An unknown alkene 'X' with molecular formula C_8H_ 16 undergoes...

Since alkene 'X' gives two moles of a single compound 'Y' on reductive ozonolysis, 'X' must be a symmetrical alkene.

Compound 'Y' does not give a silver mirror with Tollens' reagent, indicating it is a ketone, not an aldehyde. It gives a positive Iodoform test, which means it is a methyl ketone (contains a $\mathrm{-COCH_3}$ group).

Given the molecular formula of 'X' is $\mathrm{C_8H_{16}}$, compound 'Y' must have 4 carbon atoms ($\mathrm{C_4H_8O}$). The only 4-carbon methyl ketone is butan-2-one ($\mathrm{CH_3COCH_2CH_3}$).

Connecting two molecules of butan-2-one at their carbonyl carbons gives the structure of alkene 'X': $\mathrm{CH_3CH_2C(CH_3)=C(CH_3)CH_2CH_3}$, which is 3,4-dimethylhex-3-ene.

Oct-4-ene and 2,5-dimethylhex-3-ene would yield aldehydes (butanal and 2-methylpropanal respectively), which are Tollens positive. 2,4,4-Trimethylpent-2-ene would yield a mixture of two different compounds.

Answer: 3,4-Dimethylhex-3-ene