Organic Chemistry 1 · Reaction energy and kinetics
tert-Butyl chloride from 2-methylpropene: a two-step profile
Conditions: HCl
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Protonation to the tertiary cation (rate-determining)
The π bond takes the proton of HCl at the CH2 end, so the positive charge lands on the carbon carrying two methyl groups. This step is uphill, so by the Hammond postulate its transition state looks like the tertiary cation and is lowered by the same hyperconjugation that stabilizes it.
The curved arrows
- The π electrons of the alkene bond to the acidic proton on the less substituted CH2 carbon, the Markovnikov orientation.
- The H–Cl bond pair stays with chlorine, which leaves as chloride ion.
Chloride captures the cation
Chloride bonds to the cationic carbon with almost no barrier. On the diagram this second hill is small and sits just above the carbocation valley, so the first hill remains the highest point.
The curved arrows
- A chloride lone pair forms the C–Cl bond to the electron-poor tertiary carbon.
tert-Butyl chloride
The profile has two transition states and one intermediate, the tertiary carbocation. The overall reaction is downhill, and the slow protonation decides both the rate and the Markovnikov regiochemistry.
What you'll practice
Draw the two steps of HCl addition to 2-methylpropene and use the Hammond postulate to explain why the tertiary cation forms and controls the rate.
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