Question

Use the mass spectrum and either the NMR spectrum or the IR spectrum to deduce the structure of the following compound

(c) C8HO3H NMR, 300 MHz, 6.1 ppm (singlet, 2H), 6.9 ppm (doublet, 1H), and 7.3 ppm (singlet, 1H), 7.4 ppm (doublet, 1H), 9.8 ppm (singlet, 1H); significant IR absorbances at 1687, 1602, 1449, 1264, 1038, 929, and 815 cm1. 100 149 (150) 80 60 40 121 63 20 65 91 0 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 150 155 m/z Relative Abundance (percent

Answer 1

• Molecular sites or where - formula (& H603 unsaturation - (Cn+1)-( n) Cn Total number of carbons. Hn- Total no. of hydrogens. - 5.0.0. = (8+1) -(6) = 9-3 = 6 - 5.0.0. =61 IR data? 1687 cont Indicate presence of carbonyl group ! - Stretching frequency 1602 cmt Indicate cac stretching freg. - Mency of benzene ring carbon 815 cmt Indicate meta-disubstituted I benzene ring is present TH NMR :- 6.1 ppm (singler, 2H) - Indícate presence of - -cth group but it is wighly deshielded. Gig ppm ( doubler, 1H) Indi case - ch proton. of benzene ring. 7.3 ppm (singlet, 1H) → Also indicate on proton of benzene ring 7.4 ppm) (doubler, 14) Indicate presence of =ch proton of benzene ring. 9.8 ppm (singlet, 14) - Indicate presonce - of aldehydic proton le. - EH. m/2 = Mass specbum:- mass specbum shows a peak at 150, 149, 14, 1, 65, 63. Peak at 150 m/2 is due to base of compound peak

• By considering all above data structure of given organic compound is - 73(singlett _ ì 1687 cm (arbonyl) _CHE 9.88 (5) T he 7.48 ( doublet) 6.9 & doublet)