Question

On the following page, you will find a spectroscopy problem. For full credit, you must do all of the following (Show your work or you will receive Zero Credit for this Assignment) (1) Calculate the Degrees of Unsaturation (2) Completely assign all relevant peaks in the IR spectrum (3) Completely assign all peaks in the 13C NMR spectrum NOTE: DEPT was discussed last quarter and it is in the NMR PowerPoint) 4) Completely assign all peaks in the IH NMR spectrum (5) Provide the correct structure for this molecule 6) Provide the structures for peaks 29,57, and 100 in the mass spectrum Problem IR Spectrum 743 3000 2000 16001200800 100 Mass Spectrum No significant UV absorption above 220 nm м" . 174 (с 1%) CgH1404 40 80 120 160 200 240 280 m/e 13C NMR Spectrum proton decoupled 200 160 120 80 40 0 8 (ppm) H NMR Spectrum (200 MH, CDC, solution) 10 9 8 6 6 (ppm)

Answer 1

[1] double bond index or degree of unsaturation is find by the formula = [2C+2+N -X - H] /2

here C is number of carbon , N is no. of nitrogen , X is no. of halogen , H is no. of hydrogen present , no egfeef pf presence of oxygen on degree of unsaturation

here for a given formula C8 H14O4 ,

here N=0 , H=14 , C=8 , X=0

degree of unsaturation = [2x8 +2 + 0 - 0 - 14] /2 = 4/2 =2

degree of unsaturation is 2 says either two double bonds present

[2] peak at 1743 cm-1 shows ester group

peak at 1200 cm-1 shows C--O stretching frequencies

peak around 1100 cm-1 shows C--H bending motion

peak at 2900 - 3000 cm-1 shows C--H stretching

[3] peak at 10 ppm for - C H3

peak at 30 ppm for - O - C H2 - CH3

peak at 60 ppm for - C​​​​​​ H2(  C=O) - O

peak at 180 ppm for ester group - (C =O ) - O

[4] in 1 Hnmr note for hydrogen written in bold letters

triplet at 1.1 ppm due to - CH3

quartet at 2.3 ppm due to - C H2 - C=O-

triple at 4.1 ppm due to - O - C​​​​​​ H2 - CH3

[5] molecule is di ethyle succinate structure is

CH3- CH2 - (C=O) - O - CH2 - CH2 - (C=O) - O - CH2 - CH3

for DEPT nmr we see two inverted peaks for two - CH2 and one peak in above direction for - CH3

[6] peak at m/z 29 due to [- CH2 - CH3]+.

peak at m/z = 57 due to CH3 - CH2 - C=O].+

peak at m/z =100 due to =