Final answer:
Compound Y, with the formula C₆H₁₂ and a higher heat of hydrogenation compared to compound X, is likely a terminal or less substituted internal alkene, such as 1-hexene or cis-trans 2-hexene or 3-hexene. These isomers yield the same 2-methylpentane upon hydrogenation with a palladium catalyst and the same products in bromination and hydroboration/oxidation reactions.
Step-by-step explanation:
Compounds X and Y both have the formula C₆H₁₂ and can hydrogenate to form 2-methylpentane. Since the heat of hydrogenation of X is less than that of Y, it indicates that X is more stable and thus has less strain or fewer steric hindrances. Compounds that react with HBr to form the same bromoalkanes and undergo hydroboration/oxidation to give the same alcohols are likely positional isomers. Isomers are compounds with the same molecular formula, such as C₄H₁₀, but different properties and structures, like n-butane and 2-methylpropane.
In this context, since both X and Y yield 2-methylpentane upon hydrogenation, we can deduce that they must be different alkenes that can form the same alkane through the addition of hydrogen, typically facilitated by a catalyst like palladium. Considering that Y has a higher heat of hydrogenation, it is likely less substituted than X, making it more reactive and possessing more torsional strain. Therefore, Y could be a terminal alkene such as 1-hexene or a less substituted internal alkene like cis-trans isomers 2-hexene or 3-hexene.