Considering the reaction sequence given below, the correct...

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.

${NaBH}_{4}$ cannot reduce the acids.

Conc . {NH}_{4}OH$exist in the following equilibrium,</p><p>${NH}_{4}^{+} + {OH}^{-} ⇋ {NH}_{3} + {H}_{2}O