Question

Compound 2 Data Page Compound 2 Exact Mass: 164.0837 amu Compound 2 FTIR Compound 2 1H-NMR Coupling Pattern Coupling Constant, J (Hz) 10 Signal Chemical Shift Relative Integration (ppm) A 6.83 B 6. 82 1 C 6. 811 D 6 .291 1 E 6.049 1 F 5. 731 G3. 8123 J 1.830 3 Singlet Doublet Doublet Doublet of quartets Doublet of quartets Broad Singlet Singlet Doublet of Doublets Not determined Not determined Jor 15.6, Joy 1.7 Jo 15.6,JE 6.6 0 ), 6.6, 1.7

formula: C10H12O2

determine structure given the IR and H NMR

Answer 1

* Molecular formula- cio HizO2 * Degree of unscetusction:- U = 2c +2+N-H-X = (2x10+2+0-12-0 Lu= 5 *IR Data :- IR detel indicate poesence of -OH yours. IR et 3ol8ca celso indicate c-l steaching of desoretic compound. IR value above 1600 cm indicate that c=c double bond. IR value below 1600cm indicate c=c dereble best of aromatic compound, IR value in between 1000 cm to 1300 con Jodiecte presence of c-o board. HNMR :- ) 6.83 ppm TH singlet > Heromatic proton, ne coupling mose doshielded. 6.82 ppm 1H doubleta Aromatic pooten, couple - with one hydrogen 16-.81 epm TH doublet => Aromatic proton couple - with one hydrogen 6.291 ppm In doblet of =) olefinic proton couple quastet (J= 15.6,J=1.7 ) with bruns pouten and methyl group.

6.049 ppoa 11 Darblet of 3 Olefinic poston shores quartets (J = 15:6,J=66) trang_cczepling and - carpling with - "Hz -5.73 ppm yin, BS con gorrap attach to aromatic ring 3.812 , 3 H, singlet = -0CH3 group attach to casoon cetic oing. 1.83 ppm, 34 doublet =) -CHz gorrip eltelt of doublet to derible bond show eczepling with tuo hysogens. from above delta poobable stouettone is - octa Ho- 5.73 ppm 3.812 ppm -o-chot uo 1 6.83 ppm HK6.83 pm 26.219 ppm. 6.84ppm पर 6.82 ppo 6.049 ppm 1.83ppm