Find the number of rings 1 double bonds in molecules with the following compositions and draw one plausible structure for each: ${\mathrm{C}}_{11}{\mathrm{H}}_{18}{\mathrm{N}}_2{\mathrm{O}}_3$(a) (b)${\mathrm{C}}_{12}{\mathrm{H}}_{15}\mathrm{BrNPOS}$(c) fragment in a mass spectrum with the composition ${\mathrm{C}}_3{\mathrm{H}}_5^{+}$.

The total number of unsaturation contained in an organic molecule is known as the Double Bond Equivalence (DBE), and an unsaturation is defined as a double bond or a cyclic or ring system.

We must determine the structure of the provided chemicals in this challenge.

If we know the molecular formula of a molecular ion, we may calculate the number of rings and double bonds:

$R+DB=c-\frac{h}{2}+\frac{h}{2}+1...\left(1\right)$

Here,

c = The number of atoms in group 14 (make four bonds)

h = The number of halogen atoms in the H+ state (make one bond)

n = The number of atoms in group 15 (make three bonds)

Group 16 atoms, such as O and S, are not included in the calculatio

The number of atoms in group 14 is 11, the number of atoms in group H is 18, and the number of atoms in group 15 is 2.

So, to get the number of rings and double bonds, we plug data into equation 1

$R+DB=11-\frac{18}{2}+\frac{2}{2}+1=4$

One ring and three double bonds are possible in a molecule.

As a result, the structure is as follows:

The number of atoms in group 14 is 12, the number of atoms in group H+ halogen is 1 5+1, and the number of atoms in group 15 (N+P) is 1 + 1.

So, to get the number of rings and double bonds, we plug data into equation 1:

$R+DB=12-\frac{15+1}{2}+\frac{1+1}{2}+1=6$

Two rings and four double bonds are possible in a molecule.

As a result, the structure is as follows:

The number of atoms in group 14 is three, while the number of atoms in group H is five.

So, to find the number of rings and double bonds, we plug data into equation 1.

$R+DB=3-\frac{5}{2}+\frac{0}{2}+1=1.5$

One double bond is possible in the molecule. We have a fraction, and the C atom has three bonds (rather than four), thus we have 1.5.

As a result, the structure is as follows: