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小木蟲論壇-學術科研互動平臺 » 科研生活區(qū) » 專業(yè)外語 » 計算模擬 » Alcohol
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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) 烷烴 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 異丙醇 叔丁醇 醇 醇的特點是存在與碳原子相連的羥基。脂肪醇可以被認為是由烴中一個sp3雜化鍵上的氫原子被OH替換而生成的。它們也可以被認為是一種由水的一個氫原子被烷基基團所取代而生成的衍生物。烴可能包含初級、二級、三級氫原子，醇也存在同樣的分類。下面的例子說明了每一類的典型成員。 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 醇可能被認為是由在烷烴氧化生成CO2和H2O的氧化方案中產(chǎn)生的首個氧化產(chǎn)物。醇可能會被進一步氧化。已經(jīng)發(fā)現(xiàn)，只要部分氧化的碳原子上仍有氫原子（α-氫），就有進一步氧化的可能。下面的公式說明發(fā)生的結(jié)構變化，這里需要強調(diào)的是，他們沒有說明反應的機理。上述反應涉及C—H鍵。由于部分氧化的碳原子上沒有α-氫原子，它們將不會進一步反應。 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。 結(jié)合在醇上氧的質(zhì)子比結(jié)合在碳上的酸性強。這種差別很容易用氧的電負性較大，因而它極化了O—H鍵而進行解釋。舉個例子，氫氧鍵中的氫原子被鈉原子取代。這種反應類似于酸和金屬反應生成氫氣。在這種反應中醇的反應活性順序為l°>2°>3°。這種反應活性順序被認為是烷基誘導效應的結(jié)果。利用氫的相對電負性值2.1和碳的相對電負性值2.5，H—C可以表示如下：Hδ+-> Cδ-.因此這種鍵的數(shù)目越多，如在叔丁基中，則結(jié)合電子的釋放效應越大。由于結(jié)合電子的釋放效應增大，醇鹽陰離子的不穩(wěn)定性變大，從而導致所觀察到的活性順序為1°>2°>3°。 |
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