Question

A trisubstituted compound ‘A’, \(\text{C}_{10}\text{H}_{12}\text{O}_2\), gives neutral FeCl\(_3\) test positive. Treatment of compound ‘A’ with NaOH and CH\(_3\)Br gives \(\text{C}_{11}\text{H}_{14}\text{O}_2\), with hydroiodic acid gives methyl iodide, and with hot conc. NaOH gives a compound ‘B’, \(\text{C}_{10}\text{H}_{10}\text{O}_2\). Compound ‘A’ also decolourises alkaline KMnO\(_4\). The number of \(\pi\)-bond/s present in the compound ‘A’ is ______.

Hint:

For determining the number of π bonds in a compound, carefully analyze the structure, including aromatic and aliphatic double bonds. Utilize chemical reactions (e.g., FeCl₃ test, KMnO₄ decolourisation  to identify functional groups and confirm bond types.)

Answer 1

Correct Answer: 4

Step 1: Analyze the Structure of Compound ‘A’
Given the molecular formula \(\text{C}_{10}\text{H}_{12}\text{O}_2\), compound ‘A’ gives a positive neutral FeCl\(_3\) test, indicating the presence of a phenolic (\(-\text{OH}\)) group. Compound ‘A’ also decolourises alkaline KMnO\(_4\), indicating the presence of a double bond.
Step 2: Chemical Reactions of ‘A’
- Treatment with NaOH and CH\(_3\)Br forms \(\text{C}_{11}\text{H}_{14}\text{O}_2\), confirming the presence of a second hydroxyl group.
- Hydroiodic acid treatment yields methyl iodide, confirming the presence of an ether bond (\(-\text{OCH}_3\)).
- Hot NaOH treatment forms compound ‘B’, \(\text{C}_{10}\text{H}_{10}\text{O}_2\), which retains a double bond.
Step 3: Determine the Number of \(\pi\) Bonds
The structure of compound ‘A’ contains:
One aromatic benzene ring with three \(\pi\) bonds.
One aliphatic double bond outside the aromatic ring.
Thus, the total number of \(\pi\) bonds in compound ‘A’ is:
\[\text{Number of \(\pi\) bonds} = 3 \, (\text{aromatic}) + 1 \, (\text{aliphatic}) = 4.\]
Conclusion:
The number of \(\pi\) bonds present in compound ‘A’ is \(\mathbf{4}\).