91Ó°ÊÓ

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:

$\begin{array}{c}\text{¶Ù´Ç³Ü²ú±ô±ð â¶Ä‰b´Ç²Ô»å â¶Ä‰e±ç³Ü¾±±¹²¹±ô±ð²Ô³¦±ð }=\text{C}+1−\frac{\text{H}+\text{X}−\text{N}}{2}\\ =8+1−\frac{9+1−0}{2}\\ =9−5\\ =4\end{array}$

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: