The following spectroscopic data corresponds to an an unknown compound with the molecular formula C4H8O2. Deduce and draw the st
ructure of the compound that corresponds to the data. 1H NMR: δ 4.07 (quartet, 2H), 1.97 (singlet, 3H), 1.18 (triplet, 3H) ppm. 13C NMR: δ 170, 60, 20, 14 ppm.
1 answer:
Answer:
The name of the structure is ethyl acetate with one double bond. The attached document contains the structure of the compound
Step-by-step explanation:
¹H NMR =δ4.07(quartet), 1.97( singlet), 1.18(triplet)
2H 3H 3H
¹³C NMR = δ 170, 60, 20, 10 ppm
from the molecular formula,C₄H₈O₂ and spectrum study, the structure of the compound will have a double bond which is calculated as DU= 4+1-(8/2)= 1
thus, the name of the compound as drawn in the attachment is ethyl acetate
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