Discussion
In the 1H NMR spectrum, multiplicity of the signal (bold and underlined H atom) in the following species is
(I) [HNi(OPEt3)4]+
(II) Ph2Si(Me)H
(III) PH3
(IV) (Cp*)2ZrH2 (Cp* = pentamethylcyclopentadienyl)
Answer: A
The multiplicities are I- pentet, II- quartet, III- doublet and IV- singlet.
Apply the n + 1 rule to the nuclei with spin 1/2 that couple to each hydride:
- (I) The Ni-H couples to four equivalent 31P nuclei (I = 1/2, 100% abundant): 4 + 1 = 5, a pentet.
- (II) The Si-H couples to the three protons of the Si-CH3 group (three bonds apart): a quartet.
- (III) Each H of PH3 couples to the single 31P: a doublet; the three H are equivalent and do not split each other.
- (IV) The two Zr-H hydrides are equivalent, and Zr has no abundant spin-1/2 isotope (and the Cp* methyl protons are too remote): a singlet.
Each of the other options mis-counts at least one of these neighbours, for example treating PH3 as a singlet by ignoring phosphorus or giving the zirconium hydride a coupling partner it does not have.
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