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

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.

b. Propose a mechanism to show how (S)-2-bromo-2-fluorobutane reacts to give (S)-2-fluoro-2-methoxybutane. Has this reaction gone with retention or inversion of configuration?

Accepted Answer

welcome back everyone and welcome to another video. The problem states when pure three R four R three idol 34 dimethyl vaccine is heated in ethanol, it produces a mixture of two ethers and four Elkins draw a mechanism that accounts for the formation of the two eaters, discuss the retention or inversion of configuration taking place in this reaction. Now we begin this problem by drawing the original structure and we noticed that this is hexane. So we need to draw a six member chain. We understand that positions three and four have two metal substitutions. So if we count the carbon atoms, this is position number three, which has a methyl group and there's carbon number four that has a metal group. In addition position number three has iodine bonds. Is we are going to draw another bond and we are going to indicate the presence of iodine over there. We also noticed that positions three and four have configurations and therefore there carl centers, we may indicate them. These are our carol centers. So it makes sense to draw an additional implicit hygiene and disposition To clearly see the presence of four different substations at these two Carl Carbon atoms. Now let's recall the rules of con Ingle pre log and we will assign priorities to our carbon atoms. Let's start with this Cairo center, position number three, we understand that we have four different substations. One of them is iodine and then all the other ones are carbon atoms. Specifically here we have a metal group, an ethyl group and here we have another carbon atom bonded Comparing the atomic numbers. We understand that iodine has the greatest atomic number. So it gets priority number one and then we will compare the remaining carbon atoms in particular. This is the shortest elk. He'll change. So it would have the lowest priority priority number four. And we have to compare the remaining two chains which are longer than a method group. So we are comparing an ethyl substation. So we are going to separately draw a carbon atom and we will draw the atoms monitor directly. So we have two hydrogen atoms, one hygiene, another hygiene and then we have a metal group. But it's just sufficient to show that there's another carbon bonded to this specific carbon. If we look at the right substitution, there's a carbon atom that has two carbons bonded to it. So we are going to show two carbon atoms and then one hydrogen atom. Notice that if we eliminate equivalent parts, we can eliminate these hydrogen, these carbon atoms. And then we have a competition between hydrogen and carbon and since carbon is greater and atomic number, the substation on the right gets a higher priority so that we are priority to for by priority number three for the ethyl group. And we can say that we may now assign priority number two to the substation and priority number three to the whole group. Now, if we think about the con angle pre log rules, let's just automatically assume that priority for is pointing away from us. If we are wrong, we will just modify our choice. But for now let's just suppose that it's pointing away from us because if priority for is pointing away from us, we may trace our path directly. Now if we go along positions 12 and three, we noticed that the path is going clockwise so that's indeed our rotation. We have successfully identified the configuration here. That's our which works for us because we have three hour meaning. Since priority force pointing away priority number one will be pointing towards the viewer. Therefore iodine is on a wedge And similarly we will redraw the structure to perform similar steps to identify the correct configuration for position number four we are going to redraw our structure. We already know that at position number three, we have iodine on a wedge and now we are only interested in position number four. We have hydrogen, we have a muscle group because hydrogen has the lowest possible atomic number, it automatically gets priority number four. Again, the metal group would get priority number three because it's the shorter chain. I noticed that the carl carbon atoms specifically here has three neighboring carbon atoms and the shortest chain would get the lowest priority. So that would be priority number three. And then we have to determine which one would be priority one and which would be priority number two. We may use similar steps here. We will just see what kind of carbon atoms we have. So in particular this carbon atom is a carbon atom with two hydrogen atoms. And because there's a method group over here, we can just say there's another carbon atom bonded to it. On the other hand, on the left side we have a carbon atom that is wanted to iodine, there's another hydrogen atom here and then it's one, it's another carbon. If we eliminate similar parts or equivalent parts to be precise, we get rid of these carbon hydrogen atoms, these carbon carbon atoms and then we have a competition between iodine and hydrogen. Apparently iodine is greater and atomic number. So this would be the highest priority. We can state that priority. Number one goes to the substation and priority number two goes to the ethyl group. Now, according to the con Ingle pre log rules, we wish to have priority for pointing away from the viewer. So you'll just automatically assign it on a dash and we will check if that works for us to trace our path in our direction. If we trace our path from 1 to 2 to three. We noticed that indeed this is a clockwise rotation, which is our and that's perfect for us because according to the problem we have for our for our configuration and therefore position number three would have a wedge for the method group. Great job we've done a lot here. So let's move on and let's redraw our structure correctly with all the correct configurations. Starting with hexane. We understand that Position. # three has iodine on a wedge And position # four has a muscle group on a watch. Also my apologies. Don't forget that here we had a method group pointing away so we will add that and essentially this is our substrate. As a problem suggests we are forming two different eaters. We have to analyze given substrate. We call that according to the problem. We are using ethanol for this problem is our new profile or base as well as heat. Now, first of all, let's analyze our substrate, specifically the leaving group or iodine is bounded to a tertiary carbon atom. So it's a tertiary substrate. And now the nuclear file at the base is a weak nuclear file because we're dealing with a territory substrate and a weak nuclear file. We have a competition between SN one and E. One reactions specifically because we are forming two ethers. We are interested in S. N. One reaction in this problem and E one reaction essentially with produce an elimination products where these Elkins mentioned. But here we will be analyzing the mechanism of an SN one reaction the world. First step in an S. N. One reaction is the loss of the leaving group. So we will indicate that we are essentially we may essentially erase that for the purpose of drawing the mechanism because we already understand. We're using heat and ethanol and the word first step that we have is the loss of the leaving group. As we just said, this iodine would just leave. We may indicate that this is the loss of the leaving group during the loss of the leaving group. We are forming a carbo cat iron. So we may ritual what we have. There's no change in the configuration of the center because it's not participating in the reaction. At least yet. We have the loss of iodine. We still have a metal group there and we are forming a stable tertiary carbo cast iron. So that's perfect for us. And I recall that we are using ethanol as our regent. So we made sure the molecule of ethanol, the lone pair on oxygen would act as a nuclear file in this case. And we would have the nuclear attack stuff on the carbo. Karen notice that there will be a systemic mixture form because this is this is an sn one reaction and we will be forming a Cairo center. So we will have a pair of finance humors. Let's not worry about it yet. We may just say that this is the nuclear attack stop now we will form a protein ated intermediate. We made through that protein ated intermediate below. So let's redraw our structure. There's this method group and there's protein ated oxygen. So oxygen would have a formal positive charge because it's also bonded to an ethyl group. So we have our previous model group at this position. Now we have our protein ated intermediate and we may essentially d protein ate it using iodine or another molecule of ethanol. So let's suppose that for the purpose of this mechanism we release under molecule of ethanol, he may indicate another molecule of ethanol. And we have our deep rotation step in the deportation step, the lone pair attacks hydrogen and simultaneously these bonding electrons will be placed in the form of a lone pair and oxygen. And therefore we will form two different products will form a rose emmick mixture because this is a Cairo center as discussed before and we will be forming 50% of our and 50% of us in anti immerse. In other words, we may attack that carbo Karen from each side and therefore, let's draw two products, one of them will have the muscle group pointing up, so we may draw it on a wedge. And then the easter form would be the easter group would be pointing away from us. So that would be oh with an ethos substitution. So that's one of our products, let's not forget the remaining stereo center and the other product would essentially have an opposite configuration. So we made sure that as well now the muscle group would be pointing away And then the remaining one would be pointing up. Let's not forget the remaining center. We may now state that these two Ethers are our final products in this mechanism. In a regarding configuration, let's just think about it. We call that this stereo center did not even participate in the reaction. So the configuration is retained now at the remaining stereo center, the configuration is not retained simply because we are forming two different configurations. We were forming a systemic mixture, so there is only one center at which configuration was retained. The correct answer for this multiple choice question is B and if you have any questions, feel free to leave them down below. Thank you for watching this video.

Key Concepts

Nucleophilic Substitution Mechanism

Stereochemistry and Configuration

Leaving Groups and Reactivity

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