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Organic chemistry
Course: Organic chemistry > Unit 5
Lesson 6: E1 and E2 reactions- E1 mechanism: kinetics and substrate
- E1 elimination: regioselectivity
- E1 mechanism: stereoselectivity
- E1 mechanism: carbocations and rearrangements
- E2 mechanism: kinetics and substrate
- E2 mechanism: regioselectivity
- E2 elimination: Stereoselectivity
- E2 elimination: Stereospecificity
- E2 elimination: Substituted cyclohexanes
- Regioselectivity, stereoselectivity, and stereospecificity
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Regioselectivity, stereoselectivity, and stereospecificity
Reviewing the difference between regioselectivity, stereoselectivity, and stereospecificity in elimination reactions.
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- What is the difference between a cis alkene and a Z alkene?(2 votes)
- "cis" indicates that there are two identical groups on the same side of something (in the case of an alkene that something is a double bond).
"Z" indicates that the highest priority groups§ are on the same side of something.
§note: priority is determined according to the Cahn, Ingold, and Prelog system, which more or less says the most massive atom has the highest priority.
There is lots of information on this here on Khan Academy, but you might also find this helpful:
https://www.masterorganicchemistry.com/2016/11/03/alkene-nomenclature-cis-and-trans-and-e-and-z/(10 votes)
- At, I can see how the trans product is formed, but can't quite wrap my head around what the mechanism is in order to get the cis product. How is the cis product formed when what I assume is the only beta carbon is to the right of the alpha carbon? 5:20(6 votes)
- I think it is about the repulsion.
In the Beta carbon, there are two hydrogens. Let's place it on the plane, one is above and other below(just like a 2D image)
If the above Hydrogen is being protonated, double bond will be above( of the two products, in the trans double bond is above). This means, there is a hydrogen below and this can cause possible repulsion with the end carbon making the end carbon to take a geometry pointing upwards.
In the cis, the hydrogen which is protonated is the one below( Double bond is seen below). This means the remaining hydrogen will be present above. The end carbon needs to occupy a position in space to stay away from the positive hydrogen which is above. So the end carbon accepted a cis geometry in which it points to down, away from un-protonated hydrogen.
I think it's all about proper distribution of charges. Positive hydrogen and end carbon needs to be properly arranged to avoid the steric hindrance.(0 votes)
- For stereospecificity example, it would have helped if you showed the other hydrogen on the bromine carbon. Basically for anyone who didn't understand that one. In the first substrate, the hydrogen was going back and methyl was coming out, so its trans. In the second, both the methyl and hydrogen are going back so the product is cis.(3 votes)
- After, you say that we know what the product must be, but I am not sure how you came to that conclusion. How does the stereochemistry of the substrate determine the stereochemistry of the product? Is it based on sterics? 3:48(2 votes)
- It is being explained in the previous videos under the same topic
https://www.khanacademy.org/science/organic-chemistry/substitution-elimination-reactions/e1-e2-tutorial/v/e2-elimination-stereospecificity
Over here, he had already figured out the products .
You should watch it to get an idea of it(1 vote)
- how do you determine which is the major product and the minor product?(1 vote)
- Stable one will be the major product.
In regioselectivity, the more substituted product was the major product. Because of more the substitution, more the stability.
In stereoselectivity, trans was the major product. Because trans arrangement kept the bulky groups far away.
In stereospecificity, E alkene kept the bulky phenol far away and hence more stable.
You should look at the possible products and find the major product by looking at its arrangement. The arrangement which demands stable arrangement will be the major product.(3 votes)
- R there any specific types where the selectivity and will tend to or does it depend on the reaction(1 vote)
- Cis*Cis=Z
Trans*Trans=E
Cis is the same side group, while Z is a more complex and the same side group.
Trans is a different side group, while E is a more complex and different group.(1 vote)- And your question is?(1 vote)
- At, in the video he says they are regio-isomsers. The reason given by him was the different placement of the double bond. But then how is this different from position isomers? 0:35(1 vote)
Video transcript
- [Instructor] Sometimes,
definitions can be confusing, and I wanted to go through the
difference between the terms regioselectivity, stereoselectivity,
and stereospecificity. And I'm going to use examples that we've talked about in earlier videos, so if you want to know the
details of these reactions, go back and watch those earlier videos. We'll start with regioselectivity. So the reaction I've shown here is a regioselective reaction. This alcohol gets dehydrated
to form two products, the alkene on the left and
the alkene on the right. These two alkenes are regioisomers. So let me write this down here. So they are regioisomers. They're isomers of each other, but they differ in terms of the region or the location of the double bond. The isomer on the left
has the double bond here, and the isomer on the right
has the double bond here. This is a regioselective reaction. One regioisomer is favored over the other, and in this case, the
tri-substituted alkene, the one on the left here,
is the major product, whereas the di-substituted alkene, the one on the right,
is the minor product. So that's regioselectivity. Let's compare that to stereoselectivity. So for this next example, this alcohol, it's another
dehydration reaction, reacts with sulfuric acid
to give us two alkenes. The mechanism for this reaction, first we protonated the OH to form water as a good leaving group so water left and that
gave us a carbocation with a plus one formal
charge on this carbon, this benzylic carbon. So we have a benzylic
carbocation in our mechanism. Let me just go ahead
and draw that in here. So here's our benzylic carbocation, so plus one formal charge
on this carbon in magenta. And from this carbocation, we had a choice of which
proton we wanted to take. There are two hydrogens on this carbon, and depending on which proton we took, we got one of these different products. So this would be the trans product, just let me write that down here, this is the trans product, and this would be the cis product. These are stereoisomers, so let me write that down. These two are stereoisomers. That's one term that you
could use to describe them. And this reaction is stereoselective. One stereoisomer is
favored over the other. In this case, the trans products, right, this is the most stable product, so this is the major product,
this is favored over the cis. So, this would be the minor product. Now, let's look at stereospecificity. In a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry
of the product. And the E2 reaction can be a good example of a stereospecific reaction. On the left, we have our substrate, and we have these two phenyl groups here. We have a bromine, but notice the stereochemistry
at this carbon. You have a methyl group
coming out at us in space and a hydrogen going away from us. When our strong base
takes our beta proton, we end up with the E alkene, so there's stereochemistry in our product. We would have the two phenyl groups on opposite sides of the double bond. Look at this reaction now. We have the phenyl groups,
we have our bromine. Those are all the same. The difference is the
stereochemistry at this carbon. Now, we have a hydrogen coming out at us and a methyl group going away from us. Our strong base takes our proton, our beta proton in the mechanism, but this time we get the Z alkene. So the stereochemistry of
the substrate determined the stereochemistry of the product. There's no choice
because of the mechanism. You could also think about
that going backwards. If you look at this product here, the Z alkene, because you know
the product is a Z alkene, you know the stereochemistry
at this carbon, it must have this
particular stereochemistry. Same thing for the other reaction. We form only the E alkene, and because we form only the E alkene, we know the stereochemistry
at this carbon. So the stereochemical information is kept in a stereospecific reaction. Finally, let's directly
compare stereoselectivity with stereospecificity. We just said in a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry
of the product, so the stereochemical
information is preserved because of the mechanism. That's not the case in this
stereoselective reaction. If we look at the
stereochemistry of this OH here, at this carbon, the
stereochemistry is not preserved in our products. This stereochemical information is lost when we formed our carbocation. For example, we could
have started out with, let me go ahead and draw this in here, we could have started
out with the OH on a dash and we would have ended
up with the same products. So the stereochemical
information in the substrate is not preserved. Let's think about that
concept going backwards again. So for this E alkene, because this is an E alkene with the phenyl groups on opposite sides, we know the stereochemistry at this carbon in this stereospecific reaction. But if we look at the products here, if we look at this trans
product and the cis product, that does not tell us the
stereochemistry of our substrate. We do not know, we do not know
what the stereochemistry is at this carbon. Is the OH on a dash or
is the OH on a wedge? So this is not a stereospecific reaction. Instead, think about a
stereoselective reaction as being selective for, in this case, the trans isomer, the more stable isomer.