The usual choice is H 2 SO 4. The HSO 4 - ion is a relatively poor nucleophile due to the negative charge of the oxygen being distributed throughout the molecule resonance. Since the E1 and S N 1 reactions both proceed through a common carbocation intermediate, these pathways can and do compete with each other under a lot of reaction conditions. I disagree that the conjugate base of the strong acid is a competent base for the final deprotonation.
Same with Cl- or TsO- or any of the other conjugate bases of strong acids. In my class, HSO4- as a base would be marked incorrect.
But I see your point, especially regarding the last part! HSO4- is the conjugate base. Carbocation can give a proton to a base at elimination reaction.
E1 is just alternative of SN1. Am I right? I had a question that as u have mentioned in text that heat favours elimination so is always valid even in case of Hoffman products. Your email address will not be published. Save my name, email, and website in this browser for the next time I comment. Notify me via e-mail if anyone answers my comment. This site uses Akismet to reduce spam. Learn how your comment data is processed. Previous Bulky Bases in Elimination Reactions.
That means that the reaction, as written, is very unlikely to happen. Add acid! The halide ions Cl- , Br-, I- are decent nucleophiles under the reaction conditions.
So how can we stack the deck in favor of the E1 process? Use a strong acid with a conjugate base that is a poor nucleophile. Next Post: The E1 Reaction With Rearrangements Advanced References and Further Reading Since the E1 and S N 1 reactions both proceed through a common carbocation intermediate, these pathways can and do compete with each other under a lot of reaction conditions.
Ion Pairs in Elimination M. Cocivera and S. Saul Winstein UCLA was a highly influential physical organic chemist in the early 20 th century, and introduced several concepts that are now fundamental to organic mechanisms, such as anchimeric assistance, ion-pairing, internal return, and many others. And if it was a tertiary carbon, it would favor Sn1 or E1 because it would favor a stable carbocation.
The leaving group could just leave. And if this guy was bonded to another carbon, it would be very stable. But in this situation, it's a secondary carbon bonded to two carbons. It's a little bit neutral. Any of these reactions might occur. When we look at all of the other data points, they're pointing at both Sn2 or E2. We have an aprotic solvent.
It's going to be Sn2 or an E2 reaction. So let's actually draw the reactions. Let me do the Sn2 first. So let me do it in orange.
So if we were to have an Sn2 reaction, let me redraw the molecule. Let me draw the cyclopentane part. I want to make sure-- let me draw it the same way I had it drawn up there. So the pentagon is facing upwards. And then we have our bromo group right there. So we have our methoxide ion right over here. So CH3O minus. Or another way we could view it is that this oxygen has one, two, three, four, five, six, seven valence electrons with a negative charge.
One of these electrons right over here, this can attack the substrate right over there, that carbon. Right when that happens, simultaneously this bromine is going to be able to nab an electron from that same carbon.
And then we are going to be left with-- the bromine now becomes the bromide anion. It had one, two, three, four, five, six, seven valence electrons. One, two, three, four, five, six, seven. Now it nabbed one more electron, making it bromide. Now it has a negative charge. And if we were to draw the chain, it would look like this.
Well, we could draw it on this, and I might as well draw it on this side, just so it's attacking from the other side. This is the chiral substrate, so we don't have to be too particular about how we draw the connections to the carbon.
We're not actually even showing anything popping in or out. But we would have the methoxide ion, where now it's bonded, so it's no longer an ion, so it's OCH3, just like that. It has bonded to this carbon. Obviously, implicitly this carbon had another hydrogen that we are not showing. Just that quickly, that was the Sn2 reaction. That is the mechanism. Now let's think about what the E2 reaction is.
To do the E2 properly, to give it justice, we're going to have to draw some of the hydrogens. So on the E2 reaction, let me draw that in blue. Let me draw it big. Actually, over here, it's less important to draw it too big.
So let me draw the pentagon. The pentagon just like that. That is the bromine, three, four, five, six, and then it has a seventh valence electron right over here. This is the alpha carbon. That right there is the alpha carbon. And then there are two beta carbons. There are two beta carbons right over there and there. They each have two hydrogens on them. They each have two hydrogens. I know it's becoming a little hard to read. And in an E2 reaction, the strong base will react-- let me make it a little cleaner than that.
Let me get rid of the beta. The beta makes it's a little dirty. OK, so they each have two hydrogens on them. Now in an E2 reaction, the strong base-- over here, the methoxide ion was acting as a strong nucleophile. In E2, it's going to act as a strong base. It's going to nab off a hydrogen off of one of the beta carbons. And you might want to say, OK, which one? If the leaving group dissociates first, there is an equally likely chance of the nucleophile attacking SN1 as there is the base pulling off the b-hydrogen E1.
Both E1 and SN1 start the same, with the dissociation of a leaving group, forming a trigonal planar molecule with a carbocation. This molecule is then either attacked by a nucleophile for SN1 or a base pulls off a b-hydrogen for E1. This carbocation affects the progression of the reaction. Another aspect of the carbocation that must be considered is the possibility of rearrangement.
If a hydride or alkyl shift can occur to put the positive charge on a more stable carbon, it will:. So what about anti-Zaitsev E1 product?
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