What is the difference between e1 and sn1

In this paper, Prof. Winstein shows that the E1 reaction has a strong solvent dependence — as the dielectric constant or polarity of the solvent decreases, the amount of olefin obtained also decreases, since carbocations cannot be formed easily.

Mechanism of elimination reactions. Part XI. Kinetics of olefin elimination from tert. Dhar, E. Hughes, and C. Ingold J. Chemical Effects of Steric Strains. Brown provides a perfect study of how the structure of the substrate can influence the E1-S N 1 competition. Part VIII. Temperature effects on rates and product-proportions in uni- and bi-molecular substitution and elimination reactions of alkyl halides and sulphonium salts in hydroxylic solvents K.

Cooper, E. Hughes, C. Ingold, and B. MacNulty J. An increase in temperature produces a modest increase in the proportion of olefin in E1-S N 1 reactions. Part VII. Solvent effects on rates and product-proportions in uni- and bi-molecular substitution and elimination reactions of alkyl halides and sulphonium salts in hydroxylic solvents K.

Cooper, M. 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. But remember, this is mastered only by doing lots of practice problems. This only happens when the base if the mechanism is E2 or the nucleophile if S N 2 is strong — very reactive. The weak reactants are mainly going to be the water and alcohols. And choosing between E1 and S N 1 is easy — the main factor is the heat. If heat is mentioned, then it is a hint to you that E1 elimination is the main mechanism in that reaction:.

Notice now that we are already done with the left side of the flowchart and the only riddle is to choose between S N 2 and E2. Strong bases take us to the right side E2 or S N 2.

Bases can be nucleophilic and non-nucleophilic. Non-nucleophilic means most often it reacts as a base while nucleophilic base means it can react both as a base and a nucleophile and this where the problems start. For example, ethoxide and butoxide ions are relatively strong bases. The difference is that the butoxide can only take it as mostly… I know act as a base , but the ethoxide reacts as a base and a nucleophile depending on the substrate.

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.