Compound A,${\mathbf{C}}_8{\mathbf{H}}_{10}$, yields three substitution products,${\mathbf{C}}_8{\mathbf{H}}_9\mathbf{Br}$, on reaction with${\mathbf{Br}}_2$. Propose two possible structures for A. The${}^1\mathbf{H}$NMR spectrum of A shows a complex four-proton multiplet at 7.0 d and a six-proton singlet at 2.30 d. What is the structure of A?

The double bond equivalence of a molecule helps in the determination of the amount of unsaturation present in the molecule. DBE is of great use in the determination of the structure of a compound, given its molecular formula is known. The double bond equivalent is given by the following formula.

$\mathbf{DBE}\mathbf{=}\mathbf{C}\mathbf{+}\mathbf{1}\mathbf{-}\left(\frac{{\displaystyle H+X-N}}{{\displaystyle 2}}\right)$

Here, C is the number of carbon atoms, H is the number of hydrogen atoms, X is the number of halogen atoms, and N is the number of nitrogen atoms.

The double bond equivalence for the formula $C_8{H}_{10}$can be calculated as follows:

$\begin{array}{c}\mathrm{DBE}=\mathrm{C}+1-\left(\frac{\mathrm{H}+\mathrm{X}-\mathrm{N}}{2}\right)\\ =(8+1-\frac{(10+0-0)}{2})\\ =4\end{array}$

From the value of DBE, which is equal to 4, it can be deduced that the unknown compound must have had three double bonds and a ring. The molecule closest to this approximation is a benzene derivative.

The following benzene derivatives are possible for the molecular formula$C_8{H}_{10}$are

Possible compounds

From the NMR spectrum data, compound A must have two distinct sets of protons, So, compound I ( three sets of protons) and compound II (four sets of protons) get eliminated from the list. Therefore, two possible structures for A are as follows:

From the above two possible structures, the compound (II) shows a complex four-proton multiplet at $7.0\delta$and a six-proton single at $2.30\delta$. Therefore, the structure of compound A is represented as:

Compound A (o-xylene)

Therefore, the given compound is o-xylene .