Amines
Preparation, Basicity Trends, Distinguishing Tests & Fully Solved Board PYQs
1. Introduction and Classification
Amines are organic derivatives of ammonia ($NH_3$) obtained by replacing one, two, or all three hydrogen atoms by alkyl and/or aryl groups.
The nitrogen atom in amines is $sp^3$ hybridized and possesses one unshared (lone) pair of electrons, making them Lewis bases and nucleophiles. The geometry is pyramidal.
Classification
Amines are classified based on the number of alkyl/aryl groups attached directly to the nitrogen atom:
- Primary ($1^\circ$) Amines: One hydrogen atom of $NH_3$ is replaced ($R-NH_2$). Example: Ethylamine, Aniline.
- Secondary ($2^\circ$) Amines: Two hydrogen atoms are replaced ($R-NH-R'$). Example: Dimethylamine, N-Methylaniline.
- Tertiary ($3^\circ$) Amines: All three hydrogen atoms are replaced ($R-N(R')-R''$). Example: Trimethylamine, N,N-Dimethylaniline.
2. Preparation of Amines
A. Ammonolysis of Alkyl Halides (Hoffmann's Exhaustive Alkylation)
When an alkyl halide is heated with an alcoholic solution of ammonia in a sealed tube at 373 K, nucleophilic substitution occurs. This yields a mixture of $1^\circ, 2^\circ, 3^\circ$ amines and quaternary ammonium salt.
$R-X + NH_3 \xrightarrow{373 \text{ K}} R-NH_2 \xrightarrow{R-X} R_2NH \xrightarrow{R-X} R_3N \xrightarrow{R-X} R_4N^+X^-$
Note: To get primary amine as the major product, a large excess of ammonia is used.
B. Reduction Methods
- Reduction of Nitro Compounds: Nitroalkanes/Nitroarenes are reduced to $1^\circ$ amines using $Sn/HCl$, $Fe/HCl$, or catalytic hydrogenation ($H_2/Pd$).
$R-NO_2 + 6[H] \xrightarrow{Sn/HCl} R-NH_2 + 2H_2O$ - Reduction of Alkyl Cyanides (Nitriles) - Mendius Reduction: Nitriles are reduced to $1^\circ$ amines using Sodium amalgam and ethanol ($Na(Hg)/C_2H_5OH$) or $LiAlH_4$.
$R-C \equiv N + 4[H] \xrightarrow{Na/C_2H_5OH} R-CH_2-NH_2$ - Reduction of Amides: Amides on reduction with $LiAlH_4$ give $1^\circ$ amines containing the same number of carbon atoms.
$R-CONH_2 \xrightarrow{LiAlH_4, \text{ ether}} R-CH_2-NH_2$
C. Gabriel Phthalimide Synthesis (For Pure $1^\circ$ Aliphatic Amines)
This method is used exclusively for the preparation of pure primary aliphatic amines. Aromatic primary amines (like aniline) cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the phthalimide anion easily.
- Phthalimide reacts with ethanolic $KOH$ to form potassium phthalimide.
- Potassium phthalimide is heated with an alkyl halide ($R-X$) to form N-alkyl phthalimide.
- Alkaline hydrolysis (or hydrazinolysis) of N-alkyl phthalimide yields the primary amine ($R-NH_2$) and the salt of phthalic acid.
D. Hoffmann Bromamide Degradation (Step-down Reaction)
When a primary amide is heated with bromine and an aqueous or ethanolic solution of $NaOH$ or $KOH$, a primary amine is formed.
Why "Degradation"?
The amine formed contains one carbon atom less than the starting amide (the carbonyl carbon is lost as carbonate). Hence, it is a step-down reaction used to decrease the length of a carbon chain.
$R-CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$
3. Basicity of Amines (Highly Important)
Amines are basic due to the presence of a lone pair of electrons on the nitrogen atom, which can be donated to a proton ($H^+$). The basic strength is expressed in terms of $K_b$ or $pK_b$. Smaller the $pK_b$ value, stronger is the base.
A. Aliphatic Amines vs. Ammonia
Aliphatic amines are stronger bases than ammonia. Alkyl groups are electron-donating (+I effect). They increase the electron density on the nitrogen atom, making the lone pair more easily available for donation. They also stabilize the substituted ammonium cation formed after protonation.
B. Trends in Basicity of Aliphatic Amines
In Gaseous Phase
In the gas phase, there is no solvation effect (no water). The basicity strictly follows the +I effect of alkyl groups.
Order: $3^\circ > 2^\circ > 1^\circ > NH_3$
In Aqueous Solution (Board Focus)
In water, basicity depends on the interplay of three factors: +I effect, Solvation effect (hydrogen bonding), and Steric hindrance.
- For Methyl group ($-CH_3$):
$2^\circ > 1^\circ > 3^\circ > NH_3$
$((CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3)$ - For Ethyl group ($-C_2H_5$):
$2^\circ > 3^\circ > 1^\circ > NH_3$
$((C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3)$
C. Aromatic Amines (Aniline) vs. Ammonia
Aromatic amines (like aniline, $C_6H_5NH_2$) are weaker bases than ammonia.
Reason: The lone pair of electrons on the nitrogen atom in aniline is in conjugation with the $\pi$-electrons of the benzene ring (resonance). Thus, the lone pair is delocalized into the ring and is less available for protonation. Furthermore, the anilinium ion formed upon protonation has only two resonance structures, while aniline has five, making the unprotonated form more stable.
4. Chemical Reactions of Amines
A. Acylation
Primary and secondary amines react with acid chlorides or acid anhydrides to form N-substituted amides. (Tertiary amines do not react because they lack a replaceable hydrogen on nitrogen).
$CH_3-NH_2 + CH_3COCl \xrightarrow{\text{Pyridine}} CH_3-NH-COCH_3 \text{ (N-Methylethanamide)} + HCl$
B. Carbylamine Reaction (Isocyanide Test)
This test is given only by Primary ($1^\circ$) aliphatic and aromatic amines.
When a primary amine is heated with chloroform ($CHCl_3$) and alcoholic $KOH$, it forms an alkyl/aryl isocyanide (carbylamine), which has an extremely foul, pungent, and intolerable smell. Secondary and tertiary amines do not give this test.
$R-NH_2 + CHCl_3 + 3KOH(\text{alc}) \xrightarrow{\Delta} R-NC \text{ (Isocyanide)} + 3KCl + 3H_2O$
C. Reaction with Nitrous Acid ($HNO_2$)
Nitrous acid is prepared in situ from $NaNO_2$ and dilute $HCl$.
- $1^\circ$ Aliphatic Amines: React to form highly unstable aliphatic diazonium salts, which immediately decompose to yield alcohols and nitrogen gas ($N_2$ bubbles).
$R-NH_2 + HNO_2 \rightarrow [R-N_2^+Cl^-] \xrightarrow{H_2O} R-OH + N_2 \uparrow + HCl$ - $1^\circ$ Aromatic Amines (Aniline): React at low temperatures (273-278 K) to form stable arenediazonium salts (Diazotization).
D. Hinsberg's Test (Distinguishing $1^\circ, 2^\circ, 3^\circ$ amines)
Hinsberg's Reagent: Benzenesulfonyl chloride ($C_6H_5SO_2Cl$).
- Primary ($1^\circ$) Amines: React with Hinsberg's reagent to form N-alkylbenzenesulfonamide. Since it still has one acidic hydrogen attached to nitrogen, it is soluble in alkali (NaOH).
- Secondary ($2^\circ$) Amines: React to form N,N-dialkylbenzenesulfonamide. It has no acidic hydrogen on nitrogen, hence it is insoluble in alkali.
- Tertiary ($3^\circ$) Amines: Do not react with Hinsberg's reagent (no hydrogen on nitrogen).
5. Electrophilic Substitution Reactions of Aniline
The $-NH_2$ group is highly activating and ortho, para-directing due to resonance (+R effect). It makes the benzene ring extremely reactive towards electrophiles.
A. Bromination
Aniline reacts violently with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline.
To prepare mono-substituted (p-bromoaniline): The high reactivity of $-NH_2$ is reduced (protected) by acetylation using acetic anhydride. The resulting acetanilide is brominated, and then hydrolyzed back to p-bromoaniline.
B. Nitration (Important Concept)
Direct nitration of aniline with a conc. $HNO_3$ and conc. $H_2SO_4$ mixture gives a surprising result: in addition to para (51%) and ortho (2%) products, a significant amount of meta-nitroaniline (47%) is formed.
Reason:
In the strongly acidic nitration medium, a large proportion of aniline is protonated to form the anilinium ion ($C_6H_5NH_3^+$). The $-NH_3^+$ group is strongly deactivating and meta-directing. Hence, a high yield of meta-product is observed.
To get only p-nitroaniline: Again, protect the $-NH_2$ group by acetylation, nitrate, and then hydrolyze.
6. Diazonium Salts
General formula: $Ar-N_2^+X^-$. Formed by treating $1^\circ$ aromatic amines with nitrous acid at 0-5°C (Diazotization).
Replacement Reactions
- Sandmeyer Reaction: The $N_2^+$ group is replaced by $Cl^-, Br^-,$ or $CN^-$ by mixing the diazonium salt with Cuprous chloride/bromide/cyanide ($Cu_2Cl_2, Cu_2Br_2, Cu_2(CN)_2$) and corresponding acids ($HCl, HBr, KCN$).
- Gattermann Reaction: Same replacement, but uses Copper powder and $HCl$ or $HBr$ instead of cuprous salts. (Yield is lower than Sandmeyer).
- Replacement by Iodine: Simply warming the diazonium salt solution with Potassium Iodide ($KI$) yields iodobenzene.
Coupling Reactions
Arenediazonium salts react with highly activated aromatic rings (like phenols and anilines) to form brightly colored azo compounds ($-N=N-$). Used as dyes.
- With Phenol (in mildly alkaline medium, pH 9-10): forms p-Hydroxyazobenzene (Orange dye).
- With Aniline (in mildly acidic medium, pH 4-5): forms p-Aminoazobenzene (Yellow dye).
7. Solved Textbook Reasoning Questions
Question 1: Boiling Points
Why do primary amines have higher boiling points than tertiary amines of comparable molecular mass?
Answer:
Primary amines ($R-NH_2$) have two hydrogen atoms attached directly to the electronegative nitrogen atom. Thus, they can form extensive intermolecular hydrogen bonds, leading to stronger intermolecular forces and higher boiling points. Tertiary amines ($R_3N$) have no hydrogen atoms attached to nitrogen, so they cannot form intermolecular hydrogen bonds with themselves, resulting in much lower boiling points.
Question 2: Aniline vs Cyclohexylamine
Explain why aniline is a weaker base than cyclohexylamine.
Answer:
In aniline, the lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$-system of the benzene ring. Due to resonance (+R effect), the lone pair is delocalized over the ring and is less available to accept a proton ($H^+$). In cyclohexylamine, the cyclohexyl group is non-aromatic and exhibits a +I (electron-donating) effect, increasing the electron density on nitrogen. There is no resonance delocalization. Hence, the lone pair is highly available for protonation, making cyclohexylamine a much stronger base than aniline.
8. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Write the chemical name of Hinsberg's reagent. (March 2013, Oct 2017)
Answer: Benzenesulfonyl chloride ($C_6H_5SO_2Cl$).
Q2. Name the test used to distinguish primary amines from secondary and tertiary amines via a foul-smelling product. (March 2016, Oct 2021)
Answer: Carbylamine test (or Isocyanide test).
Q3. Arrange the following in increasing order of their basic strength in aqueous solution: $NH_3$, $CH_3NH_2$, $(CH_3)_2NH$, $(CH_3)_3N$. (March 2018)
Answer: $NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$. (Order: $NH_3 < 3^\circ < 1^\circ < 2^\circ$ for methyl amines in water).
2 Mark Questions (SA-I)
Q4. Write a short note on Hoffmann bromamide degradation. (March 2014, March 2020)
When a primary amide is heated with bromine and an aqueous or ethanolic solution of NaOH or KOH, it yields a primary amine. This reaction is a step-down reaction as the amine formed contains one carbon atom less than the parent amide.
$R-CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Q5. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis? (Oct 2015, March 2022)
Gabriel phthalimide synthesis involves the nucleophilic substitution ($S_N2$) of an alkyl halide by the phthalimide anion. To prepare an aromatic primary amine like aniline, an aryl halide (like chlorobenzene) must be used. However, aryl halides are extremely unreactive towards nucleophilic substitution due to partial double bond character of the C-X bond (arising from resonance). Hence, the phthalimide anion cannot displace the halogen from the benzene ring.
3/4 Mark Questions (LA)
Q6. How does Hinsberg's reagent help in distinguishing primary, secondary, and tertiary amines? (Oct 2014, March 2019, March 2023)
Hinsberg's reagent is benzenesulfonyl chloride ($C_6H_5SO_2Cl$).
- Primary ($1^\circ$) Amines: React with the reagent to form N-alkylbenzenesulfonamide. The hydrogen attached to nitrogen in this sulfonamide is highly acidic due to the strong electron-withdrawing sulfonyl group. Therefore, it is soluble in aqueous alkali (NaOH).
- Secondary ($2^\circ$) Amines: React to form N,N-dialkylbenzenesulfonamide. Since there is no hydrogen atom attached to the nitrogen atom in this product, it is not acidic and is insoluble in aqueous alkali.
- Tertiary ($3^\circ$) Amines: Do not react with Hinsberg's reagent at all, as they do not possess a replaceable hydrogen atom on the nitrogen.
Q7. Explain the Sandmeyer reaction and Gattermann reaction with suitable equations using benzene diazonium chloride. (March 2015, Oct 2020)
Sandmeyer Reaction:
When a freshly prepared solution of benzene diazonium chloride is treated with cuprous chloride ($Cu_2Cl_2$) dissolved in HCl or cuprous bromide ($Cu_2Br_2$) dissolved in HBr, the diazo group is replaced by -Cl or -Br respectively.
$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2 / HCl} C_6H_5Cl \text{ (Chlorobenzene)} + N_2 \uparrow$
Gattermann Reaction:
The diazo group can be replaced by a halogen by treating the diazonium salt solution with the corresponding halogen acid (HCl or HBr) in the presence of finely divided copper powder (instead of cuprous salts).
$C_6H_5N_2^+Cl^- \xrightarrow{Cu \text{ powder} / HCl} C_6H_5Cl \text{ (Chlorobenzene)} + N_2 \uparrow$
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