Question

How does the compound C9H12O look based on this NMR graph?

1: C3H120 S, 3H t, 3H Hz ppm 650.38 7.262 30 643.50 7.186 39 642.44 7.174 54 641.19 7.160 51 636.81 7.111 31 634.31 7.083 67 610.69 6.819 31 603.81 6.742 192 602.25 6.725 141 601.50 6.717 136 595.38 6.648 30 335.06 3.742 1000 244.69 2.733 33 237.13 2.648 101 229.63 2.564 111 221.94 2.479 42 116.19 1.298 206 108.88 1.216 472 101.06 1.129 155 m, 4H q, 2H 11 10 9 8 7 6 4 3 2 1 0 5 ppm HSP-00-080 ppm Int. 159.89 512 145.92 617 129.32 920 120.35 1000 113.80 831 110.98 970 55.04 537 29.01 791 15.52 751 200 180 160 140120 80 60 40 20 o 100 ppm CDS-00-689

Answer 1

Molecular formula = Cg Hi 2o Degree of unsaturation = 26+2+ N-X-14 - 2x9+2-12 2 = 4 2 Multiplicity = 2n1+l, where I= !, & n= No.of pootong IH NMR : Proton ax 2x2x1 +1 = 3 (triplet) CH, HT Ина a at S=1.29 bbm (3H) CH2 Proton b = 2x3x1+1= 4 (quartet) ? hoo-cha at d = 2.60 ppm (2H) Proton ( + 2x0x4 +1=1 (singlet) .. at s= 3.60 ppm (3H) The aromatic protons Cd, e, f, g) give multiplet ... at S= 6.95 ~7.18 ppm (4 H)

13C NMR: G → give singlelat 156 so bom G = give singlet at 29.01 tom I C=give singlet at 55.04ppm C4 + give singlet at 11.00 bbm *5 = give singlet at 113.80 ppm G give singlet at 120.35ppm Cz = give singlet at 129. 32 bbm coa give singlet at 145-92 pom Cg » give singlet at 159. go bbm There tere, the structure of compound is T і CM CHE — он, Lethyl-2-methoxybenzene