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 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

### Core Logic Statement (I): The -NH_2$-NH_2$ group has a lone pair of electrons on nitrogen, which undergoes resonance with the benzene ring (strong +M effect). This strongly activates the ring towards electrophilic substitution and directs incoming electrophiles to the ortho and para positions. Statement (II): Friedel-Crafts alkylation and acylation require a Lewis acid catalyst like anhydrous AlCl_3$AlCl_3$. Aniline is a Lewis base (due to the lone pair on N) and reacts with the Lewis acid AlCl_3$AlCl_3$ to form a stable salt/complex (C_6H_5overset+NH_2-AlCl_3^-$C_6H_5\overset{+}{N}H_2-AlCl_3^-$). This removes the lone pair from resonance and converts the -NH_2$-NH_2$ group into a strongly deactivating group, thereby halting the Friedel-Crafts reaction. ### Step 1: Evaluate Statements Statement I is correct. Statement II is correct. ### Pattern Recognition Aniline NEVER undergoes Friedel-Crafts because the base (NH_2$NH_2$) reacts with the catalyst (AlCl_3$AlCl_3$) before the reaction can proceed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Aniline undergoes electrophilic aromatic substitution (like bromination). The -NH_2$-NH_2$ group is a strongly activating group and directs incoming electrophiles to the ortho and para positions due to resonance electron donation (+M$M$ effect). ### Step 1: Identifying Sigma Complexes (Arenium Ions) When an electrophile (Br^+$Br^+$) attacks the ring, an intermediate arenium ion (sigma complex) is formed. - If attack occurs at the **ortho** or **para** position, the positive charge is delocalized onto the carbon atom bearing the -NH_2$-NH_2$ group. The lone pair on nitrogen can then stabilize this positive charge via resonance, forming a highly stable resonance structure (an octet-complete intermediate). - If attack occurs at the **meta** position, the positive charge delocalizes only over the remaining ring carbons and never rests on the carbon bearing the -NH_2$-NH_2$ group. Thus, it misses the extra stabilization provided by the nitrogen lone pair. Because the meta attack intermediate is less stable compared to ortho/para attack, and the -NH_2$-NH_2$ is strictly o/p directing, the meta-arenium ion is NOT a primary intermediate involved in standard bromination pathways of neutral aniline. ### Step 2: Conclusion Option 3 displays the arenium ion resulting from a meta-attack (positive charge skips the -NH_2$-NH_2$ substituted carbon).

Since -NH_2$-NH_2$ is ortho/para directing, the meta-arenium ion will not be formed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

### Core Logic Step 1: Nitration of benzene using conc. HNO_3$HNO_3$ and conc. H_2SO_4$H_2SO_4$ gives nitrobenzene. Step 2: Reduction of nitrobenzene with Sn/HCl$Sn/HCl$ yields aniline. Step 3: Aniline reacts with NaNO_2/HCl$NaNO_2/HCl$ at 0-5^circ C$0-5^\circ C$ to form benzene diazonium chloride (Product A). Step 4: Benzene diazonium chloride undergoes a coupling reaction with phenol (typically in a mildly alkaline medium) to form p-hydroxyazobenzene, an orange dye (Product B).

### Pattern Recognition Nitration rightarrow$\rightarrow$ Reduction rightarrow$\rightarrow$ Diazotization rightarrow$\rightarrow$ Coupling (Azo Dye Test). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines