NEET Amines Previous Year Questions with Solutions

Lithium aluminium hydride (LiAlH₄) is a powerful reducing agent that reduces the cyano group (-C≡N) to a -CH₂-NH₂ group. Reduction of benzonitrile produces benzylamine, which contains an extra aliphatic methylene carbon, NOT aniline.

• • Reaction A does not give aniline.

When benzamide is treated with bromine in aqueous or ethanolic potassium hydroxide, it undergoes Hoffmann bromamide degradation with loss of the carbonyl carbon as carbonate, yielding aniline containing one carbon less.

• • Reaction B produces aniline as the major product.

Sodium borohydride (NaBH₄) is a selective, mild reducing agent capable of reducing aldehydes and ketones. It lacks the reducing power to reduce aromatic nitro groups (-NO₂) to amino groups (-NH₂). (Catalytic hydrogenation or active metals like Fe/HCl or Sn/HCl are required).

• • Reaction C does not give aniline.

Acetanilide is prepared by protecting the amino group of aniline. Upon heating with aqueous mineral acid (HCl) followed by neutralization, the amide linkage undergoes nucleophilic acyl substitution/hydrolysis to regenerate free aniline and acetic acid.

• • Reaction D produces aniline as the major product.
• • Therefore, B and D only yield aniline.

In aromatic amines (benzenamine and N-methylaniline), the lone pair of electrons on the nitrogen atom is conjugated with the aromatic π-electron system and delocalized over the ortho and para positions:

• • Because lower pK_b corresponds to stronger basicity, aliphatic amines are over six orders of magnitude more basic than aromatic amines.
• • Therefore, both aliphatic amines stand ahead of both aromatic amines in basic strength.

In aqueous medium, the basic strength of alkylamines is governed by a balance of inductive (+I), steric hindrance, and hydration of the conjugate cation:

• • N-ethylethanamine (pK_b = 3.00) is a stronger base than ethanamine (pK_b = 3.29).

In N-methylaniline, the methyl group exerts an electron-releasing (+I) inductive effect, increasing electron density on the nitrogen atom and stabilizing the protonated anilinium ion compared to aniline:

• • Full decreasing order of basic strength:
• • \text{N-ethylethanamine} > \text{ethanamine} > \text{N-methylaniline} > \text{benzenamine}.
• • Matches Option (3) perfectly.

Friedel-Crafts alkylation requires an anhydrous Lewis acid catalyst like AlCl3 to generate carbocation electrophiles. Aniline has a basic lone pair on nitrogen that donates to the electron-deficient AlCl3, forming a coordinate adduct. The resulting positive formal charge on nitrogen (-NH2⁺-AlCl3⁻) exerts a potent electron-withdrawing (-I) effect, severely deactivating the benzene ring against electrophilic aromatic substitution:

Gabriel phthalimide synthesis is exclusively designed for the preparation of primary aliphatic amines. In the key step, potassium phthalimide acts as a nucleophile in an SN2 displacement of halide from an alkyl halide. Because aryl halides (like bromobenzene) have partial double bond character due to resonance and the aryl ring repels backside nucleophilic attack, they do not undergo SN2 reactions:

1-Iodopropane is a primary alkyl halide. Nucleophilic substitution with sodium cyanide (NaCN) in aqueous ethanol proceeds smoothly via an SN2 mechanism, adding one carbon to the chain to form butanenitrile:

Partial alkaline hydrolysis of nitriles in the presence of dilute base (or alkaline H2O2) stops at the primary amide stage, yielding butanamide:

Heating butanamide with bromine in aqueous sodium hydroxide results in the Hoffmann bromamide degradation. The alkyl group migrates from carbonyl carbon to nitrogen via an isocyanate intermediate, eliminating the carbonyl group as sodium carbonate and yielding propan-1-amine (propylamine):

Benzenediazonium chloride treated with cuprous bromide in hydrobromic acid (Cu₂Br₂ / HBr) undergoes replacement of the diazonium group by bromine with evolution of nitrogen gas (Sandmeyer reaction):

The resulting bromobenzene is treated with magnesium turnings in anhydrous dry ether to insert magnesium into the carbon-bromine bond, forming the Grignard reagent:

Grignard reagents are highly reactive toward protic sources. Water hydrolyzes phenylmagnesium bromide, transferring a proton to the highly basic phenyl carbon to produce benzene and basic magnesium bromide:

The final product obtained after all three sequential transformations is benzene (C₆H₆).

Reduction of acetamide by lithium aluminium hydride reduces the carbonyl group to a methylene group, producing ethanamine (1° amine):

Degradation of an aliphatic primary amide with bromine in alkaline KOH eliminates the carbonyl carbon, producing methanamine (1° amine):

Reduction of acetonitrile converts the cyano group into a primary amine with the same carbon count:

Reduction of an alkyl isocyanide (methyl isonitrile) adds hydrogens across the coordinate bond, yielding a secondary amine:

Hinsberg's reagent is benzenesulfonyl chloride ($C_6H_5SO_2Cl$). It is used to distinguish and separate primary ($1^\circ$), secondary ($2^\circ$), and tertiary ($3^\circ$) amines based on the presence of replaceable hydrogen atoms on nitrogen.

Ethylamine ($CH_3CH_2NH_2$, a primary amine) reacts with benzenesulfonyl chloride to yield N-ethylbenzenesulfonamide:

The hydrogen atom attached to the nitrogen atom in N-ethylbenzenesulfonamide is strongly acidic because of the strong electron-withdrawing inductive and resonance effects of the sulfonyl ($-SO_2-$) group. In the presence of base (alkali, such as aqueous $KOH$ or $NaOH$), it readily deprotonates to form a water-soluble potassium/sodium sulfonamide salt:

The reaction sequence shows: 1. 2,4,6-tribromoaniline undergoes diazotization with nitrous acid ($\text{NaNO}_2 + \text{HCl}$ at $0-5^\circ\text{C}$) to form 2,4,6-tribromobenzenediazonium chloride. 2. The diazonium chloride reacts with reagent 'R' to lose the diazonium function, yielding 1,3,5-tribromobenzene. This is a classic reductive deamination where the $-N_2^+Cl^-$ group is replaced by a hydrogen atom ($-H$).

According to NCERT Chapter on Amines: *"Certain mild reducing agents like hypophosphorous acid ($H_3PO_2$) or ethanol ($CH_3CH_2OH$) reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal, respectively."* The balanced chemical equation for the reduction by ethanol is:

- $H_2O$ (Option 1): Warming diazonium salts with water leads to hydrolysis, replacing $-N_2^+Cl^-$ with a phenolic $-OH$ group (forming 2,4,6-tribromophenol). - $HI$ (Option 3): Replaces $-N_2^+Cl^-$ with an iodine atom (forming 1-iodo-2,4,6-tribromobenzene). - $CuCN/KCN$ (Option 4): Sandmeyer reaction replacing $-N_2^+Cl^-$ with a cyano group ($-CN$). Therefore, reagent 'R' is uniquely ethanol ($CH_3CH_2OH$).

When aliphatic or aromatic primary amines are heated with chloroform ($$\text{CHCl}_3$$) and alcoholic potassium hydroxide ($$\text{KOH}$$), they form carbylamines or isocyanides ($$\text{R-NC}$$), which possess an intensely foul, offensive smell: $$\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-NC} + 3\text{KCl} + 3\text{H}_2\text{O}$$

Secondary ($$2^\circ$$) and tertiary ($$3^\circ$$) amines do not give this reaction because they lack the required two hydrogen atoms on the nitrogen atom to form the isocyanide ($$-\text{N}\equiv\text{C}$$) functional group.

- Option (1): Aniline ($$\text{C}_6\text{H}_5-\text{NH}_2$$) has an $$\text{-NH}_2$$ group attached to a phenyl ring, making it a primary ($$1^\circ$$) aromatic amine. - Option (2): $$N$$-methylaniline ($$\text{C}_6\text{H}_5-\text{NH}-\text{CH}_3$$) is a secondary ($$2^\circ$$) amine. - Option (3): $$N,N$$-dimethylaniline ($$\text{C}_6\text{H}_5-\text{N}(\text{CH}_3)_2$$) is a tertiary ($$3^\circ$$) amine. - Option (4): Diethylamine ($$(\text{C}_2\text{H}_5)_2\text{NH}$$) is a secondary ($$2^\circ$$) amine.

Aniline is the only primary amine among the choices, and thus it alone gives the carbylamine test.