Question

Compound B has the following 13C NMR, H NMR, IR and MS spectral properties: 13C NMR 1H NMR 6H 1H 80 60 40 20 0 DO ppm ppm Mass 102 Intensity 67.9 4.5 103 104 0.1 1000 500 400 2000 3000 2000 1500 Wavenumberlcm-11

analyze the spectral data for compounds and report their
-molecular formulae
-degree of unsaturation
-functional groups
-structures

Answer 1

13C-NMR Two types of carbons in the molecule. 55 ppm is may be Oxygen attached carbon 21 ppm is could be CH2 1H-NMR Two types of H in the molecule 1.2 ppm is for 6H; that means there are two CH3 3.6 ppm is for 1H , that means this is CH IR -Spectrum There is no broad peak at 3500 cm-1, so there is no OH group in the molecule Peak 1100 cm-1 is specific for C-O bond of ether Mass spectrum Molecular ion peak 102 m/z Since the molecule have two CH3 and CH and O (ether) CH3 CH3 Molecular Weight: 15 Molecular Weight: 15 Molecular Weight: 13 Molecular Weight: 16 CH Total but molecular ion peak is 102 m/z. So the molecule is symmetric Therefore the structure CH3 CH3 CH3 CH3 CH را CH H3C 3. 60 CH 55 O H3C CH3 1.2 21 1H-NMR 13 C-NMR CH3 CH3 CH CH H2C 0 CH3 Molecular Weight: 102 Chemical Formula: C6H140 Degree of unsaturation = No. of Carbon -(0.5 x No. of H) +1 = 6 - (0.5 x 14) + 1 = 6 - 7 + 1 = -1 +1 = 0 There is 0 site of unsaturation. Compound is saturated