Question

Deterine the following structure using the following spectroscopy (NMR spectrum analysis, chemical shifts, multiplicity):

MF: CuHi40 EE H-NMR HNMR "C-NMR 211 40 20 180 160 120 100 80 60 200 140 Fpm CDs-e5-332

Answer 1

To deduce the structure of any given unknown compound with molecular formula, first step is to determine the Degree of Unsaturation For the calculation of Degree of Unsaturation (DU) following formula can be used DU = Number of carbons +1 - (Number of Hydrogens + Number of monovalent atoms (like halogens)- Number of trivalent atoms (like N and P)/2 Oxygens can be ignored while calculating DU Since in the given questions only carbons, hydrogens and oxygens are present the formula to calculate DU can be further simplified as DU = Number of carbons +1 - (Number of Hydrogens)/2 For C1H1402 the DU is = 11+1- (14)/2 = 12-7 = 5 Hence the compound should have 5 double bonds or 5 rings or a benzene ring and a double bond in its structure. 4-tert-butylbenzoic acid Chemical Formula: C11H1402 In the IR spectrum, peaks around 1400-1600 cm is due to aromatic C=C stretching. peak around 1700 cm1 is due to carbonyl CEO stretching. ТОН From the given spectra, the unknown compound is 4-tert-butylbenzoic acid In the NMR spectrum, by seeing 2 doublet peaks at 6-8 ppm is due to para disubstituted benzene ring 1H NMR values 8.01 1.33 singlet 7.53 Y OR12 74 8.01 1.337.53 1.33 broad singlet doublet doublet - PPM PPM 13C NMR values 129.9 1 27.1.60 124. 9 127.69.3 OH 31.3 156.5 129.9 124.9 31.3 31.3 180.160.140' 120 '100' 80' 60' 40' 20 o PPM