Química Orgânica 2 · Substituição acílica e escolha do derivado
Make an acid chloride with thionyl chloride
Condições: Benzoic acid with thionyl chloride (SOCl₂) and pyridine. Follow the chlorosulfite pathway to benzoyl chloride, sulfur dioxide and pyridinium chloride. Sulfur is drawn with the usual expanded-valence S=O bonds. Pyridine acts only as a base here; its possible role as an acyl-transfer catalyst is not modeled.
Desenhe este problema você mesmo
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 6 passos escritos
Oxygen attacks sulfur
Sulfur in SOCl₂ is electron-poor: it carries an oxygen and two chlorines. The acid OH is a poor leaving group; bonding it to sulfur is the first step in changing that.
As setas curvas
- A lone pair on the hydroxyl oxygen bonds to sulfur.
- The S=O π pair becomes a lone pair on oxygen so sulfur does not gain an extra bond.
Expel chloride from sulfur
Addition to sulfur followed by loss of chloride is a substitution at sulfur. The chloride released here returns later as the nucleophile.
As setas curvas
- A lone pair on the negatively charged oxygen rebuilds the S=O π bond.
- The S–Cl pair leaves with chlorine as chloride.
Pyridine removes the O–H proton
This positively charged oxygen is a very strong acid (pKa well below 0), while pyridinium has pKa approximately 5.2, so pyridine takes the proton essentially completely. The product is an acyl chlorosulfite.
As setas curvas
- The pyridine nitrogen lone pair accepts the proton on the positively charged oxygen.
- The O–H pair stays on the oxygen bonded to carbon and sulfur, which becomes neutral.
Chloride adds to the carbonyl
The chlorosulfite group makes the carbonyl carbon more electrophilic than it was in the acid, so even chloride can add.
As setas curvas
- A chloride lone pair bonds to the carbonyl carbon.
- The C=O π pair becomes a lone pair on oxygen.
Collapse and fragment
The leaving group breaks apart as it leaves, into SO₂ gas and chloride. Loss of gaseous SO₂ makes the overall reaction effectively irreversible.
As setas curvas
- A lone pair on the negatively charged oxygen rebuilds the C=O π bond.
- The bond electrons between the acyl carbon and the oxygen bonded to sulfur form a second S=O π bond.
- The remaining S–Cl pair leaves as chloride, completing sulfur dioxide.
Acid chloride, sulfur dioxide and pyridinium chloride
The acyl carbon now carries chlorine in place of the original hydroxyl oxygen: benzoyl chloride. That oxygen left with sulfur as part of SO₂, and the remaining sulfur-bound chlorine became chloride next to pyridinium. Overall, the acid OH was replaced by Cl, with SO₂ and HCl (captured by pyridine) as byproducts.
O que você vai praticar
Convert a carboxylic acid OH into a leaving group at sulfur, then complete chloride addition and fragmentation to SO₂.
Você acabou de desenhar cada seta deste problema.
Cada um dos 5349 problemas de Orgânica 1 e 2 funciona assim: feedback em cada seta, dicas quando você travar e um modo prova sem dicas. Grátis com uma conta, sem cartão.
Travou na sua própria lição?
Tire uma foto: o primeiro scan é grátis, sem cadastro
Você escaneia toda semana? Scan Standard: por cerca de problemas