Hex-3-yne is reduced with Na/liq. NH3 to give X. X is treated with Br2/CCl4 to give Y, and one equivalent of Y is then heated with alcoholic KOH (a…

Sodium in liquid ammonia reduces an alkyne by anti addition of hydrogen, so $\mathbf{X}$ is $(E)$-hex-3-ene. Bromine adds anti across the double bond (bromonium ion opened from the opposite face); anti addition to an $(E)$ symmetrical alkene gives the meso dibromide, $(3R,4S)$-3,4-dibromohexane, $\mathbf{Y}$. For E2, $\mathrm{H}$ on C-3 and $\mathrm{Br}$ on C-4 must be anti-periplanar. In that staggered conformer of the meso isomer the remaining $\mathrm{Br}$ on C-3 sits anti to the ethyl group on C-4 (a 3D model gives a dihedral of about $170^\circ$). After elimination these two groups lie on opposite sides of the double bond, and as $\mathrm{Br}$ and the C-4 ethyl are the higher-priority groups, the alkene is $(E)$-3-bromohex-3-ene. The racemic dibromide would instead place them gauche, giving the $(Z)$ isomer. Hence meso $\mathbf{Y}$ and $(E)$ $\mathbf{Z}$.