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首页> 外文期刊>Chemistry: A European journal >Moderate and advanced intramolecular proton transfer in urea-anion hydrogen-bonded complexes
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Moderate and advanced intramolecular proton transfer in urea-anion hydrogen-bonded complexes

机译：脲-阴离子氢键复合物中分子的质子传递

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The study of the interactions of the three urea-based receptors AH, BH ~+ and CH~(2+) with a variety of anions, in MeCN, has made it possible to verify the current view that hydrogen bonding is frozen proton transfer from the donor (the urea NH fragment in this case) to the acceptor (the anion X~-). The poorly acidic, neutral receptor AH establishes two equivalent hydrogen bonds N-H?X~-, with all anions, including CH_3COO~- and F~-, in which moderate proton transfer from NH to the anion takes place. The strongly acidic, dicationic receptor CH~(2+) forms, with most anions, complexes in which two inequivalent hydrogen bonds are present: one involving moderate proton transfer (N-H?X~-) and one in which advanced proton transfer has taken place, described as N~-?HX. The degree of proton advancement is directly related to the basic tendencies of the anion. The cationic receptor BH~+ of intermediate acidic properties only forms complexes with two inequivalent hydrogen bonds (moderate+advanced proton transfer) with CH _3COO~- and F~-, and complexes with two equivalent hydrogen bonds (moderate proton transfer) with all the other anions. Moreover, [B?HF] and [C?HF]~+, on addition of a second F ~- ion, lose the bound HF molecule to give HF_2~-. Release of CH_3COOH, with the formation of [CH _3COOH?CH_3COO]~-, also takes place with the [B?CH_3COOH] complex in the presence of a large excess of anion.
机译：在MeCN中对三种脲基受体AH，BH〜+和CH〜（2+）与各种阴离子的相互作用的研究使得可以验证当前的观点，即氢键是从受体（阴离子X〜-）的供体（在这种情况下为尿素NH片段）。弱酸性的中性受体AH与所有阴离子（包括CH_3COO-和F〜-）建立两个等效的氢键N-H？X〜-，其中质子从NH转移到阴离子。与大多数阴离子形成的强酸性，阳离子型受体CH〜（2+）形成两个不等价氢键存在的配合物：一种涉及中等质子转移（NH3X〜-），一种涉及高级质子转移，描述为N〜-？HX。质子前进的程度与阴离子的基本趋势直接相关。具有中等酸性的阳离子受体BH〜+仅与CH _3COO〜-和F〜-形成具有两个不等价氢键（中度+高级质子转移）的配合物，与所有具有两个等效氢键（中度质子转移）的配合物形成。其他阴离子。而且，[B 2 HF]和[C 2 HF] +，在加入第二个F-离子后，失去结合的HF分子，得到HF_2--。在存在大量过量阴离子的情况下，[B？CH_3COOH]配合物也会释放出CH_3COOH，并形成[CH _3COOH→CH_3COO]-。

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