Question

Deduce the structure based on the information below:

Problem 52 IR Spectrum (Tiquid film) t760 U 4000 3000 1200 800 2000 V (cm 1600 ) Mass Spectrum TTTTTTTTTT % of base peak No significant UV absorption above 220 mm 87 M+* = 146 (<1%) LLLLLLLLLLLL 40 80 120 160 m/e 200 240 280 13C NMR Spectrum (20.0 MHz, CDCI, solution) solvent proton decoupled www pornowww 200 160 120 80 40 0 8 (ppm) 1H NMR Spectrum (100 MHz, CDCIsolution) 10 9 8 7 6 5 4 3 2 1 8 (ppm)

Concepts: organic spectroscopy, organic chemistry

Answer 1

Answer 1) IR= 1760cm - Rio-r' The Ir band at 1760cm indicate the presence of ester Carbonyl funcional group in the compound. 2) Mass Spectrum : mlz= 146(M+.), 87 49(10) base aj The molecular iom at M+- 146 indicate molecular weight of Compound Ps 146. b) The base peak at 43 indicate the presence of fragment icmm ch-G8 or 10. 3) 13 C-NMR (Proton decoupled) S 18, 20, 88,170 aj The cm R Spectra indicate there are four types of carbon atom in the molecule b) The peak at 170ppm indicate the presence of ester CR-COOR') carbonyl group in molecule. c) The molecule has symmetrical or equivalent group in the structure. 4) Molecular formula The molecular loeight of Compound is 146.and by wing sule of 13. (146/13 = 11). molecular formula determined. According to rule os 13 compound has expected 11 Carbon atoms. 2 The CMR show only four peak. The molecule bove equivalent carbom atoms. The symmetrical structure is possible

3) The IR show the presence of estes carbonyl (R-COOR') We consider q-ester carbonyl group is equivalent on the basis of IR, Ne mass of CMR, the possible molecular formula is - Co H1006 5) HAAR S 1.5(d, 0.5mm= 3H) → Chach- 2.1 ( S, Imm= 6H) 5 CHF $ Two Gcyl group. 6: 8019, 0.1702= 14) --O-CH-CH2 On the basis of IR Mass, em R and NMR the possible structure of Compound is 18 88 g a to d o- c - CH₂ 2.1(s) 20 1220 1.5(d) 6.8019) Mass Fragmentation m2=146 m/2=87