Identify A, B, and C.
A, B, and C`(C_6H_(10))overset(Br_2//C CI_2)rarrunderset(Br_2 "solution")("All decolourises")`
`underset(conc. H_2SO_4)overset(Cold)rarr` All are soluble
`(A)underset(AgNO_3)overset("Ammoniacal")rarr` White precipitate
`(A and B)underset(of H_2//Pt)overset("excess")rarr` Hexane
`(C)underset(H_2//Pt)overset(1 mol)rarrD(C_6H_(12))`
`(A)underset(KMnO_4)overset(Hot alk.)rarr` Pent anoic acid only
`(B)underset(KMnO_4)overset(Hot alk.)rarr` Propanoic acid only
`(C)underset(KMnO_4)overset(Hot alk.)rarr` Adipic acid only

Explanation

DU in A, B, and `C=((2xx6+2)-10)/(2)=27^@`.
Structure of (A):
(A) reacts with ammoniacal `AgNO_3` solution, which shows that it is a terminal alkyne with straight chain of six C atoms (because on hydrogenation it gives hexane). Therefore, (A) is hex-1-yne.

Structure of (B):
(B) does not react with `[Ag(NH_3)_2]^(o+)` and hence it must be internal alkyne with straight chain of six C atoms (because on hydrogenation it gives hexane).
Location of triple bond is determined by oxidative cleavage. As it gives propanoic acid (3C-atom acid), it should be symmetrical internal alkyne.
Therefore, (B) is hex-3-yne.

Structure of (C):
It absorbs 1 mol of `H_2` and the product (D) `(C_6H_(12))` after catalytic hydrogenation has 1 DU. So (D) must be a 6C-atom cyclic compound. Therefore, (C) is cyclohexene.