H3 Chemistry 9813 · Study focus: H3 Chemistry: Compare E1 and E2 profiles and rate laws
H3 Chemistry: Compare E1 and E2 profiles and rate laws
Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.
Your success criteria
- Compare E1 and E2 profiles and rate laws
- Use named chemical evidence.
- Transfer the governing reason to an unfamiliar case.
Match kinetics to elementary steps
E1 ionisation forms a carbocation before deprotonation; E2 couples proton removal, π-bond formation and leaving-group departure in one step.
Use concentration-rate data together with profile topology and curved arrows rather than identifying a mechanism from one label alone.
Rate laws and profile features
The simple E1 rate law is rate = k[RX], first order overall; its profile has two maxima separated by a carbocation minimum.
The E2 rate law is rate = k[RX][base], second order overall; its concerted profile has one maximum and no intermediate minimum.
Changing base concentration does not alter the rate-determining unimolecular ionisation in E1, although base removes a proton in the later step.
In E2, both substrate and base occur in the sole elementary step, so doubling either concentration doubles rate when the other and temperature are fixed.
Text alternative: The text enumerates every maximum and minimum, names the species at each level and states both rate laws and units.
Relate orders to mechanism
Changing base concentration does not alter the rate-determining unimolecular ionisation in E1, although base removes a proton in the later step.
In E2, both substrate and base occur in the sole elementary step, so doubling either concentration doubles rate when the other and temperature are fixed.
Extract an E2 rate constant
At 298 K, [2-bromobutane] = 0.040 mol dm−3, [ethoxide] = 0.050 mol dm−3 and rate = 1.20 × 10−4 mol dm−3 s−1.
k = rate/([RX][base]) = 0.060 dm3 mol−1 s−1.
- Recalculate k and predict the rate if [ethoxide] doubles.
Open the feedback checkpoint after attempting
- Award 0.060 dm3 mol−1 s−1 and 2.40 × 10−4 mol dm−3 s−1 at fixed substrate.
Read an E1 profile
Ionisation of tert-butyl bromide produces a carbocation minimum after the first transition-state maximum.
A second, usually smaller barrier represents base removal of β-H to form alkene; the rate law remains governed by substrate ionisation.
- Annotate reactant, two transition states, carbocation intermediate and product on the profile.
Open the feedback checkpoint after attempting
- Require two maxima, one intervening minimum and rate = k[tert-butyl bromide].
Use an initial-rate table
Runs at fixed [RX] show identical rate when [water] doubles, while doubling [RX] doubles rate.
This supports first order in substrate and zero order in the base/nucleophile for the measured rate, consistent with simple E1 ionisation.
- Infer the rate law from the stated factors and give k units.
Open the feedback checkpoint after attempting
- Credit rate = k[RX] and s−1, with a cautious ‘consistent with E1’ conclusion.
Do not count arrows as steps
Three simultaneous curved arrows in E2 depict one elementary step and one transition state.
E1 uses more than one arrow-bearing step because the carbocation is a real intermediate at an energy minimum, not a transition state.
- Repair: ‘E2 has three transition states because it has three curved arrows.’
Open the feedback checkpoint after attempting
- Reject it; the arrows are simultaneous electron movements within one concerted transition state.
Triangulate mechanism evidence
Report reaction orders, calculate k with units, count profile maxima/minima and map each curved arrow to the proposed step.
Next compare how these distinct pathways choose among available β-sites and alkene regioisomers.
- State the minimum evidence for distinguishing E1 from E2.
Open the feedback checkpoint after attempting
- Require base dependence, overall order, profile topology, presence/absence of carbocation and concerted/stepwise arrows.