Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes.
Give the structures of A and B in the following reaction :
\( \mathrm{C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow[273\ K]{HNO_2} B} \)
Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes.
\( \mathrm{C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow[273\ K]{HNO_2} B} \)
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Solution and Explanation
The reaction has two stages. First a nitro group is reduced to an amino group. Then the primary aromatic amine is treated with nitrous acid at a low temperature (diazotisation).
Step 2: Key Formula or Approach:
1. Fe with HCl reduces \( \mathrm{-NO_2} \) to \( \mathrm{-NH_2} \).
2. Nitrous acid is made in the flask from NaNO2 and HCl. At 273-278 K it converts \( \mathrm{Ar-NH_2} \) into a diazonium salt \( \mathrm{ArN_2^+Cl^-} \).
Step 3: Detailed Explanation:
Stage 1, reduction of nitrobenzene:
\[ \mathrm{C_6H_5NO_2 + 6[H] \xrightarrow{Fe/HCl} C_6H_5NH_2 + 2H_2O} \]
So A is aniline (benzenamine), \( \mathrm{C_6H_5NH_2} \).
Stage 2, diazotisation of aniline:
\[ \mathrm{C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273\ K} C_6H_5N_2^+Cl^- + 2H_2O} \]
So B is benzenediazonium chloride, \( \mathrm{C_6H_5N_2^+Cl^-} \), a benzene ring joined to the group \( \mathrm{-N{\equiv}N^+} \) with a chloride ion.
The temperature is kept near 273 K because the diazonium salt breaks down to phenol and nitrogen gas above about 278 K.
Final Answer:
A = aniline, B = benzenediazonium chloride. \[ \boxed{A = \mathrm{C_6H_5NH_2},\quad B = \mathrm{C_6H_5N_2^+Cl^-}} \]
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Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
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Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
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(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
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Top CBSE CLASS XII Importance of DiazoniumSalts in Synthesis of Aromatic Compounds Questions
- How will you obtain the following from benzene diazonium chloride ? Write chemical equations involved : (a) Phenol
- How will you obtain the following from benzene diazonium chloride ? Write chemical equations involved : (b) Iodobenzene
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- Assertion (A) : Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Reason (R) : Diazonium salts of aliphatic amines undergo resonance.
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