Draw the elimination products for each of the following E2 rea...

Question

Draw the elimination products for each of the following E2 reactions; if the products can exist as stereoisomers, indicate which stereoisomers are obtained.

c. (2R,3S)-2-chloro-3-methylpentane + high concentration of CH₃O^-

Accepted Answer

Welcome back everyone to the major elimination product of the E two reaction between three R four, R three chloro four methyl hexane and sodium and metox side. If any stereo isomers are possible, specify which stereo isomers are produced. We're given the structure of the starting material. And we have to understand that we are treating it with sodium metox side. And because sodium oxide naoch three is an ionic compound, that means we are going to dissociate it. And for an E two reaction, we are only going to include our strong base which is met. Before we continue, we're going to redraw the structure with solid lines only. And soon I will explain why we're going to do this. Now, the reason why we're doing this is just because we want to evaluate our beta hydrogens for an elimination and we have to determine the nature of the major product, whether it's of product or Hoffman product. So we have determined our alpha carbon which is directly bonded to chlorine. Now the neighboring carbons will have beta hydrogens. So we have one neighboring carbon and then we have another. So we have one beta hygiene that can participate in an elimination, let's label it in blue and then we have two more, but we are only interested in one of them because they are equivalent. So we have to carefully think about the nature of metox side. It is a small base, which is not hysterically hindered, meaning we favor the product as the major product, which is a more substituted Alcaine. So what if we go ahead and remove the proton given in blue? In that case, we will get the following product. We can simply redraw our carbon chain. So that'd be 123456 carbon atoms. Now, we're not going to show the implicit hydrogen atoms, it's just sufficient to draw the chloral group or actually because we eliminate that right, we eliminate hydrogen fall by the loss of the chloral group. So we're going to draw a double bond and then we understand that we have the metal group remaining. So now the question is how substituted is this king? And the answer is try substituted because we have three alki substi bonded to the, the sp two carbon atoms. And now if we remove the green one, we remove the proton form a double bond fall by the loss of the leaving group. So we are drawing a double bond than our remaining carbons. And then we are adding the muscle group and the alkin produced in this case is di substituted, right? Because we have two alk substituent bonded to our sp two carbons, meaning we're going to favor the blue pathway well done because it's that pathway forms a more stable try substituted ALCA. But we also have to be careful and recall the answer co planner requirement. Basically, that means the hydrogen atom and the leaving group must be in anti arrangement. And that's exactly what we're seeing. We have hydrogen pointing up, then we have our carbon carbon bond and then we have our chloral group pointing down. So they are anti to each other because hygiene is pointing up and the chloral group is pointing down. So that means not only we are able to form the states of product, but also we have the requirement fulfilled in terms of the spatial arrangement such that the proton and the living group are in an arrangement. Meaning we are ready to perform the reaction, the metox side, I will attack the proton. We're going to break our carbon hydrogen bond and form a pi bond out of the carbon hyen bonds. Electrons followed by the loss of the leaving group, which is the chloride and ion. So now we are going to redraw our six member chain. We are going to add the group because we're forming a pie bond, we're going to add that pie bond. And we are done, we got our product, we simply want to determine if it is an E isomer or a Z isomer. And to do that, a good idea is to add an implicit hygiene and consider every carbon step by step. So let's see how that works. So, first of all, let's leave our first carbon in black that is SB two hybridized, it has two substitutions and L substituent, right. So there's a carbon atom bonded to an SB two carbon. And then we have hydrogen hygiene gets a lower priority because it's lower in atomic number than carbon. So the group, we'll get priority one in E or C assignment well done. And similarly, if we look at the second carbon in blue, we have a methyl group versus an ethyl group because the two atoms directly bonded are carbon atoms, we just want to identify a longer chain. Recall that according to the Khan Ingle pre-law rules, this way of determining which alki group dominates only works when we have linear chains because both the metal group and the L group are linear, the longer one will get a higher priority. And finally, we understand that priority one on carbon, on the first double bonded carbon and priority one on the second double bonded carbon atom, they are on opposite sides of the double bond, meaning this would be an E Eimer because Z means that priorities number one are on the same side of the double bond. But in this case, that's e well done. So we have determined our final product and we know that its configuration is E. Thank you. For watching and I hope to see you in the next video.

Draw the elimination products for each of the following E2 reactions; if the products can exist as stereoisomers, indicate which stereoisomers are obtained.
c. (2R,3S)-2-chloro-3-methylpentane + high concentration of CH₃O^-

Key Concepts

E2 Mechanism

Stereochemistry

Stereoisomers

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