Draw the substitution and elimination products for the followi...

Question

Draw the substitution and elimination products for the following reactions, showing the configuration of each product:

c. 1-chloro-1-methylcyclohexane + CH₃O⁻

d. 1-chloro-1-methylcyclohexane + CH₃OH

Accepted Answer

All right. Hi, everyone. So this question is asking us to determine the product for the given reactions. Note show the correct configuration of each product. For part one, we have one bromo, one methyl cyclops reacting with oxide. And for part two, we have one bromo, one methyl cyclops reacting with ethanol. So first, let's go ahead and draw out our starting material for both parts. One and two, we have a sac pentane ring. So thats a five carbon ring just fix how it looks for a second with a bromine atom and a methyl substituent on carbon one. Now, to understand the product that each part, each reaction is going to provide, we have to distinguish between an E one E two sn one or SN two. And we can do that using our lovely flowchart. Now recall that elimination reactions that's E one and E two produce an all keen as your final product. On the other hand, sen one and Essen two reactions that represent nucleic substitution don't produce alkenes, but the nucleo replaces the leaving group with itself on the organic starting material. Now, the uni molecular reactions E one and SN one are characterized by the formation of a carbo cion. First. This is after the departure of the leaving group, the bi molecular reactions that's E two and SN two are concerted mechanisms where the intended product is created in one single step. So now let's go through our lovely flow chart and ask ourselves questions as we go along. So let's do that for part one, for part one, we have one broma one methyl cyclops reacting with a oxide just for the record. There is a Typo in the question because it should be CH three C H2O negative. But anyway, let's go ahead and go through our flow chart. The first question in our lovely flow chart is asking about the strength of the nucleo file. So for part one, the nucleo file in question being a foxy is in fact strong due to the negative charge on oxygen from there, we move to the right side. Excuse me, the left side of the flow chart towards question two, which is asking about whether or not the strong nucleo file is bulky or relatively sternly hindered. Examples of bulky bases include turp bet to side and LD. But because oxide is relatively small, the answer to this question is going to be no. So we can proceed on to part three or question three. Now, question three is asking about the type of leaving group. So let's go ahead and re examine the structure of one bromo one methyl cycling, our leaving group is bromine and bromine is attached to a tertiary carbon. So that means that our leaving group is tertiary. So question four is about whether or not a oxide is more of a nucleo or a base. So for the sake of space, I just went ahead and rearranged the order in which I wrote down these questions. But the fourth question discussing whether or not it's a nucleo file or a base is heavily dependent on what you start with. Recall, that aides hydride, NH two negative or oxides tend to be more effective as bases than nucleo files. So because a oxide is an example of an all oxide, then that's going to be more of a base than a Nile. So after all, this is being considered, then oxide for part one here is going to participate in an E two reaction because a Foxy is a more effective base than it is a nucleo file. It would rather go ahead and detonate the adjacent carbon to go ahead and create a pi bond between carbons one and two. So to predict the product of an E two mechanism, we're going to remove bromine from the carbon chain and create the most stable alkene with an adjacent carbon that has a hydrogen atom. So the most stable alkene would be inside of the ring because it would be more substituted. Keep in mind, however, that due to the symmetry of the ring, you would end up with the same product regardless of where you choose to draw the double wand. So our product for part one ends up being one methyl cyclops one E. So now lets go ahead and consider part two. So for part two, I'm going to scroll down somewhat and I am going to redraw our starting material again. But this time for part two, our nuclei is ethanol was CH three CH two. Oh So going back to our flow chart and once again, beginning at the very top here, question one is asking about the Nile this time. However, for part two, our nucleo ethanol is weak because it has a neutral charge that moves us towards the right side of the flow chart were question two is asking about the leaving group because our starting material is the same, our leaving group is once again tertiary. This leads us to the conclusion that part two is going to end up with a mixture of S and one and E one products. So not only would we end up with an alkene similar to the E two reaction in part one here, but we're also going to see an sn one reaction take place to a certain extent. So in that case, the oxy group of ethanol can behave as a nucleo and attach itself to the tertiary carbon where bromine initially was. So first I draw out my carbon chain and then install the oxy group on position one. And also I would get an all keen inside of the cyclops ring in the competing elimination reaction. Now, keep in mind here that the SN one mechanism would normally result in a reca mixture of stereo isomers. But due to the lack of stereo centers in this molecule, that semic mixture is actually going to produce the same product. So that's why I, I did not indicate any stereo chemistry, but there you have it. So with that being said, I thank you so very much for watching and I hope you found this helpful.

Draw the substitution and elimination products for the following reactions, showing the configuration of each product:
c. 1-chloro-1-methylcyclohexane + CH₃O⁻
d. 1-chloro-1-methylcyclohexane + CH₃OH

Key Concepts

Nucleophilic Substitution Reactions

Elimination Reactions

Stereochemistry

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