The ammonia derivatives of carbonyl compounds obtained as colorless precipitate are
(A) Oxime
(B) Hydrazone
(C) Phenyl hydrazone
(D) Semicarbazone
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B), (C) and (D)
Aldehydes and ketones react with ammonia derivatives, \( \ce{H2N-Z} \), in weakly acidic medium. The nitrogen adds to the carbonyl carbon (nucleophilic addition), and a molecule of water is then lost, giving a C=N bond:
\( \displaystyle \ce{R2C=O + H2N-Z -> R2C=N-Z + H2O} \)
The products are crystalline solids with sharp melting points, so they are used to identify carbonyl compounds. The oximes, hydrazones, phenylhydrazones and semicarbazones of simple aldehydes and ketones separate as colourless (white or very pale) solids. The coloured derivatives are the 2,4-dinitrophenylhydrazones, which are yellow, orange or red.
(A) Oxime — Correct. With hydroxylamine (Z = OH): \( \ce{R2C=O + H2N-OH -> R2C=N-OH + H2O} \).
(B) Hydrazone — Correct. With hydrazine (Z = \( \ce{NH2} \)): \( \ce{R2C=O + H2N-NH2 -> R2C=N-NH2 + H2O} \).
(C) Phenyl hydrazone — Correct. With phenylhydrazine (Z = \( \ce{NHC6H5} \)): \( \ce{R2C=O + H2N-NHC6H5 -> R2C=N-NHC6H5 + H2O} \).
(D) Semicarbazone — Correct. With semicarbazide (Z = \( \ce{NHCONH2} \)): \( \ce{R2C=O + H2N-NHCONH2 -> R2C=N-NHCONH2 + H2O} \).