But-$2$-yne is reacted separately with one mole of Hydrogen as shown below:

$\underset{¯}{\mathrm{B}}\underset{\mathrm{liq} {\mathrm{NH}}_3}{\overset{\mathrm{Na}}{\leftarrow }}{\mathrm{CH}}_3-\underset{+{\mathrm{H}}_2}{\mathrm{C}\equiv }\mathrm{C}-{\mathrm{CH}}_3\underset{∆}{\overset{\mathrm{Pd}/\mathrm{C}}{\to }}\underset{¯}{\mathrm{A}}$

Identify the incorrect statements from the options given below:

A. A is more soluble than $\mathrm{B}$.
B. The boiling point & melting point of $\mathrm{A}$are higher and lower than $\mathrm{B}$ respectively.

C. A is more polar than $\mathrm{B}$ because dipole moment of $\mathrm{A}$ is zero.
D. ${\mathrm{Br}}_2$ adds easily to $\mathrm{B}$ than $\mathrm{A}$.

A: Cis - But-2-ene

B: Trans - But-2-ene

More polar molecules are more soluble in ${\mathrm{H}}_2\mathrm{O}$. Hence A >B.

Molecules having higher dipole moments have higher boiling points due to larger intermolecular forces of attraction. More symmetric isomers have higher melting points due to better packing in crystalline lattice & trans isomers are more symmetric than cis. Hence, in case of boiling point it is A>B and in case of melting point it is B>A.

cis-isomer has resultant of dipoles while in trans isomer dipole moments cancel out. 

So, the correct answer is B, C and D only.