E1, E2 and E1cb: Three Ways to Lose Two Groups · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Three ways to lose two groups

Science · Chemistry · ages 19-20
Name ______________________   Date ____________

  1. What geometry does the concerted route demand?

    • The breaking bonds must sit anti periplanar, 180 degrees apart
    • The breaking bonds must sit side by side at zero degrees
    • Geometry never matters for elimination

  2. How do E2, E1 and E1cb differ in timing?

    • All three are concerted single steps
    • E2 is concerted, E1 loses the leaving group first, E1cb loses the proton first
    • E1 is concerted and E2 goes through a carbanion

  3. The Zaitsev preference favours the more substituted alkene.

    Circle one:   True   False

  4. A substrate has a poor leaving group beside an acidic proton. Which door opens?

    • The E1cb door through a carbanion
    • The E1 door through a carbocation
    • No elimination is possible

  5. A bulky strong base meets a hindered halide. Which route and product follow?

    • E1 with the Zaitsev alkene
    • E1cb with no alkene at all
    • E2 Hofmann, open proton

  6. A cyclohexyl halide holds only one proton anti to the leaving group, and it is not the one leading to the Zaitsev product. Which alkene forms?

    • The Zaitsev product, since it is always more stable
    • The alkene from the lone anti proton
    • Both alkenes in equal amounts

  7. A carbanion-like elimination gives mostly the less substituted alkene. A student calls it Zaitsev. Why is that wrong?

    • Carbanion paths lean Hofmann
    • Less substituted alkenes cannot form at all
    • Names never apply to carbanion routes

  8. A student predicts the Zaitsev product from an equatorial leaving group with no anti neighbour. What is wrong?

    • Nothing, stability always overrules geometry
    • The concerted path is shut
    • Equatorial groups always eliminate fastest
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Three ways to lose two groups W1-mt_vVq_xeZrKv-s1

  1. The breaking bonds must sit anti periplanar, 180 degrees apart · Only aligned orbitals can merge into a pi bond.
  2. E2 is concerted, E1 loses the leaving group first, E1cb loses the proton first · Order of departures defines each route.
  3. True · More alkyl groups around the double bond stabilize it.
  4. The E1cb door through a carbanion · Acidic protons deprotonate first when the leaving group lingers.
  5. E2 Hofmann, open proton · Bulk cannot reach crowded protons, so it takes the open one.
  6. The alkene from the lone anti proton · Structure dictates reactivity here, not wishes.
  7. Carbanion paths lean Hofmann · E1cb crowds lean the other way.
  8. The concerted path is shut · No alignment means no concerted elimination.
Worksheet · LightMySky