Considering the reaction sequence given below, the correct statement(s) is(are)

Propanoic acid is having  α-hydrogen hence it is halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-bromo propanoic acid. The reaction is known as Hell-Volhard-Zelinsky reaction.

Second step of the reaction involves Gabriel phthalimide synthesis Where potassium salt of phthalimide which on heating with 2-bromo propanoic acid  followed by alkaline hydrolysis produces the corresponding 2-amino propanoic acid. Aromatic primary amines cannot be prepared by this method.​​​​​​

${\mathrm{NaBH}}_4$ cannot reduce the acids.

$\mathrm{Conc}. {\mathrm{NH}}_4\mathrm{OH}$ exist in the following equilibrium,

${\mathrm{NH}}_4^{+} + {\mathrm{OH}}^{-}\leftrightharpoons {\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}$