Absorbance7. Compound 7 is a neutral, low-melting solid (mp - 22.24°C)Elemental Analysis: C-72.00%, H-6.67%, O-21.33%"C Spectral Data: 196.5ppm(s), 166.4ppm(s), 129.7ppm(s), 129.6ppm(d), 114.0ppm(d), 56.0ppm(q), 22.8ppm(q)Mass SpectrumCompound 7 StructureWInlanaltyInfrared Spectrom4000 3000 25001'H NMR20 100 110 120 130 140 150 100 170 100 100 201Wave Number, am1600 1300 1300 1100130Wavelength, microns700CHEM 241L Exp#12 NMR Spectroscopywww.................................www.Ppm, 8Rev. 052318JKO

Given C = 72.00 % ; H = 6.67 % and O = 21.33 % Elemental analysis: Element Atomic weight (a.m.u.) Mass % Moles Molar ratio Simplest Ratio C 12.0 72.00 72.00/12.0 = 6.0 6.00/1.33 = 4.5 4.5×2 = 9 H 1.00 6.67 6.67 / 1.00 = 6.67 6.67/1.33 = 5.01 5.01×2 = 10 O 16.00 21.33 21.33/16 =1.33 1.33 / 1.33 = 1 1×2 = 2Therefore emperical formula is C9H10O2 Which is equal to molecular formula. 2). Formula for degree of unsaturation of molecule CnHxNyOz = n+1-(x/2)+(y/2) Now degree of unsaturation of molecule C9H10O2 = 9 +1-10/2 = 5 It means molecule have multiple bonds / rings. Analysis of mass spectrum). Molecular ion peak at m/z 150 . It means molar mass of molecule is 150 g/molAnalysis of IR spectra: A strong band at ~1705 cm-1 due to carbonyl group of ketone. Analysis of 13CNMR data: A peak at 196.5 ppm due to carbonyl carbon. Four peaks at 166.1 ppm(s) , 129.7 ppm(s) , 129.6 ppm(d) and 111.0 ppm(d) due to benzene carbons. Two peaks at 56.0 ppm (q) and 22.8 ppm (q) due to two methyl groups, in which one attached with Oxygen atom.Analysis of 1HNMR spectrum: Label Chemical shift (ppm) Multiplicity Integration Assignment a 2.5 Singlet 3H CH3 attached with C b 3.8 Singlet 3H CH3 attached with O atom c 6.9 Doublet 2H Two benzene protons d 7.8 Doublet 2H Two benzene protonsOn behalf of analysis molecule will be 4-methoxy acetophenone. Structure will be