Question

identifying the unknown using IR, HNMR CNMR and Mass spectrums

Unknown H407 100- 18-N-9500 Relative Intensity 160 180 200 220 Order 20 40 60 80 100 120 140 m/z

Answer 1

|Answer A] Mass Speckum Data m12= 192(M+.), 147(100), 118,92 aj The mass of molecular i on peak at m2 = 192, indicate that molecular weightos Compound is 192 b) The even mass of molecular in indicate there is O sitrogen atom in Compound or even numbes ols nitrogen atoms are present B) IR Speckun: 9 The broad IR band at 8200-2600 crord indicate the presence ols Carbescyllic aid -OH group in the Compound. 3 (R-COOH) by The IR band at 1700 fra indicate the presence ols Carboxyllic aid Carboxyl group. c] molecular bormula a) The cross os molecularico indicate that rolecular weight ols Compound is 192. b) IR spectrum show that presenceols carbonyllic neid group. Thus compound have two coeygen The NMR spectra shows that there are 16 hydrogen atom are present d) The cmr show that presence os 12 carboor atom. This possible molecular bormula os Compound is m.F- (2H 1602 Degree of unsaturahim = 5 The 5- degree ols unsaturation indicate there is presence ols benzene sing in Compound.

D] +H-NMR Data Sppm 0.90(3H, t] =) Cez che 1.0 (3H, d) = CH-CH- 1.7(2H, M) = -cthy 2.3(1H, m] = CH- 3.8(1H1, d) = -9 CH CH 7.30-7:50(54 m) = 11.00 (1H, bs) 3 -4-OH E) 13C-NOR Data 8ppm 12 (9) che 19 (9] => ero 28(t) = -cth 35(2) 3 - 62(d) -c-Con 127(0) = =(-4 1291ds a 2=CH 130(d) 2=CH 136(5) 180(s) 3 -(= -4 OH

on the basis of IR, NMR, emr and mass, the Possible structure of Compound is Propel.corbes 2-3cm) o choch CH CH 0.90lt) W e also tem 1.0(d) 7.30-7.50m) 3.8(d) Mass Fragmentation ayon Tt- miz=192 I Hoo * , co miz= 147 m12= 175 соти • -0022 m12=148 to 4o yout: OF Hoy on it. .co ... . m12=136 h To m12= m12-92 = 136 corron 7to to Cornaczotto 7/23118