Which of the following amine(s) show(s) positive carbylamine test? A.

B. (CH₃)₂NH$(CH_3)_2NH$ C. CH₃NH₂$CH_3NH_2$ D. (CH₃)₃N$(CH_3)_3N$ E.

Choose the correct answer from the options given below:

Core Logic

Aniline is the common name for the simplest aromatic amine, which consists of a phenyl group attached to an amino group (C₆H₅NH₂$C_6H_5NH_2$). 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.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Statement (I): The -NH₂$-NH_2$ 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 AlCl₃$AlCl_3$. Aniline is a Lewis base (due to the lone pair on N) and reacts with the Lewis acid AlCl₃$AlCl_3$ to form a stable salt/complex (C₆H₅ +NH₂-AlCl₃^-$C_6H_5\overset{+}{N}H_2-AlCl_3^-$). This removes the lone pair from resonance and converts the -NH₂$-NH_2$ 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 (NH₂$NH_2$) reacts with the catalyst (AlCl₃$AlCl_3$) before the reaction can proceed.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Aniline undergoes electrophilic aromatic substitution (like bromination). The -NH₂$-NH_2$ 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 -NH₂$-NH_2$ 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 -NH₂$-NH_2$ 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 -NH₂$-NH_2$ 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 -NH₂$-NH_2$ substituted carbon).

Since -NH₂$-NH_2$ is ortho/para directing, the meta-arenium ion will not be formed.

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Step 1: Nitration of benzene using conc. HNO₃$HNO_3$ and conc. H₂SO₄$H_2SO_4$ gives nitrobenzene. Step 2: Reduction of nitrobenzene with Sn/HCl$Sn/HCl$ yields aniline. Step 3: Aniline reacts with NaNO₂/HCl$NaNO_2/HCl$ at 0-5^° 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 arrow$\rightarrow$ Reduction arrow$\rightarrow$ Diazotization arrow$\rightarrow$ Coupling (Azo Dye Test).

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Step 1: Bromobenzene + Mg, ether arrow Phenylmagnesium bromide$Mg, ether \rightarrow Phenylmagnesium bromide$ (Grignard reagent). Step 2: Grignard + CO₂$CO_2$ followed by H^+ arrow Benzoic acid$H^+ \rightarrow Benzoic acid$ (C₆H₅COOH$C_6H_5COOH$). Step 3: Benzoic acid + NH₃, Δ arrow Benzamide$NH_3, \Delta \rightarrow Benzamide$ (C₆H₅CONH₂$C_6H_5CONH_2$). Step 4: Benzamide + Br₂/NaOH$Br_2/NaOH$ (Hoffmann bromamide degradation) arrow Aniline$\rightarrow Aniline$ (C₆H₅NH₂$C_6H_5NH_2$).

Step 1: Tracing the product

The final product 'A' is Aniline.

Chapter Mix

Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Haloalkanes and Haloarenes