Question

Relevant i = 12 nuclei and relative natural abundances are as follows (assume all other nuclei are spin inactive): H (1 = 12, 100%) 295i (1 = 42, 4.7%) 19F (1 = 2, 100%) 31P (1 = 72, 100%) 77Se (1 = 72, 7.6%) Also, assume 1 > 2/> 34, there is no resolvable 4 coupling, and there is no observable coupling to 13C nuclei. 1. Using stick diagrams, sketch the TH NMR spectra for the following compounds. Assume structures are rigid in solution (give all isomers). Include all relative peak intensities and label all couplings. a) PBr Me b) AsFMe2 c) Se Me2 d) PF2Me e) SbCI(CH2CH3)2

Answer 1

a) Singlet will appear

Explanation: There are no coupling partners for the protons of methyl group (3H).

b) Doublet will appear

Explanation: The 6 protons of two methyl groups couple with ¹⁹F, multiplicity of the signal = 2nI + 1 = 2*1*1/2 + 1 = 2 (doublet)

c) Doublet will appear

Explanation: The 6 protons of two methyl groups couple with ⁷⁷Se, multiplicity of the signal = 2nI + 1 = 2*1*1/2 + 1 = 2 (doublet)

d) Triplet of doublet will appear

Explanation: The 3 protons of methyl group first couple with ¹⁹F, multiplicity of the signal = 2nI + 1 = 2*2*1/2 + 1 = 3 (triplet), followed by coupling with ³¹P, multiplicity of the signal = 2nI + 1 = 2*1*1/2 + 1 = 2 (doublet). Hence, each line of the triplet will further split into doublet, i.e. triplet of doublet.

e) A triplet and a quartet (two different signals) will appear.

Explanation: triplet appears for the protons of 2 methyl groups (6H), quartet appears for the protons of two methylene groups (4H).