Question

8. Fill in the table below with the information on each unique 1H-NMR signal. You may treat all aromatic protons as a single signal if that helps. Peak # Chemical Shift (8) # of H Multiplicity Proposed Fragment Structure 9. Based on your molecular formula, 'H-NMR integration and 13C-NMR, is there any symmetry in your molecule? Explain.

Answer 1

Elemental Analysis of compound is c. 53.31 ; H. 1118 and 35:51. molecular formula can be calculated by dividing Percentage with atomic mass. Then convert the values into integers by dividing least valua as shown below c H o The 53.31 11-18 35.56 Then divide least values as shown divide least values as shown Tai T T6 to get integers 4.4425 ; 1118 ze 4.4425 ; 11:18 2-2025 2.2225 2.2225 20225) 2.2225 1.999 5.03, 1 . emperital formula = CH 5o; molae mass of emperical fel- milla is 45: However from mass spectra moleculae Ion Peak is at m/= 90 so molecular formula is empecical filmul X2= CalH50)X = C4H1002 molecular formula = C4H1002 Degree 4 ansaturation : C-CHEX)+1 C= number of carbonaton, 4; H = number of hugdeogenatong = 10 X=number of halogen atoms=0 Degree of cusaturation - 4- (10-0)+1 = 4-5+1=0 I so no double bond is present. IR Spectrum * strong peak at 2990cm belongs to GH stretching A alkane. * Peaks at 1470 cm' and 1280 cmt belongs to G-H bending of Ch, and CH₃ groups. * strong peak at 1150cñ' belongs to Go Stretching frequency of ether Based upon this the compound contain ether frank tional group. THNMR spectrum * Singht at chemical shift value S 8.40ppm with integration

of SH petons belongs alkoxy methyl group. * Singht at 8 3.5 ppm belongs to alkaxy Che group since NMR has two peaks and mote molecular formula has 10 peotons. So It has double the groups as shown below CHg-o-che-CH2O-CH₃ IHNMR values > CH2-0- CH2- CH - - CH 3. (345) (3-5,5) (3:58) (3:48) The molecule has plain of symmetey as shown below. CH₃-o- CH₂ I CHO-CH3 Plain equal on both sides of aus of plain •
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