The graphic representation depicting the possible arrangement of atoms in the compounds with feasible energy and chemical geometry is known as the chemical structure.

Mass/charge ratio =72

IR absorption= 1710cm^-1

¹H-NMR spectra = 0.9 ppm (triplet, 3H), 1.2 ppm (singlet, 3H), 2.4 ppm

(quartet, 2H)

The molecular formula of the compound is evaluated from the value of mass/charge ratio.

Since 72 is completely divisible by 12, isotope carbon-13 is expected to be present in the compound.

Number of carbon atoms $=\frac{72}{13}$

=5 remainder 7

This indicates that the compound may have five C atoms and seven hydrogen atoms.

Replace one carbon with five hydrogen atoms;

Replace one CH₄ with an O atom.

The molecular formula was found to be .

The degree of unsaturation (IHD) $=\frac{\left(\right(2n+2)-x}{2}$

where

n= number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting the values,

the degree of unsaturation (IHD) $=\frac{\left(\right(2\times 4+2)-8)}{2}$

= 1

The value of IHD suggests the presence of a pi-bond.

From the NMR spectra,

a) the triplet at 1.0 ppm attributes to the 3H atoms due to the split of adjacent 2H protons;

b) the singlet at 2.1 ppm attributes to the 3H with an adjacent C with no hydrogen atoms;

c) the quartet at 2.4 ppm attributes to the 2H due to the split of adjacent 3H protons; and

d) the IR vibrational stretching at 1710 cm^-1 indicates the presence of carbonyl stretching of a ketone group.

Mass/charge ratio =88

IR absorption= 3600-3200 cm^-1

¹H -NMR spectra = 0.9 ppm (triplet, 3H), 1.2 ppm (singlet, 6H), 1.5 ppm

(quartet, 2H), 1.6 ppm (singlet, 1H)

The molecular formula of the compound is evaluated from the value of mass/charge ratio.

Number of carbon atoms $=\frac{72}{12}$

= 7 remainder 4

This indicates that the compound may have seven C atoms and four hydrogen atoms.

Replace one carbon with seven hydrogen atoms;

Replace one C atom with an O atom.

The molecular formula was found to be C₅H₁₂O .

The degree of unsaturation (IHD) $=\frac{\left(\right(2n+2)-x)}{2}$

where

n= number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting the values,

The degree of unsaturation (IHD) $=\frac{\left(\right(2\times 5+2)-12}{2}$

= 0

The value of IHD suggests the absence of pi-bond.

From the NMR spectra,

a) the triplet at 0.9 ppm attributes to the 3H atoms due to the split of adjacent 2H protons;

b) the singlet at 1.2 ppm attributes to the 6H due to two adjacent methyl groups;

c) the quartet at 1.5 ppm attributes to the 2H due to the split of adjacent 3H protons;

d) the singlet at 1.6 ppm attributes to the 1H due to the OH proton; and

e) the IR vibrational stretching at 3600-3200cm^-1 indicates the presence of hydroxyl group.

The required structure of J and K are as follows:

Structure of J

Structure of K