Read the following statements and identify the correct ones:
A. Gabriel phthalimide synthesis is preferred for secondary amines only.
B. Aniline doesn't undergo Friedel-Craft's reaction.
C. Methylamine in water doesn't precipitate ferric ions from ferric chloride.
D. Azo dyes are formed as a result of a coupling reaction of aryl diazonium salts with phenols and arylamines.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B and D only
This question tests four facts about amines. Judge each statement on its own.
A. Gabriel phthalimide synthesis is preferred for secondary amines only. — Incorrect. Gabriel synthesis prepares primary amines. Phthalimide is converted to its potassium salt, alkylated with an alkyl halide, and then hydrolysed to give \( \ce{R-NH2} \):
\( \displaystyle \text{phthalimide} \xrightarrow{\text{ethanolic KOH}} \text{potassium phthalimide} \xrightarrow{\ce{R-X}} \text{N-alkylphthalimide} \xrightarrow{\ce{NaOH(aq)}} \ce{R-NH2} \)
It cannot make aromatic primary amines, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
B. Aniline doesn't undergo Friedel-Craft's reaction. — Correct. Aniline is a Lewis base and reacts with the Lewis acid catalyst \( \ce{AlCl3} \) to form a salt. The nitrogen then carries a positive charge, which strongly deactivates the ring, so no alkylation or acylation takes place:
\( \displaystyle \ce{C6H5NH2 + AlCl3 -> C6H5NH2^{+}-AlCl3^{-}} \)
C. Methylamine in water doesn't precipitate ferric ions from ferric chloride. — Incorrect. Methylamine is basic and produces \( \ce{OH^{-}} \) ions in water. These ions precipitate iron as hydrated ferric oxide:
\( \displaystyle \ce{2FeCl3 + 6CH3NH2} + (x+3)\,\ce{H2O} \rightarrow \ce{Fe2O3.xH2O + 6CH3NH3^{+}Cl^{-}} \)
D. Azo dyes are formed as a result of a coupling reaction of aryl diazonium salts with phenols and arylamines. — Correct. Benzenediazonium chloride couples with phenol to give p-hydroxyazobenzene (an orange dye), and with aniline to give p-aminoazobenzene (a yellow dye). Both contain the azo group, –N=N–:
\( \displaystyle \ce{C6H5N2^{+}Cl^{-} + C6H5OH + OH^{-}} \rightarrow \text{p-hydroxyazobenzene} + \ce{Cl^{-} + H2O} \)
\( \displaystyle \ce{C6H5N2^{+}Cl^{-} + C6H5NH2} \xrightarrow{\text{mild acid}} \text{p-aminoazobenzene} + \ce{HCl} \)