### Core Logic Let's track sequence (1): 1) Nitrobenzene (Ph-NO_2$Ph-NO_2$) is reduced using Sn/HCl$Sn/HCl$ to form Aniline (Ph-NH_2$Ph-NH_2$). 2) Aniline undergoing diazotization with NaNO_2/HCl$NaNO_{2}/HCl$ at cold temperatures (0-5^circC$0-5^{circ}C$) creates Benzene diazonium chloride (Ph-N_2^+Cl^-$Ph-N_{2}^{+}Cl^{-}$). 3) The diazonium salt undergoes a coupling reaction with beta$\beta$-naphthol in alkaline conditions (NaOH$NaOH$) to synthesize a highly vibrant red-orange azo dye.

### Pattern Recognition The standard sequence for azo dye preparation is: Aromatic Nitro ightarrow$ ightarrow$ Primary Amine ightarrow$ ightarrow$ Diazonium Salt ightarrow$ ightarrow$ Phenol/Naphthol Coupling. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic The reaction sequence details the functional conversion of chlorobenzene down to product B (aniline, with a molar mass of 93\,mathrmg\,mol^-1$93\,\mathrm{g\,mol}^{-1}$). Following stoichiometric preservation: textmoles of chlorobenzene = textmoles of Aniline (B)$\text{moles of chlorobenzene} = \text{moles of Aniline (B)}$ Molar mass of chlorobenzene (mathrmC_6mathrmH_5mathrmCl$\mathrm{C}_6\mathrm{H}_5\mathrm{Cl}$) = 112.5\,mathrmg\,mol^-1$112.5\,\mathrm{g\,mol}^{-1}$.

textmoles = frac11.25 times 10^-3\,mathrmg112.5\,mathrmg\,mol^-1 = 10^-4\,mathrmmol$\text{moles} = \frac{11.25 \times 10^{-3}\,\mathrm{g}}{112.5\,\mathrm{g\,mol}^{-1}} = 10^{-4}\,\mathrm{mol}$ Mass of product B produced: textMass = 10^-4\,mathrmmol times 93\,mathrmg\,mol^-1 = 9.3 times 10^-3\,mathrmg = 9.3\,mathrmmg$\text{Mass} = 10^{-4}\,\mathrm{mol} \times 93\,\mathrm{g\,mol}^{-1} = 9.3 \times 10^{-3}\,\mathrm{g} = 9.3\,\mathrm{mg}$ Expressing in the specified format: 9.3\,mathrmmg = 93 times 10^-1\,mathrmmg Rightarrow x = 93$9.3\,\mathrm{mg} = 93 \times 10^{-1}\,\mathrm{mg} \Rightarrow x = 93$ ### Pattern Recognition Sees: Conversion sequence preserving a 1:1 mole ratio layout. Shortcut: Directly compute target weight via W_B = W_A cdot fracM_BM_A = 11.25 cdot frac93112.5 = 9.3$W_B = W_A \cdot \frac{M_B}{M_A} = 11.25 \cdot \frac{93}{112.5} = 9.3$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Let us resolve each structural step sequentially: 1. **Step 1:** Aniline undergoes diazotization when treated with textNaNO_2 + textHCl$\text{NaNO}_2 + \text{HCl}$ at 273-278text K$273-278\text{ K}$, forming benzene diazonium chloride [A]$[A]$ (textC_6textH_5textN_2^+textCl^-$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$). 2. **Step 2:** Warming benzene diazonium chloride with potassium iodide (textKI$\text{KI}$) substitutes the diazonium group with iodine, producing iodobenzene [B]$[B]$ (textC_6textH_5textI$\text{C}_6\text{H}_5\text{I}$). 3. **Step 3:** Treating iodobenzene with sodium metal in dry ether causes a Fittig coupling reaction, dimerizing two phenyl radicals into biphenyl [C]$[C]$ (textC_6textH_5-textC_6textH_5$\text{C}_6\text{H}_5-\text{C}_6\text{H}_5$).

### Pattern Recognition Shortcut: Aniline ightarrow textNaNO_2/textHCl ightarrow$ ightarrow \text{NaNO}_2/\text{HCl} ightarrow$ Diazonium salt ightarrow textKI ightarrow$ ightarrow \text{KI} ightarrow$ Iodobenzene. The final sodium metal treatment triggers a symmetrical radical dimer homocoupling (Fittig reaction) to yield a biphenyl product. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Haloalkanes and Haloarenes

### 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