Question

need help on quetion 6 & 7

6 HAMK 1 CNMR 240 220 200 180 140 160 120 100 60 40 20 ppm IR MS 404 3000 2000 Propose a compound for that would furnish all of the above spectra. (1 pt) go zo arbitrary intensty

Answer 1

6 Feom mass spectrum molecular Ion Peak at m/2108 and m+1 Peak at log. HNMR has 8 Pestons and 13C NMR Shows 7C so GHz. total 92. remaining atom is oxygen. so molecular formula of the compound is CHgO Degree of cunsaturation - C-(H)+1= 7-(8)+1=4 C= Numbero carbon atom=1, H = number of lydeogenalom = 8 Degree of unsaturation of the compound = 4 IR specteum has following peaks * Bicad peak around 3450 to 3200 cm belongs to-OH stitching * Peak at 3080 an belongs c=C-H, CH stretching of benzene. * Peaks at 2900 and 2870 um belongs to alkane cistrutching * Peaks at 1600cm, and 1500cm belongs to benzene c=csteetching * Peak at 138oom! belongst bending * Peak at 1170cm belongs to c-o stretching frequency IHNMR specteum * teiplet 87.1 pp mb with it belongs aromatic hydeogen - * doublets at 868 Pemand 866 ppm with it integration are Aromatic Prydrogeny and singlet atset ppm with integration based upon the splitting Pattren Compound is meta disubstituted . * Singht at S2:40pm CH₂ group attached to phenyl group. Peakat s 155 ppm belongs to aromatic cartron attached to oxygen. Peak at s 140 ppm belongs to substituted aromatic carbon. Peaksat s 130 122, 116, 112 PPm belongs to alomatic carbony Peak at S 22ppm belongs to aliphatic carbon Based upon the data structure of the compound is shoron below oH IHNMR 6-10 H 466.9%CH (58,9) (1162 CH3(22) 13C NMR 130 La G.1+)# TILH:1,8) C10 cogat CH3 (2:4,5) (155)

From mass specteum molecular Ion Peak at m/2 120 • IHNMR has total 12 Peotons, pestans at aromatie region. 13 C specteren has 3 Peaks.cach Peaks Corresponds to 3 carbons. So totalt 9 coorben molecular formula of compound is cq Hizi negree of unsaturation - c# C-(4-9-13+=4m IRSRdeum * Peak at sosoom belongs to coromatic c-H stretching * Peaks at 2980 and 920an belongs to aliphatic GH stretching * Peaks at 1600 and 1420an belongs to alomatic cac stretching * Peak at 1850 on belongs to C-H bonding. IHNMR spectrum * Singlet at 87.00 ppm with integration 3H belongs to 3 hydrogens of same chemical Environment. * singlet at 52.4 ppm with at belongs to 3 CH3 groups attached to benzene ring. 13C NMR spectrum - * Peaks at s 138 ppm and 1280pm belongs to aromatic carbons * Peak at 8 22ppm belongs to aliphatic carbon. Based upon the spectrums data structure of the Compound is 1 H NMR Chemical shift values 3 (NMR 143 G0S). CH3 (2:4,3) H670,5) (1297 628) CHA C H3(9:48) >(138) SH3(22) (28) CH3 - 02:4,3) 167.0,8) 22) The mdecule loo is symmetric. So we have on Peaks in HNMR due to same equivalent Peotons. (38) have only two