Question

Problem 3 IR Spectrum (quam 4000 3000 2000 1600 V (cm) 1200 800 Mass Spectrum So base per 107/109 152/54 M 180/182 CgH9O2Br 40 80 120 160 mle 200 240 280 13C NMR Spectrum (50.0 CDC, on DEPT CHIC CHA over proton decoupled 200 160 120 80 400 8 (ppm) 'H NMR Spectrum (200 M COCI, solution 40 pm 10 9 8 7 6 5 4 3 2 8 (ppm)

Answer 1

Answer D IR Spectrum 17406ro: The IR band at 1740 cm' indicate the presence of ester Carbonyl funch'cmal group in the molecule. (R-GARI 2) Molecular Formula - CsHgO, Br Degree of Unsaturation = 1. B) Mass Spectrum m12= 182, 180, 154, 152, 109, 107- 4) '-NMR spectrum a) Pooten decoupled - 6 13, 19, 40,62,170 The Pootcos de coupled (-13 NMR show the five signals. Thus Five carbons present in molecule is non-equivalent b) DEPT is 13, 19, 40 up (1) CH(CH3 is 62 clown (d) - CH-0 The down Lield chemical shilst indicate et group attached to electronegah've seygen . 5) IH NMB Spectrum 1.2 (t, 0.9mm = 3H) = Chechy 1.81d, 0.9mm = 3H) = City -CH- 4.169, 0.6mm = 2H) >O-CH, CH3 4.419, 0-3mm = 1H) -CH-CH3

On the basis ols IR. EMR and NMR the possible Structure ols Compound is 62 19 13. i Bro 13 H CH 6o-Crh CH 3 1.8(d) 4.469 | 4.1(9) 1.2ct) ΤΟΝ Mass -Fragmentation 20 a your tee you 1/2 = 180, 182 ọc 11 CH. b) I Mclakbestys Fragmentation B6 Br m12= 152, 150 86 B76 112=107, 109