### Related Formula textBasicity propto textAvailability of lone pair of electrons on Nitrogen atom$\text{Basicity} \propto \text{Availability of lone pair of electrons on Nitrogen atom}$ textBasicity propto +textI, +textM groups quad textBasicity propto frac1-textI, -textM groups$\text{Basicity} \propto +\text{I}, +\text{M groups} \quad \text{Basicity} \propto \frac{1}{-\text{I}, -\text{M groups}}$ ### Core Logic 1. 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. 2. Among aliphatic amines (aqueous standard configurations implicit): Secondary amine (textCH_3)_2textNH$(\text{CH}_3)_2\text{NH}$ is a stronger base than primary textCH_3textNH_2$\text{CH}_3\text{NH}_2$ due to combined inductive effect (+textI$+\text{I}$) and solvation fields. Hence, textE > textD$\text{E} > \text{D}$. 3. Among substituted aromatic amines: - **(B) p-Methoxyaniline**: -textOCH_3$-\text{OCH}_3$ exerts a strong electron-donating resonance effect (+textM$+\text{M}$), maximizing ring density and electronic availability on N. - **(A) Aniline**: Baseline reference value with no extra substitutions. - **(C) p-Nitroaniline**: -textNO_2$-\text{NO}_2$ acts as an intensive electron-withdrawing field (-textM, -textI$-\text{M}, -\text{I}$), pulling electron clouds heavily and quenching basicity. ### Step 1: Consolidating Rankings Combining both structural domains cleanly provides: textE > textD > textB > textA > textC $\text{E} > \text{D} > \text{B} > \text{A} > \text{C} $ ### Pattern Recognition Basicity hierarchy shortcut: Aliphatic secondary >$>$ Aliphatic primary >$>$ Aromatic with EDG (+textM$+\text{M}$) >$>$ Unsubstituted aniline >$>$ Aromatic with EWG (-textM$-\text{M}$). This immediately gives textE > textD > textB > textA > textC$\text{E} > \text{D} > \text{B} > \text{A} > \text{C}$ without deep arithmetic. ### Evaluation Rubric / Model Answer null ### 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: 1. Treating Aniline with conc. mathrmH2SO_4$\mathrm{H}{2}SO_{4}$ at high temperature (453-473text K$453-473\text{ K}$) yields Sulfanilic acid due to para sulfonating preference. 2. Acetylation with mathrmAc_2O$\mathrm{Ac}_{2}O$ protects the amine as an acetanilide functionality to moderate activation power and prevent over-bromination. 3. Electrophilic substitution using mathrmBr_2$\mathrm{Br}_{2}$ selectively places bromine at the position ortho to the protected acetamido group (the only available activated site since para is occupied). 4. Acidic/thermal desulfonation via mathrmH_2O(Delta)$\mathrm{H}_{2}O(\Delta)$ cleaves the para-sulfonic acid group. 5. Alkaline hydrolysis with mathrmNaOH$\mathrm{NaOH}$ 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 ightarrow$ ightarrow$ Protection ightarrow$ ightarrow$ Halogenation ightarrow$ ightarrow$ Desulfonation ightarrow$ ightarrow$ Deprotection. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Related Formula The Carbylamine reaction is specific to primary amines: R-NH_2 + CHCl_3 + 3KOH xrightarrowDelta R-NC + 3KCl + 3H_2O$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O$ ### Core Logic Evaluating each amine statement: - (A) Primary aromatic amines form stable diazonium salts with NaNO_2/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 (C_6H_5SO_2Cl$C_6H_5SO_2Cl$) 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^circ$1^\circ$) amine groups. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Related Formula textMoles = fractextMasstextMolar Mass$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$ textMass of HCl consumed = n_textamine cdot M_textHCl$\text{Mass of HCl consumed} = n_{\text{amine}} \cdot M_{\text{HCl}}$ ### Core Logic Step 1: Identify the most basic amine Benzylamine (mathrmC_6mathrmH_5mathrmCH_2mathrmNH_2$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{NH}_2$) 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 mathrmC_6H_5CH_2NH_2 + mathrmHCl ightarrow mathrmC_6H_5CH_2NH_3^+ Cl^-$\mathrm{C_6H_5CH_2NH_2} + \mathrm{HCl} ightarrow \mathrm{C_6H_5CH_2NH_3^+ Cl^-}$ Molar Mass of Benzylamine (mathrmC_7mathrmH_9mathrmN$\mathrm{C}_7\mathrm{H}_9\mathrm{N}$): M = (7 cdot 12) + (9 cdot 1) + 14 = 84 + 9 + 14 = 107 \, mathrmg/mol$M = (7 \cdot 12) + (9 \cdot 1) + 14 = 84 + 9 + 14 = 107 \, \mathrm{g/mol}$ Moles of Benzylamine in 1.0text g$1.0\text{ g}$ : n = frac1.0107 simeq 0.009346 \, mathrmmol$n = \frac{1.0}{107} \simeq 0.009346 \, \mathrm{mol}$ Since 1 mole of benzylamine reacts with 1 mole of mathrmHCl$\mathrm{HCl}$ : textMoles of HCl consumed = 0.009346 \, mathrmmol$\text{Moles of HCl consumed} = 0.009346 \, \mathrm{mol}$ textMass of HCl = 0.009346 cdot 36.5 = 0.3411 \, mathrmg = 341.1 \, mathrmmg ightarrow 341$\text{Mass of HCl} = 0.009346 \cdot 36.5 = 0.3411 \, \mathrm{g} = 341.1 \, \mathrm{mg} ightarrow 341$ ### Pattern Recognition Aliphatic localized clusters (like the -mathrmCH_2mathrmNH_2$-\mathrm{CH}_2\mathrm{NH}_2$ segment in benzylamine) always show higher basicity than aromatic ring-conjugated arrays. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Aniline is the common name for the simplest aromatic amine, which consists of a phenyl group attached to an amino group (C_6H_5NH_2$C_6H_5NH_2$). According to IUPAC nomenclature, the amino group attached to a benzene ring can also be called aminobenzene. ### Step 1: Statement Validation Statement I: True. Aminobenzene is just the systematic IUPAC name for aniline. Statement II: False. They refer to the exact same molecule. ### Pattern Recognition Common names for simple aromatic compounds are often accepted as IUPAC names. Aniline = Benzenamine = Aminobenzene. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines