Question

bined Spectra Problems All Together": within mbined Spectra Problems friends, the moment has come... pre puring the past two semesters, we have talked about infrared spectros- (IR), 'H-NMR spectroscopy, "C-NMR spectroscopy, and Mass spec trometry (MS). We've also talked about using information gathered from the mass spectrum to generate a molecular formula and the index of hydro- gen deficiency which helps us determine information about our compound when there isn't any available. The moment has come in which we piece all of these together. There are 8 problems in this chapter, each problem comes with an IR, both H- and C-NMR spectra, and a mass spectrum. With this information, you need to do the following for each problem: • Identify all major functional groups in the IR spectra • Identify all peaks in 'H-and "SC-NMR spectra • Identify the molecular mass of the unknown from the mass spectrum • Determine the pattern of fragmentation from the mass spectrum • Generate a molecular formula • Determine the Index of Hydrogen Deficiency • Determine the structure of the unknown compound. You will need to do your work on a separate sheet of paper. While you can write on the spectra, your TAs will grade your written work. Good luck!!

Answer 1

IR:

They have already clearly mentioned there is no actual OH & C=O groups.

2850-3000 cm-1, strong peaks, sp3 C-H stretch

1350-1480 cm-1, medium, sp3 C-H bending

1150 cm-1, strong, C-O stretch in ether

1HNMR:

4.75 ppm, 4H, triplet - indicates two symmetrical -CH₂- groups are interconnected by one -CH₂- group

2.7 ppm, 2H, doublet of triplet - indicates above group (one -CH₂- group is in between two -CH₂- groups)

13CNMR:
72.6 ppm, double intensity than other peak means strong electron withdrawing atom is present and connected to two -CH₂-

22.5 ppm, is actually one carbon atom that is -CH₂-

Mass:

Molecular ion at m/z 58 ==> molar mass of compound is 58 g/mol

base peak is at m/z 28 ==> H₂C=CH₂

we already found there are 6 "H" atoms present in unknown from 1HNMR, 3 "C" atoms from 13CNMR

6 + (3 x 12) = 42

molar mass = 58, so 58 - 42 = 16 ==> "O" atom present

Molecular formula is: C₃H₆O

IHD = [ 2C + 2 + N - H - X] / 2 = [(2 x 3) + 2 - 6] / 2 = 1

IHD = 1, ==> one ring or one double bond

Conclusion: The unknown compound is oxetane ( 1,3-EPOXYPROPANE )

Assign. ppm 72.6 22.5 Assign. Shift (ppm) 4.752 2.704 Ι Ι Ι Ι Ι Ι Ι Ι Ι Ι Ι 180 Ι 160 Ι 140 Ι 120 Ι 80 Ι 60 Ι 40 Ι 20 Ι Ι Ο Ι 200 Ι Ι Ι Ι Ι 100 ppm Ι Ι Ι Ι Τ 11 Ι 10 Ι 9 Ι 8 Ι 7 Ι 6 Ι Ι 5 ppm Ι 4 Ι 3 Ι 2 Ι 1 Ι 0

Hope this helped you!

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