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

The stability of arenediazonium salts is dictated by the ability of substituents on the benzene ring to disperse the positive charge on the diazonium group (-N₂^+$\mathrm{-N_2^+}$).

Electron-donating groups (EDG) increase the stability by resonance (+M effect) and inductive (+I) effects, dispersing the positive charge. Electron-withdrawing groups (EWG) decrease the stability by intensifying the positive charge (-M, -I effects).

Let's evaluate the given substituents at the para position: (A) -OCH₃$\mathrm{-OCH_3}$: Strong +M effect. Highly stabilizing. (C) Plain benzene ring: Neutral baseline. (D) -CN$\mathrm{-CN}$: Strong -M and -I effect. Destabilizing. (B) -NO₂$\mathrm{-NO_2}$: Very strong -M and -I effect. Most destabilizing.

Step 1: Final Conclusion

Arranging them in decreasing order of stability: (A) -OCH₃$\mathrm{-OCH_3}$ (+M$+M$) > (C) Baseline > (D) -CN$\mathrm{-CN}$ (-M$-M$) > (B) -NO₂$\mathrm{-NO_2}$ (Stronger -M$-M$). Thus, the correct order is A > C > D > B$\text{A} > \text{C} > \text{D} > \text{B}$.

Pattern Recognition

Stability of carbocations and diazonium ions follows: +M > +I > no effect > -I > -M$+M > +I > \text{no effect} > -I > -M$. Look directly at the para group to rank.

Chapter Mix

Class 12 Chemistry: Amines Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Related Formula

Core Logic

Given mass of aniline (C₆H₅NH₂$C_6H_5NH_2$) = 9.3 gm Molar mass of aniline (MW$MW$) = 93 g/mol Moles of aniline (ntheoretical$n_{\text{theoretical}}$) = (9.3)/(93) = 0.1 moles$\frac{9.3}{93} = 0.1 \text{ moles}$

Given mass of dry acetanilide (C₆H₅NHCOCH₃$C_6H_5NHCOCH_3$) obtained = 11 gm Molar mass of acetanilide (MW$MW$) = 135 g/mol Moles of acetanilide obtained (nactual$n_{\text{actual}}$) = (11)/(135) 0.08148 moles$\frac{11}{135} \simeq 0.08148 \text{ moles}$

Step 1: Calculate Percentage Yield

Pattern Recognition

A 1:1 molar ratio dictates that theoretical yield in moles of product directly matches moles of starting material. Converting experimental mass to moles removes the need to calculate theoretical mass, speeding up the final ratio.

Chapter Mix

Class 12 Chemistry: Amines Class 11 Chemistry: Some Basic Concepts of Chemistry

Core Logic

Statement I: While it is true that the dipole moment of R-CN is greater than R-NC, the hydrolysis of isocyanide (R-NC) under acidic conditions yields a primary amine and formic acid, not a carboxylic acid (R-COOH) with the same alkyl chain intact. Reaction: R-NC H₃O^+ R-NH₂ + HCOOH$\mathrm{R-NC} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{R-NH}_2 + \mathrm{HCOOH}$ Thus, Statement I is False.

Statement II:

• Hydrolysis of Cyanide: R-CN H₃O^+ R-COOH$\mathrm{R-CN} \xrightarrow{\mathrm{H}_3\mathrm{O}^+} \mathrm{R-COOH}$
• With SOCl₂$\mathrm{SOCl}_2$: R-COOH SOCl₂ R-COCl$\mathrm{R-COOH} \xrightarrow{\mathrm{SOCl}_2} \mathrm{R-COCl}$
• With NH₃$\mathrm{NH}_3$: R-COCl NH₃ R-CONH₂$\mathrm{R-COCl} \xrightarrow{\mathrm{NH}_3} \mathrm{R-CONH}_2$ (Compound x)
• Hoffmann Bromamide / Hypochlorite degradation: R-CONH₂ NaOCl/NaOH R-NH₂$\mathrm{R-CONH}_2 \xrightarrow{\mathrm{NaOCl/NaOH}} \mathrm{R-NH}_2$
• Carbylamine Reaction: R-NH₂ CHCl₃ + KOH R-NC$\mathrm{R-NH}_2 \xrightarrow{\mathrm{CHCl}_3 + \mathrm{KOH}} \mathrm{R-NC}$

The sequence correctly produces R-NC. Thus, Statement II is True.

Step 1: Final Conclusion

Statement I is false but Statement II is true.

Pattern Recognition

Isocyanide hydrolysis breaks the N-C bond, leaving the R group attached to Nitrogen (forming a 1°$1^{\circ}$ amine). The remaining C becomes formic acid.

Chapter Mix

Class 12 Chemistry: Amines

Step 1: Evaluate Reaction (1)

Nitration of acetanilide produces primarily the para-isomer because the -NHCOCH₃$-NHCOCH_3$ group is strongly ortho/para directing and sterically hinders the ortho position. The option incorrectly suggests the meta product. (Incorrect)

Step 2: Evaluate Reaction (2)

Gabriel phthalimide synthesis does not work with aryl halides because aryl halides do not readily undergo nucleophilic substitution. Thus, bromobenzene cannot form aniline this way. (Incorrect)

Step 3: Evaluate Reaction (3)

Reaction of secondary amine with CHCl₃/KOH$CHCl_3/KOH$ does NOT yield an isocyanide. The carbylamine test is strictly for primary amines. (Incorrect)

Step 4: Evaluate Reaction (4)

Hoffmann bromamide degradation: H₃C-CH₂-CO-NH₂ + Br₂ + 4KOH Δ H₃C-CH₂-NH₂ + 2KBr + K₂CO₃ + 2H₂O$H_3C-CH_2-CO-NH_2 + Br_2 + 4KOH \xrightarrow{\Delta} H_3C-CH_2-NH_2 + 2KBr + K_2CO_3 + 2H_2O$ This cleanly forms ethylamine, completely accurately represented. (Correct)

Pattern Recognition

Carbylamine test = 1° amines only. Gabriel Phthalimide = aliphatic 1° amines only (no aryl halides). Hoffmann degradation strictly lowers carbon chain by 1 (CO$CO$ removal).

Chapter Mix

Class 12 Chemistry: Amines

Core Logic

Step 1: Empirical Formula calculation: Moles of C = (61.01)/(12) ≈ 5.08$C = \frac{61.01}{12} \approx 5.08$ Moles of H = (15.25)/(1) ≈ 15.25$H = \frac{15.25}{1} \approx 15.25$ Moles of N = (23.74)/(14) ≈ 1.69$N = \frac{23.74}{14} \approx 1.69$ Ratio C : H : N = (5.08)/(1.69) : (15.25)/(1.69) : (1.69)/(1.69) ≈ 3 : 9 : 1$\frac{5.08}{1.69} : \frac{15.25}{1.69} : \frac{1.69}{1.69} \approx 3 : 9 : 1$ Empirical formula = C₃H₉N$C_3H_9N$.

Step 2: Chemical tests deduction. The compound is treated with HNO₂/H₂O$HNO_2/H_2O$ and produces a nitrogen-free compound Y. This implies X is a primary aliphatic amine (C₃H₇NH₂$C_3H_7NH_2$), giving an alcohol Y (C₃H₇OH$C_3H_7OH$).

Compound Y upon controlled oxidation forms Z, which gives positive iodoform test. For Z to give an iodoform test, it must be a methyl ketone. Thus Z is acetone (CH₃COCH₃$CH_3COCH_3$).

For Y to oxidize to acetone, Y must be Isopropyl alcohol (CH₃CH(OH)CH₃$CH_3CH(OH)CH_3$). Therefore, X must be Isopropylamine: (CH₃)₂CHNH₂$(CH_3)_2CHNH_2$.

Step 1: Reaction Sequence

Step 2: Final Conclusion

The structure of 'X' is Isopropylamine.

Pattern Recognition

Iodoform test on an oxidized alcohol means the alcohol was a secondary alcohol with a terminal methyl group. Combining this with HNO₂$HNO_2$ reacting with primary amines gives the exact functional groups.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Amines Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids