Sn1 Sn2 E1 E2 Mechanism Guide

SN1, SN2, E1, E2: Mechanism Guide

SN1, SN2, E1, E2: How to Choose the Mechanism

Substitution (SN1/SN2) and elimination (E1/E2) compete for alkyl halides/alcohol derivatives. Predicting outcomes hinges on substrate, nucleophile/base strength, solvent, and temperature.

Mechanism Snapshots

  • SN1: Two-step via carbocation; rate = k[sub]; racemization; rearrangements possible; favored by 3° (some 2°), good leaving group, polar protic solvent, weak/neutral nucleophile.
  • SN2: One-step backside attack; inversion; no rearrangement; rate = k[sub][Nu]; favored by 1° (methyl) substrates, strong nucleophile, polar aprotic solvent; hindered substrates slow/blocked.
  • E1: Two-step via carbocation; rate = k[sub]; competes with SN1; Zaitsev alkene; possible rearrangements; favored by 3° (some 2°), polar protic solvent, heat.
  • E2: One-step, strong base abstracts β-H as LG departs; anti-periplanar geometry required; rate = k[sub][base]; often with 2°/3° + strong (often bulky) base; Hofmann possible with bulky base; no rearrangements.

SN1: 2-bromo-2-methylbutane in water

A tertiary alkyl bromide ionizes first, then water adds. The product is 2-methylbutan-2-ol.

Curved arrow moving the C–Br bonding electrons of 2-bromo-2-methylbutane onto bromine
1. The leaving group leaves. Bromide is a good leaving group: it is a weak base and carries the negative charge easily. The carbon is tertiary, so it can stabilize the positive charge that is left behind.
A lone pair on water attacks the tertiary carbocation
2. Water adds. Water is only a weak nucleophile, but the carbocation is so electron-poor that a lone pair on oxygen forms a bond to the positively charged carbon.
A second water molecule removes a proton from the oxonium ion
3. Deprotonation. A nearby water molecule removes a proton from the newly attached water, giving the neutral alcohol.
2-methylbutan-2-ol, the SN1 product
Product. 2-Methylbutan-2-ol.

SN2: 1-chlorobutane with NaCN

A primary alkyl chloride and a strong nucleophile react in a single step. The product is pentanenitrile.

Cyanide attacks the carbon bearing chlorine from the back as the C–Cl bond breaks
1. One concerted step. The halogen is a good leaving group and cyanide is a strong nucleophile. Cyanide attacks the carbon from the side opposite the leaving group (backside attack) at the same time as chloride leaves.
Pentanenitrile, the SN2 product
Product. Pentanenitrile. At a stereocenter, backside attack inverts the configuration.

E1: 2-chloro-2-methylbutane in hot water

E1 reactions happen in two steps and share their first step with SN1. The product is 2-methylbut-2-ene, the more substituted (Zaitsev) alkene.

Curved arrow moving the C–Cl bonding electrons of 2-chloro-2-methylbutane onto chlorine
1. The leaving group leaves. A good leaving group, such as a halogen, on a carbon that can stabilize a positive charge, such as a tertiary carbon, leaves the molecule and forms a carbocation.
The tertiary carbocation intermediate
2. The carbocation. This is where a hydride or alkyl shift would happen if it gave a more stable cation. Here the cation is already tertiary, so nothing moves.
Water removes a proton next to the carbocation and the C–H electrons form the double bond
3. Loss of a proton. Once the carbocation has formed, a base (water in this example) removes a proton from a neighboring carbon. The electrons of that C–H bond form the double bond to the carbocation carbon.
2-methylbut-2-ene, the E1 product
Product. 2-Methylbut-2-ene, the elimination product.

E2: (1-chloropropyl)cyclobutane with NaOMe and heat

A strong base removes a β-hydrogen as the leaving group departs, all in one step. The product is propylidenecyclobutane.

(1-chloropropyl)cyclobutane and sodium methoxide before reaction
1. Line up the geometry. The β-hydrogen that is removed must be anti-periplanar to the C–Cl bond.
Methoxide removes the beta hydrogen while the C–H electrons form the double bond and chloride leaves
2. One concerted step. The leaving group (Cl in this example) leaves at the same time as the base (methoxide) pulls the proton off the adjacent carbon, and the electrons of that C–H bond form the new C=C bond. No carbocation or carbanion is formed along the way.
Propylidenecyclobutane, the E2 product
Product. Propylidenecyclobutane, the more substituted alkene.

How to Choose

  • Substrate: 3° blocks SN2; 1° blocks SN1/E1; 2° is contextual. Allylic/benzylic can do both SN1/SN2; vinyl/aryl generally neither.
  • Nucleophile vs base: Strong, unhindered Nu → SN2 (especially 1°); strong bulky base → E2; weak/neutral in protic → SN1/E1 (with 2°/3°).
  • Solvent: Protic stabilizes carbocations/ions → SN1/E1; aprotic boosts nucleophiles → SN2/E2.
  • Temperature: Higher T often favors elimination (E1/E2) over substitution when pathways compete.

Summary

  • SN1/E1 share a carbocation first step (rearrangements, racemization, Zaitsev alkene).
  • SN2/E2 are concerted (no rearrangements); SN2 inverts stereochemistry, E2 needs anti geometry.
  • Decide by weighing substrate hindrance, nucleophile/base strength, solvent, and temperature.