The study of boranes (boron hydrides) has played an especially important role in understanding broad — Analytical Chemistry — Spectroscopy Chemistry Question
Structures of boron hydrides and NMR spectroscopy
The study of boranes (boron hydrides) has played an especially important role in understanding broad structural principles. Work in this area began in 1912 with the classic research of Alfred Stock (1876–1946), and chemists soon learned that the boranes had unusual stoichiometries and structures and an extensive and versatile reaction chemistry. William Lipscomb (1919–2011) received the Nobel Prize in 1976 “for his studies of boranes which … illuminated problems in chemical bonding.”
Predict the most likely structure for the BH4 – ion.
Model Answer
The BH4 – ion is tetrahedral. That is, the H atoms of the borohydride ion are tetrahedrally arranged and are equivalent.
The 1H NMR spectrum of the BH4 – ion is illustrated below. It consists of a 1:1:1:1 multiplet along with a smaller 7-line multiplet. (The nuclear spin for 1H is ½, for 11B it is 3/2 and for 10B it is 3.) Interpret this spectrum.
Model Answer
In NMR spectroscopy, the number and spins of the attached nuclei determine the number of lines or multiplicity and pattern of the signal for the observed nucleus. The multiplicity is given by 2nI + 1, where n is the number of attached nuclei and I is their nuclear spin.
The nuclear spin of the 11B nucleus is 3/2, so the 1H spectrum has
2nI + 1 = 2 (1)(3/2) + 1 = 4 lines of equal intensity
The nuclear spin of the 10B nucleus is 3, so the 1H spectrum has
2nI + 1 = 2 (1)(3) + 1 = 7 lines of equal intensity
The multiplet for the splitting by the 10B isotope has a smaller area than the lines for splitting by the 11B isotope because 10B is only 20 % abundant, whereas 11B is 80 % abundant.
Explain why the 11B NMR spectrum of the BH4 – ion is a 1 : 4 : 6 : 4 : 1 quintet with JB-H = 85 Hz.
Model Answer
In the 11B spectrum, interaction of boron with 4 equivalent 1H atoms leads to the 1 : 4 : 6 : 4 : 1 quintet.
2 n I + 1 = 2(4)(1/2) + 1 = 5 lines
The molecular structure of Al(BH4)3 is symmetrical, with all B atoms and the Al atom being in one plane and 120o angles between the three Al–B lines. Each BH4 – ion is bonded to aluminum through Al–H–B bridge bonds, and the line through the bridging H atoms is perpendicular to the AlB3 plane. The reaction of Al(BH4)3 with additional BH4 – ion produces [Al(BH4)4]–. The 11B NMR spectrum of the ionic compound [Ph3MeP][Al(BH4)4] (Ph = phenyl; Me = methyl) in solution has a well-resolved 1:4:6:4:1 quintet (with J = 85 Hz). At 298 K, the 1H NMR spectrum has a multiplet at 7.5–8.0 ppm, a doublet at 2.8 ppm (J = 13 Hz), and a broad signal at 0.5 ppm. The broad signal remains broad on cooling to 203 K. Interpret this spectrum. (Note that the nuclear spin for 11B is 3/2 and for 31P is ½.)
Model Answer
[Ph3MeP][Al(BH4)4]. Structure published by D. Dou, et al., in Inorg. Chem., 33, 5443 (1994).
The [Al(BH4)4]– ion has 4 BH4 – ions bound to the Al3+ ion. The quintet observed for the 11B NMR indicates coupling to 4 equivalent 1H atoms. In the solid state, the H atoms are in different environments; in each BH4 – ion two of the H atoms bridge to Al and two are terminal atoms. That the NMR spectrum indicates they are equivalent means that the structure is dynamic in solution, the H atoms switching between bridging and terminal environments on a time scale faster than the NMR experiment. This is also indicated by the 1H NMR spectrum, where only a broad signal at 0.5 ppm is observed for the protons of the borohydride ions.
Finally, the doublet at 2.8 ppm and the multiplet at 7.5–8.0 ppm arise from the protons of the Ph3MeP+ cation. The doublet at 2.8 ppm represents the methyl group protons (coupling with the 31P nucleus, I = ½). The multiplet much further downfield is assigned to the protons of the phenyl groups.