Give the structure that corresponds to the following molecular formula and H1 NMR spectrum: The proton N M R shows a triplet at 0.9 p p m integrating to 11238 and a quartet at 1.4 integrating to 7473. THere is a singlet at 1.35 p p m integrating to 1275 that disappears after a D 2 O shake. The unknown compound has a molecular formula of C 7 H 16 O. The magenta numbers in the spectrum are the relative integrals in arbitrary units. Draw the structure.

Kindly check the attached picture for the diagram of the chemical compound.

Explanation:

So, the following parameters were given from the question above;

=> A triplet at 0.9 ppm and a quartet at 1.4 ppm, a singlet at 1.35 ppm. Now, the unknown compound has a molecular formula of C7H16O.

For a triplet at 0.9ppm, there are nine (9) atoms of hydrogen, for the quartet at 1.4ppm there are six(6) atoms of hydrogen and for the singlet at 1.35 ppm, the number of hydrogen atoms is one(1). Hence, the total number of hydrogen atoms = 16.

Therefore, number of bondings = [(2 × number of carbon atoms) + 2 - number of hydrogen atoms present on the compound)/2 .

Thus number of bonds =[( 2× 7) + 2 - 16] ÷ 2 = 0.

Hence, there is no double bond or ring in the compound.

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