Química Orgânica 1 · Eliminação e caminhos competitivos
Choose the Zaitsev proton after ionization
Condições: 2-Methylbutan-2-ol heated with concentrated sulfuric acid (H₂SO₄). Follow the E1 dehydration to the major product, 2-methylbut-2-ene; 2-methylbut-1-ene is the minor product and no ratio is claimed. H₂SO₄ is drawn in full, and the HSO₄⁻ it forms is the base in the final step. Carbons are numbered as in the name (C1–C4 along the main chain); the explicit H is attached to the CH₂ carbon of the ethyl group.
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Arraste dos elétrons que se movem até onde eles vão, ou toque em um e depois no outro. Cada seta é verificada; Dica ajuda quando você travar. Não precisa de conta.
O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 5 passos escritos
Protonate the hydroxyl group
Sulfuric acid is a very strong acid, and protonation converts the poor –OH leaving group into –OH₂⁺, which can leave as water.
As setas curvas
- A lone pair on the alcohol oxygen bonds to the acidic H of sulfuric acid.
- The O–H pair stays on the sulfate oxygen, giving hydrogen sulfate, HSO₄⁻.
Lose water
Ionization is the slow step. The cation is planar at C2, and every C–H bond on a neighboring carbon can now line up with its empty p orbital.
As setas curvas
- The C2–O pair leaves with oxygen as neutral water, leaving a tertiary carbocation at C2.
Which β proton?
Three carbons are attached to C2: C1, the 2-methyl carbon and C3. Removing an H from any of them makes a C=C bond to C2, but the alkenes differ in how many carbon groups sit on the double bond.
Remove the C3 proton
The weak base HSO₄⁻ is enough to remove a proton next to the cation. H₂SO₄ is regenerated.
As setas curvas
- A lone pair on the negatively charged oxygen of HSO₄⁻ bonds to the drawn hydrogen on the ethyl group.
- The C3–H pair becomes the C2=C3 π bond, neutralizing C2.
2-Methylbut-2-ene, water and sulfuric acid
The double bond carries three carbon groups: two methyls at one end and one methyl at the other. It has no E/Z isomers because one alkene carbon bears two identical methyl groups. The oxygen left as water, and H₂SO₄ is recovered unchanged, so it is a catalyst. Concentrated acid also binds the water formed, which helps drive the dehydration.
O que você vai praticar
Decide which of two kinds of β hydrogen an E1 base removes from a tertiary cation, and form the more substituted alkene.
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