Given below are some nitrogen containing compounds.

Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ________ mg of HCl. (Given molar mass in g mol ^-1$^{-1}$ C:12, H : 1, O : 16, Cl : 35.5)

### Core Logic When benzene diazonium chloride is treated with ethanol (textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$), it undergoes a reduction reaction (deamination). Ethanol acts as a reducing agent and gets oxidized to ethanal (textCH_3textCHO$\text{CH}_3\text{CHO}$), while the diazonium group is replaced by hydrogen to yield pure **benzene**, not phenetole (ethoxybenzene).

### Step 1: Review of Alternative Choices Reactions (2), (3), and (4) show standard correct transformations: hypophosphorous acid reduction to benzene, potassium iodide substitution to iodobenzene, and cuprous cyanide substitution to benzonitrile. ### Pattern Recognition Shortcut: Remember that textH_3textPO_2$\text{H}_3\text{PO}_2$ and textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$ are standard classic reducing agents that reduce textArN_2^+textCl^-$\text{ArN}_2^+\text{Cl}^-$ directly down to textArH$\text{ArH}$ (benzene). They do not undergo nucleophilic ether substitution paths. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Let's track the chemical transformations sequentially: 1. **Step 1 (Sn + HCl$Sn + HCl$):** Nitrobenzene is cleanly reduced to yield Aniline (C_6H_5NH_2$C_6H_5NH_2$). 2. **Step 2 (Ac_2O + textPyridine$Ac_2O + \text{Pyridine}$):** Protecting step. Aniline undergoes acetylation to form Acetanilide (C_6H_5NHCOCH_3$C_6H_5NHCOCH_3$). This tempers the highly activating -NH_2$-NH_2$ group to prevent poly-bromination. 3. **Step 3 (Br_2 + AcOH$Br_2 + AcOH$):** The -NHCOCH_3$-NHCOCH_3$ amide group safely directs electrophilic bromination to the less-hindered **para** position, yielding p-bromoacetanilide. 4. **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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### 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