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The following spectroscopic data corresponds to an an unknown compound with the molecular formula C4H8O2. Deduce and draw the structure 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.

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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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