TpPtMe ((H) ((2)):なぜメチル群のH/D乱雑化があるが,メタンの損失はないのか?
Mark A Iron1, H Christine Lo, Jan M L Martin
1Department of Organic Chemistry, Weizmann Institute of Science, 76100 Rehovot, Israel.
Journal of the American Chemical Society
|June 13, 2002
まとめ
ヒドリドトリス (((ピラゾリル) ボラートプラチナ複合体TpPtMe (((H) ((2) は,硬いリガンド構造によりメタンの損失に抵抗します. しかし,デュテラートメタノールによる可逆的なH/Dスクランブリングを容易に行う.
科学分野:
- 有機金属化学 有機金属化学
- コンピューティング・ケミストリー
- 反応メカニズム 解明 解明
背景:
- ヒドリドおよびメチルリガンドとのプラチナ複合体の反応性は,触媒作用において極めて重要です.
- 複合的な安定性と反応経路に対するリガンド効果を理解することは鍵となる.
- TpPtMe ((H) ((2) は,メタンの損失に対する安定性のユニークなケースを示しています.
研究 の 目的:
- TpPtMe (((H) (((2) の反応性を,特にメタンの損失に対する耐性を調査する.
- デュテラート溶媒で観察されたH/Dスクランブリングのメカニズムを解明する.
- 観察された反応性における硬質のTpリガンドの役割を決定する.
主な方法:
- 反応を研究するために,密度関数理論 (DFT) の計算が採用されました.
- 計算はmPW1k/LANL2DZ+P//mPW1k/LANL2DZの理論レベルで行われました.
- 比較分析には,モデルシステムである[(NH(3)) (((3) PtMe(H) ((2)) ](+) が使用されました.
主要な成果:
- Tpリガンドの硬さは,メタンの除去に必要なトランス幾何学を阻害し,エネルギー的に不利になります.
- メチルおよびヒドリドリガンドの可逆的なH/Dスクランブルは,エタ (2-CH) -CH (4) の中間体を通して容易に発生します.
- メタンの損失は観察されず,Tpリガンドの安定作用が確認されました.
結論:
- Tpリガンドの硬さは,TpPtMeのメタン損失を抑制する主要な要因である.
- H/Dスクランブルは,明確な中間経路を通って発生する簡単なプロセスです.
- 同様のシステムの精密なエネルギー計算には,mPW1k/SDB-cc-pVDZなどの特定のDFTレベルが必要です.
関連する概念動画
Hybridization of Atomic Orbitals I
The mathematical expression known as the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals. The new orbitals that...
Proton (¹H) NMR: Chemical Shift
Organic molecules primarily contain carbon and hydrogen atoms. While all the hydrogen isotopes are NMR-active, protium or hydrogen-1 is the most abundant. It has a significant energy separation between its nuclear spin states due to its large gyromagnetic ratio. As per Boltzmann's distribution, an increase in the energy separation implies a greater excess population of nuclei available for excitation, resulting in a strong NMR absorption signal.
Absorption signals of all the protium nuclei in a...
Absorption signals of all the protium nuclei in a...
Inductive Effects on Chemical Shift: Overview
The protons in unsubstituted alkanes are strongly shielded with chemical shifts below 1.8 ppm. Methine, methylene, and methyl protons appear at approximately 1.7, 1.2 and 0.7 ppm, while the proton signal from methane appears at 0.23 ppm. An electronegative substituent, such as chlorine, withdraws the electron density from the protons, increasing their chemical shift. Progressive substitution of the hydrogens in methane by chlorine shifts the proton signals increasingly downfield, to 3.05 ppm in...
¹H NMR of Labile Protons: Deuterium (²H) Substitution
This lesson illustrates the role of deuterium substitution in simplifying the NMR spectrum of compounds comprising labile protons. One method employed is the use of deuterium. Amongst the three isotopes of hydrogen, deuterium (2H) has a nucleus composed of one proton and one neutron. When the D2O solvent is added to a pure dry ethanol solution, its labile proton is substituted with deuterium.
¹³C NMR: Distortionless Enhancement by Polarization Transfer (DEPT)
When proton-coupled carbon-13 spectra are simplified by a broadband proton decoupling technique, structural information about the coupled protons is lost. Distortionless enhancement by polarization transfer (DEPT) is a technique that provides information on the number of hydrogens attached to each carbon in a molecule. While the DEPT experiment utilizes complex pulse sequences, the pulse delay and flip angle are specifically manipulated. The resulting signals have different phases depending on...
¹H NMR Chemical Shift Equivalence: Homotopic and Heterotopic Protons
Protons in identical electronic environments within a molecule are chemically equivalent and have the same chemical shift. The replacement test is a useful tool to identify chemical equivalence and predict NMR spectra. A substituent replaces each of the protons being examined and the resulting molecules are compared. If the same molecule is obtained, the protons are equivalent or homotopic. Replacement of any hydrogens in ethane by chlorine yields chloroethane because all six protons are...


