Question

9. Identify the compound with give molecular formula that gives the IR and 'H NMR spectra shown. Label the IR peaks with the type of functional group. Draw the structure of the compound and label the signals on the structure using letters (a, b, c, etc.) then label the signals on the 'H NMR spectrum with the corresponding letters on the structure. a. C6H1202 Waar 8 (ppm) -frequency

   For part B also calculate degree of unsaturation and use the MS and 13CNMR spectra to identify the molecule. Label the graphs with the information gained.

Answer 1

3 0 calculation of double boriol equalent (Dee) for C6H1203 for ce Saturated blydton gens = 140 m four c l'ine.color furtum lor Hydrogens: 126 DBE - 141-1941 JCE = 1 * It Rag one double bonde.) - From tr Denta at 1940 сril Indicauley Cao Stretching. For the given NHR spectru, the probable solutie is H₂c-c-c-o-CH3 It con88888 of Two types of protons. Helly1 - 22-Diretky, propanoate H26-¢-C-0-CH3 Integration ratio (a:b) 3:9 01:3 Type of proton | Integration multiplicity 1 C Prry Singler - Cth . (a,b). qu. 3.8 PPM I 9H singlet | IPPM (b) 33/33

of (B) calculation DBE for cutting: forcy saturated Hydrogeng = lott rentanter Cyho for Cy Molecular torrula Hydrogens = 10th DBE - 10-10 DBE z o (It is a saturated Molecule) t. From IR Data at 3300 cil Indicates -OH Stretching

- Fron 4' NER the probable structure in ats C H₂L 'CH-Cute-0- H C H-CH₂-0-4 1t3c Iso butyl alcohou (a) H3C 3 G (d) It consists of a types of protons (a, b, c, d) Type of Integration multiplicity / Proton Svakve cepe a) 14 doublet oogpor Multiplet . 117 PPM 3.4 ppm doubler singlet frosso 3.9 ppni HCC - E13 NAR: 3 signals (a, bel a, bel CH-CH₂-04 H3C/ as. mass spectrun Hill CH-CH2-OH Foly H3 CH-CH₂H e) Hzc nolecularion H34, CH-CH2-OH CH 3 CH Case peak h = 43 HBC EHoH) Hzc at н- си, to Сычу % % (- near the 43 74 3