Question

for the following 2 compounds, please calculate, and show the calculations for, the degree of unsaturation, assign the IR spectrum peaks, assign the 13C NMR peaks, assign the 1H NMR peaks, and draw the structure for the unknown compound.

CHIM 245 Spectroscopy Problem Set #2 In this problem set there are two unknown compounds. You are provided with the formula, IR spectrum, "C NMR spectrum, and 'H NMR spectrum for each compound. Each unknown is worth 10 points, with an emphasis on correctly determining the unknown structure and assigning the protons in the 'H NMR spectrum. The points for each unknown are allocated as shown below: Degrec of unsaturation (show calculation)-1 point IR spectrum assigned peaks- I point C NMR spectrum assigned peaks - 1 point H NMR spectrum assigned peaks -3 points Structure of unknown -4 points The best way to present your NMR analysis is to assign carbons using numbers and protons using lett For CNMR you need to assign chemicals shifts as well as make assignments. For H NMR you need to assign chemical shifts, integrations, and multiplicity as well as make assignments ters. Please include all of your responses in tabulated form. See examples on Canvas. You can turn in the entire packet or just the answer sheets. Be sure to include your name, TA name, section, and instructor. Submit as a pdf filc to the Problem Set 2 Assignment page in Canvas, or bring a hard copy to lab the week the assignment is duc. Late submissions will not be accepted. Unknown #1: Molecular formula for Compound #l is: C-H,NO2 13C NMR for Compound #1 (7 peaks at 148.4. I 39.9. 135.4. I 29.2. 123.8. I 20.7 and 21.2 ppm) 135.4 129.2 120.7 123 21.2 39.9 148.4 200 180 160 140 120 100 80 0 40 20 'H NMR for Compound #1 (Multiplicity and Chemical shifts: singlet at 7.98, doublet at 7.96, doublet at 7.43, triplet at 7.36, and singlet at 2.47 ppm) 3H 1H 1H 1H 1H 10 9 6 4 ppm IR for Compound #1 LDD 2927 3072 2960 1097 1527 1350 500 L000 Compound #1 Answer Sheet: Note: You may not need every row in the IR table. Emphasis in grading is on the proposed structure and how it matches up with 'H NMR table. You may need to do a little research to complete the IR table for this compound. C NMR Table: Chemical Shift (ppm) Assignment CH H NMR Table: Chemical Shift (ppm) Integration Multiplicity Assignment IR Table: Frequency (cm) Assignment Labeled Structures: Unknown #2: Molecular formula for Compound #2 is: C6H100 13C NMR of Compound #2 (6 peaks at 194.8. 149.4, 146.4, 17.0, 14.5 and 130 ppm) 149.4 170 13.0 146.4 194.8 200 180 160 140 120 100 6 40 20 0 pprm 1H NMR of Compound #2 (Multiplicity and Chemical Shifts: singlet at 94, quartet at 6.6, quartet at 2.3, doublet at 2.0, and triplet at 1.0) зн зн 1H 2H 1H 6 11 10 9 8 ppm IR of Compound #2 2717 3003 2971 1464 1212 1066 1649 s0 1693 800 500 Compound #2 Answer Sheet: Note: You may not need every row in the IR table. Emphasis in grading is on the proposed structure and how it matches up with H NMR table. 3C NMR Table: Chemical Shift (ppm) CH Assignment HNMR Table: Chemical Shift (ppm) Integration Assignment IR Table: Frequency (em) Labeled Structures:

Answer 1

To deduce the structure of any given unknown compound with molecular formula, first step is to determine the Degree of unsaturation (DU) For the calculation of DU following formula can be used DUNumber of carbons +1 (Number of Hydrogens +Number of monovalent atoms (like halogens)- Number of trivalent atoms (like N and P))/2 Oxygens can be ignored while calculating DU For C7H7NO2 the DU is -7+1-(7-1)/2 8-3 5 Hence the unknown compound should have 5 double bonds or 5 rings or a benzene ring and 2 double bonds in its structure From the given spectra, the unknown compound is 1-methyl-3-nitrobenzene In the IR spectrum, peak at 3027 cm1 is due to aromatic C-H stretching peaks at 2960, 2927 cm1 are due to aliphatic C-H stretching Strong peak at 1527 cm1 is due to N-O stretching NO2 1-methyl-3-nitrobenzene Chemical Formula: C7H7N02 1H NMR peaks 13C NMR peaks 478123.7 129.5139.3 8.09 8.10 7.56 31.5 20.3 7.43 2.30

For C6H1o0 the DU is 6+1. (10)12 7-5-2 Hence the unknown compound should have 2 double bonds or 5 rings or a benzene ring and 2 double bonds in its structure From the given spectra, the unknown compound is 2-ethylbut-2-enal 2-ethylbut-2-enal Chemical Formula: C6H10O In the IR spectrum, peak at 1693 cm1 is due to C-O stretching of conjugated aldehyde peaks at 2971, 2927 is due to aliphatic C-H stretching 1H NMR peaks 13C NMR peaks 6.37 145.3 H O 144.7 194.0 15.7 10.18 13.8 2.35 14.5 2.44 0.89