When 2-chloro-2-methylbutane is treated with a variety of stro...

Question

When 2-chloro-2-methylbutane is treated with a variety of strong bases, the products always seem to contain two isomers (A and B) of formula C₅H₁₀. When sodium hydroxide is used as the base, isomer A predominates. When potassium tert-butoxide is used as the base, isomer B predominates. The ¹H and ¹³C NMR spectra of A and B are given below.

(a) Determine the structures of isomers A and B.

(b) Explain why A is the major product when using sodium hydroxide as the base and why B is the major product when using potassium tert-butoxide as the base.

Accepted Answer

All right. Hi everyone. So for this question it says that when two chlor oh 23 dimethyl butane is treated with different strong bases, then elimination causes the formation of two is A and B. With the formula C six H 12. When sodium hydroxide is used as a base, then the major product is A. And when turbo oxide is used as a base, then the barbie is formed in a higher quantity. So now let's begin by determining the structures of the two ice members using the NMR spectrum given below. So let's get started by first drawing out our starting material. Right? So our starting material is two chloral 23 dimethyl butane, which is gonna be a four carbon parent chain With methyl groups attached to positions two and 3. And we're also going to have chlorine on position two. So let's begin and recall for a second that the question is explicitly telling us that elimination is going to occur. So with an elimination reaction, we're talking about um the departure of a leaving group after a beta hydrogen has been removed. So what does that mean? Right, our chlorine here on position two is going to be our leaving group because it's it's stable by itself. So because carbon too has chlorine attached to it, which is our leaving group, it's considered the alpha carbon and the beta carbons therefore are those ones that are directly attached to the alpha carbon. So in this case it's going to be carbons one and three. You can consider the methyl group as well attached to carbon too, but it's equivalent to carbon one. So really they're the same. So now um beta hydrogen is therefore are going to be attached to beta carbons. And if you look here positions one and 3 each have a hydrogen that can be removed during an elimination. So now we have to consider the different possibilities, right? We have to consider the products for both eliminations and that's gonna be our customers A and B. So for S. M. R. A. If we look at the NMR spectrum, we notice that there's only one peak corresponding to 12 hydrogen. So that's kind of a big number, right? But if we go ahead and we divide our 12 hydrogen By three, then that's going to correspond to four methyl groups. That's gonna correspond to four methyl groups associated to our structure. So that must mean that we have a symmetrical all keen. Right. And we're going to achieve that if we go ahead and D protein eight, Um the beta hydrogen on position three. So, I summer A is going to be This symmetrical walkin with a double bond in between Carbons two and 3. So this is A And here we have our four method groups which is gonna add up to 12 hydrogen. So now let's talk about ice and maybe And S. M. R. B. must therefore be the result of removing a beta hydrogen from carbon one. So we know that it must look something like this, Right? We're going to have the double bond now in between Carbons one and 2. So this is going to be I Somerby and we can justify it with our NMR spectrum. Right? So first of all this peak here on the right side of the spectrum corresponding to six hydrogen, if we go ahead and we divide six hydrogen by three, then we know that we have two methyl groups two MG for short because I'm running out of space here, we have two methyl groups um that are equivalent to each other And that's exactly the case here with the two methyl groups that we see attached to carbon three. So that's one peak and our next peak here which is slightly more downfield or more to the left side of the spectrum than our six H. That must correspond to one additional methyl group. So we have one additional method group here attached to carbon to and that's going to be more downfield because recall that it's in the vin Illich position, right? It's directly attached to an all keen, so it's gonna be more D shielded and therefore more downfield and then Our one H here, it's a multiplayer. And if we look at our hydrogen here on position three on S. M. R. B, then we noticed that it's going to be attached to our two methyl groups. Right? So it's going to appear as a multiplayer and there's only one proton associated with it. So now we've justified our structures and now we can move on to part B. So part B is asking us to explain why I summer A predominates what sodium hydroxide is used while I simmer B predominates when ter beat oxide is used. So before I begin, let's recall for a second that when we're talking about elimination reactions, we have what's called the sights and the Hoffman rules. So if we go ahead and we look at our starting material Right are two chloral to three dimensional butane. We have two possibilities right? We have A and B. So recall that the decisive product is going to be the more substituted. All keen. It's going to be the more substituted. All keen. Which in this case is going to be A. Because it's tetra substituted meaning we have four R groups associated to it, whereas barbie is di substituted, there are only two R groups attached to it. So if the sites of product is the more substituted all keen than the Hoffmann product is going to be the less substituted. Okay, so we have our sights in our Hoffman products and so the major product for a given elimination reaction actually depends on the base that we use so relatively small. Unhindered bases if you prefer the formation of the sites of product. So if we have any O. H. For example any O. H. Is small. Right? It's unhindered. So that means that it's going to favor the formation of the sites of product which in this case is a. Meanwhile on the other hand relatively large hysterically hindered bases are going to favor the formation of the Hoffmann product. So if we consider the fact that we're comparing sodium hydroxide to turk detox side. So T. EBU oh minus this is a very large L. Kill group. Right? It's a turd bugle group. It's very big and therefore hysterically hindered. So that means if we look at our starting material here, if it were to go ahead and attack or rather remove the beta hydrogen on carbon three here it would face significant starik hindrance with the two methyl groups associated to carbon three. But if we look at carbon one then it's free game right? Because um it's relatively unhindered. It's at the very end of the parent chain. So terp detoxified can very easily go ahead and D. Protein eight carbon one as opposed to carbon three. And therefore the Hoffmann product is going to be favored when we have hysterically hindered um relatively large base being used in our Hoffmann product like we discussed before is assumed to be. So with that being said, we went ahead and We cracked the code right? We solved for R. two I. Summers. And we were able to successfully explain why I said where A. Is favored when you add any O. H. And why I. Somerby is favored when you add to it a toxic. So with that being said, if you made it to the end, thank you very much for watching, and I hope you found this helpful.

Key Concepts

Elimination Reactions

Zaitsev's Rule

NMR Spectroscopy

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