a hydrocarbon A ( molecular formula C5H10) yields 2- methylbutane on catalytic hydrogenation. a adds hbr compound B which on reaction with silver hydroxide forms an alcohol C (C5H12). alcohol C on oxidation gives a ketone D. deduce the structure of A, B, C and D.
The molecular formula of A is C5H10. As it fits into general formula CnH2n therefore it could be an alkene or cycloalkane.
As it undergoes hydrogenation therefore it is an alkene.
Since on hydrogenation we get 2-methylbutane therefore a double bond may be present on any of the carbon but the basic skeletal of the compound would be same as that of 2-methylbutane.
As C i.e. Alcohol on oxidation give ketone therefore carbon bearing -OH group should be secondary.
Hence, the compound which satisfies all the above conditions is 3-methylbut-1-ene.
The scheme for above reactions is as follows:
Hence, A is 3-methylbut-1-ene, B is 2-Bromo-3-methylbutane, C is 3-methylbutan-2-ol (please note the molecular formula that you have provided is incorrect, it should be C5H12O) and D is 3-methylbut-2-one.
During the conversion of alkyl bromide to alcohol the reagent mentioned in you question is not clear, you may use aqueous KOH or NaOH for this reaction.