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Please help this is super hard... Spectra posted below are in this order: IR, H

ID: 489635 • Letter: P

Question

Please help this is super hard...

Spectra posted below are in this order: IR, H NMR, C and DEPT NMR, HETCOR correlations, COSY correlations, HMBC correlations.

Please perdict a structure using the following spectra and be as elaborate as possible.The molar mass of the compound is 316! Please elaborate on each spectra. Thanks!

The molar mass of the compound is 316

NO NEED FOR MASS SPECTRA. PLEASE REFER TO THE FIRST IMAGE (mass spectra info is on the left side of the IR graph)

C and DEPT NMR carbons: 187, 172, 151, 146, 135, 134, 130, 127, 118, 114, 75, 58, 52, 39, 38, 19, 17, 13

Chem 318319 Beauchamp Name- Problem 70 Predict a reasonable structure from the spectral information provided below. As much as possib match the spectral information to the part of the structure that it explains. Show all of your work. IR Spectrum: Interpret as fully as possible from structure. Not every peak is interpretable. (units cm MS data exact mass Exact Mass: 316.18 1630 M+ 316, 18 (100.0%), 1380 M+1 317-18 20.6%), 3030 M 2-318.19 (1.9%) so 1700 1660 1470 1230 2960-2850 4000 3500 3000 2500 2000 Proton NMR: interpret data (calculate chemical shifts to confirm they match actual values, N neighbors) 2.85 3H, s 1H,brd s 6.79 7.36 3.69

Explanation / Answer

Since Mass is even ( 316)

No Nitrogen present ( According to Nitrogen Rule )

Peaks due to IR shows that Alcohol ( 3300-3400) and ketonic group (1690) present .

M+1 peak abundance is 1.08 , Therefore percent abundance tells that there are 19 Carbons present.

1H NMR signal shows the presence of Aromatic Group ( 6 to 8 ppm ).13 C NMR at 190 ppm shows presence at ketone group .

COSY tells the similar nature of protons.

Hence the formula would be C19H24O4

Hence the Structure would be a naphthalene ring with ketone and alcohol Group.

(( Double bond Equivalence is 8 )

Double bond Equivalence is No of double bonds Present in the ring including No of rings .

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