Question

Mass Spectrum (not shown): \([\mathrm{M}]=193(100 \%) \mathrm{m} / 2\)

IR Spectrum (not shown): \(3300(\mathrm{~m}), 3200(\mathrm{~m}) 3060,2985,1725,1600,1495 \mathrm{~cm}^{-1}\) (all listed are strong (s) unless otherwise indicated)

- \(\mathrm{H}\) NMR Spectrum \(\left(400 \mathrm{MHz}, \mathrm{CDCl}_{3}, 25^{\circ} \mathrm{C}\right)\)

The answer is not this: 

Pretty sure the formula is C11H15NO2.

Answer 1

O mass spectrum data = 193 (mto) @ # molecular weight of compound is 193 . # The odd mass of molecularico indicate the presence of odd number or cme nitrogen atom in Compound. (N-Rule) FR spectrum data. 83010 costa 3200CR 3 –Nithe group (i gmine present ) 3060 -CH bond 2985 as =ch band $ 3 -&o- estz carbony! poesent 4600 I ceas chi' -cec in benzene ring present molecular koomula Using Rule of 13 i š n= 14, 13 8= 11 Cn Hnto=C4H25 But compound have one N and two czuygen, then demove C3H10. Isoom Sule ob 13 koomula. Then roolecular Isormula ols Compound is - Cui HI5NoR Degree ob Unsaturahcm = 2(0)+9+N-H © = 26113*2+175- 25 Compound have 5-degree ols unsatesahom. This indicate the presence of benzene ring in structure

4) HNMR data CH3 s 1.4ppm (GH, d) = CH CH is present 2-3 ppm (3H, S) 3 cl As present 3.6 ppm (bs, 2H) = NH- group os amine 5. Pệm ( , IR) = (-CH-o- 6.3 ppm (dd, 5=8, 2H2, IH) 3 6.9 ppm (d, S=286) 2 P 1H 31 To7 To 8ppm (d, J=846, 14). OZY H present 5) 13C NMR data - structure 2-3(5) lold) - CH3 o clean ty 4o-ch-cth 1 1 367 L 5.31m) UN H.- 19 ppm =) two equivalent-Ch 22ppm> ag-ar 68 ppm -CH-o 6-9(d) 116ppm =64 so² 116.5ppm ac-n sp 120ppm 3 =ç- spac 131ppm y =(-H sp²c 140 ppm-> =ęH sp2c 152ppm =( spac 167 ppm 3 – so eslez carbonyl 7-8(d) 3.6(65) - 6.3(dd)