Top 50 Most Important
Subjective Questions
Amines. Master Basicity Trends, Preparation Methods, Diazonium Salts, and Organic Conversions.
How to use this module:
Click on any question to reveal the meticulously formatted answer. When you are ready to study offline, click the download button at the bottom—the system will automatically generate a perfectly formatted, watermark-stamped PDF.
Classification & Preparation
Amines are classified based on the number of alkyl or aryl groups attached directly to the nitrogen atom.
- Primary ($1^\circ$) Amines: One group attached to N (e.g., Methylamine, $CH_3NH_2$).
- Secondary ($2^\circ$) Amines: Two groups attached to N (e.g., Dimethylamine, $(CH_3)_2NH$).
- Tertiary ($3^\circ$) Amines: Three groups attached to N (e.g., Trimethylamine, $(CH_3)_3N$).
The nitrogen atom in amines is $sp^3$ hybridized. Out of the four $sp^3$ hybrid orbitals, three form sigma bonds with carbon or hydrogen, and the fourth contains an unshared lone pair of electrons. Due to the presence of this lone pair, the geometry is pyramidal (like ammonia), and the bond angle is slightly less than the standard tetrahedral angle of 109.5° (typically around 108°).
Ammonolysis is the cleavage of a C-X bond by an ammonia molecule. An alkyl halide reacts with an ethanolic solution of ammonia at 373 K to form a primary amine.
Disadvantage: The primary amine formed acts as a nucleophile and can further react with the remaining alkyl halide to form secondary, tertiary, and eventually a quaternary ammonium salt. This yields a complex mixture of amines which is difficult to separate, making it unsuitable for pure $1^\circ$ amine synthesis.
To maximize the yield of a primary amine and prevent further successive alkylations (to $2^\circ$ and $3^\circ$ amines), Ammonia must be taken in large excess. This statistically ensures that the alkyl halide molecules are overwhelmingly more likely to collide with fresh ammonia rather than the newly formed primary amine.
It is a premier method to prepare pure primary aliphatic amines.
1. Phthalimide reacts with ethanolic KOH to form Potassium phthalimide.
2. This is heated with an alkyl halide ($R-X$) to undergo $S_N2$ substitution, forming N-alkylphthalimide.
3. Alkaline hydrolysis of N-alkylphthalimide yields purely the primary amine ($R-NH_2$) and the salt of phthalic acid.
To prepare aniline, one would have to use an aryl halide (like chlorobenzene) to react with the potassium phthalimide nucleophile in Step 2. However, aryl halides do not undergo nucleophilic substitution ($S_N2$) easily under normal conditions due to partial double bond character. Hence, the nucleophilic attack fails.
An amide ($R-CONH_2$) is treated with Bromine ($Br_2$) in an aqueous or ethanolic solution of Sodium Hydroxide ($NaOH$). This yields a primary amine.
$R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
Main Feature: It is a step-down (degradation) reaction. The amine formed has one carbon atom less than the parent amide.
Nitriles ($R-CN$) are reduced to primary amines ($R-CH_2NH_2$) using Lithium Aluminium Hydride ($LiAlH_4$), or by catalytic hydrogenation ($H_2/Ni$).
Synthesis Utility: It is exceptionally useful for "stepping up" a carbon chain. An alkyl halide ($R-X$) reacts with $KCN$ to form a nitrile ($R-CN$), adding a carbon. Reduction then yields an amine with one carbon atom more than the starting halide.
Nitro compounds ($R-NO_2$) are reduced to amines ($R-NH_2$) using active metals in acidic medium, such as $Sn/HCl$ or $Fe/HCl$.
Industrial Preference: $Fe$ scrap and $HCl$ is highly preferred because $FeCl_2$ formed gets hydrolyzed to release $HCl$ during the reaction. Thus, only a tiny initial amount of $HCl$ is needed to initiate the process.
- $CH_3-CH(NH_2)-CH_3$: The principal chain is propane, and the amine group is on the 2nd carbon. Name: Propan-2-amine.
- $(CH_3)_3N$: The principal chain is methanamine. The nitrogen has two methyl substituents. Name: N,N-Dimethylmethanamine.
Physical Properties & Basicity
Primary amines ($R-NH_2$) have two hydrogen atoms available on the nitrogen, allowing them to form strong, extensive intermolecular hydrogen bonds. Tertiary amines ($R_3N$) have absolutely no hydrogen atoms attached to the nitrogen, making them incapable of intermolecular hydrogen bonding. Thus, $1^\circ$ amines boil at significantly higher temperatures.
In aliphatic amines, the alkyl groups ($R-$) exert a strong electron-donating inductive effect (+I effect). This strongly pushes electron density onto the nitrogen atom, making its lone pair of electrons much more readily available to be donated to a proton ($H^+$) compared to ammonia ($NH_3$). Additionally, it stabilizes the conjugate acid formed.
In aniline ($C_6H_5NH_2$), the lone pair of electrons on the nitrogen atom heavily delocalizes into the $\pi$-electron system of the benzene ring through resonance. Consequently, the lone pair is fundamentally far less available for protonation. Also, the $sp^2$ hybridized carbon of the ring exerts an electron-withdrawing (-I) effect.
Order: $3^\circ > 2^\circ > 1^\circ > NH_3$.
In the gaseous phase, there are no solvent interactions (solvation effects). The basicity is decided purely by the +I effect of the alkyl groups. The tertiary amine has three electron-donating alkyl groups, making its lone pair maximally available.
Order (for Methyl group): $2^\circ > 1^\circ > 3^\circ > NH_3$.
In aqueous solution, three competing factors decide basicity: +I effect, Steric hindrance, and Solvation (hydration) energy of the conjugate acid. While $3^\circ$ has the max +I effect, its conjugate acid lacks hydrogens to form stabilizing hydrogen bonds with water, and bulky groups block solvent molecules. The $2^\circ$ amine hits the optimal balance of all three factors.
- EDG (e.g., $-CH_3, -OCH_3$): They pump electrons into the ring, increasing electron density on nitrogen and increasing basicity.
- EWG (e.g., $-NO_2, -X, -CN$): They drain electron density from the ring, making the nitrogen lone pair even less available, drastically decreasing basicity.
Remember: Higher $pK_b$ means Weaker Base.
Basicity order: N-methylaniline (strongest) > p-toluidine > Aniline (weakest).
Therefore, $pK_b$ order is exactly opposite: N-methylaniline < p-toluidine < Aniline.
Lower amines are soluble because they can easily form hydrogen bonds with water molecules. In higher amines, the massive hydrophobic (water-repelling) alkyl chains dominate the molecule, blocking hydrogen bond formation. Aniline is insoluble because the large, hydrophobic benzene ring overwhelms the hydrogen-bonding capacity of the small $-NH_2$ group.
Friedel-Crafts reactions require an anhydrous Lewis acid catalyst like Aluminum Chloride ($AlCl_3$). Aniline is highly basic, while $AlCl_3$ is a strong Lewis acid. They instantly undergo an acid-base neutralization reaction to form an insoluble complex salt. This salt physically ties up the nitrogen lone pair, deactivating the ring entirely and destroying the catalyst before any electrophilic substitution can occur.
Being basic, all amines rapidly react with mineral acids (like $HCl$) to form stable, water-soluble ammonium salts ($R-NH_3^+Cl^-$).
Utility: This salt formation is the foundation for separating and extracting basic amines from neutral or acidic non-water-soluble organic compounds.
Chemical Reactions
When aliphatic or aromatic primary amines are heated with chloroform ($CHCl_3$) and ethanolic potassium hydroxide ($KOH$), they form incredibly foul-smelling and poisonous compounds called isocyanides (carbylamines).
$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$
Detection: It is a highly specific qualitative test exclusively for primary ($1^\circ$) amines. Secondary and tertiary amines do not respond.
- Primary Aliphatic Amines: React with $HNO_2$ (prepared in situ from $NaNO_2 + HCl$) to form highly unstable aliphatic diazonium salts, which instantly decompose to yield alcohols and release nitrogen gas ($N_2$) quantitatively.
- Primary Aromatic Amines (Aniline): React at cold temperatures (273-278 K) to form stable benzenediazonium salts, which do not immediately release $N_2$ gas.
Amines are treated with Hinsberg's reagent (Benzenesulphonyl chloride) and then subjected to aqueous alkali (KOH).
- $1^\circ$ Amine: Forms a sulfonamide that has an acidic hydrogen. Thus, it strictly dissolves in the alkali to form a clear solution.
- $2^\circ$ Amine: Forms a sulfonamide with NO acidic hydrogen. Thus, it forms a precipitate that remains insoluble in alkali.
- $3^\circ$ Amine: Completely fails to react with the reagent itself (remains as the original amine layer).
The $-NH_2$ group is an incredibly powerful activating and ortho/para directing group. It activates the benzene ring so aggressively that the reaction cannot be stopped at monosubstitution. It instantly undergoes tri-substitution to yield a dense white precipitate of 2,4,6-Tribromoaniline.
To prevent tri-substitution, the excessive activating power of the $-NH_2$ group must be "tamed" (protected).
1. Aniline is first acetylated with acetic anhydride to form acetanilide. The acetyl group engages the nitrogen lone pair in resonance, heavily reducing its availability to the ring.
2. Bromination ($Br_2/CH_3COOH$) now strictly yields the para-isomer (p-bromoacetanilide).
3. Acidic or basic hydrolysis removes the acetyl group to finally yield p-Bromoaniline.
Direct nitration requires a highly acidic mixture (conc. $HNO_3$ + conc. $H_2SO_4$). In this strongly acidic medium, a large portion of aniline molecules immediately get protonated to form the anilinium ion ($-NH_3^+$). The anilinium ion is a strongly deactivating and strictly meta-directing group, which directly causes the massive formation of the meta isomer.
Aniline reacts with concentrated $H_2SO_4$ upon intense heating (453-473 K) to form p-aminobenzenesulfonic acid (Sulfanilic acid).
Unique property: Sulfanilic acid contains both a basic amino group ($-NH_2$) and an acidic sulfonic group ($-SO_3H$) within the same molecule. It undergoes internal acid-base neutralization to exist strictly as a dipolar ion called a Zwitterion.
Aliphatic and aromatic primary and secondary amines react with highly electrophilic acid chlorides or anhydrides. The amine acts as a nucleophile, attacking the carbonyl carbon to form an amide. The reaction is typically carried out in the presence of a strong base like pyridine to instantly neutralize the $HCl$ byproduct, pushing the reaction forward.
When an amine (primary, secondary, or tertiary) is treated with a large excess of methyl iodide ($CH_3I$), it undergoes repeated and continuous alkylation until all the hydrogen atoms attached to nitrogen are completely replaced by methyl groups, eventually culminating in the final formation of a Quaternary Ammonium Iodide salt.
Aniline (a primary amine) reacts with benzaldehyde via nucleophilic addition-elimination (condensation). A water molecule is eliminated between the $-NH_2$ group of aniline and the $=O$ of benzaldehyde to form a substituted imine containing a $C=N$ bond. This product is famously known as a Schiff Base (Benzylideneaniline).
Diazonium Salts
Diazotization is the chemical process of converting a primary aromatic amine (like aniline) into a diazonium salt.
Conditions: It requires treating the amine with freshly prepared nitrous acid (mixture of $NaNO_2$ and dilute $HCl$) at strictly cold temperatures, precisely between 273 K to 278 K (0°C to 5°C), because the salts are highly unstable at higher temperatures.
Aromatic diazonium salts ($Ar-N_2^+ X^-$) possess a positive charge on the terminal nitrogen which is powerfully dispersed and stabilized over the entire benzene ring through resonance. Aliphatic diazonium salts ($R-N_2^+$) completely lack resonance stabilization and decompose explosively instantly even at low temperatures to release $N_2$ gas.
When the $-N_2^+$ group leaves, it departs as completely neutral, exceptionally stable molecular nitrogen gas ($N_2$). The formation of this highly stable, triple-bonded molecule provides a massive thermodynamic driving force, allowing nucleophiles to easily displace the diazonium group from the aromatic ring.
A freshly prepared cold aqueous solution of benzene diazonium chloride is rigorously mixed with specific Cuprous salts (Cuprous chloride $Cu_2Cl_2$ in $HCl$, Cuprous bromide $Cu_2Br_2$ in $HBr$, or Cuprous cyanide $CuCN$ in $KCN$). The diazo group is replaced by $-Cl, -Br$, or $-CN$ to form the respective aryl compounds.
Both achieve the same goal of replacing the diazonium group with a halogen. While Sandmeyer uses Cuprous halides ($Cu_2X_2$), the Gattermann reaction utilizes freshly divided copper powder mixed directly with the corresponding halogen acid ($HCl$ or $HBr$). Sandmeyer is universally preferred due to superior product yield.
No copper catalyst is required. The iodide ion ($I^-$) is such a powerful nucleophile that simply warming the benzene diazonium chloride solution with aqueous Potassium Iodide ($KI$) easily and efficiently displaces the nitrogen to form Iodobenzene.
Benzene diazonium chloride is first treated with fluoroboric acid ($HBF_4$) to precipitate an insoluble, relatively stable intermediate complex called benzene diazonium fluoroborate ($C_6H_5N_2^+BF_4^-$). This dry salt is then strongly heated, causing it to decompose into Fluorobenzene, $BF_3$, and $N_2$ gas.
The diazonium salt can be successfully reduced back to Benzene by treating it with mild reducing agents.
Reagents: Hypophosphorous acid ($H_3PO_2$) with water (which oxidizes to phosphorous acid), OR Ethanol ($C_2H_5OH$) (which oxidizes to ethanal).
1. Diazotization: Treat Aniline with $NaNO_2$ and $HCl$ at 273-278 K to form benzene diazonium chloride.
2. Hydrolysis: Simply warm the aqueous solution of the diazonium salt (or treat with dilute $H_2SO_4$). It furiously hydrolyzes to yield pure Phenol and $N_2$ gas.
The diazonium salt is first treated with fluoroboric acid ($HBF_4$) to precipitate the stable benzene diazonium fluoroborate salt. Unlike the Balz-Schiemann reaction (which heats the dry salt alone), this salt is heated with aqueous Sodium Nitrite ($NaNO_2$) in the presence of Copper powder to yield Nitrobenzene.
Coupling & Conversions
Coupling reactions occur when diazonium salts (acting as weak electrophiles) react with highly activated aromatic rings (like phenol or aniline) at their para position to form azo compounds containing the $-N=N-$ linkage.
They are intensely colored because the two aromatic rings are joined by the $-N=N-$ bond, creating an exceptionally large, extended conjugated pi-system that absorbs visible light.
Benzene diazonium chloride reacts with phenol in a mildly alkaline medium (pH 9-10). The electrophile attacks the para position of phenol to form p-Hydroxyazobenzene, which is an intensely colored Orange dye.
Benzene diazonium chloride reacts with aniline in a mildly acidic medium (pH 4-5). The electrophile attacks the para position of aniline to form p-Aminoazobenzene, which is an intensely colored Yellow dye.
Avoid direct ammonolysis. Use the Gabriel Phthalimide Synthesis. The bulky phthalimide group protects the nitrogen atom from multiple alkylations. After the single alkylation step, hydrolysis cleaves off the protecting group, ensuring 100% pure primary aliphatic amine.
1. Nitration: Treat benzene with concentrated $HNO_3$ and concentrated $H_2SO_4$ to form Nitrobenzene.
2. Reduction: Treat the nitrobenzene with $Sn / HCl$ (or $Fe / HCl$) to cleanly reduce the nitro group to an amine, yielding Aniline.
1. Diazotization: Aniline + $NaNO_2/HCl$ (0-5°C) $\rightarrow$ Benzene diazonium chloride.
2. Sandmeyer: Treat with $CuCN/KCN$ $\rightarrow$ Cyanobenzene (Benzonitrile).
3. Hydrolysis: Boil with $H_3O^+$ $\rightarrow$ Benzoic acid.
4. Reduction: React with strong $LiAlH_4$ followed by water $\rightarrow$ Benzyl alcohol.
1. Nitration: Treat chlorobenzene with conc. $HNO_3/H_2SO_4$. The Cl is ortho/para directing. Separate the major product, p-nitrochlorobenzene.
2. Reduction: Treat p-nitrochlorobenzene with $Sn/HCl$ to specifically reduce the $-NO_2$ group, yielding p-Chloroaniline.
During the acylation of amines with acid chlorides, a strong acid, Hydrogen Chloride ($HCl$), is produced as a byproduct. Pyridine is a base that is specifically added to react with and neutralize the $HCl$. By constantly removing the byproduct, Le Chatelier's principle drives the equilibrium aggressively forward, significantly increasing the yield of the amide.
They act as universal chemical stepping stones. Direct introduction of groups like $-F, -I, -CN$, and $-OH$ onto a benzene ring is highly difficult or impossible via standard electrophilic substitution. However, Diazonium salts contain the $-N_2^+$ group (an ultimate leaving group), allowing any of these groups to be effortlessly substituted onto the ring.
1. Protection: Acetylate aniline using acetic anhydride/pyridine to form acetanilide, preventing the formation of the anilinium ion.
2. Nitration: Treat acetanilide with conc. $HNO_3 / H_2SO_4$. The bulky acetyl group directs the nitro group exclusively to the para position, yielding p-nitroacetanilide.
3. Deprotection: Hydrolyze with dilute acid/base to remove the acetyl group, yielding pure p-Nitroaniline.
Chapter 9 Mastered!
You have just conquered the 50 most critical subjective questions for Class 12 Chemistry, Chapter 9: Amines. Your nitrogen chemistry is now perfectly synthesized!
No comments:
Post a Comment