Question: Identify the structures of isomers A and B (molecular formula ${\mathbf{\text{C}}}_{\mathbf{\text{8}}}{\mathbf{\text{H}}}_{\mathbf{\text{9}}}\mathbf{\text{Br}}$)

In benzene, the protons on the aromatic ring are magnetically and chemically equivalent and are in the downfield region.

When any of these protons is substituted with a different group or atom, some protons become magnetically inequivalent.

All sets of protons couple to each other and have a corresponding J value based on the type and number of bonds.

Calculating double bond equivalence:

$\begin{array}{c}\text{Double\hspace{0.17em}\hspace{0.17em}bond\hspace{0.17em}\hspace{0.17em}equivalence\hspace{0.17em}}=\text{C}+1-\frac{\text{H}+\text{X}-\text{N}}{2}\\ =8+1-\frac{9+1-0}{2}\\ =9-5\\ =4\end{array}$

Since double bond equivalence is 4, the structure contains a benzene ring

A.

• One triplet at 1.2 ppm - 3 hydrogens - 3 H’s adjacent to 2 H’s
• One quartet at 2.6 ppm - 2 hydrogens - 2 H’s adjacent to 3 H’s
• Two signals at 7.1 and 7.4 - 2+2 hydrogens - para disubstituted benzene

Based on the spectra and molecular formula, the most probable structure for A is:

B.

• One triplet at 3.1 ppm - 2 hydrogens - 2 H’s adjacent to 2 H’s
• One quartet at 3.5 ppm - 2 hydrogens - 2 H’s adjacent to 2 H’s
• Multiplet at 7.1 and 7.4 - 5 hydrogens - monosubstituted benzene

Based on the spectra and molecular formula, the most probable structure for B is: