Organic Lecture 7: Sn1, Sn2, E1, E2 (All you need to know)

Most organic reactions occur between nucleophiles and electrophiles. Nucleophiles are species that have unshared pairs of electrons or pi bonds and frequently have a negative charge. They are "nucleus seeking" or "nucleus loving" and electron donors (aka Lewis Base).

Nucleophilicity is a measure of how "strong" a nucleophile is. Here are some general trends:
1. Nucleophilicity increases as negative charge increases. NH2- is more nucleophilic than NH3.
2. Nucleophilicity increases going down the periodic table within a particular group. This is related to polarizability (how easy it is for electrons surrounding an atom to be distorted--therefore, larger atoms are more polarizable and more nucleophilic). F < Cl< Br< I
3. Nucleophilicity increases going left on the periodic table. NH2 is more nucleophilic than OH-. This is related to electronegativity. The more electronegative the atom is, the better it is able to support its negative charge. Therefore, the less electronegative an atom is, the higher its nucleophilicity.

Electrophiles are electron-deficient species. They have a full or partial positive charge and "love electrons". Since electrophiles are electron pair acceptors, they are also known as Lewis acids.

Here's a brief synopsis of the substitution and elimination mechanisms:

Sn2: nucleophile attacks an electrophile (generally an alkyl halide); this is a backside attack. Alkyl halide acts as a leaving group and the nucleophile forms a bond with the electrophile. There is only one step in this mechanism--therefore, it's a concerted mechanism. Inversion occurs when the substrate is chiral (this means that if you have an absolute configuration of S, then it goes to R, and vice versa). This is a bimolecular reaction because the rate is determined by the concentrations of both the nucleophile and the electrophile.

Sn2 reaction rate = k[nucleophile][electrophile]

**Less substituted substrates react faster than more substituted ones
**To favor Sn2, avoid protic solvents (i.e. water and alcohols) becasue these strongly solvate the nucleophile and therefore hinder the backside attack
**Use aprotic (non-hydrogen bonding) solvents--acetone, DMF, DMSO

Sn1: occurs in two distinct steps. In the first step, a carbocation is formed--this occurs when the leaving group falls off. This is the slow step of the mechanism (rate limiting step). In the final step, racemization occurs as the nucleophile attacks on either side of the carbocation. Racemic mixture results. The rate of the reaction depends only upon the concentration of the electrophile

Sn1 rate = k[electrophile]

**The more substituted the carbocation intermediate the better. This stablizes it and therefore makes the reaction proceed faster.
**Use protic solvents (water and alcohols) to help stablize the forming carbocation and solvate the leaving group.

E1: leaving group falls off and then a weak base removes a proton, leaving behind its electrons to form a C=C double bond; E1 works best with tertiary substrates because they best support the positive charge after the leaving group leaves; most highly substituted alkene is formed; this mechanism is favored by a protic solvent

E2: proceeds via a 1-step mechanism; a strong base removes the beta hydrogen while the leaving group leaves; the carbon-carbon double bond forms at the same time; this mechanism only works with alkyl halides; must have ANTIPERIPLANAR geometry (H is anti to leaving group); small bases yield the most substituted alkene while bulky bases favor the least substituted one (because of steric hindrance)

I found this on SDN. It's just a different way to explain what I have above. Very good though.

1) Number of R Groups on the Substrate
This is a very important factor. Keep in mind that primary substrates (like alkyl halides) will not react via Sn1 or E1, because the carbocation formed would be too unstable. Likewise, tertiary substrates will not react via Sn2, because they are too hindered. To sum:
-Primary substrates may undergo Sn2 or E2, but not Sn1 or E1
-Tertiary substrates may undergo Sn1, E1, or E2, but not Sn2.
-Secondary substrates are tricky and may react by any of the four mechanisms, depending on other factors. Keep reading.
2) Strength of the Base/Nucleophile
This is another very important factor. Keep in mind that carbocations are very powerful Lewis acids. You may not be used to thinking of them this way, because the Bronsted-Lowry definition of an acid doesn't include carbocations. However, the same general principle applies to carbocations that applies to all strong acids: when you mix them with strong bases, you get a neutralization reaction. That is not what we are trying to accomplish in organic reactions! Thus, E1 and Sn1 reactions, both of which have carbocation intermediates, will not occur in the presence of a strong base. In contrast, E2 requires the presence of a strong base, and a strong base will promote E2, particularly if the base is also bulky. To sum:
-A species that is BOTH a good nucleophile and a strong base (ex. hydroxide, Grignard reagents, and alkoxide ions) will tend to promote E2 or Sn2, depending on other factors like substrate hindrance and base bulkiness
-A species that is a good nucleophile but NOT a strong base (ex. cyanide, sulfur nucleophiles, azide) will tend to promote Sn2 if the substrate is not too hindered.
-A species that is a strong base but is NOT nucleophilic (ex. NaH) will often just deprotonate an acidic hydrogen on the substrate.
-A species that is BOTH a poor nucleophile and a weak base (ex. water, alcohols) will tend to promote Sn1 and E1, assuming that the substrate is hindered enough.
3) Solvent Type
This is another important factor that is often overlooked by students. Generally the solvent will be polar, and it can either be protic or aprotic, which affects the mechanism. Thus, it is essential that you learn how to correctly identify whether a solvent is protic or aprotic. (Read the Intermolecular Interactions post in the General Chemistry Explanations thread about protic versus aprotic solvents if you are not familiar with these terms.) A protic solvent stabilizes charged species. For mechanisms that form carbocations (Sn1 and E1), this is a good thing, because it lowers the potential energy of the carbocations, making them easier to form. However, for mechanisms where the base or nucleophile is charged, a polar protic solvent will stablize the nucleophile and make it less reactive. In contrast, a polar aprotic solvent will destabilize it and make it more reactive. (This is sometimes referred to as making a "naked nucleophile.") To sum:
-Polar protic solvents favor E1 and Sn1 by lowering the energy of the carbocation intermediate and making the RDS occur more readily.
-Polar aprotic solvents favor Sn2 by increasing the energy of the nucleophile and making it more reactive.
-E2 reactions are often carried out in polar aprotic solvents or in a polar protic solvent that is the conjugate acid of the strong base (ex. t-butanol for t-butoxide).
4) Heat and Hindrance
It is relatively easy to force the reaction to go by E2 over Sn2 by manipulating the choice of base. A large, bulky base like t-butoxide or LDA (lithium diisopropylamide) is too hindered to serve as a nucleophile, and the presence of either of those bases should immediately tell you that the correct mechanism is E2. Forcing a reaction to go by E1 instead of Sn1 is very difficult, because these two mechanisms have the same RDS (formation of the carbocation). The one thing you can do to promote elimination over substitution is to heat the reaction. In organic chemistry, we use the Greek symbol delta (looks like a little triangle over the reaction arrow) to designate that the reaction is being heated. (Note that the delta symbol means "change in" for the physical sciences, and often precedes a variable: eg. delta G = delta H - T delta S.) To sum:
-A large, bulky base (t-butoxide or LDA) should immediately clue you in that the mechanism is E2.
-A heat symbol (the Greek letter delta) over the reaction arrow tells you that the reaction will probably go by elimination rather than substitution.
5) None of the Above
Don't get so caught up in trying to decide which mechanism will predominate for a given reaction that you don't consider the possibility that no reaction will occur at all. There are two common situations that you should watch out for where there will be no reaction.

One is if you have a primary substrate and no good nucleophile or strong base is present. You can't have E1 or Sn1 in this scenario because the substrate is primary. You can't have E2 because there isn't a strong base. And you can't have Sn2 because there isn't a good nucleophile. The correct response here is that no reaction will occur.

Another situation where you must be cautious is if you are asked about how a cyclohexane or other hindered sytem will react, and you see that you have E2-only conditions. Remember that for an E2 to occur, the base, proton, both carbons forming the new double bond, and the leaving group must all be in what is called the antiperiplanar geometry. This means that all five atoms must lie in the same plane of space, and that the proton and leaving group must be 180 degrees apart in orientation. If the ring cannot orient itself in such a way that the proton and the leaving group are coaxial (both in axial positions on the ring with one up and one down), then an E2 cannot occur. Again, the correct response here is that there will be no reaction.