Chapter 13: Amines
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What are amines? How are they classified based on the number of alkyl/aryl groups attached to nitrogen?
Amines are organic derivatives of ammonia ($NH_3$) in which one, two, or all three hydrogen atoms are replaced by alkyl or aryl groups.
Classification:
- Primary ($1^\circ$) Amines: One hydrogen of $NH_3$ is replaced by an alkyl/aryl group ($R-NH_2$).
- Secondary ($2^\circ$) Amines: Two hydrogens of $NH_3$ are replaced ($R-NH-R'$).
- Tertiary ($3^\circ$) Amines: All three hydrogens of $NH_3$ are replaced ($R_3N$).
2. Distinguish between aliphatic and aromatic amines with examples.
- Aliphatic Amines: The nitrogen atom is directly bonded to one or more alkyl groups (and no aryl group is directly attached to N).
Example: Ethylamine ($CH_3CH_2NH_2$). - Aromatic Amines (Arylamines): The nitrogen atom is directly bonded to one or more aromatic rings.
Example: Aniline ($C_6H_5NH_2$).
3. Discuss the structure and hybridization of the nitrogen atom in trimethylamine.
In trimethylamine, $(CH_3)_3N$, the nitrogen atom is $sp^3$ hybridized. It has three bond pairs (with three methyl groups) and one lone pair of electrons in the fourth $sp^3$ hybrid orbital.
Due to the presence of the lone pair, the geometry is pyramidal (not perfectly tetrahedral), and the $C-N-C$ bond angle is slightly less than $109.5^\circ$ due to lone pair-bond pair repulsion (approx $108^\circ$).
4. Describe the preparation of amines by the reduction of nitro compounds. Mention the preferred reagent.
Nitro compounds can be reduced to primary amines by passing hydrogen gas in the presence of finely divided Raney Ni, Pt, or Pd, or by reduction with metals in acidic medium (like $Sn/HCl$ or $Fe/HCl$).
$$R-NO_2 + 6[H] \xrightarrow{Fe/HCl} R-NH_2 + 2H_2O$$
Preferred Reagent: $Fe$ scrap and $HCl$ is preferred because $FeCl_2$ formed gets hydrolyzed to release $HCl$ during the reaction. Thus, only a small amount of $HCl$ is required to initiate the reaction.
5. Explain Hoffmann's ammonolysis of alkyl halides. What is its major disadvantage?
When an alkyl halide ($R-X$) is heated with an alcoholic solution of ammonia ($NH_3$) in a sealed tube at 373 K, nucleophilic substitution occurs. The $C-X$ bond is broken by ammonia (ammonolysis) to form a primary amine.
$$R-X + NH_3 \xrightarrow{\Delta} R-NH_2 + HX$$
Disadvantage: The primary amine formed acts as a nucleophile and can further react with the alkyl halide to form a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. This yields a complex mixture of amines, which is difficult to separate.
6. How can the yield of primary amine be maximized in Hoffmann's ammonolysis?
The yield of primary amine can be maximized by taking a large excess of ammonia. This ensures that the alkyl halide molecules are more likely to collide with ammonia molecules rather than with the newly formed primary amine molecules, minimizing further alkylation.
7. Describe the reduction of Nitriles (Cyanides) to prepare amines. What is Mendius reduction?
Nitriles on reduction with Lithium aluminum hydride ($LiAlH_4$) or catalytic hydrogenation ($H_2/Ni$) produce primary amines containing the same number of carbon atoms.
$$R-C \equiv N + 4[H] \xrightarrow{LiAlH_4 \text{ or } H_2/Ni} R-CH_2-NH_2$$
When this reduction is carried out using Sodium amalgam and ethyl alcohol ($Na/C_2H_5OH$), it is specifically known as Mendius reduction.
8. How are primary amines prepared by the reduction of amides?
Amides on reduction with strong reducing agents like Lithium aluminum hydride ($LiAlH_4$) yield primary amines containing the same number of carbon atoms as the parent amide.
$$R-CONH_2 \xrightarrow{1. \ LiAlH_4 \quad 2. \ H_2O} R-CH_2-NH_2$$
9. Describe Gabriel Phthalimide Synthesis. What type of amines can be prepared by this method?
It is used for the preparation of pure aliphatic primary amines.
- Phthalimide is treated with ethanolic KOH to form potassium phthalimide.
- This salt is heated with an alkyl halide ($R-X$) to form N-alkyl phthalimide (nucleophilic substitution).
- Alkaline hydrolysis of N-alkyl phthalimide with aqueous NaOH yields the primary amine ($R-NH_2$) and the sodium salt of phthalic acid.
10. Why cannot aniline (an aromatic primary amine) be prepared by the Gabriel phthalimide synthesis?
In Gabriel synthesis, the crucial step involves nucleophilic substitution ($S_N2$) of an alkyl halide by the phthalimide anion.
To prepare aniline, one would need to use an aryl halide (like chlorobenzene). However, aryl halides do not undergo nucleophilic substitution easily under normal conditions due to the partial double bond character of the C-X bond (resonance). Hence, the phthalimide anion cannot displace the halogen from the benzene ring.
11. Describe Hoffmann Bromamide Degradation reaction.
When a primary amide is treated with bromine and an aqueous or ethanolic solution of sodium hydroxide, it undergoes degradation to form a primary amine. This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon to the nitrogen atom.
$$R-CONH_2 + Br_2 + 4NaOH \xrightarrow{\Delta} R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$$
12. What is the synthetic utility of the Hoffmann Bromamide Degradation reaction?
The key utility of this reaction is that the primary amine formed contains one carbon atom less than the parent amide (the carbonyl carbon is lost as $Na_2CO_3$). It is a very useful reaction for stepping down a homologous series in organic synthesis.
13. Discuss the boiling points of $1^\circ, 2^\circ$, and $3^\circ$ amines of comparable molecular mass. Explain the trend.
Order of boiling points: Primary ($1^\circ$) > Secondary ($2^\circ$) > Tertiary ($3^\circ$).
Reason: Primary amines have two N-H bonds and can form more extensive intermolecular hydrogen bonding. Secondary amines have only one N-H bond, so their H-bonding is weaker. Tertiary amines lack N-H bonds entirely and cannot form intermolecular hydrogen bonds (only weak dipole-dipole forces), resulting in the lowest boiling points among the three.
14. Compare the boiling points of amines with alcohols and alkanes of comparable molecular mass.
Order: Alcohols > Amines > Alkanes.
Alcohols have higher boiling points than amines because Oxygen is more electronegative than Nitrogen. Thus, the O-H bond is more polar than the N-H bond, resulting in stronger intermolecular hydrogen bonding in alcohols compared to amines. Alkanes, being non-polar, have the weakest forces and lowest boiling points.
15. Are amines soluble in water? Explain.
Lower aliphatic amines (up to 3 carbon atoms) are completely soluble in water because they can form hydrogen bonds with water molecules (using the lone pair on nitrogen and the hydrogens of water).
However, solubility rapidly decreases with an increase in molar mass because the large hydrophobic (water-repelling) alkyl part hinders hydrogen bonding. Higher amines are essentially insoluble in water.
16. Explain the basic nature of amines.
Amines act as Lewis bases because the nitrogen atom possesses an unshared (lone) pair of electrons which it can easily donate to form a coordinate bond with a proton ($H^+$) or an electrophile.
$$R-NH_2 + H_2O \rightleftharpoons R-NH_3^+ \text{ (Alkylammonium ion)} + OH^-$$
The generation of $OH^-$ ions in an aqueous solution proves their basic character.
17. What is $K_b$ and $pK_b$? How do they relate to basic strength?
$K_b$ is the base dissociation constant. A larger $K_b$ value indicates a higher concentration of $OH^-$ ions, and therefore a stronger base.
$pK_b$ is the negative logarithm of $K_b$ ($pK_b = -\log K_b$).
Therefore, a smaller $pK_b$ value corresponds to a stronger base.
18. Why are aliphatic amines stronger bases than ammonia?
In aliphatic amines ($R-NH_2$), the alkyl group ($R$) has an electron-donating inductive effect (+I effect). It pushes electron density towards the nitrogen atom.
This increases the electron density on nitrogen, making the lone pair more readily available for donation to a proton. It also disperses the positive charge on the resulting alkylammonium cation, stabilizing it. Ammonia lacks this +I effect, so it is a weaker base than aliphatic amines.
19. Discuss the order of basicity of $1^\circ, 2^\circ,$ and $3^\circ$ aliphatic amines in the gaseous phase.
In the gaseous phase, there is no solvation effect (no water). The basicity depends entirely on the +I (inductive) effect of the alkyl groups.
A tertiary amine has three electron-donating alkyl groups, secondary has two, and primary has one.
Therefore, the order of basicity in the gas phase is exactly as predicted by the +I effect:
$3^\circ \text{ amine} > 2^\circ \text{ amine} > 1^\circ \text{ amine} > NH_3$.
20. Discuss the order of basicity of methylamines in an aqueous solution. Why does it differ from the gas phase?
In an aqueous solution, basicity is determined by a combination of the +I effect, steric hindrance, and solvation (hydration) effect. The protonated amine is stabilized by hydrogen bonding with water. The $1^\circ$ amine has 3 hydrogens for H-bonding, while the $3^\circ$ amine has only 1, so $1^\circ$ is better solvated.
The interplay of these factors gives the experimental order for methylamines in water:
$2^\circ \text{ amine} > 1^\circ \text{ amine} > 3^\circ \text{ amine} > NH_3$
($(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$)
21. Discuss the order of basicity of ethylamines in an aqueous solution.
For ethylamines, the ethyl group is bulkier than the methyl group. This increases the +I effect but also increases steric hindrance, particularly affecting the $3^\circ$ amine's solvation.
The order in aqueous solution for ethylamines is:
$2^\circ \text{ amine} > 3^\circ \text{ amine} > 1^\circ \text{ amine} > NH_3$
($(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$)
22. Why is Aniline less basic than Ammonia?
In aniline ($C_6H_5NH_2$), the lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$-electron system of the benzene ring. It gets delocalized into the ring via resonance (+R effect).
Because the lone pair is "busy" in resonance, its availability for donation to a proton ($H^+$) is significantly decreased. Hence, aniline is a much weaker base than ammonia (where the lone pair is localized on nitrogen).
23. Explain the effect of substituents on the basicity of aniline.
- Electron Donating Groups (EDG) (like $-CH_3, -OCH_3, -NH_2$) attached to the ring increase electron density, making the lone pair more available. They increase the basicity.
- Electron Withdrawing Groups (EWG) (like $-NO_2, -X, -CN$) pull electron density away from the ring and nitrogen, making the lone pair less available. They decrease the basicity.
24. What happens when an amine reacts with an acid?
Due to their basic nature, amines react with mineral acids (like $HCl, H_2SO_4$) to form water-soluble ammonium salts.
$$R-NH_2 + HCl \rightarrow R-NH_3^+Cl^- \text{ (Alkylammonium chloride)}$$
These salts react with strong bases (like NaOH) to regenerate the free amine, which is a useful method to purify amines from non-basic organic compounds.
25. Describe the alkylation of amines.
Amines react with alkyl halides via $S_N2$ mechanism. The amine acts as a nucleophile. A primary amine reacts to form a secondary amine, which reacts further to form a tertiary amine, and finally a quaternary ammonium salt.
$R-NH_2 \xrightarrow{R'-X} R-NH-R' \xrightarrow{R'-X} R-N(R')_2 \xrightarrow{R'-X} [R-N(R')_3]^+X^-$
26. Describe the acylation of amines. Why do tertiary amines not undergo acylation?
Primary and secondary amines react with acid chlorides, acid anhydrides, and esters by nucleophilic acyl substitution to form amides. This reaction is carried out in the presence of a stronger base like pyridine to remove the $HCl$ formed.
$R-NH_2 + CH_3COCl \xrightarrow{\text{Pyridine}} R-NH-COCH_3 \text{ (N-alkylacetamide)} + HCl$
Tertiary amines do not undergo acylation because they lack a replaceable hydrogen atom attached to the nitrogen.
27. Describe the Carbylamine reaction (Isocyanide test). What is its utility?
When an aliphatic or aromatic primary amine is heated with chloroform ($CHCl_3$) and ethanolic potassium hydroxide ($KOH$), it forms an isocyanide or carbylamine, which is a substance with an extremely foul, foul-smelling odor.
$$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC \text{ (Alkyl isocyanide)} + 3KCl + 3H_2O$$
Utility: Secondary and tertiary amines do not show this reaction. Therefore, it is used as a specific test to detect primary amines.
28. How do primary aliphatic amines react with nitrous acid ($HNO_2$)?
Nitrous acid is unstable and prepared in situ from $NaNO_2$ and $HCl$. Primary aliphatic amines react with cold nitrous acid to form aliphatic diazonium salts. These are highly unstable and immediately decompose to form primary alcohols with the brisk evolution of nitrogen gas.
$R-NH_2 + HNO_2 \xrightarrow{NaNO_2/HCl} [R-N_2^+Cl^-] \xrightarrow{H_2O} R-OH + N_2 \uparrow + HCl$
Quantitative evolution of $N_2$ gas is used to estimate amino acids and proteins.
29. How do primary aromatic amines (aniline) react with nitrous acid?
Primary aromatic amines react with nitrous acid at low temperatures (273-278 K) to form stable aromatic diazonium salts. This process is called diazotization.
$$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278K} C_6H_5N_2^+Cl^- \text{ (Benzene diazonium chloride)} + NaCl + 2H_2O$$
30. How do secondary amines react with nitrous acid?
Both aliphatic and aromatic secondary amines react with nitrous acid to form N-nitrosoamines, which separate as yellow oily liquids.
$$R_2NH + HO-N=O \rightarrow R_2N-N=O \text{ (N-Nitrosodialkylamine)} + H_2O$$
31. What is Hinsberg's reagent?
Hinsberg's reagent is Benzenesulfonyl chloride ($C_6H_5SO_2Cl$). It is used to distinguish between primary, secondary, and tertiary amines.
32. Describe the reaction of a primary amine with Hinsberg's reagent.
A primary amine reacts with benzenesulfonyl chloride to form an N-alkylbenzenesulfonamide.
$C_6H_5SO_2Cl + R-NH_2 \rightarrow C_6H_5SO_2NHR + HCl$
The product still has one hydrogen attached to the highly electron-withdrawing sulfonamide nitrogen. This hydrogen is strongly acidic. Therefore, the product is soluble in aqueous alkali (like NaOH) to form a clear solution.
33. Describe the reaction of a secondary amine with Hinsberg's reagent.
A secondary amine reacts with benzenesulfonyl chloride to form N,N-dialkylbenzenesulfonamide.
$C_6H_5SO_2Cl + R_2NH \rightarrow C_6H_5SO_2NR_2 + HCl$
Since the product has no acidic hydrogen attached to the nitrogen atom, it is insoluble in aqueous alkali and separates as a solid precipitate.
34. How do tertiary amines react with Hinsberg's reagent?
Tertiary amines do not have any hydrogen atoms attached to the nitrogen. Therefore, they do not react with benzenesulfonyl chloride. However, because they are basic, they dissolve in the acidic by-products if the medium is acidified later.
(Note: Modern tests often use p-toluenesulfonyl chloride, but the principle remains identical).
35. Is the $-NH_2$ group in aniline activating or deactivating? What is its directing nature?
The $-NH_2$ group has a strong +R (resonance) effect. It donates its lone pair of electrons into the benzene ring, vastly increasing the electron density. Hence, it is a strongly ring activating group.
The electron density increases specifically at the ortho and para positions. Therefore, it is an ortho-para directing group for electrophilic substitution.
36. What happens when aniline reacts with bromine water?
Because the $-NH_2$ group is highly activating, aniline reacts rapidly with bromine water at room temperature to undergo polyhalogenation. It yields a white precipitate of 2,4,6-tribromoaniline.
$$C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2(NH_2)Br_3 \downarrow + 3HBr$$
37. How can you prepare mono-brominated aniline (e.g., p-bromoaniline) from aniline?
To get a monosubstituted product, the high reactivity of the $-NH_2$ group must be reduced. This is done by protecting it via acetylation with acetic anhydride.
- Aniline + $(CH_3CO)_2O \rightarrow$ Acetanilide (the lone pair on N now resonates with the carbonyl group, reducing ring activation).
- Bromination of acetanilide with $Br_2$ in acetic acid yields mainly p-bromoacetanilide.
- Hydrolysis of p-bromoacetanilide with acidic or basic water removes the acetyl group to yield p-bromoaniline.
38. Direct nitration of aniline yields a significant amount of m-nitroaniline, even though $-NH_2$ is an o/p directing group. Explain why.
Direct nitration involves strongly acidic conditions ($HNO_3 + H_2SO_4$). In this strongly acidic medium, the basic aniline is protonated to form the anilinium ion ($C_6H_5NH_3^+$).
The $-NH_3^+$ group has a strong -I effect and no lone pairs for resonance. Thus, it acts as a strongly deactivating and meta-directing group. This causes a substantial amount (~47%) of the meta-isomer (m-nitroaniline) to form alongside the para (51%) and ortho (2%) products.
39. How do you synthesize pure p-nitroaniline from aniline?
To prevent protonation and meta-direction, the amino group must be protected by acetylation.
- React aniline with acetic anhydride to form acetanilide.
- Nitrate the acetanilide using conc. $HNO_3$ and conc. $H_2SO_4$ to form p-nitroacetanilide (major product).
- Hydrolyze the p-nitroacetanilide with aqueous acid or base to regenerate the amino group, yielding p-nitroaniline.
40. Describe the sulfonation of aniline. What is Zwitter ion formation here?
Aniline reacts with concentrated $H_2SO_4$ to form anilinium hydrogen sulfate. Upon strong heating (453-473 K), it rearranges to form p-aminobenzenesulfonic acid, commonly known as Sulfanilic acid.
Sulfanilic acid contains an acidic group ($-SO_3H$) and a basic group ($-NH_2$). The acidic proton transfers to the basic nitrogen, forming an internal salt with a positive and a negative charge. This dipolar ion is called a Zwitter ion ($H_3N^+-C_6H_4-SO_3^-$).
41. Why does aniline not undergo Friedel-Crafts reaction?
Friedel-Crafts reactions require a Lewis acid catalyst like anhydrous Aluminum chloride ($AlCl_3$). Aniline is a Lewis base due to the lone pair on nitrogen.
Instead of acting as a catalyst for the ring, $AlCl_3$ reacts violently with the $-NH_2$ group to form an insoluble salt complex ($C_6H_5NH_2^+ \to AlCl_3^-$). This converts the activating amino group into a strongly deactivating group, shutting down the Friedel-Crafts substitution completely.
42. Write the formula of Benzenediazonium chloride. Why are aromatic diazonium salts more stable than aliphatic ones?
Formula: $C_6H_5-N \equiv N^+ \ Cl^-$
Stability: The aromatic diazonium ion ($C_6H_5N_2^+$) is stabilized by resonance. The positive charge on the nitrogen can be delocalized into the $\pi$-system of the benzene ring. Aliphatic diazonium ions lack this resonance stabilization, making them highly unstable even at low temperatures.
43. Describe Sandmeyer's reaction for diazonium salts.
In Sandmeyer's reaction, the diazonium group ($-N_2^+X^-$) is replaced by $-Cl, -Br$, or $-CN$ by treating the diazonium salt solution with the corresponding cuprous salt ($Cu_2Cl_2, Cu_2Br_2, Cu_2(CN)_2$). Nitrogen gas is evolved.
$$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl \text{ (Chlorobenzene)} + N_2 \uparrow$$
44. Describe Gattermann's reaction for diazonium salts. How does it differ from Sandmeyer's?
Gattermann's reaction is a modification of Sandmeyer's reaction. Instead of using a cuprous salt, the diazonium salt solution is warmed with the corresponding halogen acid ($HCl$ or $HBr$) in the presence of copper powder as a catalyst.
$$C_6H_5N_2^+Cl^- \xrightarrow{Cu / HCl} C_6H_5Cl + N_2 \uparrow$$
Sandmeyer's reaction generally provides a better yield than Gattermann's.
45. How is Fluorobenzene prepared from benzene diazonium chloride? (Balz-Schiemann reaction).
When benzene diazonium chloride is treated with fluoroboric acid ($HBF_4$), benzene diazonium fluoroborate precipitates. Upon heating, this salt decomposes to give fluorobenzene, boron trifluoride, and nitrogen gas.
$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- \downarrow + HCl$
$C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F \text{ (Fluorobenzene)} + BF_3 + N_2 \uparrow$
46. How do you replace the diazonium group with Hydrogen (Deamination)?
The diazonium group can be replaced by hydrogen (converting the compound to benzene) by using mild reducing agents like Hypophosphorous acid ($H_3PO_2$) or Ethanol ($C_2H_5OH$).
$$C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_6 \text{ (Benzene)} + N_2 \uparrow + H_3PO_3 + HCl$$
47. What are Coupling Reactions? Give the reaction of diazonium chloride with phenol.
Coupling Reaction: Diazonium salts react with highly activated aromatic rings like phenols and aromatic amines to form highly colored azo compounds (containing $-N=N-$ bond). These are widely used as dyes. It is an electrophilic aromatic substitution where the diazonium ion acts as the electrophile.
With Phenol (in mildly alkaline medium, pH 9-10): The coupling occurs at the para position.
$C_6H_5-N_2^+Cl^- + H-C_6H_4-OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4-OH \text{ (p-Hydroxyazobenzene, Orange Dye)} + HCl$
48. Give the coupling reaction of diazonium chloride with Aniline.
With Aniline (in mildly acidic medium, pH 4-5): The coupling occurs at the para position of aniline.
$C_6H_5-N_2^+Cl^- + H-C_6H_4-NH_2 \xrightarrow{H^+} C_6H_5-N=N-C_6H_4-NH_2 \text{ (p-Aminoazobenzene, Yellow Dye)} + HCl$
49. How will you convert Aniline to Chlorobenzene?
This is a two-step process utilizing Sandmeyer's reaction.
- Diazotization: Treat aniline with $NaNO_2 + HCl$ at 273-278 K to form benzene diazonium chloride.
$C_6H_5NH_2 \xrightarrow{NaNO_2, HCl, 0^\circ C} C_6H_5N_2^+Cl^-$ - Sandmeyer's Reaction: Treat the diazonium salt with cuprous chloride ($Cu_2Cl_2$) and $HCl$.
$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2, HCl} C_6H_5Cl \text{ (Chlorobenzene)} + N_2$
50. How will you convert nitrobenzene to aniline?
Nitrobenzene is reduced to aniline using a metal and an acid, preferably Iron scrap and Hydrochloric acid.
$$C_6H_5NO_2 + 6[H] \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (Aniline)} + 2H_2O$$
Alternatively, catalytic hydrogenation ($H_2/Pd$ or $H_2/Ni$) or Tin and HCl ($Sn/HCl$) can also be used.
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