Question

CH 203 Organic Chemistry Infrared Spectroscopy Experiment Name 10 points Based on the following IR and NMR spectra and using the given molecular formula, formulate the structure of the compound. Identify regions of your structure that correspood to the peaks that you see in the graph below. Show your organic compound structure here on this perd). No additional write-up is needed for this laboratory class. PS. If your NMR protons are shown in fruction you can round them to a whole number by multiplying all protos by a simplesmeltiple such as a 2 or 3. PROBLEM 23 WW 162 *C 74.0 6.2 WE GoHoo2 LITE 21 1994 Solid sample 2H 1H IH TTTTTTTTTTTTTTTTTTT 7.5 7.6 6.5 5.6 5.6 5.6 4.5 TTTTTTTTTTTTTTTTTTTTTTT 11.0 10.5 10.0 9. PP Pratun 11

Answer 1

The structure of the compound was given in the attached image based on systematic elucidation from the given spectroscopic data.

The given Formula is clothio O2 the double bond equivalent (DBE) = ļ[(2n+2) -m ] for a compound having formula C Hmo. Hence, DBE for Cohoo2 = Ž J (2x10'+2) -10] = 1 [22-10] = 6 DBE 6' is indicative of Aromatic ring, and two double bond. The IR spectrum clearly shows Crostr bond at an 1700 cm which indicates carbonyl group of carboxylic acid. This is supported by presence ok - Hshy band ~ 3000 em, Also, the O-H proton appeared as broad signal at 8 - 10.5 in 1 H NMR. In 1H NMR there are SH im aromatic region (8772) indicates, monosubstituted Aromatic comp ound. Anothere 24 are visible ans-5.5, which indicates proton due to presence of double bond. Im aliphatic region, 24 can be seen at dn 3.3 which indicates a CH2 group. Togethere all the information from the spectral dală points to woveds the following structure Styrylacetic acid 4-Phenyl-3- butenoic acid IR region (ent O-H ) ~ 3000 N 3000 cm in 3ooooml Toocmil 800 -- 1400 m