An organic compound (A) of molecular formula C2H5NO reacts with Br2/NaOH to give compound (B) of molecular formula CH5N. (A) is reduced by LiAlH4 to give cmpound (C) of formula C2H7N. Identify (A),(B) and (C).
Here Br2/NaOH acts as Hoffmann reagent since it gives the product with one carbon less than starting material. Hence (A) is an amide and (B) is an amine.
When amide reacts with LiAlH4, it gives amine with the same number of carbon atoms.
CH3CONH2(A)Br2/NaOH−−−−−−−→CH3NH2methylamine(B)+CO2
CH3CONH2LiAlH4−−−−→CH3CH2NH2ethylamine(C)+H2O
Hence, A= acetamide
B= methylamine
C=ethylamine