Influence of the Lewis Acid/Base Pairs on the Reactivity of Geminal E‐CH₂‐E′ Frustrated Lewis Pairs

Abstract

The influence of the nature of the acid/base pairs on the reactivity of geminal frustrated Lewis pairs (FLPs) (Me2E‐CH2‐E′Ph2) has been computationally explored within the density functional theory framework. To this end, the dihydrogen‐activation reaction, one of the most representative processes in the chemistry of FLPs, has been selected. It is found that the activation barrier of this transformation as well as the geometry of the corresponding transition states strongly depend on the nature of the E/E′ atoms (E=Group 15 element, E′=Group 13 element) in the sense that lower barriers are associated with earlier transition states. Our calculations identify the geminal N/Al FLP as the most active system for the activation of dihydrogen. Moreover, the barrier height can be further reduced by replacing the phenyl group attached to the acidic atom by C6F5 or 3,5‐(CF3)2C6H3 (Fxyl) groups. The physical factors controlling the computed reactivity trends are quantitatively described in detail by means of the activation strain model of reactivity combined with the energy decomposition analysis method.

A Zwitterionic Phosphonium Stannate(II) via Hydrogen Splitting by a Sn/P Frustrated Lewis‐Pair and Reductive Elimination

“…The transition state for hydrogen‐splitting (TS) is located 58 kJ mol −1 in energy above the vdW complex. Lower activation energies for hydrogen splitting have been calculated for geminal nitrogen/triel systems [17] . In this late transition state, the H−H bond length rises to 1.05 Å, comparable to the respective distance of 0.99 Å in the transition state of hydrogen activation by the geminal FLP Ph 2 BCH 2 P( t Bu) 2 [18] .…”

“…Conversions weredetermined using 1 HNMR integrals. 1.800(10), P(1)-C(6)1.830(8),P(1)-C(10) 1.831 (9), Sn(1)-C(1) 2.238 (13), Sn(1)-C(3) 2.347 (17),S n(1)-C(5) 2.275(5);P(1)-C(5)-Sn(1) 118.6(4), C(5)-P(1)-C(6) 118.5(12), C(5)-P(1)-P(10) 105.5 (13), C(6)-P(1)-P(10)1 18.1(4), C(1)-Sn(1)-C 392.3 3, C(1)-Sn(1)-C(5) 93.9(7), C(3)-Sn(1)-C(5) 84.4 (9). which meanst he neutral phosphane function is more basic than the anionic stannate(II) function.…”

“…Lower activation energies for hydrogen splittingh aveb een calculated for geminaln itrogen/triel systems. [17] In this late transition state, Scheme3.Theoretical equilibrium between the tautomers 3 and 5. Figure 3.…”

Understanding chemical reactivity using the activation strain model

“…The ASM can be applied to all unimolecular and bimolecular reactions in both homogeneous and heterogeneous systems and has been used routinely by theoretical and experimental chemists [28][29][30][31][32][33][34][35][36][37][38][39][40][41][42] . We provide specific examples of the ASM being applied to understand inorganic, organic, and supramolecular chemistries, namely, the transition metal-mediated oxidative addition of C-X bonds in cross-coupling reactions, the reactivity of cycloalkynes in 1,3-dipolar cycloadditions, the reactivity of dihalogen-catalyzed Michael addition reactions, and the bonding mechanism in hydrogen-bonded systems.…”

Understanding the Reactivity of Neutral Geminal Group 14 Element/Phosphorus Frustrated Lewis Pairs

“…Therefore, both key donor−acceptor orbital interactions occur along the reaction path but at different stages of the transformation, which confirms our previously reported cooperative and asynchronous mode of action of FLPs in related dihydrogen activation reactions. 32,34 Interestingly, the data gathered in Figures 4 and 5 clearly indicate that both orbital interactions are significantly stronger for the process involving the heavier FLP 1-Sn than for the analogous process involving its silicon counterpart 1-Si. For instance, at the same consistent P•••C forming bond distance of 2.7 Å, ΔE(ρ 1 ) = −12.6 kcal/mol and ΔE(ρ 2 ) = −5.1 kcal/mol for the 1-Sn + PhNCO reaction, whereas much less stabilizing (i.e., less negative) values were computed for the analogous reaction involving 1-Si (ΔE(ρ 1 ) = −8.4 kcal/mol and ΔE(ρ 2 ) = −2.8 kcal/mol).…”