Chemistry - Hydrocarbon Question with Solution | TestHub
An organic compound A has molecular formula and it is non-resolvable. A does not decolourize brown colour of bromine water solution. A on treatement with yields B as major product. B on treatment with yields which on treatment with yields three monochloro derivative. Also B on boiling with acidic permanganate solution yields . C on heating with sodalime yields . D on reducing with followed by heating the product with concentrated yields as major product. E on treatment with ozone followed by work-up with yields 6-Ketononanal.
Compound B is :
Options:
Answer:
Solution:
Question Explanation: Identify Compound formed via elimination from Compound A .
Concept: Elimination Reaction and Stability of Alkenes
Solution:
Final product, 6-Ketonoanal (a 9-carbon chain with a ketone at and an aldehyde at ). Compound E ⟶ 6-Ketonoanal (Ozonolysis)
D → E (Dehydration/Elimination)
(Reduction)
B → C (Hydrolysis/Acidic )
(Substitution/Rearrangement)
A is a compound with . Since there's no (no water decolorization), it must have two rings.
Reaction : A reacts with base/heat (elimination) or via a carbocation ( side reaction) to yield . Elimination of HBr from 1-bromodecalin forms a bicyclic alkene.
Product B Identification: Following Saytzeff's rule, the most stable (most substituted) alkene is formed. Elimination from 1-bromodecalin yields an octalin (bicyclic alkene with one double bond). Final Answer:
