Chapter 13: Amines
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The IUPAC name is N-Methylethanamine. (It is a secondary amine where the ethyl group forms the parent chain).
Answer: The chemical name is Benzenesulfonyl chloride and its formula is $C_6H_5SO_2Cl$.
Answer: Only Primary ($1^\circ$) Amines (both aliphatic and aromatic) give this test, known as the Carbylamine test.
Answer: $NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$.
(Note: For methyl amines in aqueous solution, the order is $NH_3 < 3^\circ < 1^\circ < 2^\circ$ due to combined inductive, solvation, and steric effects).
2 Mark Questions (Short Answer-I)
Answer:
When a primary amide is heated with bromine and an aqueous or ethanolic solution of sodium hydroxide (NaOH) or potassium hydroxide (KOH), it yields a primary amine. This reaction is called Hoffmann bromamide degradation.
It is called a "degradation" (step-down) reaction because the primary amine formed contains one carbon atom less than the parent amide (the carbonyl carbon is lost as carbonate).
Example (Preparation of Methylamine from Acetamide):
$CH_3-CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} CH_3-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
(Acetamide) (Methylamine)
Answer:
Nitrous acid is generated in situ by reacting $NaNO_2$ with dilute $HCl$ at 273-278 K.
- (a) On Ethylamine ($1^\circ$ Aliphatic amine): It forms a highly unstable aliphatic diazonium salt, which immediately decomposes by reacting with water to yield ethanol and brisk effervescence of nitrogen gas.
$C_2H_5NH_2 + HNO_2 \xrightarrow{NaNO_2/HCl} [C_2H_5N_2^+Cl^-] \xrightarrow{H_2O} C_2H_5OH + N_2 \uparrow + HCl$ - (b) On Aniline ($1^\circ$ Aromatic amine): It reacts at low temperatures (273-278 K) to form a relatively stable aromatic diazonium salt (Benzenediazonium chloride). This reaction is called diazotization.
$C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278 \text{ K}} C_6H_5N_2^+Cl^- + 2H_2O$
Answer:
In aniline ($C_6H_5NH_2$), the unshared (lone) pair of electrons on the nitrogen atom is in conjugation with the $\pi$-electron system of the benzene ring. Due to this resonance (+R effect), the lone pair is delocalized over the benzene ring, making it less available for donation to a proton ($H^+$).
In contrast, in ammonia ($NH_3$), the lone pair on the nitrogen atom is localized and readily available for protonation. Furthermore, the anilinium ion formed after protonation has only two resonance structures, whereas aniline has five, making the unprotonated form of aniline more stable. Hence, aniline is a weaker base than ammonia.
3 Mark Questions (Short Answer-II / Mechanisms)
Answer:
Hinsberg's reagent is Benzenesulfonyl chloride ($C_6H_5SO_2Cl$). It reacts differently with $1^\circ$, $2^\circ$, and $3^\circ$ amines:
- Primary ($1^\circ$) Amines:
They react with Hinsberg's reagent to form N-alkylbenzenesulfonamide. The hydrogen atom attached to the nitrogen in this sulfonamide is highly acidic because of the strong electron-withdrawing nature of the sulfonyl group. Therefore, the product is soluble in aqueous alkali (NaOH).
$C_6H_5SO_2Cl + R-NH_2 \rightarrow C_6H_5SO_2NHR + HCl$ - Secondary ($2^\circ$) Amines:
They react to form N,N-dialkylbenzenesulfonamide. Since there is no hydrogen atom attached to the nitrogen in this product, it is not acidic. Therefore, it is insoluble in aqueous alkali.
$C_6H_5SO_2Cl + R_2NH \rightarrow C_6H_5SO_2NR_2 + HCl$ - Tertiary ($3^\circ$) Amines:
They do not possess a replaceable hydrogen atom on the nitrogen. Therefore, they do not react with Hinsberg's reagent at all.
Answer:
Gabriel Phthalimide Synthesis:
This is a highly useful method for preparing pure primary aliphatic amines. It involves three steps:
- Phthalimide is treated with ethanolic potassium hydroxide (KOH) to form the potassium salt of phthalimide.
- The potassium phthalimide is then heated with an alkyl halide ($R-X$). The nucleophilic phthalimide anion undergoes an $S_N2$ reaction to form N-alkyl phthalimide.
- Alkaline hydrolysis (or hydrazinolysis) of N-alkyl phthalimide yields the pure primary aliphatic amine ($R-NH_2$) and the salt of phthalic acid.
Reason for failure with Aromatic Amines:
Primary aromatic amines (like aniline, $C_6H_5NH_2$) cannot be prepared by this method. To prepare aniline, one would need to react potassium phthalimide with an aryl halide (like chlorobenzene). However, aryl halides do not undergo nucleophilic substitution reactions easily due to the partial double bond character of the C-X bond (arising from resonance). Hence, the phthalimide anion cannot displace the halogen from the benzene ring.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 2-mark name reaction combined with a 2-mark reasoning question).
(b) Write a short note on the Sandmeyer reaction. [2 Marks] March 2014, March 2018, Oct 2022
Answer (a): Carbylamine Reaction
The Carbylamine reaction (or isocyanide test) is a specific test for primary amines (both aliphatic and aromatic). When a primary amine is heated with chloroform ($CHCl_3$) and an alcoholic solution of potassium hydroxide ($KOH$), it forms an isocyanide (carbylamine) which has an extremely pungent, foul, and intolerable smell.
Reaction with Ethylamine:
$C_2H_5-NH_2 + CHCl_3 + 3KOH(\text{alc.}) \xrightarrow{\Delta} C_2H_5-NC + 3KCl + 3H_2O$
(Ethylamine) (Ethyl isocyanide)
Answer (b): Sandmeyer Reaction
The Sandmeyer reaction is used to synthesize aryl halides or aryl cyanides from an aromatic diazonium salt. When a freshly prepared solution of benzene diazonium chloride is treated with cuprous chloride ($Cu_2Cl_2$) dissolved in HCl, cuprous bromide ($Cu_2Br_2$) in HBr, or cuprous cyanide ($Cu_2(CN)_2$) in KCN, the diazonium group ($-N_2^+Cl^-$) is replaced by $-Cl$, $-Br$, or $-CN$ respectively, liberating nitrogen gas.
$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2 / HCl} C_6H_5Cl \text{ (Chlorobenzene)} + N_2 \uparrow$
$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2(CN)_2 / KCN} C_6H_5CN \text{ (Benzonitrile)} + N_2 \uparrow$
(b) How is an azo dye prepared from benzene diazonium chloride? Write the chemical equation. [2 Marks] Oct 2016, March 2021
Answer (a): Nitration of Aniline
The amino group ($-NH_2$) is strongly ortho, para-directing. However, direct nitration is carried out using a strongly acidic nitrating mixture (conc. $HNO_3$ + conc. $H_2SO_4$). In this highly acidic medium, a large proportion of aniline accepts a proton to form the anilinium ion ($C_6H_5NH_3^+$).
The $-NH_3^+$ group is strongly electron-withdrawing and is meta-directing. As a result, the incoming electrophile ($NO_2^+$) is directed to the meta position, resulting in a surprisingly high yield (about 47%) of meta-nitroaniline alongside the expected ortho and para products.
Answer (b): Coupling Reaction (Azo Dye)
Azo dyes are prepared by the coupling reaction of benzene diazonium chloride with highly activated aromatic compounds like phenols or anilines. The diazonium ion acts as an electrophile and attacks the para position of the phenol or aniline ring to form a brightly colored compound containing the azo linkage ($-N=N-$).
Example: Reaction with Phenol (Orange Dye)
$C_6H_5-N_2^+Cl^- + H-C_6H_4-OH \xrightarrow{\text{mildly alkaline (pH 9-10)}} C_6H_5-N=N-C_6H_4-OH + HCl$
(Benzene diazonium chloride) (Phenol) (p-Hydroxyazobenzene - Orange Dye)
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