For each of the following compounds, draw the product that for...

Question

For each of the following compounds, draw the product that forms in an E2 reaction and indicate its configuration:

b. (1S,2R)-1-bromo-1,2-diphenylpropane

Accepted Answer

Hey, everyone. And welcome back, determine the elimination product or products of the E two reaction between one R two S one Chloro 12 diphenyl butane and sodium oxide, assign an appropriate configuration to the product or products formed. OK. So let's not worry about one R two S, let's just draw butane. So butane would be four member chain. OK? And at position number one, we have our Chloro substituent at positions one and two, we have two fey substituent. So those would be our benzene rings. We're going to use an abbreviation for now. Wonderful. So let's label carbon number one and carbon number two to assign the appropriate configurations one R two S. We're going to start with carbon number one and we're going to think about the substitutions bonded to it. So we have our Chloro substituent, we have an implicit hydrogen atom, we have our fennel substituent and then we have our long group. So that would be C H, right, which is then bonded to another carbon. So you can just say that later on it's bonded to another carbon and it's also bonded to the fennel substituent. So let's assign priorities chlorine gets the highest atomic number, priority number one, hydrogen has the lowest atomic number, priority number four. And now we have to consider our p substitute against our kiel group. So if we carefully consider phenyl substituent, we can show it as carbon being double bonded to another carbon in being single bonded to the other carbon, which essentially is not really equal to. But according to the Khan Engel prelaw rules, we can say that those are three carbon carbon, single bonds. Now, for the other one, we have c bonded to H bonded to el which essentially corresponds to carbon atom to begin with and then another carbon. OK. And if we exclude equivalent atoms C C C C C H versus C C, that means the final substituent gets a higher priority. So we can state that priority number two goes to the final ring. Priority. Number three goes to our long longer al group. OK. And now assuming that hydrogen is pointing down, if we assign our rotation, we get clockwise rotation, which is r exactly what we need, right? So what we want to do here, we want to make sure that our groups are shown. OK. So we are going to show them, we're going to try to assign hydrogen pointing down. And we understand that we have priority one, priority 23 and four. So 123, right? That are meaning we assign our configuration properly according to the pre rules. Priority number four must be pointing down to assign a rotation directly well done. And now we are going to consider our next position. So the other carbon atom car number two, it basically has several substitutions L group, then it has an L group. So C H two C H three on the left side, it has carbon chlorine bond, right. Uh we are interested in chlorine because it has the greatest atomic number. And then we also noticed that there is an implicit hydrogen. Once again, we assign our priorities, we have three carbon atoms. So that means hydrogen gas, priority four. It's the lowest an atomic number. Now, for the remaining ones, Fey group, that's three cc bonds versus C bonded to C L, right? So if we think about the other substi that's carbon bonded to chlorine bonded to hydrogen bonded to another carbon from the fenny group. Now, if we exclude similarities C C, we have hydrogen versus carbon, but chlorine is higher in atomic number than carbon, right? So we are going to say that our priority number one goes to C C L followed by our fennel ring because we said it has three carbon carbon, single bonds according to the rules. And then we have our carbon bonded to one carbon only and two hydrogens. So that'd be priority number three. And if we assign a rotation directly, that would be R and we want to get s so that means high should actually be pointing up so that we can reverse it. Meaning if we go back to our structure, let's suppose that we assign hydrogen on a wedge, then fennel group pointing up my apologies, then fennel group pointing down, we will once again assign our priorities. Or now we have number one for the group, for the final group, that's priority number two. And then priority number three or the ethyl substituent. And if we trace our path 123, that's R but because hydrogen is pointing up instead of down, that's actually right, because priority number four is pointing up well done. So we understand how the structure looks like we are now ready to actually redraw it in a more readable form and think about the reaction itself. So if we redraw it carefully, we are going to add our fennel substitute hyen pointing down glory and pointing up at the adjacent position. Our final group is winding down hygiene pointing up and that's it. Now we are using sodium oxide, sodium is an ion spectator. So it's sufficient to think about oxide itself. The problem says that we have an E two reaction that makes sense, right? We are using a strong base and because it's a small base, we're going to find our major product as the of product because the base is not hysterically hindered. So we identify our alpha carbon which has the leaving group. And we want to think about the hyen atoms that are beta hydrogens. So the beta carbons, the right has hydrogen. On the left, we have our phenyl group that, that that does not have beta hydrogen with respect to our alpha carbon. But the problem is that our chloral group and our hydrogen they are in arrangement. And we need to recall the N C co planner requirement. So we want to make sure that Hodgen is pointing down. If chlorine is pointing up, do that, we are going to perform the bond rotation. So if we do that, we are essentially going to rotate it and we can just redraw what we get. So we are going to draw our parent right? And now nothing really changes at the very beginning. We are just going to add hydrogen pointing down chlorine pointing up because we are performing a rotation. We want to make sure that hyen is pointing down. So let's add Hyen pointing down. Meaning now, if we are rotating it, the metal group will be pointing up and the fennel group will be in plain on a solid line. So we are just going to shorten our chain because now it's the fennel group that is in playing and the ethel group is now pointing up. OK. Well done. So now we have our anti co planner requirement fulfilled and we are, we can just move the original structure and think about the actual one where we have an appropriate arrangement between the beta hygiene and the leaving group. So, oxide comes in, we are removing hydrogen forming our pie bond and we get rid of the leaving group. So as we notice, there's only one product formed and we are ready to try it. So we have our final group, then we have our chain with a double bond formed another fennel group. And now we end up with an ethyl group right in plain, because we have an S B two carbon and an implicit hygiene atom. So now we simply can redraw it, right? If we simply expand our final groups, we can just add our six member rings. OK? One of them and the other and everyone just signed a configuration. Let's remember that Al Keynes, we, we have to analyze the priorities, right? So for the first double bonded carbon atom, we have two substitutions. One of them is carbon, the other one is our implicit hyen. So carbon gets priority number one versus hyen because it has the greater atomic number. And then we are comparing the other two substitutions. As we said, fenny can be considered as C C C, right? Specifically that carbon has a double bond and a single bond versus the ethyl substitute, which has carbon carbon bond and two hydrogens. So if we exclude similarities, we have C C versus C H, right? In each case, that means once again, the final group gets a higher priority. And because priorities number one are on opposite sides of the double bond. That's an E eamer. So we are just going to neatly redraw it without any information that we do not want to have in the final structure and state that we have formed an E E or that king. Thank you for watching and I hope to see you in the next video.

For each of the following compounds, draw the product that forms in an E2 reaction and indicate its configuration:
b. (1S,2R)-1-bromo-1,2-diphenylpropane

Key Concepts

E2 Reaction Mechanism

Stereochemistry and Configuration

Substituent Effects on Elimination Reactions

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