Zaitsev's Rule: Which Alkene Wins?

One substrate, two possible alkenes. Zaitsev's rule tells you the more substituted one on the right of the pair usually wins.

When a leaving group and a neighboring hydrogen depart in an elimination, the substrate often has more than one hydrogen it could lose — and each choice gives a different alkene. Zaitsev's rule (also spelled Saytzeff) is the shortcut that tells you which one dominates: the more substituted, more stable alkene. Below we build the rule from the ground up, learn to spot the winner by counting, and cover the bulky-base exception that reverses it.

1. The more substituted alkene is the major product

Consider 2-bromobutane treated with sodium ethoxide, a small, strong base. The bromide can leave while a hydrogen is removed from either the C1 methyl on one side or the C3 methylene on the other. Removing the C3 hydrogen places the double bond between C2 and C3, giving 2-butene, which carries an alkyl group on each alkene carbon. Removing a C1 hydrogen puts the double bond at the end of the chain, giving 1-butene with alkyl substitution on only one carbon.

2-bromobutane + NaOEt gives 2-butene as the major (Zaitsev) product — a disubstituted alkene beats the monosubstituted alternative.

Experimentally, 2-butene is the major product. That observation, generalized across countless eliminations, is Zaitsev's rule: the alkene with more alkyl groups attached to the double-bonded carbons forms preferentially.

2. "More substituted" means counting alkyl groups on the C=C

To apply the rule you only need to count. Look at the two carbons of each candidate double bond and count how many carbon (alkyl) groups are attached directly to them. A terminal alkene like 1-butene is monosubstituted (one alkyl group). An internal alkene like 2-butene is disubstituted (one alkyl group on each carbon). Trisubstituted and tetrasubstituted alkenes carry three and four alkyl groups, respectively, and rank even more stable.

Substitution ladder: the more alkyl groups hanging off the double bond, the more stable — and the more "Zaitsev" — the alkene.

The alkene highest on this ladder that a given substrate can actually form is the Zaitsev product. Everything less substituted is called a Hofmann product.

3. More substitution lowers the alkene's energy

Why should extra alkyl groups matter? Two reasons. First, hyperconjugation: electrons in adjacent C–H and C–C sigma bonds delocalize into the empty pi* orbital of the double bond. Each alkyl group brings more such neighboring bonds, spreading electron density and stabilizing the alkene. Second, alkyl groups are electron-donating and relieve strain relative to a terminal =CH2 group.

The proof is quantitative. Heats of hydrogenation measure how much energy is released when an alkene is reduced to the alkane — a more stable alkene sits lower to begin with, so it releases less. 1-butene releases about 127 kJ/mol, while 2-butene releases only about 120 kJ/mol. Since both give the same butane, the ~7 kJ/mol difference is exactly how much more stable the disubstituted alkene is. The trend continues up the ladder: each additional alkyl substituent shaves several more kJ/mol off the alkene's energy, which is why a trisubstituted alkene beats a disubstituted one, and tetrasubstituted sits lowest of all. More substitution, lower energy, favored product.

4. The transition state has partial double-bond character

Zaitsev's rule works because the stability of the product alkene is already felt in the transition state. In an E2 reaction, as the base pulls off the proton and the leaving group departs, the two carbons rehybridize and a pi bond is partly formed at the transition state. That developing double bond enjoys the same hyperconjugative stabilization the finished alkene does, so the pathway leading to the more substituted alkene has the lower-energy transition state and the faster rate.

2-bromo-2-methylbutane with a small base gives the trisubstituted 2-methyl-2-butene, not the disubstituted 2-methyl-1-butene.

The same logic governs E1, where a carbocation loses a proton in the product-determining step: the base removes whichever proton yields the most stable alkene. E/Z geometry follows too — when the Zaitsev alkene can be trans (E) or cis (Z), the trans isomer usually dominates because its bulky groups sit on opposite sides, avoiding steric strain.

5. The exception: bulky bases give the Hofmann alkene

Zaitsev is a strong default, not a law. Swap the small ethoxide base for a big one — potassium tert-butoxide (KOtBu) or LDA — and the outcome flips. The more substituted hydrogen sits in a crowded position, and a bulky base is too large to reach it. It instead grabs a more exposed hydrogen at the end of the chain, giving the less substituted Hofmann alkene as the major product.

Same substrate, bulky base: KOtBu steers 2-bromo-2-methylbutane to the terminal 2-methyl-1-butene (Hofmann product) instead.

So the base's size, not just the substrate, decides the regiochemistry. Small strong base gives Zaitsev; bulky base gives Hofmann. We unpack the sterics in detail in bulky bases in elimination.

6. Summary

Zaitsev's rule predicts that eliminations favor the more substituted alkene because extra alkyl groups stabilize the double bond through hyperconjugation, giving both a lower-energy product (seen in heats of hydrogenation) and a lower-energy, product-like transition state. Identify the Zaitsev product by counting alkyl groups on each candidate C=C and picking the highest on the substitution ladder; when trans/cis is possible, the trans (E) isomer usually leads. The one big exception is a bulky base (KOtBu, LDA), which can only reach the exposed terminal hydrogen and delivers the less substituted Hofmann alkene. Master both the default and its exception and you can call the major product of nearly any elimination.