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E1 Reactions &
Competing Pathways

Understand unimolecular elimination, its two-step mechanism, and how it competes with \(S_N1\) and E2 pathways in real reactions.

Organic Chemistry 1

What Is an Elimination Reaction?

An elimination reaction removes atoms or groups from adjacent carbons to form a new pi bond (typically a \(C=C\) double bond). The substrate loses a leaving group and a proton (\(H^+\)) from the beta carbon.

\(C_\alpha\)
\(C_\beta\)
 LG H C C eliminate alkene product

The E1 Mechanism: Two Steps

E1 stands for Elimination, Unimolecular. The rate depends only on the substrate concentration: \(\text{Rate} = k[\text{substrate}]\). The mechanism has two distinct steps.

Step 1: Ionization
The leaving group departs, forming a carbocation intermediate. This is the slow, rate-determining step.
Step 2: Deprotonation
A weak base removes a \(\beta\)-hydrogen, and the electron pair forms the new \(C=C\) pi bond.

Energy Diagram for E1

Energy Reaction Coordinate R-LG
\(R^+\)
 alkene TS1 TS2
\(E_{a1}\)
 Step 1 (slow) Step 2 (fast)

Key Conditions for E1

1
Tertiary or secondary substrate: must form a stable carbocation (\(3^\circ \gg 2^\circ \gg 1^\circ\)).
2
Weak or neutral base/nucleophile: water, alcohols, or no base added. Strong bases favor E2.
3
Polar protic solvent: stabilizes the carbocation intermediate and the departing anion.
4
Heat (elevated temperature): higher temperatures favor elimination over substitution.

E1 vs E2: Key Distinctions

E1 E2
Steps Two (carbocation intermediate) One (concerted)
Rate Law \(\text{Rate} = k[\text{sub}]\) \(\text{Rate} = k[\text{sub}][\text{base}]\)
Base Strength Weak / neutral Strong, bulky
Substrate \(3^\circ > 2^\circ\) \(3^\circ > 2^\circ > 1^\circ\)
Stereochemistry Not anti-periplanar required Anti-periplanar required
Rearrangements? Yes (carbocation) No
Regioselectivity Zaitsev (more sub. alkene) Zaitsev or Hofmann

Competing Pathways: The Big Picture

A tertiary alkyl halide in a polar protic solvent can undergo four possible reactions simultaneously. Your job is to predict which pathway dominates.

R-LG (substrate)
\(S_N1\)
\(S_N2\)
E1
E2
 weak nuc, polar protic
strong nuc, \(1^\circ\) sub
 weak base, heat strong base

E1 vs \(S_N1\): Same Step 1

E1 and \(S_N1\) share the same rate-determining step: carbocation formation. They diverge only in Step 2.

\(S_N1\) Path
A nucleophile attacks the carbocation, forming a new \(C-Nu\) bond: substitution product.
E1 Path
A base removes a \(\beta\)-H, electrons form a \(C=C\) pi bond: elimination product (alkene).

Zaitsev's Rule in E1

In E1 elimination, the most substituted alkene is the major product. The base preferentially removes the \(\beta\)-hydrogen that leads to the more stable, more substituted double bond.

\(C^+\)
\(CH_3\)
\(CH_3\)
\(CH_2\)
\(CH_3\)
\(H_a\)
\(H_b\)
 Major: trisubstituted (Zaitsev product) Minor (less substituted)

Carbocation Rearrangements in E1

Since E1 forms a carbocation intermediate, it can rearrange via hydride shifts (\(1,2\)-H shift) or methyl shifts (\(1,2\)-\(CH_3\) shift) to become more stable before elimination occurs.

\(2^\circ\) carbocation
 (less stable) 1,2-H shift
\(3^\circ\) carbocation
 (more stable) alkene Elimination occurs *after* rearrangement, leading to an unexpected product structure.

How to Predict the Pathway

1
Classify the substrate: Is it \(1^\circ\), \(2^\circ\), or \(3^\circ\)? Primary substrates cannot do E1 or \(S_N1\).
2
Identify the base/nucleophile: Strong and bulky? Likely E2. Weak or neutral? E1 or \(S_N1\).
3
Check solvent: Polar protic favors E1/\(S_N1\). Polar aprotic favors E2/\(S_N2\).
4
Consider temperature: High temperature (\(\Delta\)) favors elimination (E1 over \(S_N1\)).
5
Check for rearrangements: If a carbocation forms, can it rearrange to become more stable?
6
Apply Zaitsev's Rule: The most substituted alkene is the major E1 product.

Example 1: Identifying E1

Problem: 2-bromo-2-methylpropane is dissolved in ethanol and heated. What is the major product?

\((CH_3)_3C\text{-}Br\)
\(3^\circ\) alkyl halide
 EtOH, heat
\((CH_3)_2C=CH_2\)
 E1 product (major)
+ \((CH_3)_3C\text{-}OEt\)
\(S_N1\) product (minor)

Example 2: Zaitsev Product

Problem: 2-bromo-2-methylbutane undergoes E1 elimination. Predict the major alkene product.

2-bromo-2-methylbutane
\(CH_3\)
 C
\(CH_2\)
\(CH_3\)
\(CH_3\)
 Br
\(H_a\)
\(H_b\)
 2-methylbut-2-ene (major, trisubstituted) 2-methylbut-1-ene (minor, disubstituted)

Example 3: Rearrangement Trap

Problem: 2-bromo-3-methylbutane undergoes E1 in ethanol with heat. What happens?

1
Ionization: Br leaves from C2, giving a \(2^\circ\) carbocation at C2.
2
1,2-hydride shift: H migrates from C3 to C2, giving a \(3^\circ\) carbocation at C3 (more stable!).
3
Elimination: Base removes \(\beta\)-H to form 2-methylbut-2-ene (trisubstituted alkene).

Common Mistakes to Avoid

✕ Ignoring Rearrangements
Forgetting that E1 goes through a carbocation that can undergo 1,2-shifts to become more stable before elimination.
✕ E1 on Primary Substrates
Primary carbocations are too unstable. \(1^\circ\) substrates undergo E2 or \(S_N2\), never E1 or \(S_N1\).
✕ Confusing E1 and E2
E1 requires no strong base and has no stereochemical requirement. E2 needs a strong base and anti-periplanar geometry.

Summary: All Four Pathways

\(S_N1\) \(S_N2\) E1 E2
Steps 2 1 2 1
Rate Law \(k[\text{sub}]\) \(k[\text{sub}][\text{nuc}]\) \(k[\text{sub}]\) \(k[\text{sub}][\text{base}]\)
Substrate \(3^\circ > 2^\circ\) \(\text{Methyl} > 1^\circ > 2^\circ\) \(3^\circ > 2^\circ\) \(3^\circ > 2^\circ > 1^\circ\)
Nuc / Base Weak Strong Weak Strong / bulky
Solvent Polar protic Polar aprotic Polar protic Any
Rearrange? Yes No Yes No
Favored by heat? No No Yes Yes

Question 1 of 5

What is the rate-determining step of the E1 mechanism?

Question 2 of 5

Which set of conditions most favors E1 over \(S_N1\)?

Question 3 of 5

2-bromo-3-methylbutane undergoes E1. A 1,2-hydride shift occurs. What is the major product?

Question 4 of 5

Why does E1 not occur with primary (\(1^\circ\)) alkyl halides?

Question 5 of 5

Zaitsev's Rule states that E1 elimination preferentially forms: