Question

what is the organic compounds structure represented im these graphs? please explain how you figured it out so I know di the correct steps. thank you!

80 sbsbre 1379. 19 %Transmittance 629.30 3079.17 992.79 1456.54 1641.93 2874.69 2960 31 2927.63 98808 4000 3500 1500 2500 2000 Wavenumbers (cm-1) 500 50 Thu Mar 12 08:41:58 2020 (GMT-04:00) FIND PEAKS: Spectrum: cbshr8 Region: 4000.00 400.00 Absolute threshold: 49.603 Sensitivity: Peak list: Position: 404.53 Position: 434.94 434.94 Position: 629.30 Position: 909.36 Position: 992.79 Position: 1379.19 Position: 1456.54 Position: 1641.93 Intensity: Intensity Intensity: Intensity: Intensity: Intensity Intensity: Intensity: -1,629 - 1.786 44.080 6.184 19.905 27.567 19.554

Answer 1

Using MS data, We have a molecular ion peak at 84 m/z

i.e. we may have at least 6 carbons (84/12 = 7, C6+ other elements)

Therefore, probable molecular formula is C6H12

This molecular formula is also supported by 13C NMR and 1H NMR data as there are 6 Carbons peaks (in 13C NMR) and 12 Hydrogen peaks (from integration values i.e. 1+2+2+2+2+3 = 12 from left to right in the 1 HNMR septrum).

Now, to calculate double bond equivalance or degree of unsaturation, we have

H N DBE=C +

DBE = 6 - +

This means there is at least one unsaturations, that is also supported by IR peak due to C=C (alkene) at 1641 cm-1.

Probable Strature is ] m/z= 84 hou holky @ @ @ @ Maz-Hy-che-CH₂-CH=lte peaks HNMR (ppm) 0.89 1.3 is 1.9 Sb 4.8 Peak 3 (NMR (ppm) in 23 32 34 140 116 as (unit) 2800-3000 due to sp3 Hybridized (6-4) Ms peaks CH₂ =44- CH2 CH=CH-Cha-Cho mu (m22 41) (no 255)