Complete the following reactions `:`
`a.`
`b.`
`c.`
`d.`
`e.`
`f.`
`g.` ltbr `h.`
`i.`
`j. `
`k`.`underset((A))(CH_(2)=C=CH_(2))overset(KNH_(2))underset(underset(underset(underset(C_(2)H_(5)OH)(+))(C_(2)H_(5)ONa))(or))rarr(B)`

Explanation

a.
b.
Although product `(Z)` is formed by Saytzeff's elimination, but the product `(Y)` is formed in major amount because product `(Y)` is a more stable conjugated diene than `(Z)`, isolater diene.
c.
d.
e.
f.

`g.`
If `1 mol` of `H_(2)O` is eliminated, a more stable Saytzeff product is formed, which undergoes tautomerism to give ketone.

So, compound `(X)` is not used for the preparation of buta `-13-`diene because a more stable product butan`-2-`one is formed.
`h.`
`i.`
`j.`
`k.` Base `(overset(ddot)overset(Θ)(B),e.g., overset(ddot)overset(Θ)(N)H_(2)" or "C_(2)H_(5)overset(ddot)overset(Θ)(O))` removes a proton, leaveing the anion, which is resoanace stabilised, as shown below.
Hence propyne is more stable than cumulative diene `(A)`. The heat of hydrogenation of allene `(A)` is `-298 kJ mol^(-1)`, whereas that of propyne is `-290 k J mol ^(-1)`.
The more negative is the value of hydrogenation, the less stable is the diene.