Chemistry - Nomenclature Question with Solution | TestHub
A trisubstituted compound ' A ', gives neutral test positive. Treatment of compound ' A ' with NaOH and gives , with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound . Compound ' A ' also decolorises alkaline . The number of it bond present in the compound ' A ' is 。
Answer:
Solution:
Explain Question:
The question asks for the total count of π bonds in a trisubstituted compound based on its molecular formula and specific chemical test results.
Concept:
Degree of Unsaturation (DU), FeCl₃ Test (Phenols), KMnO₄ Test (Alkenes), Ether Cleavage.
Solution:
Calculate DU: For C10H12O₂, the DU is calculated as:
This means the total of (rings + π bonds) is 5.
Identify Aromaticity: A positive neutral FeCl₃ test indicates a phenol. A benzene ring accounts for DU = 4 (1 ring + 3π bonds).
Identify Aliphatic Unsaturation: Decolorizing alkaline KMnO₄ confirms an aliphatic double bond (C=C). This accounts for the remaining DU = 1 (1π bond).
Other Groups: The HI test confirms a methoxy group (-OCH₃), which contains no π bonds.
Total π bonds = 3 (from benzene ring) + 1 (from aliphatic double bond) = 4.
Final Answer: 4.