Question
Write the molecular formulas from the mass spec, IR, and H-NMR of each of the following. (You can ignore the C-NMR spectra)
A)
1818 1751 IR Spectrum 1200 v (cm 100 60 No significant UV absorption above 220 nm 74 M 130 19) C6H10O3 2 40 80 120 160 200 240 280 m/e 13C NMR Spectrum (50.0 MHz, CDC, solution) 160 400 H NMR spectrum (200 MHz, CDC, solution) 10 9 8 7 6 5 4 3 2 1 0 6 (ppm)
B)
media%2F6ee%2F6ee40787-e71c-44de-9135-c0
C)
media%2Fc6f%2Fc6f75cdb-7dbe-43c1-b146-18

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Answer #1

A)

in this Ir shows at anhydride 1818 and at 1750 is for ester carbonyl

from proton coupled spectra 13 C value at near 1bout 160 ppm is for carbonyl and and splitin n+1 for ch2 is triplet and for ch3 is quartet

from nmr spectra it is clear that quartet at downfield is observed due to attachment of ch2 group to electron withdrawing carbonyl and it gives quartet at 2.5 ppm, ch3 gives triplet at 1.2 ppm value.

hence str is 1818 cm-1 1750 cm-1 C) Triplet at 1.3 ppm2)

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