A unifying idea in chemistry is the theory of acid-base behavior put forward by G. N. Lewis (1875–19 — Organic Chemistry Chemistry Question
Lewis Acid-Base Chemistry
A unifying idea in chemistry is the theory of acid-base behavior put forward by G. N. Lewis (1875–1946) early in the 20th century. That is, acids are electron-pair acceptors, whereas bases are electron-pair donors. There are thousands of molecules that can be classified as Lewis acids or bases, and hundreds of studies of the quantitative aspects of Lewis acid-base chemistry were carried out in the 20th century. One person deeply involved in such work was H. C. Brown (1912–2004), who received the Nobel Prize (1979) for his work using Lewis base complexes of the Lewis acid borane (such as C4H8O–BH3) in synthetic organic chemistry.
Trisilylamine, N(SiH3)3, like all amines, is potentially a Lewis base. This question explores this function with this interesting compound.
The NSi3 framework of the compound is essentially planar. Account for this observation.
Model Answer
The usual explanation is that the N atom lone pairs can interact with a 3d orbital on Si, giving a pπ–dπ bond. Because this interaction is most efficient if the molecule is planar, this is the thermodynamically stable form of the molecule. (Another explanation involves hyperconjugation involving the N lone pair and an antibonding orbital centered on Si–H.)
In Purcell and Kotz, Inorganic Chemistry (page 494, Saunders, 1977), the authors say that: “With respect to the VSEPR model of structures, you might initially think that trisilylamine would have a pyramidal geometry about nitrogen. … Many believe that the planarity is good evidence for the existence of a pπ–dπ interaction of the N atom lone pair with the d orbitals of silicon (recall the near planarity of organic amides).”
Housecraft and Sharpe (Inorganic Chemistry, Pearson-Prentice Hall, 3rd edition, 2008, page 398) say that “A stereoelectronic effect also contributes to N(SiH3)3 being planar. The polarity of the N–Si bonds … is such that there are significant long-range electrostatic repulsions between the SiH3 groups. These are minimized if the NSi3- skeleton in N(SiH3)3 adopts a trigonal planar, rather than pyramidal, geometry.”
Consider the following reaction enthalpies, ∆rHo, for acid-base reactions of trimethylborane [B(CH3)3] with given Lewis bases.
[VISUAL]
i) Using N(CH3)3 as the reference base, explain why the other Lewis bases have smaller or larger values of the reaction enthalpy.
ii) Explain why trisilylamine does not form a stable complex with trimethylborane.
Model Answer
i) In forming a Lewis acid-base complex, several factors must be taken into account when predicting the overall enthalpy change for the reaction. Both the Lewis acid and Lewis base will undergo some change in geometry, which contributes to the overall enthalpy change. Severe changes in geometry will lead to a smaller overall enthalpy change. The enthalpy change involved in donor-acceptor bond formation will be affected by electronic considerations and by steric effects of the acid, the base, or both. As this question involves the same Lewis acid [B(CH3)3] throughout, geometry changes and electronic effects of this acid are constant. The different reaction enthalpies will depend, therefore, only on the base.
N(C2H5)3 versus N(CH3)3: The enthalpy change for N(C2H5)3 is smaller than for N(CH3)3 owing to steric effects. The ethyl groups of the amine can interact sterically with the B-methyl groups, decreasing the energy of the N–B interaction relative to N(CH3)3.
NH3 versus N(CH3)3: Although there may be less steric strain in NH3, the inductive effect of the N-methyl groups will lead to a strong N–B bond in the (CH3)3B–N(CH3)3 complex.
C7H13N versus N(CH3)3: Quinuclidine is a strong Lewis base because of its geometry. The N–C bonds are already in a geometry that bares the N atom lone pair. In addition, there is little steric strain with the acid.
ii) As explained above, the silyl amine is stable as a planar molecule, so energy is required to fold it back into a geometry for N–B bond formation. Further, there is likely to be significant steric strain. These two factors apparently completely outweigh any energy gain from N–B bond formation, and the acid-base complex is not stable.
Gaseous (CH3)3NB(CH3)3 is introduced into an evacuated vessel at 100.0 °C to give the initial pressure of 0.050 bar. What is the equilibrium pressure of B(CH3)3 at this temperature? (For the dissociation of (CH3)3NB(CH3)3: ∆dissocHo = 73.7 kJ·mol–1 and ∆dissocS° = 191 J·K–1·mol–1.)
Model Answer
Based on the thermodynamic data in this question we have:
∆G at 100.0 °C = +2.46 kJ mol –1
K = 0.45
Pressure at equilibrium of B(CH3)3 = 0.046 bar