For each of the following reactions, draw the major eliminatio...

Question

For each of the following reactions, draw the major elimination product; if the product can exist as stereoisomers, indicate which stereoisomer is obtained in greater yield.

a. (R)-2-bromohexane + high concentration of CH₃O⁻

b. (R)-3-bromo-3-methylhexane + CH₃OH

Accepted Answer

All right. Hello everyone. So this question is asking us to draw the major products formed in the following elimination reactions. If the products formed can exist as stereo isomers identify the stereo isomer that will be in, that will be higher in amount and label it as the major product. Now, for part one, we have R two chloro heptane and a high concentration of oxide. And for part two, we have R three chloro three methyl octane with ethanol. Now towards the bottom of the screen, I've gone ahead and drawn both starting materials. So recall first and foremost, that elimination reactions can proceed via an E one or an E two mechanism. A few different factors are considered when deciding which one applies to a specific reaction. First is the substrate. Recall that tertiary substrates tend to favor E one, whereas primary substrates tend to favor E two. And also we tend to consider the nucleo file E two reactions tend to favor or actually require strong N files generally with negative charges. Whereas E one reactions can take place with a weak Nile. So this leads us to the first reaction we have R two chloro hept which is a secondary ocho Halide reacting with an eth oxide ion which has a negative charge and is therefore considered a strong nucleic base. This means that reaction one is going to precede via an E two mechanism. Recall that E two mechanisms are concerted. So everything is happening in one single step, the negatively charged nucleo behaving as a base. Here is going to eliminate a proton from a carbon atom next to the carbon that contains the leaving group. Those positions are called beta carbons, which is why elimination is also referred to as beta elimination. Now, in addition to this E two requires anti coplanar stereo chemistry. So because chlorine here is on a wedge to depict the R configuration, the hydrogen atom that's being removed on the next carbon over must be on a dash that's that anti coplanar stereo chemistry. But I want to talk about why I went ahead and selected a proton from carbon three as opposed to carbon one because both beta carbons are non equivalent, they're not the same as each other. They're going to produce distinct keen products. We have the zeit product which is more substituted and we have the Hoffman product which is less substituted. Now, oxide, in addition to being strong is relatively small in size, its definitely not sternly hindered in the way that something like TP COCC side very much is. So because of this, the Zs of product is going to predominate So the pi bond of the final product would be in between carbons two and three. So now that we know that the sites of product is going to predominate here recall also that the Zs of product can exist as cis trans isomers. So our two products are actually going to be the cist trans isomers. So first, I drew the trans isomer where hydrogens here are on opposite sides and we also have the cis isomer. So now that we have the structures of both products for reaction, one recall, the trans isomers oops tend to be more energetically favorable than their cis counterparts. It's thermo dynamically more stable because the R groups are on opposite sides of the py bond which reduces steri hindrance. So even though we have this trans isomers for part one, it's trans hep tone that is the major or predominant product. So now let me go ahead and scroll down somewhat before we can talk about reaction to. Now reaction two features a tertiary o hide and our base or nucleo file is ethanol which notably is neutral. So it is quote unquote, a weak Nile or base. This means that an E one reaction is going to proceed. Now, unlike E two reactions, e one reactions are not stereo specific. This is because the first step is characterized by the formation of a Carbocaine. So to understand the products that form for reaction too, let's talk about the mechanism for illustrative purposes. The first step of an E one mechanism is characterized by the leaving group. In this case, chlorine detaching itself from carbon to create a carbo cion. So now we have a crumble cion on position three. And keep in mind that carbo C ions are cleaner, they are sp two hybridized. So that's why I'm not drawing any stereo chemistry on carbon three there. So the next step of the mechanism involves the nucleic attack of ethanol. And once again, our Nile or base would attack a beta hydrogen, a hydrogen on a carbon next to the carbo cion. Now recall that earlier, we had a conversation about the Zaitsev versus the Hoffman product. There's a distinction to make between a more or a less substituted, all key. But notice here how both beta carbons in the given molecule would produce equally substituted all kinds. So we have two possible positions for the pi bond and both of them can produce cis trans isomers. So, in reality, we end up with four distinct products for part two here. So let's go ahead and showcase those products. And here I'm going to go ahead and erase ethanol. So the first two products that I'm going to draw feature a double bond in between carbons, two and three of the parent chain. So first, we have an E I so and subsequently, we end up with a, the isomer. So here are two products with the double bond in between carbons two and three. So now let's go ahead and draw out our other two products with the double bond in between carbons three and four. Here's re isomer and then we have our Z isomer. So to go ahead and finish off this part of the video for the same reasons discussed earlier, the e isomers would predominate since they're more thermo dynamically stable due to reduced STAIC hindrance between our groups and there you have it. So now we have our products here or reactions one and two. And in the case of both compounds, the trans or E isomer is more favorable. So it's going to predominate. So with that being said, if you watch this video all the way through, thank you so very much. I do appreciate it and I hope you found this helpful.

For each of the following reactions, draw the major elimination product; if the product can exist as stereoisomers, indicate which stereoisomer is obtained in greater yield.
a. (R)-2-bromohexane + high concentration of CH₃O⁻
b. (R)-3-bromo-3-methylhexane + CH₃OH

Key Concepts

Elimination Reactions

Stereochemistry

Regioselectivity

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