13C NMR: Draw out the molecule associated with this graph:

Please explain how to solve the following NMR:

A compound with the molecular formula C5H₁002 has the following ¹H NMR spectrum: Proton NMR 5.0 4.5 4.0 3.5 3.0 2.5 6.33 Integration Values 19.4 37.9 Determine the number of proton(s) giving rise to the signal at 5 ppm. 0000 10 3 6 1 2.0 1.5 1.0 4 0.5 ppm

Identify the molecule for the 13C-NMR spectrum below. 120 100 PPM 40 20

Q2: The proton NMR spectrum is shown for a compound with formula C5H9NO4. The infrared spectrum displays strong bands at 1750 and 1562 cm-1 and a medium-intensity band at 1320 cm-1. The normal carbon- 13 and the DEPT experimental results are tabulated. Draw the structure of this compound. Normal Carbon DEPT-135 DEPT-90 14 ppm Positive No peak No peak 16 Positive 63 Negative No peak 83 Positive Positive 165 No peak No peak Proton spectrum C;H,NO4 0.92 2.01 3.00 3.00 5.0 4.5 4.0 3.5 3.0 25 2.0 1.5 10 0.5 0.0

Assign the resonances in this C-NMR spectrum.

2. (Chapter 13 - Q58b) The compound whose H NMR spectrum is shown has the molecular formula C7H7B1. Follow the following questions to predict the unknown structure. Chem. shift Rel. area 2.31 1.50 701 1.00 7.35 1.00 TMS 10 7 6 O ppm 4 3 2 Chemical shift (8) e20s Cenge leaming 2(a) Degree of the unsaturation of this compound is= 2(b) The two distinct peaks in the aromatic region of the H NMR indicate that compound is .disubstituted = 2(c) The splitting pattern of the peak at 2.31 ő is = 2(d) The group that corresponds to the splitting pattern in 1(c) is = 2(e) This compound has the plane of symmetry Yes or No = 2(f) The name of the unknown compound = Intensity

3. (Chapter 13 - Q58c) The compound whose H NMR spectrum is shown has the molecular formula C3H9B1. Follow the following questions to predict the unknown structure. Chem. Rel. shift area 3.16 1.00 3.56 1.00 1.00 1.50 Z18 729 TMS 10 O ppm Chemical shift (8) ng 3(a) Degree of the unsaturation of this compound = 3(b) The two peaks at 3.16 8 and 3.56 ő indicate two triplets Yes or NO = 3(c) The compound has 2 CH2 groups Yes or No = 3(d) Depending on the two triplets' chemical shifts one of the CH2 may have attached to the Br, Yes or No = 3(e) The multiplet at 7.18 õ to 7.29 õ indicate that compound is. disubstituted benzene - 3(f) The name of the unknown compound is = Intensity