Identification of Chemical Properties and Isomeric Relationship of Product Y

To determine the correct statement regarding product $\mathbf{Y}$, let us analyze the nature of products $\mathbf{X}$ and $\mathbf{Y}$, their functional groups, and their chemical properties:

1. Functional Groups and Isomeric Relationship

• Product $\mathbf{X}$ is an unsaturated ether (an alkoxyalkene) with the molecular formula $\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}$, containing an ether linkage ($-\mathrm{O}-$) and an olefinic double bond ($\mathrm{C}=\mathrm{C}$). Its degree of unsaturation (DU) is: $\mathrm{DU} = 7 + 1 - \frac{14}{2} = 1$
• Product $\mathbf{Y}$ has the same molecular formula $\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}$ ($\mathrm{DU} = 1$), but possesses a ketone functional group, specifically a methyl ketone of the type: $\mathrm{CH}_3-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{R}$
• Since compounds $\mathbf{X}$ and $\mathbf{Y}$ share the identical molecular formula ($\mathrm{C}_7\mathrm{H}_{14}\mathrm{O}$) but have different functional groups (an ether-alkene vs. a carbonyl ketone), they are functional isomers of each other.
• They are not geometrical isomers, as geometrical isomerism requires the same connectivity of atoms with different spatial orientations across a restricted rotation site (such as cis/trans alkenes).

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2. Chemical Tests for Product $\mathbf{Y}$

1. Tollens' Test:

   • Tollens' reagent ($\left[\mathrm{Ag}(\mathrm{NH}_3)_2\right]^+\mathrm{OH}^-$) acts as a mild oxidizing agent that readily oxidizes aldehydes and $\alpha$-hydroxy ketones to give a silver mirror.
   • Simple aliphatic ketones, such as $\mathbf{Y}$, are not oxidized by Tollens' reagent and therefore give a negative Tollens' test.

2. Iodoform Test:

   • Compounds containing the methyl carbonyl moiety ($\mathrm{CH}_3-\mathrm{C}=\mathrm{O}$) react with iodine in the presence of base ($\mathrm{I}_2 / \mathrm{NaOH}$) to undergo haloform cleavage, producing a characteristic yellow precipitate of iodoform ($\mathrm{CHI}_3$): $\mathrm{CH}_3-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{R} + 3\mathrm{I}_2 + 4\mathrm{NaOH} \longrightarrow \mathrm{R}-\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{O}^-\mathrm{Na}^+ + \mathrm{CHI}_3\downarrow + 3\mathrm{NaI} + 3\mathrm{H}_2\mathrm{O}$
   • Because $\mathbf{Y}$ contains the $\mathrm{CH}_3-\mathrm{CO}-$ group, it gives a positive iodoform test.

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Conclusion:

Product $\mathbf{Y}$ gives a positive iodoform test and is a functional isomer of $\mathbf{X}$.

Therefore, the correct statement is given by option (C).