Pinnick oxidation Pinnick 氧化
（重定向至 Pinnick 氧化）
| Pinnick oxidation Pinnick 氧化 | |
|---|---|
| Named after 以命名 | Harold W. Pinnick 哈罗德·W·皮尼克 |
| Reaction type 反应类型 | Organic redox reaction 有机氧化还原反应 |
The Pinnick oxidation is an organic reaction by which aldehydes can be oxidized into their corresponding carboxylic acids using sodium chlorite (NaClO2) under mild acidic conditions. It was originally developed by Lindgren and Nilsson.[1] The typical reaction conditions used today were developed by G. A. Kraus.[2][3] H.W. Pinnick later demonstrated that these conditions could be applied to oxidize α,β-unsaturated aldehydes.[4] There exist many different reactions to oxidize aldehydes, but only a few are amenable to a broad range of functional groups. The Pinnick oxidation has proven to be both tolerant of sensitive functionalities and capable of reacting with sterically hindered groups. This reaction is especially useful for oxidizing α,β-unsaturated aldehydes, and another one of its advantages is its relatively low cost.[4][5]
Pinick 氧化是一种有机反应，通过使用亚氯酸钠（NaClO）在温和的酸性条件下，醛可以被氧化成相应的羧酸。这项反应最初由 Lindgren 和 Nilsson 开发。Kraus 通过 G. A. Kraus 开发了今天常用的典型反应条件。H.W. Pinnick 后来证明，这些条件可以应用于氧化α,β-不饱和醛。存在许多不同的氧化醛的反应，但只有少数对功能团具有广泛的适应性。Pinick 氧化已被证明对敏感功能团具有耐受性，并能够与空间位阻的基团反应。这种反应特别适用于氧化α,β-不饱和醛，另一个优点是其相对较低的成本。
PinnickOxidationReaction
Mechanism 机制
[edit]The proposed reaction mechanism involves chlorous acid as the active oxidant, which is formed under acidic conditions from chlorite.
提出的反应机制涉及亚氯酸作为活性氧化剂，它在酸性条件下从亚氯酸盐形成。
- ClO2− + H2PO4− ⇌ HClO2 + HPO42−
First, the chlorous acid adds to the aldehyde. Then resulting structure undergoes a pericyclic fragmentation in which the aldehyde hydrogen is transferred to an oxygen on the chlorine, with the chlorine group released as hypochlorous acid (HOCl).[6]
首先，氯酸加入到醛中。然后，生成的结构经历一个环状断裂过程，在这个过程中，醛中的氢原子转移到氯原子上的氧原子上，氯团释放为次氯酸（HOCl）。
Side reactions and scavengers
副反应和清除剂
[edit]The HOCl byproduct, itself a reactive oxidizing agent, can be a problem in several ways.[6] It can destroy the NaClO2 reactant:
副产品 HOCl 本身是一种反应性氧化剂，以多种方式造成问题。 [6] 它可以破坏 NaClO 2 反应物：
- HOCl + 2ClO2− → 2ClO2 + Cl− + OH−
HOCl + 2ClO → 2ClO + Cl + OH
making it unavailable for the desired reaction. It can also cause other undesired side reactions with the organic materials. For example, HOCl can react with double bonds in the organic reactant or product via a halohydrin formation reaction.
使其无法用于所需反应。它还可能导致与有机材料的其他不希望的副反应。例如，HOCl 可以通过形成卤代缩醛反应与有机反应物或产物中的双键反应。
To prevent interference from HOCl, a scavenger is usually added to the reaction to consume the HOCl as it is formed. For example, one can take advantage of the propensity of HOCl to undergo this addition reaction by adding a sacrificial alkene-containing chemical to the reaction mixture. This alternate substrate reacts with the HOCl, preventing the HOCl from undergoing reactions that interfere with the Pinnick reaction itself. 2-Methyl-2-butene is often used in this context:
为了防止 HOCl 的干扰，通常会在反应中添加清除剂，以消耗正在形成的 HOCl。例如，可以利用 HOCl 倾向于进行这种加成反应的倾向，向反应混合物中添加含有烯烃的牺牲化学品。这种替代底物与 HOCl 反应，防止 HOCl 进行干扰 Pinnick 反应本身的反应。2-甲基-2-丁烯通常在这种情况下使用：
Resorcinol and sulfamic acid are also common scavenger reagents.[6][7]
酚醛树脂和硫铵酸也是常用的清除剂试剂。
Hydrogen peroxide (H2O2) can be used as HOCl scavenger whose byproducts do not interfere in the Pinnick oxidation reaction:
过氧化氢（H2O2）可以用作 HOCl 清除剂，其副产品不会干扰 Pinnick 氧化反应：
- HOCl + H2O2 → HCl + O2 + H2O
HOCl + H2O → HCl + O2 + H2O
In a weakly acidic condition, fairly concentrated (35%) H2O2 solution undergoes a rapid oxidative reaction with no competitive reduction reaction of HClO2 to form HOCl.
在弱酸性条件下，相对集中的（35%）H 2 O 2 溶液与无竞争性的 HClO 2 发生快速氧化反应，形成 HOCl。
- HClO2 + H2O2 → HOCl + O2 + H2O
Chlorine dioxide reacts rapidly with H2O2 to form chlorous acid.
氯 dioxide 与 H 2 O 2 快速反应形成亚氯酸。
- 2ClO2 + H2O2 → 2HClO2 + O2
Also the formation of oxygen gives good indication of the progress of the reaction. However, problems sometimes arise due to the formation of singlet oxygen in this reaction, which may oxidize organic materials (i.e. the Schenck ene reaction). DMSO has been used instead of H2O2 to oxidize reactions that do not produce great yields using only H2O2. Mostly electron rich aldehydes fall under this category.[7] (See Limitation below)
形成氧也很好地指示了反应的进程。然而，在这个反应中由于形成了单态氧，有时会出现问题，单态氧可能会氧化有机材料（即 Schenck ene 反应）。在仅使用 H 2 O 2 无法产生大量产物的反应中，使用 DMSO 代替 H 2 O 2 来氧化反应。大多数电子丰富的醛类属于这一类别。 [7] （请参见下方的限制）
Also, solid-supported reagents such as phosphate-buffered silica gel supported by potassium permanganate and polymer-supported chlorite have been prepared and used to convert aldehydes to carboxylic acid without having to do conventional work-up procedures. The reaction involves the product to be trapped on silica gel as their potassium salts. Therefore, this procedure facilitates easy removal of neutral impurities by washing with organic solvents.[8]
此外，已经制备并使用了诸如由高锰酸钾支持的磷酸缓冲硅胶和聚合物支持的氯酸盐等固体支持试剂，将醛转化为羧酸，而无需进行常规的后处理步骤。该反应涉及将产物作为其钾盐捕获在硅胶上。因此，此过程便于通过用有机溶剂清洗轻松去除中性杂质。
Scope and limitations 范围和限制
[edit]The reaction is highly suited for substrates with many group functionalities. β-aryl-substituted α,β-unsaturated aldehydes works well with the reaction conditions. Triple bonds directly linked to aldehyde groups or in conjugation with other double bonds can also be subjected to the reaction.[7][9] Hydroxides, epoxides, benzyl ethers, halides including iodides and even stannanes are quite stable in the reaction.[7][9][10][11] The examples of the reactions shown below also show that the stereocenters of the α carbons remain intact while double bonds, especially trisubsituted double bonds do not undergo E/Z–isomerization in the reaction.
反应非常适合于具有多种官能团的底物。β-芳基取代的α,β-不饱和醛在反应条件下工作良好。直接连接到醛基的三键或与其他双键共轭的三键也可以接受反应。 [7] [9] 羟基、环氧基、苯基醚、卤化物包括碘化物甚至锡烷在反应中相当稳定。 [7] [9] [10] [11] 下面所示的反应示例还表明，在反应中，α碳的立体中心保持不变，特别是三取代双键不发生 E/Z 异构化。
Scope
Lower yields are obtained for reactions involving aliphatic α,β-unsaturated and more hydrophilic aldehydes. Double bonds and electron-rich aldehyde substrates can lead to chlorination as an alternate reaction. The use of DMSO in these cases gives better yield. Unprotected aromatic amines and pyrroles are not well suited for the reactions either. In particular, chiral α-aminoaldehydes do not react well due to epimerization and because amino groups can be easily transformed to their corresponding N-oxides. Standard protective group approaches, such as the use of t-BOC, are a viable solution to these problems.[12]
对于涉及脂肪族α,β-不饱和和更亲水性醛的反应，得到的产率较低。双键和电子丰富的醛基底物可能导致氯化作为替代反应。在这种情况下使用 DMSO 可以得到更好的产率。未保护的芳香胺和吡咯也不适合这些反应。特别是，手性α-氨基酸由于异构化和氨基可以很容易地转化为相应的 N-氧化物而不易反应。标准的保护基方法，如使用-BOC，是解决这些问题的可行方案。
Thioethers are also highly susceptible to oxidation. For example, Pinnick oxidation of thioanisaldehyde gives a high yield of carboxylic acid products, but with concomitant conversion of the thioether to the sulfoxide or sulfone.[7]
硫醚同样高度易氧化。例如，Pinnick 氧化硫代安息香醛可以得到羧酸产物的高产率，但同时硫醚会转化为硫醚或砜。
See also 也请参见
[edit]References 参考文献
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维基媒体共库包含与 Pinick 氧化相关的媒体。