Elimination Reactions and the Substitution-Elimination Competition: every key term you need (+ practice quiz)
25 flashcard terms for Organic Chemistry I Topic 6, written to match the course framework. Study them here, then drill them as interactive flashcards, or test yourself with the 15-question quiz — free, no account needed.
Loss of two groups from adjacent carbons to create a pi bond. In the common case a proton and a leaving group depart from neighboring positions to give an alkene.
Beta hydrogen
A hydrogen on a carbon adjacent to the one bearing the leaving group. Its removal is what creates the new pi bond, so a substrate lacking one cannot eliminate.
Bimolecular elimination
A single concerted step in which base removal of a beta hydrogen, pi bond formation, and leaving group departure happen together. Its rate depends on both substrate and base.
Unimolecular elimination
A stepwise route where the leaving group departs first to give a carbocation and a weak base then removes an adjacent proton. Its rate is independent of base concentration.
Anti-periplanar requirement
In a concerted elimination the beta hydrogen and the leaving group must lie in the same plane on opposite sides, so their orbitals align to form the new pi bond.
Syn-periplanar arrangement
The alternative coplanar geometry with both groups eclipsed on the same side. It is far less favorable and appears only when the anti arrangement is geometrically impossible.
Zaitsev product
The more highly substituted and therefore more stable alkene, formed when a small base removes the proton that leads to the most alkyl-bearing double bond.
Hofmann product
The less substituted alkene, favored when a bulky base can only reach the most exposed beta hydrogen or when the leaving group is unusually large.
Alkene stability order
Stability rises with the number of alkyl groups on the double bond, and for a given substitution count the trans arrangement beats the cis one because of reduced crowding.
Potassium tert-butoxide
A strong, sterically demanding alkoxide base used specifically to push a reaction toward elimination and toward the less substituted alkene.
Ethoxide as a base
A strong but small alkoxide that acts as both base and nucleophile, so with secondary substrates it gives a mixture of elimination and substitution products.
Ring elimination geometry
On a six-membered ring the leaving group must occupy an axial position before a concerted elimination can occur, since only then is a neighboring hydrogen anti-periplanar.
Substrate class as the first filter
Deciding among the four pathways begins with whether the reacting carbon is primary, secondary, or tertiary, because that alone rules out several options immediately.
Reagent strength as the second filter
After the substrate, ask whether the reagent is a strong base, a strong nucleophile that is a weak base, or a weak neutral species, since each points to a different route.
Temperature effect on elimination
Heating favors elimination over substitution because breaking one molecule into two raises entropy, and that entropy term grows in importance as temperature rises.
Bulky base effect
Steric demand suppresses attack at carbon while leaving proton removal easy, so hindered bases convert substitution-prone conditions into clean elimination conditions.
Weak base and weak nucleophile conditions
Heating a tertiary substrate in a protic solvent alone gives ionization followed by a mixture of solvent capture and proton loss from the cation.
Dehydrohalogenation
Removal of hydrogen halide from an alkyl halide using a strong base, the standard laboratory route from a halide to the corresponding alkene.
Acid catalyzed dehydration
Heating an alcohol with concentrated sulfuric or phosphoric acid protonates the hydroxyl, water leaves, and loss of a neighboring proton gives the more substituted alkene.
Rearrangement during dehydration
Because a cation forms, the skeleton can shift before the proton is lost, so an alcohol may yield an alkene whose double bond sits on a rearranged framework.
Deuterium kinetic isotope effect
Replacing the beta hydrogen with deuterium slows a concerted elimination noticeably, which is direct evidence that the carbon-hydrogen bond breaks in the rate determining step.
Regiochemistry of elimination
Which beta hydrogen is removed decides where the double bond appears. Base size, leaving group size, and substrate geometry together settle the outcome.
Stereospecific alkene formation
Because the concerted route demands a specific coplanar geometry, a single diastereomer of the starting halide gives a single geometric isomer of the alkene.
Competing product mixture
Secondary substrates with small strong bases are the classic ambiguous case, since substitution and elimination proceed at comparable rates and give a blend of products.
Twofold elimination to an alkyne
Treating a dihalide with excess strong base removes hydrogen halide twice, first giving a vinyl halide and then the triple bond after a second, harsher deprotonation.