The major product (A) formed in the following reaction sequence is:

### Core Logic Step 1: Bromobenzene + Mg, ether rightarrow Phenylmagnesium bromide$Mg, ether \rightarrow Phenylmagnesium bromide$ (Grignard reagent). Step 2: Grignard + CO_2$CO_2$ followed by H^+ rightarrow Benzoic acid$H^+ \rightarrow Benzoic acid$ (C_6H_5COOH$C_6H_5COOH$). Step 3: Benzoic acid + NH_3, Delta rightarrow Benzamide$NH_3, \Delta \rightarrow Benzamide$ (C_6H_5CONH_2$C_6H_5CONH_2$). Step 4: Benzamide + Br_2/NaOH$Br_2/NaOH$ (Hoffmann bromamide degradation) rightarrow Aniline$\rightarrow Aniline$ (C_6H_5NH_2$C_6H_5NH_2$).

### Step 1: Tracing the product The final product 'A' is Aniline. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Haloalkanes and Haloarenes

### Core Logic The reaction sequence represents the classic dye test for aromatic primary amines. Step 1 (Diazotization): Aniline (Ph-NH_2$Ph-NH_2$) reacts with nitrous acid (generated in situ from NaNO_2 + HCl$NaNO_2 + HCl$) at low temperature (0-5^circ C$0-5^\circ C$) to form benzene diazonium chloride (Ph-N_2^+Cl^-$Ph-N_2^+Cl^-$). Thus, Reagent A is NaNO_2 + HCl$NaNO_2 + HCl$ at 0-5^circ C$0-5^\circ C$. ### Step 2: Coupling Reaction Step 2: The diazonium salt undergoes an electrophilic substitution (coupling reaction) with an electron-rich aromatic ring to form an azo dye. The formation of a 'scarlet red dye' is specifically the result of coupling benzene diazonium chloride with beta$\beta$-naphthol in a weakly basic medium (NaOH).

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic 1) Sulphanilic acid reacts with NaNO_2$NaNO_2$ and CH_3COOH$CH_3COOH$ to form a diazonium salt (X). 2) The diazonium salt (X) then reacts with N,N-dimethylaniline (given in the coupling step image). The coupling takes place at the para position of the highly activated N,N-dimethylaniline ring. 3) This coupling yields Methyl Orange, an azo dye. Its structure is p$p$-dimethylaminoazobenzenesulphonic acid.

### Step 1: Final Identification The final product (Y) matches option (4) structurally, containing the sulphonic acid group on one ring, the azo linkage, and the N,N-dimethylamine group on the para position of the other ring. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Statement I: Aniline reacting with concentrated H_2SO_4$H_2SO_4$ gives anilinium hydrogensulphate, which on heating at 453-473 K produces sulphanilic acid (p-aminobenzene sulphonic acid). Because sulphanilic acid contains both Nitrogen and Sulphur, it gives a blood-red colouration in Lassaigne's test due to the formation of thiocyanate ion SCN^-$SCN^-$ which reacts with Fe^3+$Fe^{3+}$ to form [Fe(SCN)]^2+$[Fe(SCN)]^{2+}$. Thus, Statement I is true. Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst AlCl_3$AlCl_3$ reacts with the lone pair on the nitrogen atom of aniline to form a salt. This generates a positive charge on the nitrogen, transforming the -NH_2$-NH_2$ group from a strong activating group into a strong deactivating group, thus preventing the Friedel-Crafts reaction from occurring. Thus, Statement II is true.

### Step 1: Final Conclusion Both Statement I and Statement II are true. Option (4) is correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula R-NH_2 + CH_3COCl rightarrow R-NH-COCH_3 + HCl$$R-NH_2 + CH_3COCl \rightarrow R-NH-COCH_3 + HCl$$ ### Core Logic During the acetylation of an amino group, one hydrogen atom (mass = 1text g/mol$1\text{ g/mol}$) is replaced by an acetyl group (-COCH_3$-COCH_3$, mass = 43text g/mol$43\text{ g/mol}$). Gain in molecular weight for every one -NH_2$-NH_2$ group acetylated = 43 - 1 = 42text g/mol$43 - 1 = 42\text{ g/mol}$. ### Step 1: Calculating Number of Groups Total increase in molecular weight = Final mass - Initial mass = 192 - 108 = 84text g/mol$192 - 108 = 84\text{ g/mol}$. textNumber of amino groups = fractextTotal mass increasetextMass increase per group = frac8442 = 2$$\text{Number of amino groups} = \frac{\text{Total mass increase}}{\text{Mass increase per group}} = \frac{84}{42} = 2$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines