Predict the product(s) of the following substitution or elimin...

Question

Predict the product(s) of the following substitution or elimination reactions, paying close attention to the stereochemical outcome of the reactions.

(h)

Accepted Answer

All right. Hello everyone. So this question says that the given reaction could undergo substitution or elimination. Here we have two bromo, 234 Tropane reacting with sodium amide in the presence of D MS O. What is its product note? Pay attention to stereo chemistry. Now recall first and foremost that there are a few different reactions possible when we discuss substitution or elimination. Because for substitution, we have either SN one or SN two. On the other hand, for elimination, we have E one and E two. Now recall first and foremost that the uni molecular reactions discussed here, that's SN one and E one, our book characterized by the formation of a carbo cion, even though their end product is of course different in an sn one mechanism. A carbo cion is formed which the Nile then proceeds to attack in an E one mechanism, an alkene is produced after a carbo on intermediate forms. Now, on the other hand, the bi molecular reactions that's SN two and E two are concerted mechanisms. In other words, everything is happening in one single step in an E two reaction, the base removes the hydrogen that is anti complainer to the leaving group as it leaves. So everything is happening at the same time. But in an sn two reaction, the Nile is attacking the carbon that's connecting to our leaving group at the same time that the leaving group is departing, right? It's being removed. Now, to differentiate between each of these or reactions, we have a very handy flow chart to help us decide right. Our first question that we have to ask ourselves is the strength of the Nile or base. If you're talking about elimination, our nuclei or base is sodium amide. Now sodium is a cion when it dissociates in water. And NH two here is in fact negative, it has a negative charge. So this means that our nuclei or our base is strong because of this negative charge. This automatically means that SN two and E two are favored because SN two and E two require stronger, negatively charged nucleo files or bases. SN one and E one prefer a weak or a neutral nuclei or base. So after answering the first question in our very handy flow chart, the second question is whether or not the base or the nucleo file is bulky or large in size. Now, examples of bulky bases include turp oxide and LDA. And in this case, NH two negative is neither of those, right? And it's actually relatively small. So the answer to the second question is no, right? Because our base here is not that large. So this leaves us to our third question. What type of living group are we dealing with in this case? So to enter the to answer this question, we have to analyze the carbon that's connecting to our leaving group, which in this case is bromine, right? In this case, we have an o hide with bromine as our leaving group. And that atom of bromine is connecting to a tertiary carbon. So our answer for the type of lead group is tertiary. However, this is where it can get a little bit tricky, right? Because recall the tertiary leaving groups are not favorable for S and two reactions because of the ster hindrance that they pose on the nuclei when it approaches. So earlier, I did mention that a strong nucleo file or base is going to favor SN two or E two. But in this case, that requires a little bit more nuanced because of the fact that we have a tertiary leaving group. So this leaves us to our fourth question is sodium amide more of a Nile or more of a base. In this case, recall that it just so happens that sodium amide is more basic than it is nucleic. So because it is better suited as a base as opposed to a nucleo file, this means that E two is going to be our primary reaction mechanism according to our incredible flow chart. So now that that's established, we are now aware that this is going to be an elimination reaction where an alkene is produced. Now, the double bond is going to form in between the alpha carbon, which is the carbon that contains our leaving group here and a beta carbon or a carbon directly adjacent to the alpha position. In this case, we have to be a little bit more careful because this particular tertiary carbon happens to have two non equivalent beta positions. This means that two elimination products are possible. But the question is which of those two is more favorable? Well, let's go ahead and draw them first to go ahead and draw both of my products. I'm going to first go ahead and draw out my carbon skeleton and I'm going to do so twice. So for each product that I've drawn here, I'm going to go ahead and draw my double bond in the two different positions that are possible. The first possibility is an alkene within the carbon chain. In other words, in between carbons two and three. And my second possibility is in between carbons one and two. Notice how bromine is notably missing because it has since left after beta elimination or the removal of a beta hydrogen. Now, here we have two different alkenes, one of which is more substituted than the other, right, the double bond in between carbons two and three of the chain is more substituted than the one in between carbons one and two. This means that the alkene in between carbons two and three is the sides of product. Whereas the double bond in between carbons one and two is the Hoffman product. Because sodium amide is an a base that is not large in size, the zits of product is going to be favored. Which means that our final answer is going to be our major product, which is 234 triethyl P two E that is our Zaitsev product and therefore our major product in this case and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Predict the product(s) of the following substitution or elimination reactions, paying close attention to the stereochemical outcome of the reactions.
(h)

Key Concepts

Nucleophilic Substitution Reactions

Elimination Reactions

Stereochemistry

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