Question

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

Short answer

Answer

Structure of isomer A

Structure of isomer B

Step-by-step solution

Aromatic compound NMR

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.

Elucidating the structure

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: