Question

For this compound:

100 95 90 3000 2000 4000 1000 Wavenumbers (em) AL 1 43 741 711 100 Exact mass 102.0678 50 597 877 1021 90 40 50 60 70 80 100 m/z 011 1103 -188 S-1265 -1365 -1442 -1759 -2970 aourusueL%

1. Molecular weight
2. Elements present in the molecule
3. Molecular formula
4. Functional groups from IR

5. SODAR

6. Draw structure and label the protons in your spectrum with their chemical shifts.

7. Fill in the tables by labeling the carbons in the carbon spectra with letters of the alphabet from left to right. List the the chemical shifts and information from the DEPT spectra as to CH3, CH2, CH, Q including any special structural feature associated with this particular carbon i.e. carbonyl, vinyl etc.

Answer 1

17590 Answer 1) TR Spectrum Data 2970ca', 1759 Co 1442, 1365,1265,1188,1103, 1011cm 297001' = -(-H bond :streteching vibratooi Frequency W -quo- ester Carbony absorphicor band 1442 cm = -(-1 bcod ecissoring vibrolion 13652 1265ý 3-sto- boond absorption band 1188 chai 1103 cm l = -(-or bood absorphica band. To|| Cô J The IR spectrum represent the given molecule is ester Carbonyl Compound. (R G-OR') 3 In molecular formula tho meygen atoms are present. 2) Mass Spectrum - Data 2)/2 = 102 (M+), 87, 74, 71, 59, 43. The mto at 102 indicate that the molecular weight of Compound is 102. 3) 13e-NMR Spectrum a) Protoor decoupled - S 13,18,36, 52, 174 ppm i) The 5-CMR signal indicate given Compound hos Five Carbon atom in it's structural Isormula and molecular formula ii) The S 174 ppm indicate the presence of ester corbonyl in Compound. b) DEPT-135 s 13, 52 (1) up = CH CH₂ 18, 36 64. clown => -cth (both are 016-) 114 absent 3 i+-&-0 quarternary carbon ( No hydrogen alom)

C) DEPT-90 - No Peaks DEPT-90 indicate that there is no <-H group present in molecule on the basis DEPT-135, DEPT-90 13ppm (9) CH- on the basis (mR 18(+) -CH1 CH2-CH=CH-C-0-11h 36(+) 3-0 52(9) 5 (10- 174(5) -go- 4) Molecular Foomula i) The IR show that the compound has estes Carbonyl group There love, it has two Gocygen alom in molecular formula. ii) The Cmr show 5 Signal it indicate Compound have 5- Carbon atoms. Hii) Thus on the basis of IR, mass of emr, the Possible molecular formula i's C5 Hro02 5) SODAR.Ox Degree of Unsaturali cn = 1. 6) IH NAR 8 0.9 (t, 1.5mm = 3H) => (15-CH2 1.6 (sextet, imm= 24) => c-CH=C113 2.35't, Imm,= 21) - -5-C12-ci 3.6(S, 1-513m) = 331) = -0-clla on the basis IHNMR possible sloucture is Cle-ciech Co-chz

on the basis IR, NMR CMR the structure os Compound is 13 18 36 74 52 Clly Cly-ch o -ch, TL 235() 3.6(5) 2-35(+) 0.9(+) 1-6(sextet) Mass Foagmentation Tto _ o + eo 5 No. m2 102 7+. Nior - meo, viens -60, A miz- YU m2=43 T + H- 977. 2- 74 No T* -me, ovo 7/2=87 riot Motor 19)/2=59

1. Molecular Weight - 102 2. Element present in the molecule -> CHO 3. Molecular Formula - C5 H1OO2 4. Funch mal groups from IR - Ester (R-TOR') 5. SOPAR = 1 e 6. structure CH CH CH Co-CH₂ 2-35(+) 3:6(5) 0.9(+) 1-G(sectet T. Answer Coz CH2 Ch 4-o-cha 52 13 18 36 174 13C stilit | DEPTINKO Structural Features 13 ppm CH₂ 18ppni Cth Chh 36ppm 01516- 174ppm . (8-0- 52 ppm CH3