Compounds X has the formula C7H15Cl; Y is C7H15Br. X undergoes base-promoted E2 elimination to give a single alkene product Z. Y likewise reacts under similar conditions to give a single alkene product that is isomeric with Z Catalytic hydrogenation of Z affords 3-ethylpentane. X readily reacts in SN2 fashion with sodium iodide in acetone. Y does not undergo a similar SN2 reaction. Propose structures for X and Y.

See explanation and image attached

Explanation:

Let us examine the statements in the question carefully. First of all, we will discover that the products of the E2 elimination of the both compounds are isomeric. However Y does not undergo SN2 reaction as  X does.

The fact that SN2 reaction does not occur in Y confirms that the bromine atom is attached to a tertiary carbon atom and SN2 reaction does not occur due to steric hinderance. Since X undergoes SN2 reaction in aprotic solvent, the chlorine atom must be attached to a secondary carbon atom.

However, E2 reactions does occur with tertiary alkyl halides when strong bases such as OH^- or RO^- are used.

The question also stated that the catalytic hydrogenation of Z affords 3-ethylpentane.

Putting all these together, the structures of X and Y have been suggested in the image attached to this answer.