Question

Deduce the structure based on the information below:

Problem 69 2288 IR Spectrum (liquid film) 4000 3000 1200 800 2000 V (cm 1600 ) Mass Spectrum 88 TTTTTTTTTTT % of base poak No significant UV absorption above 220 nm M+ = 94 < 1%) 40 80 160 200 240 280 120 m/ 13C NMR Spectrum (20.0 MHz, CDCI, solution) TMS solvent prolon decoupled 200 160120 80 40 0 8 (ppm) 1H NMR Spectrum (100 MHz, CDCI, solution) 10 9 8 7 6 5 4 3 2 1 8 (ppm)

Concepts: organic spectroscopy, organic chemistry

Answer 1

Answer DIR Data: 2268 cro! The absorphion band at 2268 cm indicate the presence of-Can functional group in compound. es Mass Data m/2= 94 66,29 aj The molecular ion mass 94 indicate that the molecular Weight of compound is 94. b) TR show the presence ols Nitrogen atom ECEN) in boomula ols Compound. Th. 1800 talt c)! The molecular weight even number, thus according to nitrogen rule, there is presence of two or even number of nitrogen atom in molecular formula of Compound. 3) 13 C-NMR Data ( Pootes de coupled ) S 10, 22,25, 112 ppm aj The CMR spectoa o show 4. Signal. Thus compound has four types of Carbon atom . b) The eme Signal at S 112 ppror indicate the presence ols – CEN. molecular Formula on the basis of IR and mass the possible molecular Lormula is C5H6N2 Degree of Unsaturalcm = 4 5) IH NMR data The NMR spectra lor compound has three signal Thus Compound have three types ols by drogen atoms, S 1-2 ( t, 1.6mm = 3H) → CityCih 2.01 quintet, I. Imm= 24) = CH-CH2-CH 3.8 ( triplet, 0.5mm = TH) -CH=CH-C=N

On the basis of IR, Mass, cmr and Nm R the possible structure ols Compound is data 1,2(+) 2.0 3.8(t) 112 CH₂ CH₂ CH-CEN CEN 22 25 112 Mass Fragmentahion VN to a) Y tee - ON CN m12 = 94 ON -EN ON +. CH m12= 29 HON + Mclabberty Fragmentation ch m12= 66 CN