Question

For problems 11-12 answer the questions on the Mass Spectroscopy problem set worksheet.

Problem 11 Problem (z m that IR Spectrum liquid film IR Spectrum oldim 4000 3000 2000 1600 1200 800 4000 3000 2000 V (cm 1800 ) 1200 800 Vom) 0.0 100F Mass Spectrum Mass Spectrum of base peak No significant UV absorption above 220 mm of base 107/109 1015486162 UV spectrum 530 mg/10mis path length: 100 cm solvencyclohexane 120 40 80 120 160 m/e 200 240 280 200 240 250 1.5h 200 350 250 300 2 (nm) 13C NMR Spectrum (50.0 M . CDC, wolution 13C NMR Spectrum (50.0 MHz, CDC, solution to signals at higher field DEPT CH CH CH DEPT CHICC the signalisatgerfield proton decoupled 200 160 120 5 (ppm) 200 160 120 80 400 (ppm) TH NMR Spectrum (200 MHz, CDCI, solution) TH NMR Spectrum 200 M CDC, con om 10 9 8 4 3 2 6 5 4 3 2 1 1 6 (ppm)

Answer 1

1) IR Spectrum duta 1740 cm 3 R-GOR Ester Carbonyl group is present 1200 cm = =c-o- bend os ester e mass spectrum da mi2= 180(m), 182(m+2), 152, 154, 107, 109, 29. # The molecular iom at ml2= 180, Compound is 180 Tom at m12= 180, indicate molecular weight ols # # The mand m+2 peauks at 180 and 182 have equal relahve intensity. This indicate Br.is present in Compound molecular Loomulg - Using rule of 13 13 = n=19, r=ll entento= C13H24. 180 But Compound have two Oseygen and one Bromine, then Pemove C7H15, then molecular Lormula of compound is mit - C5 Hg 02 BG . DUE 2C+2-4- = 205) +2-9-) 2 . 2 = 1 3) 13C NMR data Pooks decoupled S 14, 22, 40, 61, 170 ppm chair show five peak · Thees all five carbuor in structure are noso equwdleut DEPT 1669) GI(I) CHE 2269) S-CH 40(1) --- 2 170Cabremos -80-

4) HINAR data 1.28( t, 3H) = (CH chhi present 1.90(d, 3H) 3 4.5(9, 1H) che present (1 h presut 4.3 (9, 2H) 39-0-CH2chy prezut Structure 22 GI 70 I Go 14 Clb c o ch - Br the - l-gold) 4.5(9) 1-28143 4-319) 5) mass Fragmentahan roy to Y teo- roll HOY L Br Hol 1. Br - ch=1th M/2=152 m/2=180 182 154 Vol 360 m12=107 log