Question

its all one problem on spectra please help

1. to identify fhe compounds of each spectra

2.assign proton , carbon and IR spectra

3 200 100 160 140 120 1000 6040200

Answer 1

Chemical Formula: C3H7NO2 Proposed Structure Name: 1-nitropropane Exact Mass: 89.0477 2 NO₂ / 3 Degree of Unsaturation=C+1 -" + =1 CMR Data: FTIR Chemical Shift 13C Signal C marked 1 | C marked 2 C marked 3 21 59 Stretch (cm-1) 2985 1550 1370 1350 Bond C-H stretching Aliphatic) N-O asymmetric stretch C-H rock -O symmetric stretch N 1H NMR IH Signal Integration Multiplicity Chemical Shift (ppm) 1.1 2.1 4.4 H marked 1 H marked 2 | H marked 3 3 triplet sextet triplet 2

Mass Fragments: NO2 m/z: 89 m/z: 43 m/z 41 m/z: 27 Multiplicity Rule: The multiplicity of any peak depends on the number of neighboring protons. If a protons contain n number of neighboring protons then the splitting will be n+1. Suppose a proton contains 2 neighboring H then it will give a triplet (2+1). However sometimes the splitting changes due to germinal vicinal etc. coupling in certain cases. Elemental analysis: C = 40.4%, H = 7.8 %, 0 = 36% so the total comes 84.2% so the rest 15.8% is for some other element If we divide all C, H, O % by their atomic mass we get a ratio of almost C:H:0 = 3:7:2, that gives us the formula C3H702 and the molecular mass becomes 75. The given molecular mass from mass spectroscopy is given 89 so we are left with (89-75)= 14. So it has to be nitrogen (rest 15.8%). So the chemical formula becomes C3H702N. From the 1H NMR we see three peaks means there are three different sets of protons. The pattern triplet, sextet and triplet if characteristics for a propyl group. And when it is attached to a electron withdrawing nitro (NO2) group, the value for the peaks also matches. The 13C NMR also shows three sets of carbon. They also supports the structure. Presence of NO2 also confirmed by the peak at 1550 cm in FTIR (characteristic for N-O asymmetric stretch).