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Elimination Products:
Zaitsev vs Hofmann

Learn how base size and substrate structure determine which alkene forms in E2 elimination reactions.

Organic Chemistry 1

What is Elimination?

In an elimination reaction, a base removes a proton from a carbon adjacent to the leaving group. The leaving group departs, and a new \(\pi\) bond (alkene) forms between the \(\alpha\)-carbon and the \(\beta\)-carbon.

\(\text{H}\)
\(\beta\)-C
\(\alpha\)-C
\(\text{LG}\)
Base
\(\beta\)-C
\(\alpha\)-C
Alkene formed

The \(\beta\)-Hydrogen Choice

When a substrate has multiple \(\beta\)-carbons bearing hydrogens, the base can abstract from different positions, leading to different alkene products.

\(\alpha\)-C
\(\text{LG}\)
\(\beta_1\)-C(H)
\(\beta_2\)-C(H)
\(\beta_3\)-C(H)

Zaitsev's Rule

Zaitsev's rule states that elimination preferentially forms the more substituted alkene (the alkene with more alkyl groups on the double bond carbons). This is the thermodynamic product.

Why more stable?
Hyperconjugation from adjacent \(C–H\) bonds donates electron density into the \(\pi^*\) orbital, stabilizing the alkene.
When it applies
Most E2 reactions with small, strong bases like \(\text{NaOEt}\), \(\text{NaOH}\), or \(\text{NaOMe}\) on unhindered substrates.

Hofmann's Rule

Hofmann's rule states that elimination preferentially forms the less substituted alkene. This is the kinetic product, favored when steric effects dominate.

Why less substituted?
A bulky base cannot easily reach the more hindered \(\beta\)-hydrogen, so it abstracts the more accessible one instead.
When it applies
E2 with bulky bases like \(\text{KOtBu}\) on hindered substrates, or when the leaving group is a charged species (\(\text{NR}_3^+\)).

Alkene Substitution Patterns

Monosubstituted
\(\text{R}\)
\(\text{H}\)
\(\text{H}\)
\(\text{H}\)
Least stable
Disubstituted
\(\text{R}\)
\(\text{H}\)
\(\text{R}\)
\(\text{H}\)
Trisubstituted
\(\text{R}\)
\(\text{R}\)
\(\text{R}\)
\(\text{H}\)
Tetrasubstituted
\(\text{R}\)
\(\text{R}\)
\(\text{R}\)
\(\text{R}\)
Most stable
Increasing stability

Factors that Determine the Product

Base Size
Small bases (\(\text{EtO}^-\)) reach any \(\beta\)-H. Bulky bases (\(\text{tBuO}^-\)) are sterically blocked from hindered positions.
Leaving Group
Charged leaving groups (\(\text{NMe}_3^+\)) create a larger transition state, favoring Hofmann due to steric strain.
Substrate Bulk
Highly branched substrates make the more-substituted \(\beta\)-H harder to access, tipping toward Hofmann.

Zaitsev vs Hofmann: Key Distinction

Zaitsev Hofmann
Major product More substituted alkene Less substituted alkene
Favored by Small bases Bulky bases
Driving force Thermodynamic stability Steric accessibility
Typical base \(\text{EtO}^-\), \(\text{OH}^-\) \(\text{tBuO}^-\), \(\text{LDA}\)
Leaving group effect Neutral LG (Br, Cl) Charged LG (\(\text{NR}_3^+\))

How to Predict the Product

1
Identify the \(\alpha\)-carbon (the one bearing the leaving group) and all adjacent \(\beta\)-carbons with hydrogens.
2
Draw all possible alkene products by removing the LG and an H from each \(\beta\)-position.
3
Rank by substitution: count how many alkyl groups are on each double bond (mono-, di-, tri-, tetra-).
4
Check the base: Small base → Zaitsev (more substituted). Bulky base → Hofmann (less substituted).
5
Check the leaving group: Charged LG (\(\text{NR}_3^+\)) tips toward Hofmann regardless of base size.

Example 1: 2-Bromobutane with \(\text{NaOEt}\)

\(\text{CH}_3\)
\(\text{CH}\)
\(\text{Br}\)
\(\text{CH}_2\)
\(\text{CH}_3\)
\(\beta_1\)
\(\beta_2\)
NaOEt (small base)
\(\text{CH}_2=\text{CH}–\text{CH}_2\text{CH}_3\)
1-butene (Hofmann)
+
\(\text{CH}_3\text{CH}=\text{CH}\text{CH}_3\)
2-butene (Zaitsev) MAJOR

Example 2: 2-Bromobutane with \(\text{KOtBu}\)

\(\text{CH}_3\)
\(\text{CH}\)
\(\text{Br}\)
\(\text{CH}_2\)
\(\text{CH}_3\)
\(\beta_1\)
\(\beta_2\)
KOtBu (bulky base)
\(\text{CH}_2=\text{CH}–\text{CH}_2\text{CH}_3\)
1-butene (Hofmann) MAJOR
+
\(\text{CH}_3\text{CH}=\text{CH}\text{CH}_3\)
2-butene (Zaitsev) minor

Example 3: Charged Leaving Group \(\text{NMe}_3^+\)

When the leaving group is a quaternary ammonium salt (\(\text{NMe}_3^+\)), the Hofmann product dominates even with a small base. The bulky, positively charged leaving group increases steric strain in the transition state.

\(\text{CH}_3\)
\(\text{CH}\)
\(\text{NMe}_3^+\)
\(\text{CH}_2\text{CH}_3\)
NaOH
\(\text{CH}_2=\text{CH}\text{CH}_2\text{CH}_3\)
Hofmann: MAJOR
(not \(\text{CH}_3\text{CH}=\text{CH}\text{CH}_3\))

Energy Diagram: Thermodynamic vs Kinetic

Energy
Reaction Coordinate
Substrate
TS (Hofmann)
lower barrier
TS (Zaitsev)
higher barrier
Hofmann
less stable
Zaitsev
more stable
\(\Delta E\)

Don't Forget E/Z Geometry

When the Zaitsev product has two different groups on each carbon of the double bond, you must also consider E/Z stereochemistry. The trans (E) isomer is generally more stable due to less steric strain.

Trans (E)
Large groups on opposite sides of the \(C=C\). Less steric strain. Usually major among stereoisomers.
Cis (Z)
Large groups on the same side. More steric strain between adjacent substituents. Usually minor.

Common Mistakes

✕ Ignoring Base Size
Assuming Zaitsev always wins. A bulky base like \(\text{KOtBu}\) reverses the selectivity.
✕ Forgetting Charged LG
With \(\text{NR}_3^+\) as the leaving group, Hofmann dominates even with small bases.
✕ Miscounting Substitution
Counting H atoms on the double bond instead of counting alkyl groups. Focus on R groups attached to the \(C=C\).

Quick Reference Summary

Condition Product Rule
Small base + neutral LG More substituted Zaitsev
Bulky base + neutral LG Less substituted Hofmann
Any base + charged LG Less substituted Hofmann
Bulky base + branched substrate Less substituted Hofmann (strongly)
Small base + linear substrate More substituted Zaitsev (strongly)

Question 1 of 5

When 2-bromobutane is treated with \(\text{NaOEt}\) (sodium ethoxide), what is the major product?

Question 2 of 5

Which base would you use to favor the Hofmann product from 2-bromobutane?

Question 3 of 5

A substrate with a \(\text{NMe}_3^+\) leaving group undergoes E2 with \(\text{NaOH}\). Which rule predicts the major product?

Question 4 of 5

Why is the Zaitsev product considered the thermodynamic product?

Question 5 of 5

Treating 2-bromo-2-methylbutane with \(\text{KOtBu}\) in E2 gives a major product. How many alkyl groups are on the double bond of this major product?