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.
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:
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.
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.
Some exceptions to SN1/E1 reactions are:
Carbocation rearrangement: if the positive charge can rearrange to a more stable carbon, it will, causing a hydride or alkyl shift.
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.
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.