SN1/E1 Reactions

SN1 and E1 Reactions have very similar mechanisms, the final result just depends on whether the nucleophile or the base is attacks first. Compared to second order SN2 and E2 reactions (see “SN2 Reactions” and “E2 Reactions”), SN1/E1 are first order, the rate of the reaction depends only on the substrate.

SN1 and E1 are grouped together because they always occur together. 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). This means that any reaction that proceeds through the SN1/E1 mechanism will produce 50% SN1 product and 50% E1 product, and you will have to include both for full credit.

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. If SN1 occurs, there will be an inversion of stereochemistry due to the backside attack and the other three bonds “turning inside out” around the central carbon.

This carbocation affects the progression of the reaction. In SN2 and E2 we saw that 3° carbons had a much slower rate of reaction than 1° carbons due to the steric hindrance preventing either the nucleophile attacking or the base pulling off the b-hydrogen. With SN1 and E1 however, this relationship is reversed, and 3° carbons react much faster than 1° carbons. This is due to the stability of the carbocation intermediate–3° carbocations can stabilitize the excess charge much better than 1° carbocations, so the intermediate step is more likely to occur, allowing the nucleophile or base to attack during the second step of the mechanism.

  The general mechanism for SN1/E1 reactions

The general mechanism for SN1/E1 reactions

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:

OCHEM - SN1 + E1 Reactions 2.png

As you can see the carbocation changed the SN1 product, but didn’t change the E1 product. This is because Zaitsev’s Rule still applies for E1 as it does for E2. So what about anti-Zaitsev E1 product? For an anti-Zaitsev product, you must use tert-butoxide because its bulky. However, tBuO- is a strong base, and carbocations can’t exist in the presence of a strong base, therefore E1 can’t proceed if tBuO- is present, so there are no anti-Zaitsev products of E1.

Predicting SN1/E1

First, as mentioned above, if a strong base is present, SN1/E1 can’t happen. Second, assuming you have a weak base, SN1/E1 will always take place on 3° carbons, and never on 1° or methyl carbons (SN2 will occur). To determine what mechanism will take place for 2° carbons, you must look at the nucleophile: a good nucleophile will follow SN2 (a better nucleophile is more likely to attack first), and a bad nucleophile will follow SN1/E1.

OCHEM - SN1 + E1 Reactions STEPS (Figma Out).png

Some exceptions to SN1/E1 reactions are:

  1. Carbocation rearrangement: if the positive charge can rearrange to a more stable carbon, it will, causing a hydride or alkyl shift.
  2. Tert-butoxide: if tBuO- is present in solution, SN1/E1 can’t occur! The reaction will proceed through SN2 or E2 depending on the substrate.
  3. Bredt’s Rule: this you may not see very often. The rule states that a double bond on a ring and bridge molecule cannot occur at the “bridgehead,” i.e. a double bond can’t be on the tertiary carbon of a bridged molecule. Therefore, if the leaving group is attached to this carbon or the adjacent one (don’t forget carbocation rearrangement), E1 can’t occur, so only SN1 products will be produced.
 - E1 can’t occur as it violates Bredt’s Rule. - Results in 100% SN1 product.

- E1 can’t occur as it violates Bredt’s Rule.
- Results in 100% SN1 product.

E2 Reactions

Elimination reactions are helpful in many situations, especially in synthesis when you have an alkane and need an alkene. Similar to substitution reactions, eliminations produce similar products but depend on different factors. Most notably, E1 reactions have a carbocation intermediate while E2 reactions do not (for more information on E1 reactions, see “SN1/E1 Reactions”). 

While SN2 reactions require a nucleophile, E2 reactions require a base. This base will pull off a b-hydrogen which opens up two electron. These electrons form another bond between the carbons, creating a double bond, and expelling the leaving group in order for the second carbon to maintain octet:

  Note: Only the b-hydrogen shown.

Note: Only the b-hydrogen shown.

In order for E2 to occur, the hydrogen and the leaving group must be antiperiplanar. This just means that the hydrogen and leaving group have to be on the same plane, but in opposite directions, forming a “Z” shape with the two carbons involved. With linear molecules this usually isn’t a problem, as sigma bonds can freely rotate to accommodate. However, ring structure can prove to be problematic. In the case of cyclohexane, E2 can only happen if the hydrogen and leaving group are both in the axial position on adjacent carbons. If one (or both) are equatorial, E2 can’t happen. It may help to build a molecule with your modeling kit if you’re having trouble visualizing this.

Elimination reactions also introduce another rule: Zaitsev’s Rule. This is similar to Markovnikov’s Rule in that the more substituted option will be the major product, and the less substituted option the minor product. And just like Markovnikov’s Rule, there is an anti-Zaitsev’s exception: tert-butoxide. tBuO- is too bulky to get close to the more substituted carbon, so instead it pulls a hydrogen off from the less substituted b-carbon, causing the less substituted option to be the major product.

  Zaitzev's Rule

Zaitzev's Rule

  Anti-Zaitzev's Rule

Anti-Zaitzev's Rule

Predicting E2

Unlike SN2, E2 only depends on the base present. A strong base is required to pull off a b-hydrogen, so if you don’t have a strong base you cannot proceed through E2. Additionally, if you have a strong base, SN1/E1 can’t occur, so with a strong base you know you’ll be doing either SN2 or E2.

OCHEM - E2 Reactions 4 STEPS.PNG


1. Non-antiperiplanar b-hydrogens: E2 requires the b-hydrogen and leaving group to be antiperiplanar, if they aren’t then E2 can’t occur. This is most problematic in ring structures due to the rigidity of the molecule.

2. tBuO-: Again here’s tert-butoxide causing problems. When you see it, circle it, and remember that this molecule causes the major product to be the less substituted option—anti-Zaitsev’s Rule.