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Alcohol
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Alcohol Alcohols are characterized by the presence of OH (hydroxyl group) attached to a carbon atom. Aliphatic alcohols may be considered to be derived from hydrocarbons in which an sp3 bonded hydrogen atom has been replaced with OH: R—H (Alkane) 烷烃 R—OH (Alcohol) 醇 They may also be considered as derivatives of water in which one of the hydrogens has been replaced with an alkyl group, R—. H—O—H (Water) R—0—H (Alcohol) Since hydrocarbons may contain primary, secondary, or tertiary hydrogens, the same classes of alcohols are capable of existence. The following examples illustrate the typical members of each class: CH3CH2OH (CH3)2CHOH (CH3)3COH l°Alcohol 2°Alcohol 3°Alcohol (Ethyl alcohol) (Isopropyl alcohol) (t-butyl alcohol) 乙醇 异丙醇 叔丁醇 醇 醇的特点是存在与碳原子相连的羟基。脂肪醇可以被认为是由烃中一个sp3杂化键上的氢原子被OH替换而生成的。它们也可以被认为是一种由水的一个氢原子被烷基基团所取代而生成的衍生物。烃可能包含初级、二级、三级氢原子，醇也存在同样的分类。下面的例子说明了每一类的典型成员。 Oxidation Reaction of C—H Bond Alcohols may be considered to be the first product of oxidation of the alkanes in the oxidation scheme, which eventually produces carbon dioxide and water. Alcohols might be expected to be subject to further oxidation .This has been found to be the case as long as a hydrogen atom (a -hydrogen) remains bonded to the carbon atom which has already been partially oxidized. The following equations illustrate the structural changes which occur; it should be emphasized that they do not illustrate the mechanistic steps of the reactions. The above reaction involve the C—H bond. Since 3°alcohol have no α-hydrogen atom bonded to the partially oxidized carbon atom, they resistant to further oxidation. 1. 醇的氧化 醇可能被认为是由在烷烃氧化生成CO2和H2O的氧化方案中产生的首个氧化产物。醇可能会被进一步氧化。已经发现，只要部分氧化的碳原子上仍有氢原子（α-氢），就有进一步氧化的可能。下面的公式说明发生的结构变化，这里需要强调的是，他们没有说明反应的机理。上述反应涉及C—H键。由于部分氧化的碳原子上没有α-氢原子，它们将不会进一步反应。 2. Reaction of O—H Bond The proton bonded to oxygen of an alcohol is much more acidic than protons bonded to carbon. This difference can be readily accounted for on the greater electronegativity of oxygen, which polarizes the 0—H bond." The hydrogen on oxygen is replaceable by sodium for example 2RCH2OH+2Na—>2RCH2C-Na+H2. This reaction is analogous to the liberation of hydrogen by reaction of an acid and a metal. The order of reactivity of alcohols in this reaction is : l°>2°>3°.The reason for this order of reactivity is believed to be the result of the inductive effects of alkyl groups. Utilizing the relative electronegativity value of 2.1 for hydrogen and 2.5 for carbon, the H—C may be represented as follows: Hδ+-> Cδ-. Thus, the larger the number of such bonds, as in the i-butyl group, the greater the combined electron release effect would be expected. As the combined electron release effect increases, the greater the destabilizing effect on the alkoxide anion, which thus accounts for the observed order of reactivity: 1°>2°>3°. 2. 醇的弱酸性。 结合在醇上氧的质子比结合在碳上的酸性强。这种差别很容易用氧的电负性较大，因而它极化了O—H键而进行解释。举个例子，氢氧键中的氢原子被钠原子取代。这种反应类似于酸和金属反应生成氢气。在这种反应中醇的反应活性顺序为l°>2°>3°。这种反应活性顺序被认为是烷基诱导效应的结果。利用氢的相对电负性值2.1和碳的相对电负性值2.5，H—C可以表示如下：Hδ+-> Cδ-.因此这种键的数目越多，如在叔丁基中，则结合电子的释放效应越大。由于结合电子的释放效应增大，醇盐阴离子的不稳定性变大，从而导致所观察到的活性顺序为1°>2°>3°。 |
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