Given below are two statements: Statement I:

can be synthesized from

in the order i) Acidic KMnO₄$KMnO_4$, ii) Ammonia, iii) Bromine and alkali Statement II:

In the light of the above statements, choose the correct answer from the options given below

Core Logic

Let's track the chemical transformations sequentially:

• Step 1 (Sn + HCl$Sn + HCl$): Nitrobenzene is cleanly reduced to yield Aniline (C₆H₅NH₂$C_6H_5NH_2$).
• Step 2 (Ac₂O + Pyridine$Ac_2O + \text{Pyridine}$): Protecting step. Aniline undergoes acetylation to form Acetanilide (C₆H₅NHCOCH₃$C_6H_5NHCOCH_3$). This tempers the highly activating -NH₂$-NH_2$ group to prevent poly-bromination.
• Step 3 (Br₂ + AcOH$Br_2 + AcOH$): The -NHCOCH₃$-NHCOCH_3$ amide group safely directs electrophilic bromination to the less-hindered para position, yielding p-bromoacetanilide.
• Step 4 (NaOH(aq)$NaOH_{(aq)}$): Basic hydrolysis removes the protecting acetyl group, restoring the free amine function to yield the final product: p-bromoaniline.

Pattern Recognition

Acetylation of aniline followed by halogenation and subsequent hydrolysis is the standard synthetic pathway to produce mono-substituted para-haloanilines.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

Core Logic

• Aliphatic amines vs Aromatic amines: In aromatic amines (A, B, C), the lone pair on nitrogen is delocalized into the benzene ring via resonance, decreasing basicity compared to aliphatic amines (D, E) where electron pairs are localized.
• Among aliphatic amines (aqueous standard configurations implicit): Secondary amine (CH₃)₂NH$(\text{CH}_3)_2\text{NH}$ is a stronger base than primary CH₃NH₂$\text{CH}_3\text{NH}_2$ due to combined inductive effect (+I$+\text{I}$) and solvation fields. Hence, E > D$\text{E} > \text{D}$.
• Among substituted aromatic amines:
• (B) p-Methoxyaniline: -OCH₃$-\text{OCH}_3$ exerts a strong electron-donating resonance effect (+M$+\text{M}$), maximizing ring density and electronic availability on N.
• (A) Aniline: Baseline reference value with no extra substitutions.
• (C) p-Nitroaniline: -NO₂$-\text{NO}_2$ acts as an intensive electron-withdrawing field (-M, -I$-\text{M}, -\text{I}$), pulling electron clouds heavily and quenching basicity.

Step 1: Consolidating Rankings

Combining both structural domains cleanly provides:

Pattern Recognition

Basicity hierarchy shortcut: Aliphatic secondary >$>$ Aliphatic primary >$>$ Aromatic with EDG (+M$+\text{M}$) >$>$ Unsubstituted aniline >$>$ Aromatic with EWG (-M$-\text{M}$). This immediately gives E > D > B > A > C$\text{E} > \text{D} > \text{B} > \text{A} > \text{C}$ without deep arithmetic.

Chapter Mix

Class 12 Chemistry: Amines Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Core Logic

To direct selective monobromination ortho to the amino functionality while utilizing the masking capability of the sulfonic acid group:

• Treating Aniline with conc. H2SO₄$\mathrm{H}{2}SO_{4}$ at high temperature (453-473 K$453-473\text{ K}$) yields Sulfanilic acid due to para sulfonating preference.
• Acetylation with Ac₂O$\mathrm{Ac}_{2}O$ protects the amine as an acetanilide functionality to moderate activation power and prevent over-bromination.
• Electrophilic substitution using Br₂$\mathrm{Br}_{2}$ selectively places bromine at the position ortho to the protected acetamido group (the only available activated site since para is occupied).
• Acidic/thermal desulfonation via H₂O(Δ)$\mathrm{H}_{2}O(\Delta)$ cleaves the para-sulfonic acid group.
• Alkaline hydrolysis with NaOH$\mathrm{NaOH}$ removes the acetyl protecting group to regenerate the pristine primary amine structure yielding ortho-bromoaniline.

Step-by-Step Mechanism

The reaction mechanism progresses linearly through the designated strategic intermediates:

Pattern Recognition

When dealing with aniline conversions requiring blocked para positions followed by a removal step, look for the sequence tracking: Sulfonation arrow$\rightarrow$ Protection arrow$\rightarrow$ Halogenation arrow$\rightarrow$ Desulfonation arrow$\rightarrow$ Deprotection.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

The Carbylamine reaction is specific to primary amines:

Core Logic

Evaluating each amine statement:

• (A) Primary aromatic amines form stable diazonium salts with NaNO₂/HCl$NaNO_2/HCl$ at low temperatures, making this statement false.
• (B) Both aliphatic and aromatic primary amines undergo the carbylamine test to produce foul-smelling isocyanides. This is correct.
• (C) Secondary and tertiary amines do not undergo the carbylamine reaction, making this statement false.
• (D) Benzenesulfonyl chloride (C₆H₅SO₂Cl$C_6H_5SO_2Cl$) is the definition of Hinsberg's reagent. This is correct.
• (E) Tertiary amines do not possess an acidic hydrogen on nitrogen and do not react with Hinsberg's reagent under standard analytical testing conditions, making this statement false.

Step 1: Selecting Correct Entries

Statements (B) and (D) are verified as true.

Pattern Recognition

Hinsberg's reagent and the carbylamine test are key analytical methods used to differentiate primary, secondary, and tertiary amines. The carbylamine test is strictly positive only for primary (1^°$1^\circ$) amine groups.

Chapter Mix

Class 12 Chemistry: Amines

Related Formula

Core Logic

Step 1: Identify the most basic amine Benzylamine (C₆H₅CH₂NH₂$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{NH}_2$) is the most basic compound here because its nitrogen lone pair is localized and not involved in aromatic resonance. This stands in contrast to aniline or amides, which delocalize their lone pairs into the ring or carbonyl group .

Step 2: Neutralization Stoichiometry

Molar Mass of Benzylamine (C₇H₉N$\mathrm{C}_7\mathrm{H}_9\mathrm{N}$):

Moles of Benzylamine in 1.0 g$1.0\text{ g}$ :

Since 1 mole of benzylamine reacts with 1 mole of HCl$\mathrm{HCl}$ :

Pattern Recognition

Aliphatic localized clusters (like the -CH₂NH₂$-\mathrm{CH}_2\mathrm{NH}_2$ segment in benzylamine) always show higher basicity than aromatic ring-conjugated arrays.

Chapter Mix

Class 12 Chemistry: Amines