.

a. Use mechanisms to show which three fluoroalkenes are formed.

Accepted Answer

everyone and welcome back in this problem. We are stated that when pure three R four R three idol 34 dimethyl hexane is heated with sodium with oxide in ethanol, it produces a mixture of three Elkins draw a mechanism that accounts for the formation of the three al kins. Let's begin by correctly drawing our substrate which is hexane, that's our parents. So let's just draw a six member chain. That's hexane, positions number three and four have two metal substitutions 123, so that would have a methyl group four. There's another metal group and position number three also has iodine, balance it. So we are going to add iodine and we are also going to add hydrogen and disposition because three and four are Cairo centers. So it's important to see all of the four different groups that are bonded to each Cairo carbon atom. Now let's recall kon Ingle prelaw groups to assign priorities. And we will start with carbon number three in particular. We are looking at this carbon atom based on the atomic number of our substitutions. We noticed that there are three carbon atoms, there's one here, another one and the third one. So iodine is has the greatest atomic number, meaning it gets priority number one. Yeah. Method would be priority number four because if you look at the remaining carbon chains method is the shortest chain and therefore therefore it would have the lowest priority between the remaining chains. We're just going to draw them out. We will start with this carbon atom which has two hydrogen atoms bonded. Sit there are implicit and it's also bonded to another metal group but it's just sufficient to show that it's bonded to another carbon atom. On the other hand, the supposition under ride this carbon atom has a hydrogen atom and then it has bonded to a method group. So another carbon atom and also one more carbon atom over here. If we eliminate equivalent parts ch ch C C C C. We have a competition between carbon, carbon and carbon hydrogen. So since carbon is greater in atomic number, the supposition underwrite gets a higher priority. So we may indicate that this position on the right gets priority number two and that's a position on the left gets priority number three. Based on the rules, we wish to have priority number four pointing away or having a dash. So we are going to assign dash to priority number four. And we will check if we may trace our path to get the our configuration is required. If we trace the path here going along submissions towards number three. We noticed that we get our rotation because that's clockwise. So we are good. We have to see that our assumption that priority number four was on a dash worked. This is indeed are So if priority number four is on a dash, that would imply that priority number one is on a watch. We have successfully assigned the configuration for position number 33. R that works for us. And now let's look at position number four. For complete clarity, let's just redraw our substrate. We already know that iodine is on a wedge and the method group is on a dash here. So we are primarily interested in these substitutes. We have a method group a hydrogen atom. We may use a similar approach and we may understand that first of all, hydrogen would have the lowest priority because it has the lowest possible atomic number. So we may immediately state that priority number four would go to hydrogen than the metal group is shorter than any of the to any of the three al kill chains. So since it's the shortest, it would have priority number three. And then to indicate which one is priority one and which one is priority number two, we would essentially just draw out the atoms for the ethyl substitution. We may just say that Carbon is bonded to two hydrogen atoms and another carbon atom which is a method group. And similarly, if we look at the substitution on the left, we would have a carbon atom bonded to two carbon atoms and iodine carbon and the red carbon, carbon and iodine. If we eliminate equivalent parts notice that we get rid of carbon carbon and then that's it. No matter at which you compare. You have ch versus C. C. So C C wins and ch again loses against ci because iodine and carbon, they're both higher in atomic number than hydrogen is. And therefore we may conclude that the substitutions on the left is higher in priority. That would be the highest priority number one. And that would be priority number two. If we assume that priority number four gets a dash pointing away from the viewer based on con Ingle pre log rules, we may try to assign rotation and trace the path for our substitutions going along substitutions 12 and three. We notice that such a rotation as R. And we are happy to see that because four needs in our configuration, meaning our assumption was right. And now we are signing priority number three on a wedge. So we are going to add a wedge for priority number three. Great job. So we are going to redraw our structure with correct configurations. So we have hexane At position # two. We have iodine pointing up and missile pointing down. Now, if you look at these two substitutions, we have a muscle group pointing up and hydrogen pointing down. Now the problem suggests that we are using sodium s oxide and ethanol. The problem also tells us that we are forming Elkins. So it implies an elimination reaction. However, let's just think why this is an elimination reaction. And let's just confirm, first of all this is an ionic substance and when it dissociates in water, it produces these oxide as our base or our nuclear file. And also notice that the solvent that we are using is a polar product solvent now because it's a strong base strong nuclear file. We have a competition between essence to and E two. However, look at the solvent in particular the solvent that we are using as a polar product solvent, which favors SN one reactions and actually doesn't favor sN two reactions. So as a result, it favors E two reactions primarily because the solvent is a polar product solvent. It favors elimination reaction in this case and we are dealing with the formation of three different al canes. So it pretty much implies the same. We will form elimination products. Now recall that an elimination reaction requires us to identify beta hydrogen. But to do that. First of all notice that this is our alpha carbon atom because it has the leaving group bonnets, It and the beta hydrogen in particular are hydrogen. At these positions we have beta, another beta and another beta. We may essentially exclude that beta from our analysis because we are using a strong base that is not hysterically hindered. It's a relatively small based whenever we think about the dockside and I on So we are not trying to for many Hoffman products. We are only trying to form cities of products. So we are only considering this beta in this beta carbon. Now these beta carbons will have beta hydrogen. So let's label those hydrogen. Now that we see everything. There will be two hydrogen over here. One pointing up in another pointing down and we actually see that there's one hygiene indisposition. So we don't need to leave but anything. This is the only hygiene that we have. And now let's think about the elimination mechanism. We will draw the elimination mechanism by redrawing our substrate. So the substrate looks like this. We already know that we need to expand several bonds. So that's the method group that's iodine the leaving group, our hydrogen atoms Marcel hydrogen recall that an elimination reaction. It too requires us to have anti per plane arrangement arrangement of our hydrogen with respect to the leaving group. In other words, if you're seeing a dash here, you want to make sure that our design is pointing on a watch so that they're in anti per plane arrangements. So that's good for us. We have this hydrogen in anti arrangement and this hygiene in anti arrangement as well. So this is how we will get our first two products. We will just draw it oxide, the negative charge and oxygen will attack the proton. We will form a double bond at that position and simultaneously we will have the loss of the leaving group. So the mechanism is pretty simple as you see, and we will form our first L king. We may redraw our chain. Now we noticed that essentially we will not show these implicit hydrogen atoms but at position number three, we will have a planer carbon atom or basically sp two hybridized carbon atom. And in particular, we are only left with a metal group, we have the formation of the double bond in this position and there is still this metal group remaining. So that's our first product. And now let's think about the removal of the other hydrogen essentially, we may now redraw the structure for complete clarity to indicate what we're doing. We will only show one of the two high genes that we're interested in and the one that we're interested in should be anti to iodine. So it must be on a dash. That's hydrogen and that's iodine as well as a muslim group in particular here, nothing really changes. We still have that method group and again, we're using our strong base which is s oxide. The negative charge attacks the beta hydrogen. We are forming a double bond and simultaneously the living group leaves. So we get our second product and we are ready to draw it. Now the position of the double bond is over here between Carbons two and 3. There's still that muscle group as well as that method group. There's no change in configuration because it did not even participate in that reaction. So we get our second product. Now the third product is a bit more complex because we have to think about the remaining hydrogen. What if we perform bond rotation here to make this hydrogen be anti to iodine. So if we perform bond rotation, we will draw a new structure below. And that structure would essentially have the method group pointing down instead of up after we performed bond rotation. So that one of the hydrogen is that we haven't yet used or in particular this hydrogen would be anti to iodine. So let's draw that new structure. We may some place start by drawing the method group pointing down and then everything pretty much remains the same. There will be he must have group pointing away iodine pointing towards the viewer. But now regarding these high regions, one of them will be anti to iodine. So we may indicate that the hydrogen notice that in particular we are referring to this hydrogen because after bond rotation you will see it pointing away rather than towards the viewer since we're rotating the bond. And that is good for us because if we're using our strong base in this case the stock side, the negative charge would d protein data structure. We would form a double bond at that position and simultaneously we would have the loss of the leaving group. So this would be the formation of the next product or the final product because notice that this is indeed our third al king. So we are forming a double bond at this position. Then what is that? This position still has a method group also, let's not forget that there was a metal group at this position. We will draw it. Nothing changes here because again it doesn't participate in the in the elimination mechanism, so that will be our additional structure. We now have all of our three products. We may essentially label these three Elkins, and as you can see, the correct answer for this multiple choice question is C. However, if you have any questions for this problem, feel free to leave them down below. I hope to see in the next video. Thank you.

Pure (S)-2-bromo-2-fluorobutane reacts with methoxide ion in methanol to give a mixture of (S)-2-fluoro-2-methoxybutane and three fluoroalkenes. a. Use mechanisms to show which three fluoroalkenes are formed.

Key Concepts

Nucleophilic Substitution Mechanisms

Elimination Reactions

Stereochemistry and Product Configuration

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