Question

IR/HNMR/CNMR Unknown #101 MW=142 AMU Czol D2 | M 1 ppm 200 150 160 140 120 100 ao so 10 20 rpm

I need help figuring out what the molecular formula and structure look like based on the molecular weight 142 AMU and also based in the info given in the IR, HNMR, and CNMR.

Answer 1

From the IR Spectra, the peak at -1700 cm indicate the presence of carbonyl group from the ketone. no stretching at 1500-1600 cm indicate aliphatic ketone, i.e., no C=C groups present. The 'H NMR indicate that three type of protons. The doublet at 0.95 ppm corresponding to 6H indicate the two H₃Cm methyl group attached to the carbon carrying single proton H₃d The sharp doublet at -2.25 ppm corresponding to 2H indicate the presence of methylene group attached to the carbon carrying single proton. The chemical shif at 2.25 ppm could be because of adjescent carbonyl group O C H.CH The multiplet at 2.1 ppm correspinding to 1H confirms the CH group which is attached to two mehyl and one methylene group CH The 13C NMR indicate only four type of carbon including the CO carbon atom at -210 ppm which confirms the presence of ketone as functional group As the fragment corresponding to (CH3)2CHCH CO does not satiesfy the molecular weight, thereis a possibility of symmetric ketone. Thus, pacing the (CH) CHCH2 group next to the fragment (CH3)2CHCHCO can satiesfy the molecular weigt, 'H and C spectra, indicating the structure as shown below 25 ppm CH2.1 ppm CH, - 0.95 ppm C - 21 ppm 2.25 ppm 'H NMR assignment of peaks 53 ppm 13C NMR assignment of peaks