Química Orgânica 2 · Enolatos: alquilação, condensações e adições conjugadas
2-Bromobutanoic acid by the Hell–Volhard–Zelinsky reaction
Condições: Br2, PBr3 (cat.); then H2O
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Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
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The acid OH attacks PBr3
Catalytic PBr3 first converts a little butanoic acid into butanoyl bromide. The O–H oxygen lone pair attacks phosphorus and bromide leaves.
As setas curvas
- An O–H lone pair bonds to phosphorus.
- A P–Br pair leaves as bromide.
Bromide adds to the carbonyl
Bromide attacks the carbonyl carbon, now activated by the positive oxygen next to it.
As setas curvas
- Bromide forms a bond to the carbonyl carbon.
- The C=O π pair moves onto oxygen.
Collapse to butanoyl bromide
The oxyanion re-forms C=O and expels HOPBr2, an excellent leaving group, giving the acyl bromide.
As setas curvas
- An oxygen lone pair re-forms C=O.
- The C–O pair leaves with HOPBr2.
HBr protonates the acyl bromide
The HBr formed in the reaction (assumed present) protonates the carbonyl oxygen. Acyl bromides enolize far more readily than the acid itself, which is why PBr3 is needed.
As setas curvas
- The carbonyl oxygen takes the HBr proton.
- The H–Br pair stays on bromide.
Form the enol
Bromide removes the α-hydrogen; the C–H pair forms C=C and the C=O⁺ π pair returns to oxygen, giving the enol of butanoyl bromide.
As setas curvas
- Bromide takes the α-proton.
- The C–H pair forms the enol C=C.
- The π pair returns to oxygen.
The enol attacks Br2
The enol C=C bonds to bromine with a push from the O–H lone pair, and bromide leaves: bromination happens on the α-carbon.
As setas curvas
- An oxygen lone pair forms C=O⁺.
- The C=C π pair forms the new C–Br bond.
- The Br–Br pair leaves as bromide.
Deprotonate the carbonyl
Bromide removes the O–H proton, giving neutral 2-bromobutanoyl bromide and regenerating HBr.
As setas curvas
- Bromide takes the O–H proton.
- The O–H pair stays on oxygen.
HBr protonates the acyl bromide again
In the aqueous workup the acyl bromide is hydrolysed. HBr, the strongest acid present, first protonates its carbonyl oxygen so that water can add.
As setas curvas
- The carbonyl oxygen takes the HBr proton.
- The H–Br pair stays on bromide.
Water adds to the carbonyl
Water attacks the activated carbonyl carbon; the C=O⁺ π pair moves onto oxygen, giving a neutral O–H.
As setas curvas
- A water lone pair bonds to the carbonyl carbon.
- The π pair moves onto the protonated oxygen.
Water removes the oxonium proton
A second water molecule takes the proton from the oxygen that just added.
As setas curvas
- A water lone pair takes the proton.
- The O–H pair stays on oxygen.
Lose bromide
An O–H lone pair forms C=O⁺ and pushes bromide out of the acyl carbon.
As setas curvas
- An oxygen lone pair forms the C=O⁺ bond.
- The C–Br pair leaves as bromide.
Water removes the last proton
Water deprotonates the protonated carbonyl, giving 2-bromobutanoic acid.
As setas curvas
- A water lone pair takes the proton.
- The O–H pair stays on the carbonyl oxygen.
Why PBr3?
Carboxylic acids barely enolize; acyl bromides enolize readily.
2-Bromobutanoic acid
Bromine is now on the α-carbon: 2-bromobutanoic acid (racemic, from the planar enol). HBr and HOPBr2 are by-products.
O que você vai praticar
Draw the HVZ sequence on butanoic acid: conversion to the acyl bromide with PBr3, enolization, α-bromination by Br2 and hydrolysis to 2-bromobutanoic acid.
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