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7 E1 vs E2
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Flashcards in this deck (47)

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  • How does alkene stability vary with degree of substitution?

    • More substituted alkenes are more stable; order: tetra > tri > di > mono > unsubstituted
    alkene stability

  • What stereochemical requirement for E2 eliminations is mentioned?

    • Anti‑periplanar
    e2 stereochemistry

  • Which product-prediction trends or rules are mentioned for elimination reactions?

    • Zaitsev's rule
    • Hofmann elimination
    elimination rules

  • Why is the most substituted alkene the most stable?

    • Hyperconjugation: sigma bonds overlap with the pi system, stabilizing the alkene.
    alkene stability hyperconjugation

  • Which alkene substitution level is most stable?

    • Tetra-substituted alkene: four carbon chains bonded to the sp2 carbons; this is the most stable.
    • Tri-substituted is less stable and higher in energy.
    alkene substitution stability

  • What is an elimination reaction in organic chemistry?

    An elimination reaction is losing two sigma bonds from adjacent atoms to form one pi bond.

    elimination definition

  • What two features are required for an elimination reaction to occur?

    • A leaving group
    • A beta hydrogen
    elimination requirements

  • Rank alkene stability by substitution according to the text (most to least stable).

    • Di-substituted (trans more stable than cis due to less steric strain)
    • Mono-substituted
    • Unsubstituted (ethylene) (highest energy)
    alkenes stability

  • Why is a trans di-substituted alkene more stable than the cis isomer?

    Because the trans isomer has less steric strain.

    alkenes steric

  • What does Zaitsev's rule predict in elimination reactions?

    • It helps determine the major product when multiple elimination products are possible
    elimination zaitsev

  • How is Zaitsev's rule alternatively spelled?

    • Saytzeff's
    nomenclature zaitsev

  • What does Zaitsev's rule state about which alkene is formed in an elimination?

    • Preferential formation of the more substituted alkene (elimination from the β-carbon with fewer hydrogens) because it is more stable.
    zaitsev alkenes stability

  • Why are trans alkenes generally more stable than cis alkenes?

    • Trans isomers are generally more stable than cis isomers due to steric factors.
    stereochemistry stability

  • What is an elimination that yields the least substituted alkene called?

    • Anti-Zaitsev (Hofmann) elimination
    elimination hofmann zaitsev

  • What does Zaitsev's rule typically predict?

    • The major product
    zaitsev elimination

  • When can the Hofmann product be favored in eliminations?

    • Under specific conditions with certain E2 reactions
    hofmann e2

  • Why is E2 elimination described as 'bimolecular'?

    Because both the base and substrate are involved in the rate-determining step.

    e2 kinetics

  • How does E2 differ from SN2 regarding the attacking species?

    • SN2: uses a nucleophile
    • E2: uses a base
    comparison e2

  • What type of base is typically required for E2 and give an example?

    • Type: a strong base (often with a negative charge on oxygen)
    • Example: sodium hydroxide (NaOH)
    bases e2

  • What is the E2 mechanism in elimination reactions?

    • Concerted single-step: base (OH-) deprotonates a beta-hydrogen; electrons from the C-H bond form a new pi bond; the leaving group (e.g., bromide) departs simultaneously.
    mechanism elimination e2

  • What is the rate law for the E2 mechanism?

    \(\text{Rate} = k[\text{Base}][\text{Substrate}]\)

    kinetics e2

  • What is the rate law for the reaction of NaOH with 2-bromobutane?

    \(Rate = k[\mathrm{NaOH}][\text{2-bromobutane}]\)

    kinetics organic

  • What stereochemical requirement must be met for E2 eliminations?

    Anti-periplanar geometry: the beta-hydrogen and the leaving group must be \(180^\circ\) apart.

    stereochemistry e2 organic

  • What does anti-periplanar mean in the context of elimination reactions?

    • Anti-periplanar means the hydrogen and leaving group are positioned opposite each other, nearly 180 degrees apart, though slight deviations are acceptable.
    stereochemistry definition

  • Why is the anti-periplanar geometry important for the E2 mechanism?

    • This geometry is essential for the E2 mechanism to proceed efficiently.
    e2 mechanism

  • How is the anti-periplanar arrangement commonly visualized in molecular drawings?

    • It's often visualized by having one group on a wedge and the other on a dash.
    visualization stereochemistry

  • What relative arrangement is required between the beta-hydrogen and the leaving group?

    • Anti-periplanar
    stereochemistry

  • If the beta-hydrogen and leaving group are not anti-periplanar, what may the molecule need to do?

    • Rotate to achieve the anti-periplanar arrangement
    conformation

  • Why is the dashed hydrogen in the drawing described as key?

    • Because it is positioned correctly relative to the leaving group
    visual

  • What stereochemical arrangement does an E2 reaction require for the electron movement?

    An anti-periplanar arrangement between the hydrogen and the leaving group.

    e2 stereochemistry

  • If a molecule lacks a hydrogen and leaving group in the anti-periplanar orientation, what may be necessary?

    Rotation around single bonds may be necessary to bring them into the correct orientation.

    conformation e2

  • What is the consequence if a beta carbon has no hydrogens for elimination?

    • No elimination possible
    elimination beta

  • For the second substrate, what substituents are on the beta carbon and how are they oriented?

    • One hydrogen (on a dash)
    • One methyl group (on a wedge)
    stereochemistry beta

  • What stereochemical arrangement between the beta‑hydrogen and the leaving group enables the E2 elimination in the second substrate?

    They are anti‑periplanar: the beta‑hydrogen is on a dash and the bromine (leaving group) is on a wedge.

    e2 stereochemistry elimination

  • What is the effect of this anti‑periplanar arrangement on the E2 elimination's progress in the example?

    It allows the E2 elimination to proceed readily.

    e2 reactivity

  • Which alkene stereochemical product forms from this E2 elimination when the substituents are trans?

    The E product (the alkene with substituents trans).

    product stereochemistry

  • What stereochemical arrangement is required for an E2 elimination to occur from a given conformation?

    • The beta-hydrogen must be anti-periplanar to the bromine; if rotation places the beta‑H out of anti-periplanar alignment, the E2 reaction will not occur from that conformation.
    organic e2 stereochemistry

  • What stereochemical requirement must be met for an elimination reaction to occur?

    • The beta-H and leaving group must be anti-periplanar.
    stereochemistry elimination

  • How do you determine which beta hydrogens can participate when multiple are present on a substrate?

    • Only beta hydrogens that are anti-periplanar to the leaving group can participate in the elimination.
    elimination mechanism

  • What stereochemical arrangement between bromine and a beta hydrogen allows elimination to form the E product?

    • Anti‑periplanar
    stereochemistry elimination

  • If the conformation shown does not have the hydrogen anti‑periplanar to the bromine, what is required?

    • Rotation is needed
    conformation elimination

  • Which conformation of a beta-hydrogen relative to bromine allows its removal?

    • Anti-periplanar — the beta-hydrogen that is anti-periplanar to the bromine can be removed.
    stereochemistry conformation

  • What alkene stereochemical product forms after removal of the anti-periplanar beta-hydrogen?

    • Z product — the substituents are cis.
    elimination product

  • Which beta-hydrogen will be removed in an elimination reaction?

    • Only the beta-hydrogen that can achieve anti-periplanar alignment with the leaving group will be removed.
    stereochemistry elimination

  • When multiple beta-hydrogens are available, what dictates which one is removed?

    • Zaitsev's rule often dictates removal based on alkene stability, but the anti-periplanar stereochemical requirement must always be met first.
    zaitsev stereochemistry elimination

  • When might E/Z isomerism be not observable if a beta carbon has two identical groups?

    • E/Z may not be observable when the other substituents on the double bond are also identical.
    stereochemistry alkene

  • What condition allows formation of both E and Z products for an alkene?

    • Both E and Z products can form if the alkene has distinct substituents on the double-bond carbons and appropriate anti‑periplanar hydrogens are available.
    stereochemistry alkene
Çalışma Notları

Quick overview

  • Focus: distinguishing E1 vs E2 eliminations, predicting products (Zaitsev vs Hofmann), and the anti‑periplanar stereochemical requirement for E2.
  • Key idea: more substituted alkenes are more stable (drives product distribution unless steric/base effects favor Hofmann).

Alkene stability (why substitution matters)

  • Trend: tetra‑substituted > tri‑ > trans‑di > cis‑di > mono‑ > unsubstituted in stability.
  • Reason: hyperconjugation and alkyl electron donation stabilize the π system.
  • Consequence: elimination usually forms the more substituted alkene (Zaitsev product) when allowed.

Zaitsev (Saytzeff) vs Hofmann

  • Zaitsev's rule: elimination favors formation of the more substituted (more stable) alkene.
  • Hofmann (anti‑Zaitsev): gives the less substituted alkene; favored with bulky bases, poor leaving groups, or sterically hindered substrates.
  • Stereochemistry: when E/Z matters, trans (E) is generally more stable than cis (Z) due to lower steric strain.

E2 mechanism (detailed)

  • Definition: bimolecular, concerted elimination where base and substrate participate in the rate‑determining step.
  • Mechanism steps: base removes a β‑H → C–H electrons form the C=C π bond → leaving group departs, all in one concerted step.
  • Rate law: \(\text{Rate} = k[\text{Base}][\text{Substrate}]\) (second order overall).
  • Base requirements: strong base (often negatively charged oxygen bases like OH–, OR–) favors E2.
  • Stereochemical requirement: anti‑periplanar alignment of the β‑H and leaving group is required for efficient orbital overlap; they should be ~180° apart.
  • If not aligned, rotation about single bonds may bring them into anti alignment (if conformationally possible).
  • In cyclic systems, only certain β‑H positions meet anti‑periplanar geometry, restricting which alkene forms.
  • Product control: Zaitsev usually, but Hofmann product can dominate with bulky base or steric constraints preventing Zaitsev formation.

E2 examples & implications

  • Example: 2‑bromobutane + NaOH → E2 with \(\text{Rate} = k[\text{NaOH}][\text{2‑bromobutane}]\); product is the more substituted 2‑butene (Zaitsev), but conformation (anti arrangement) determines E vs Z stereochemistry.
  • If a β‑carbon has no hydrogens, elimination from that carbon is impossible.
  • When multiple β‑H are present, only those that can adopt anti‑periplanar orientation to the leaving group can be removed.
  • If rotation gives an anti β‑H that leads to a cis arrangement, the reaction can form the Z isomer instead.

E1 mechanism (brief comparison)

  • Definition: unimolecular elimination via carbocation intermediate; rate depends only on substrate.
  • Rate law: \(\text{Rate} = k[\text{Substrate}]\) (first order).
  • Favored by: weak bases, polar protic solvents, tertiary substrates, and heat.
  • Key differences vs E2: no anti‑periplanar requirement, carbocation rearrangements possible, product distribution driven by carbocation stability and subsequent deprotonation (usually Zaitsev).

How to distinguish E1 vs E2 (practical checklist)

  1. Substrate: primary favors E2 (if strong base); tertiary favors E1 or E2 (if strong base then E2, if weak base/solvent then E1).
  2. Base strength: strong base → E2; weak base/solvent → E1.
  3. Rate law / kinetics: experimentally, second order suggests E2, first order suggests E1.
  4. Stereochemistry needed: if anti‑periplanar requirement explains product distribution, E2 is likely.
  5. Rearrangements observed: rearrangements imply an E1 carbocation intermediate.
  6. Bulky base: favors Hofmann product via E2.

Common pitfalls & tips

  • Don’t assume Zaitsev always wins: check base size, sterics, and conformational accessibility for anti‑periplanarity.
  • In cyclic systems, only β‑H that are trans‑diaxial (anti‑periplanar equivalent) lead to elimination.
  • If a β‑carbon lacks hydrogens, that β position cannot undergo elimination.
  • When predicting E or Z, determine which β‑H is anti to the leaving group in a low‑energy conformation.

Quick study checklist

  • Know the stability order of alkenes and why (hyperconjugation).
  • Memorize Zaitsev vs Hofmann conditions.
  • Be able to apply the anti‑periplanar rule to 3D structures and cyclic systems.
  • Use the distinguishing checklist (substrate, base, kinetics, stereochemistry, rearrangements) to assign mechanism.