Question

Each problem contains the formula of the compound, the IR spectrum and the H NMR spectrum.

For this problem you must:

A. Calculate the degree of unsaturation

B. Assign IR abosrbtion bands above 1500cm-1

C. Draw the structure of the Compound.

Problem 2: C3H30 TTTTTTTTTTTTTTTT 10 Proton NMR 9 2 PPN 2. integral ratios

Answer 1

Problem 2 C q Hgo & Degree of unsaturation. Molecular formulo - C&HgO. saturated analogue for this molecular formula is CHIS. C8H18 -C8H8O HO 10 * Hydrogen deficient. 9.5 sites of unsaturation Degree of unsaturation - 05. su - 20+2 +N-H-X - 2x8+2-8-0. 18-8 SW = I sin = os 205. sites of unsaturation) - os degree of unsaturation B* Assign IR absorption bands above 1sono co 0 1577 cont - coco It indicate cre streching frequency in the ring, & 1604 cont - streching frequency for ring carbon. (② 1703 on this frequency indicates carbonyl frequency of a, B unsaturated aldehyde.

4 2733 cm CH sreching frequency of aldehyde ② 2828 cont - CH streching frequency of aldehyde 6 2922 cm. spiccn streching frequency of ch 1500-2600 cont in that range aromatic cre streching frequency 2700-2850 cmt this fange for caldehyde ch streching Frequency. 1703 - daß unsaturated aldehyde. Draw the structure of the compound. 'H-NMR. 2.4 - CH3 , s. 3810 d 25.3 7.3 024 b Aromatic proton purusubstituted bennene 7.8. do 23. 3 0 .24 9.96, S. 11 i aldehyde proton Intensity - 38.0 +25.3 +23.3 +11. 1 2 9717 = 12.2 204 ppon ,S, BH 3H CH3, singlet ridicates reno proton on neighbouring carbon chemical shirt indicate attach to aromatic .. chemical shifl 2.4 ppm,

7.3 8 7.8, doublet, indicates parabubstituted. benene. g. 96 ppm chemical shift for aldehydee proton. By analysis of above data we conclude that giren organic compounds structure is the H3 H3 H 1500- 1600 cm A 1703 en thye g.96 ppm TH Har 7.8 ppm. Mb 2700- 2850on! 2.4 ppm 7-3 ppm 2922 om