Question

Relevant I 2 nuclei and relative natural abundances are as follows (assu me all other nuclei are spin inactive): H(I 100%) 19F (I, 100%) 77Se (I2, 7.6%) 29Si (I 4.7%) 31P (I 2, 100%) Also, assume 'J 2J> 3J, there is no resolvable 4J coupling, and there is no observable coupling to 13C nuclei. Using stick diagrams, sketch the 'H 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) PBr2Me b) AsFMe2 c) SeMe2 d) PF2Me e) SbCl(CH2CHs)2 1

Answer 1

Que 1 H gAsi 1F I = Y2 t Y2 1se I = Ya PBvM Br CH3 Br Valuu of Bromine S (50 69) Natural abundana No ob Signal appeau in p NNR 1 No ot signal appeau in H NMR 1 Mo ot Signal appear in 19 NMR 1 &1 B MMR:1 in Mo oSignal apear 404

No w the sputing in H NHR No oL peakss Pa gluen by tormula n+1) m no d Nucle Nucleau maqnton Quttum No Total Nomuber ot peaksTo lal Gpli thing du to P Nuclei Total spit 31 Dur to Bromina Nudi (enIH) anI+1) xIt1)(2x2+1) - (2)0) 2 14 Total peaks obseuve 14 (2) As As F Mu H H HW 35Couping tates pau with Hyduogen ot anotur methy Couping uoil be but tus is not ettective Coupling 6oth muthyl guout auu Same

gsomurs AS F и No ob peaks obtainsdl in H MR = @nt+1) = (2 4) = doubut will be obsewed 1:1 gntensity foapowbu 1:1 Ce) bent chabe сиg CH3 Senglel ui be ob Se is NNR active Huclei Se =LT an Coubl with Se thmeugh 8J Coubung H wil

ot Peaks nT + 1 No IKI Doublet eotl observed with 1: 1 3ntensity 1: 1 TMS н Both Flowunu cuu equivalen to yduogunt No ot peaks in H NAR Sperkum: (xaxl+1) ( = (3) (2) Sexlet is obtainedh