Percentage composition of compound $$X$$ is C = 61.01 %, H = 15.25 %, N = 23.74 %.

First, determine the empirical formula.
Moles of each element (using atomic masses C = 12, H = 1, N = 14):
$$n_C = \frac{61.01}{12} = 5.084$$
$$n_H = \frac{15.25}{1} = 15.25$$
$$n_N = \frac{23.74}{14} = 1.695$$

Divide by the smallest value (1.695) to obtain the simplest ratio:
$$\frac{n_C}{1.695} = 3.00,\; \frac{n_H}{1.695} = 9.00,\; \frac{n_N}{1.695} = 1.00$$

Empirical formula = $$\mathrm{C_3H_9N}$$.

All isomers of $$\mathrm{C_3H_9N}$$ are primary amines: $$\mathrm{CH_3CH_2CH_2NH_2}$$ (1-aminopropane) and $$\mathrm{(CH_3)_2CHNH_2}$$ (2-aminopropane, isopropylamine).

Reaction of $$X$$ with $$HNO_2/H_2O$$: a primary aliphatic amine is converted into the corresponding alcohol with loss of nitrogen.
$$\mathrm{RCH_2NH_2 + HNO_2 \rightarrow RCH_2OH + N_2 \uparrow + H_2O}$$

Therefore compound $$Y$$ is the alcohol having the same carbon skeleton as $$X$$.

The experiment shows that the oxidation product $$Z$$ does give the iodoform test; hence Case 2 is correct.

Therefore $$X$$ must be 2-aminopropane (isopropylamine): $$\mathrm{(CH_3)_2CHNH_2}$$.

Option B which is: isopropylamine.

Moles of given compound = mass / molar mass
= $$\ \frac{\ 137\times\ 10^{-3}}{137}\ =\ 1\times\ 10^{-3}$$
= $$0.001$$ mol

All reactions (reduction → acetylation → bromination) are 1:1, so moles of product = 0.001 mol

Mass of product = moles × molar mass
= 0.001×228
= 0.228 g = 228 mg

Satement 1

Statement 2

• $$NO_{2}$$ is strongly electron-withdrawing (-I and -M) → decreases ring electron density → deactivating
• Ortho/para attack puts + charge near -NO₂ (unstable)
• Meta avoids this instability → meta-directing

Statement I is correct because imidazole is a significantly stronger base than aniline. Therefore, its $$pK_b$$ value is lower (around 7.0 for imidazole compared to 9.4 for aniline).

Imidazole's pyridine-like nitrogen has a lone pair residing in an $$sp^2$$ orbital that is not part of the aromatic ring's cloud, making it highly available for protonation.

In contrast, aniline's lone pair is heavily delocalized into the benzene ring, which dramatically reduces its basicity.

Statement II is incorrect because the basicity order specified in the statement represents methyl-substituted amines in an aqueous medium rather than ethyl-substituted ones.

For ethyl-substituted amines, the correct decreasing order of basic strength in an aqueous solution is:

$$\mathrm{(C_2H_5)_2NH>(C_2H_5)_3N>C_2H_5NH_2>NH_3}$$

This sequence occurs due to a unique balance of the electron-donating inductive effect, steric hindrance, and hydration stability.

Correct option: (B)

We need to evaluate the correctness of two organic reaction sequences presented in Statement I and Statement II.

Statement I:

The first sequence outlines the conversion of tetralin (a bicyclic hydrocarbon where a benzene ring is fused to a 6-membered aliphatic ring) into 1,2-diaminobenzene:

• Step (i) $$\text{KMnO}_4 / \text{H}^+$$ (Oxidation): Vigorous oxidation cleaves the aliphatic ring at the benzylic positions, converting both benzylic carbons into carboxylic acid groups. This yields phthalic acid (benzene-1,2-dicarboxylic acid).
• Step (ii) $$\text{NH}_3$$ (Ammonolysis & Heating): Reaction of phthalic acid with ammonia followed by heating forms phthalamide, which undergoes cyclization to give phthalimide.
• Step (iii) $$\text{Br}_2 + \text{KOH}$$ (Hoffmann Bromamide Degradation): Phthalimide undergoes Hoffmann degradation, converting the carbonyl-linked nitrogen groups into primary amines, yielding 1,2-diaminobenzene.

$$\implies \text{Statement I is True}$$

Statement II:

The second sequence outlines the synthesis of 3,5-dibromotoluene starting from 4-methylaniline (p-toluidine):

• Step (i) $$\text{Br}_2 / \text{H}_2\text{O}$$ (Electrophilic Aromatic Substitution): The strongly activating and ortho/para-directing $$-\text{NH}_2$$ group directs bromine atoms to both of its vacant ortho positions (positions 2 and 6), producing 2,6-dibromo-4-methylaniline.
• Step (ii) $$\text{NaNO}_2 + \text{HCl, } 0{-}5^\circ\text{C}$$ (Diazotization): The primary aromatic amine group ($$-\text{NH}_2$$) is converted quantitatively into a diazonium salt group ($$-\text{N}_2^+\text{Cl}^-$$).
• Step (iii) \text{aq. } $$\text{H}_3\text{PO}_2$$ (Deamination): Hypophosphorous acid reduces the diazonium group, replacing it with a hydrogen atom ($$-\text{H}$$), which results in the formation of 3,5-dibromotoluene.

$$\implies \text{Statement II is True}$$

Conclusion:

Both reaction sequences proceed successfully as described to form their respective major products.

Answer: Option C — Both Statement I and Statement II are true

Reaction A (Correct)

• Reactants: Benzyl bromide ($$\text{C}_6\text{H}_5\text{CH}_2\text{Br}$$) + Silver cyanide ($$\text{AgCN}$$)
• Mechanism: Silver cyanide ($$\text{AgCN}$$) is covalent in nature. Because the $$\text{Ag-C}$$ bond is strong, the lone pair on the nitrogen atom acts as the nucleophile rather than the carbon atom. This nucleophilic substitution ($$\text{S}_\text{N}2$$) yields an isocyanide.
• Product: Benzyl isocyanide ($$\text{C}_6\text{H}_5\text{CH}_2\text{NC}$$)

Reaction B (Correct)

• Reactants: Benzylamine ($$\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$$) + Chloroform ($$\text{CHCl}_3$$) + aq. $$\text{NaOH}$$
• Mechanism: This is the Carbylamine reaction (Hofmann's isocyanide test). The primary amine reacts with chloroform in a basic medium via a dichlorocarbene ($$:\text{CCl}_2$$) intermediate to form an isocyanide.
• Product: Benzyl isocyanide ($$\text{C}_6\text{H}_5\text{CH}_2\text{NC}$$)

Reaction C (Incorrect)

• Reactants: Bromobenzene ($$\text{C}_6\text{H}_5\text{Br}$$) + Silver cyanide ($$\text{AgCN}$$)
• Mechanism: Bromobenzene does not easily undergo nucleophilic substitution under normal conditions due to the resonance-stabilized, partial double-bond character of the $$\text{C-Br}$$ bond. Even under drastic conditions, it would yield phenyl isocyanide ($$\text{C}_6\text{H}_5\text{NC}$$), which lacks the required $$-\text{CH}_2-$$ carbon spacer.

Reaction D (Incorrect)

• Reactants: Aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$) + Chloroform ($$\text{CHCl}_3$$) + aq. $$\text{NaOH}$$
• Mechanism: While this also undergoes the Carbylamine reaction, the starting material is an aromatic primary amine without a side-chain carbon.
• Product: Phenyl isocyanide ($$\text{C}_6\text{H}_5\text{NC}$$)

Reaction E (Incorrect)

• Reactants: 2-Phenylethyl bromide ($$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{Br}$$) + Potassium cyanide ($$\text{KCN}$$)
• Mechanism: Potassium cyanide ($$\text{KCN}$$) is ionic, providing free $$\text{CN}^-$$ ions where the negatively charged carbon acts as the nucleophile. Furthermore, the starting alkyl chain already has two carbon atoms instead of one.
• Product: 3-Phenylpropanenitrile ($$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CN}$$)

Correct Option: A and B Only

1. Structure A (Hofmann Degradation)

• Reaction Conditions: Compound A has the molecular formula$$C_{8}H_{9}ON$$ and undergoes a Hofmann bromamide degradation when treated with aqueous $$Br_{2}/HO^{-}$$. This indicates that 'A' is a primary amide attached to a substituted benzene ring.
• Product B: The amide group ($$-CONH_{2}$$) is converted into a primary amine ($$-NH_{2}$$), yielding 3-methylaniline (m-toluidine), which is basic and soluble in dilute acid.

2. Intermediate Conversions (B → C → D → E)

• Diazotization (B → C): Treatment of 3-methylaniline with$$NaNO_{2}/HCl$$ at $$0\text{--}5^{\circ}C$$ converts the amine group into a diazonium salt ($$-N_{2}^{+}Cl^{-}$$).
• Sandmeyer Reaction (C → D): Reacting the diazonium salt with$$CuCN/NaCN$$ replaces the diazonium group with a nitrile group ($$-CN$$), forming 3-ethylbenzonitrile.
• Hydrolysis (D → E): Complete acid hydrolysis of the nitrile group converts it into a carboxylic acid ($$-COOH$$), producing 3-methylbenzoic acid. This matches the product obtained from the direct hydrolysis of the amide group in 'A'.

3. Final Oxidation (E → F)

• Reaction Conditions: 3-methylbenzoic acid is oxidized using acidified$$KMnO_{4}$$ with heating. The side-chain methyl group ($$-CH_{3}$$) undergoes complete oxidation into a carboxylic acid group ($$-COOH$$).
• Product F: The reaction yields benzene-1,3-dicarboxylic acid (isophthalic acid). Due to its symmetrical meta-substitution, the structure contains two distinct types of aromatic hydrogen atoms.

1. Step (i): Amide Formation $$(\text{NH}_3 / \Delta)$$
   Propanoic acid $$(\text{CH}_3\text{CH}_2\text{COOH})$$ reacts with ammonia and heat to undergo a dehydration reaction, converting the carboxylic acid group into an amide:
   $$\text{CH}_{3}\text{CH}_{2}\text{COOH}\xrightarrow{\text{NH}_{3},\ \Delta }\text{CH}_{3}\text{CH}_{2}\text{CONH}_{2}\quad \text{(Propanamide)}$$
   
2. Step (ii): Hoffmann Bromamide Degradation $$(\text{NaOH} / \text{Br}_2)$$
   Propanamide reacts with bromine in the presence of sodium hydroxide. This causes degradation by losing the carbonyl $$(\text{C}=\text{O})$$ group, yielding a primary amine with one less carbon atom:
   $$\text{CH}_{3}\text{CH}_{2}\text{CONH}_{2}\xrightarrow{\text{NaOH}/\text{Br}_{2}}\text{CH}_{3}\text{CH}_{2}\text{NH}_{2}\quad \text{(Ethanamine)}$$
   
3. Step (iii): Diazotization and Hydrolysis $$(\text{HNO}_2 / \text{H}_2\text{O}, 0^\circ\text{C})$$
   Ethanamine reacts with nitrous acid $$(\text{HNO}_{2})$$ at a low temperature to form an unstable aliphatic diazonium salt, which instantly undergoes nucleophilic substitution with water $$(\text{H}_2\text{O})$$ to form ethanol:
   $$\text{CH}_{3}\text{CH}_{2}\text{NH}_{2}\xrightarrow{\text{HNO}_{2}/\text{H}_{2}\text{O},\ 0^{\circ }\text{C}}\text{CH}_{3}\text{CH}_{2}\text{OH}\quad \text{(Ethanol)}$$
   
4. Step (iv): Reduction of Benzenediazonium Chloride $$(\text{C}_6\text{H}_5\text{N}_2\text{Cl}, 0^\circ\text{C})$$
   When ethanol is mixed with benzenediazonium chloride $$(\text{C}_6\text{H}_5\text{N}_2\text{Cl})$$, it acts as a reducing agent. Ethanol reduces the diazonium salt to Benzene while itself getting oxidized to ethanal (acetaldehyde):
   $$\text{C}_{6}\text{H}_{5}\text{N}_{2}\text{Cl}+\text{CH}_{3}\text{CH}_{2}\text{OH}\xrightarrow{0^{\circ }\text{C}}\text{C}_{\mathbf{6}}\text{H}_{\mathbf{6}}\mathbf{\ (Benzene)}+\text{N}_{2}+\text{CH}_{3}\text{CHO}+\text{HCl}$$

Correct Option
Benzene

In the Hofmann bromamide degradation reaction, a primary amide is converted into a primary amine containing one carbon atom less.

For benzamide,

$$C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2.$$

Thus, benzamide produces aniline and not benzylamine ((C_6H_5CH_2NH_2)). Therefore, Statement I is false.

During the nitration of aniline using the nitrating mixture ((HNO_3/H_2SO_4)), the amino group is protonated in the strongly acidic medium to form the anilinium ion ((-NH_3^+)). The anilinium ion is electron-withdrawing and acts as a meta-directing group.

As a result, nitration under these conditions produces (m)-nitroaniline in a higher proportion than (o)-nitroaniline.

Therefore, Statement II is true.

Hence, the correct answer is Statement I is false but Statement II is true.

To determine the major product [C], we analyze the reaction sequence step by step starting from aniline.

Aniline undergoes rapid electrophilic aromatic substitution with bromine water because the $$-NH_2$$ group strongly activates the benzene ring. Bromination occurs at both ortho positions and the para position, producing 2,4,6-tribromoaniline as product [A].

On treatment with $$NaNO_2/HCl$$ at $$273\text{-}278\ \text{K}$$, the amino group of 2,4,6-tribromoaniline is converted into a diazonium group through the diazotization reaction, giving 2,4,6-tribromobenzenediazonium chloride as intermediate [B].

The diazonium salt then reacts with $$HBF_4$$ to form the corresponding diazonium fluoroborate. Subsequent treatment with $$NaNO_2$$ and copper replaces the diazonium group with a nitro group, yielding the final product

$$\text{2,4,6-tribromobenzenediazonium salt} \longrightarrow \text{1,3,5-tribromo-2-nitrobenzene}.$$

Thus, the major product [C] is 1,3,5-tribromo-2-nitrobenzene.

Hence, the correct answer is 1,3,5-tribromo-2-nitrobenzene, which corresponds to option B.

For substituted anilines, two electronic effects decide the stability of the protonated form (the anilinium ion):
1. Inductive effect (−I or +I): An electron-withdrawing (−I) group pulls electron density away from the positively charged nitrogen, dispersing the charge and stabilising the cation.
2. Resonance (−M or +M): A group that can withdraw electrons by resonance (−M) also delocalises and stabilises the positive charge.

Let us examine each substituent in the para position:

Case A: Aniline (no substituent)
No extra stabilising effect beyond the benzene ring. So its conjugate acid serves as the reference.

Case B: p-Methoxyaniline (−OCH3)
−OCH3 has a strong +M (electron-donating) effect, pushing electron density toward the ring. This increases electron density on nitrogen in the neutral base; once the base is protonated, that extra electron density destabilises the positive charge. Hence the conjugate acid is weaker than that of aniline.

Case C: p-Nitroaniline (−NO2)
−NO2 is a very powerful −I and −M group. Both effects withdraw electron density from the ring and ultimately from the protonated nitrogen. This delocalises and stabilises the positive charge on nitrogen, making the conjugate acid highly stabilised and therefore very strong.

Case D: p-Methylaniline (−CH3)
−CH3 is weakly +I and weakly +M (hyperconjugation). These donating effects slightly increase electron density, which destabilises the cation relative to aniline. So its conjugate acid is weaker.

Comparing all four, the conjugate acid of p-nitroaniline receives the maximum stabilisation due to the strong −I and −M effects of the nitro group. Hence it is the strongest conjugate acid among the given options.

Therefore, the correct choice is:
Option C which is: p-Nitroaniline

The reaction described is an azo-coupling reaction (electrophilic aromatic substitution) with a benzenediazonium ion. For this reaction to occur efficiently, the benzene ring must be highly activated with high electron density, primarily provided by the strong mesomeric ($$+M$$) donating effect of the nitrogen's lone pair.

• Compound P (N,N-dimethylaniline): There is no steric hindrance around the $$-\text{NMe}_2$$ group. The lone pair on the nitrogen atom is fully coplanar with the benzene ring, allowing for maximum resonance ($$+M$$ effect). This strongly activates the ring, making it the most reactive.
  
• Compound Q (2-methyl-N,N-dimethylaniline): The single bulky methyl group at the ortho-position introduces steric crowding, pushing the $$-\text{NMe}_2$$ group slightly out of the benzene ring's plane. This decreases the overlap of the nitrogen lone pair with the $$\pi$$-system (partial SIR effect), lowering its reactivity compared to P.
  
• Compound R (2,6-dimethyl-N,N-dimethylaniline): With two bulky methyl groups flanking the $$-\text{NMe}_2$$ group on both sides, the steric crowding is extreme. To minimize repulsion, the $$-\text{NMe}_2$$ group rotates completely out of the plane of the benzene ring. Because it is perpendicular, resonance is entirely inhibited (Steric Inhibition of Resonance / SIR effect). The lone pair cannot activate the ring at all, making it the least reactive.

As the number of ortho-methyl groups increases, resonance decreases, dropping the electron density on the ring and reducing its reactivity toward electrophiles:

$$\text{P} > \text{Q} > \text{R}$$

Therefore, the correct choice is [A] P > Q > R.

Formation of Compound A

Formation of Compound B (Reduction)

• Reagent: $$\text{Sn} / \text{HCl}$$
• Transformation: This is a selective reduction reaction that converts the nitro group ($$-\text{NO}_2$$) into a primary amine group ($$-\text{NH}_2$$).
• Compound B: 3-bromo-4-ethylaniline.

Step 3: Formation of Compound C (Diazotization)

• Reagent: $$\text{NaNO}_2 / \text{HCl}$$ at $$273\text{--}278\text{ K}$$ ($$0\text{--}5^\circ\text{C}$$).
• Transformation: The primary aromatic amine group ($$-\text{NH}_2$$) reacts to form a stable aromatic diazonium salt ($$-\text{N}_2^+\text{Cl}^-$$).
• Compound C: 3-bromo-4-ethylbenzenediazonium chloride.

Formation of Compound D (Deamination)

• Reagent: $$\text{C}_2\text{H}_5\text{OH}$$ (Ethanol) or $$\text{H}_3\text{PO}_2$$.
• Transformation: Ethanol acts as a mild reducing agent that replaces the diazonium group ($$-\text{N}_2^+\text{Cl}^-$$) with a hydrogen atom ($$-\text{H}$$), effectively removing it from the ring.
• Compound D: 1-bromo-2-ethylbenzene (o-bromoethylbenzene).

Formation of Compound E (Side-chain Oxidation)

• Reagent: (i) $$\text{KMnO}_4 / \text{KOH}$$, (ii) $$\text{H}_3\text{O}^+$$
• Transformation: Potassium permanganate under basic heating conditions oxidizes any alkyl side chain containing benzylic hydrogens completely down to a carboxylic acid group ($$-\text{COOH}$$). Acid workup ($$\text{H}_3\text{O}^+$$) protonates the resulting carboxylate salt.
• Compound E: 2-bromobenzoic acid.

Option B

In benzylamine, the lone pair on nitrogen is localized and is not involved in resonance with the benzene ring. In aniline, the lone pair is delocalized into the aromatic ring through resonance, making it less available for protonation. Therefore, benzylamine is more basic than aniline. Hence, statement A is correct.

Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines and proceeds through an $$S_N2$$ mechanism. Primary aromatic amines such as $$p$$-methoxyaniline cannot be prepared by this method because aryl halides do not undergo $$S_N2$$ substitution. Hence, statement B is incorrect.

In Hoffmann bromamide degradation, an amide is converted into a primary amine containing one less carbon atom. Thus,

$$\mathrm{PhCH_2CONH_2 \xrightarrow{Br_2/NaOH} PhCH_2NH_2}$$

The product, benzylamine, is a primary aliphatic amine since the $$-NH_2$$ group is attached to an $$sp^3$$ hybridized carbon and not directly to the benzene ring. Hence, statement C is incorrect.

Treatment of $$p$$-nitroaniline with $$NaNO_2/HCl$$ at $$0^\circ\mathrm{C}$$ forms the corresponding diazonium salt, which on heating with water undergoes hydrolysis to give $$p$$-nitrophenol. Since $$p$$-nitrophenol is acidic, it dissolves in aqueous $$NaOH$$ forming its sodium salt. Hence, statement D is correct.

Therefore, the incorrect statements are B and C only.

Stability Factors

• -OCH₃ (Group A): A strong electron-donating group (+M effect). It provides the highest stability by shifting electron density toward the diazonium group.
• No substitution (Group C): Acts as the baseline for comparison.
• -CN (Group D): An electron-withdrawing group (-M, -I effect). It destabilizes the cation.
• -NO₂ (Group B): A very strong electron-withdrawing group (-M, -I effect). It causes the greatest destabilization.
  

Hofmann bromamide degradation converts a primary amide into a primary amine containing one carbon atom less.

$$RCONH_2\xrightarrow{Br_2/KOH}RNH_2.$$

Therefore, only primary amides can undergo this reaction.

Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines. It is not suitable for the preparation of aromatic amines because aryl halides do not undergo nucleophilic substitution with potassium phthalimide under these conditions.

Hence, a compound must satisfy both of the following conditions:

• It should be a primary amide ((RCONH_2)).
• The amine obtained after Hofmann degradation should be a primary aliphatic amine.

Evaluating the given compounds:

• (C_6H_5CONH_2): Primary amide, but gives aniline (aromatic amine), so it does not qualify.
• (C_6H_5CH_2CONH_2): Primary amide, gives benzylamine, so it qualifies.
• (CH_3CONH_2): Primary amide, gives methylamine, so it qualifies.
• (C_6H_{11}CONHCH_2CH_3): Secondary amide, so it does not qualify.
• ((CH_3)_3CCONHCH_3): Secondary amide, so it does not qualify.
• (C_6H_{11}CONH_2): Primary amide, gives cyclohexylamine, so it qualifies.

Thus, 3 compounds satisfy both conditions.

Hence, the correct answer is 3.

Hoffmann bromamide degradation:

$$RCONH_2 + Br_2 + 4NaOH$$ $$\rightarrow$$ $$RNH_2 + 2NaBr + Na_2CO_3 + 2H_2O$$

Analysis of the statements:

For Statement A:

It states that 4 moles of NaOH and 2 moles of $$Br_2$$ are consumed per mole of amide.

From the balanced equation:

$$1 \text{ mole of amide consumes } 1 \text{ mole of } Br_2$$ and $$4 \text{ moles of } NaOH$$

Therefore, $${\text{False}}$$

For Statement B:

Hoffmann bromamide degradation is given by:

$$RCONH_2$$ (primary amides) including alkyl amides such as: $$CH_3CONH_2$$

Hence, $${\text{False}}$$

For Statement C:

The reaction converts a primary amide into a primary amine having one carbon less.

$$RCONH_2$$ $$\rightarrow$$ $$RNH_2$$

Therefore, $${\text{True}}$$

For Statement D:

Secondary amides $$RCONHR'$$ do not undergo Hoffmann degradation to give secondary amines.

The reaction is characteristic of primary amides.

Therefore, $${\text{False}}$$

For Statement E:

From the balanced equation, the by-products are:

$$Na_2CO_3,\; NaBr,\; H_2O$$

Hence, $${\text{True}}$$

Therefore, the true statements are: $${C \text{ and } E}$$

Step 1: Bromination (Electrophilic Aromatic Substitution)

Reagent: Br₂ / FeBr₃
Reaction name: Halogenation

• NO₂ is meta-directing
• Bromine enters meta position

Product: m-bromonitrobenzene

Step 2: Reduction

Reagent: Sn / HCl
Reaction name: Reduction of nitro group

• NO₂ → NH₂

Product: m-bromoaniline

Step 3: Neutralisation

• Converts salt to free amine

Step 4: Bromination in water

Reagent: Br₂ / H₂O
Reaction name: Bromination of aniline

• NH₂ is strongly activating → ortho & para substitution
• Already one Br present → total becomes 3 Br added

Product: tribromoaniline derivative

Step 5: Diazotization

Reagent: NaNO₂ / HBr (0-5°C)
Reaction name: Diazotization

• NH₂ → N₂⁺

Step 6: Sandmeyer Reaction

Reagent: CuBr / NaBr
Reaction name: Sandmeyer reaction

• N₂⁺ → Br

Final Product (P):

• Initially 1 Br
• After step 4 → +3 Br = 4 Br
• After Sandmeyer → +1 Br

Total bromine atoms = 5

Final Answer:

Number of bromine atoms = 5

Mol. wt. = $$6\times\ 12+(6\times\ 1)+(2\times\ 14)+(2\times\ 16)=138$$

% N =$$\frac{\ 28}{138}\times\ 100$$ = 20.29 %

Nearest integer = 20%

When diazoaminobenzene is warmed to $40\text{-}45^\circ\text{C}$ in the presence of hydrochloric acid ($\text{HCl}$) and excess aniline, it rearranges to form p-aminoazobenzene (a well-known yellow dye).

The product formed from the reaction 

Using the provided atomic masses ($$\text{C} = 12$, $\text{H} = 1$, $\text{N} = 14$$):

$$\text{Molar Mass} = (12 \times 12) + (11 \times 1) + (3 \times 14)$$
$$\text{Molar Mass} = 144 + 11 + 42 = 197\text{ g mol}^{-1}$$

The total mass contributed by the 3 nitrogen atoms in one mole of the compound is:

$$\text{Mass of Nitrogen} = 3 \times 14 = 42\text{ g mol}^{-1}$$

Now, find the mass percentage:

$$\text{Percentage of Nitrogen} = \left( \frac{42}{197} \right) \times 100 \approx 21.32\%$$

Rounding to the nearest integer gives 21%.

Step 1 : Reduction of the nitro group
$$4\text{-methyl-}NO_2\text{-benzene} \xrightarrow[\;OH^-]{Sn/HCl} 4\text{-methyl aniline }(p\text{-toluidine})$$

Step 2 : Acetylation of the amino group (protection)
$$p\text{-toluidine}\xrightarrow{(CH_3CO)_2O} p\text{-methyl acetanilide }( -NHCOCH_3 )$$

Step 3 : Electrophilic bromination
In acetanilide the $$-NHCOCH_3$$ group is ortho-/para-directing. The para position is already occupied by $$-CH_3$$, so bromine enters one of the two ortho positions; steric factors give the single major product.

Major brominated product after step 3:
$$\text{2-bromo-4-methyl acetanilide}$$

Step 4 : Hydrolytic deprotection
$$\text{2-bromo-4-methyl acetanilide} \xrightarrow{H_3O^+} \text{2-bromo-4-methyl aniline}$$

Thus the final compound (P) is 2-bromo-4-methyl aniline.

Step 5 : Molecular formula and molar mass of (P)

Ring carbons: 6, extra methyl carbon: 1 ⇒ $$C_7$$
Ring hydrogens: 3 (after three substitutions) + methyl H: 3 + amino H: 2 ⇒ $$H_8$$
One nitrogen and one bromine ⇒ $$N_1Br_1$$
Molecular formula: $$C_7H_8BrN$$

Molar mass (g mol$$^{-1}$$): $$7(12) + 8(1) + 80 + 14 = 84 + 8 + 80 + 14 = 186$$

Step 6 : Carius (halogen) analysis
Each mole of compound has one Br atom, giving one mole of $$AgBr$$.
Molar mass of $$AgBr = 108 + 80 = 188\text{ g mol}^{-1}$$

Moles of (P) in 1.0 g sample:
$$n = \frac{1.0}{186} = 5.38\times10^{-3}\;\text{mol}$$

Mass of $$AgBr$$ formed:
$$m = n \times 188 = (5.38\times10^{-3})\times188 \approx 1.01\text{ g}$$

Nearest integer = 1 g.

Final Answer: 1

We are asked to find the mass of benzanilide obtained from the benzoylation of 5.8 g of aniline with 82% yield.

The benzoylation of aniline (Schotten-Baumann reaction) produces benzanilide according to the equation $$\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{CONHC}_6\text{H}_5 + \text{HCl}$$ and since one mole of aniline yields one mole of benzanilide, the stoichiometry is 1:1.

The molar mass of aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$) is calculated as $$6(12) + 5(1) + 14 + 2(1) = 72 + 5 + 14 + 2 = 93 \, \text{g/mol}$$ and the molar mass of benzanilide ($$\text{C}_6\text{H}_5\text{CONHC}_6\text{H}_5$$) is $$13(12) + 11(1) + 14 + 16 = 156 + 11 + 14 + 16 = 197 \, \text{g/mol}$$.

The number of moles of aniline is given by $$\text{Moles of aniline} = \frac{5.8}{93} = 0.06237 \, \text{mol}$$.

Since the reaction is 1:1, the theoretical moles of benzanilide are also 0.06237 mol, and the theoretical mass is $$\text{Theoretical mass} = 0.06237 \times 197 = 12.287 \, \text{g}$$.

Applying the 82% yield gives an actual mass of $$\text{Actual mass} = \text{Theoretical mass} \times \frac{\text{yield}}{100} = 12.287 \times \frac{82}{100}$$ and thus $$\text{Actual mass} = 12.287 \times 0.82 = 10.075 \, \text{g}$$.

Finally, rounding to the nearest integer gives $$\approx 10 \, \text{g}$$.

The answer is 10.

When an ammonium salt contains the oxidising anion of the same oxidation state as the reducing cation (NH$$^{+}_{4}$$), gentle heating often brings about an internal redox (disproportionation) in which nitrogen changes its oxidation state and water is expelled. The exact product depends on the anion present.

Case 1 : Heating NH$$_{4}$$NO$$_{2}$$ at 60-70 $$^\circ$$C
The anion here is nitrite, $$NO_{2}^{-}$$. When warmed, ammonium nitrite undergoes complete internal oxidation-reduction forming molecular nitrogen and water:

$$\mathrm{NH_{4}NO_{2} \;\xrightarrow[\;60\;-\;70^\circ C\;]\; N_{2} + 2\,H_{2}O}$$

Thus compound X is $$\mathbf{N_{2}}$$.

Case 2 : Heating NH$$_{4}$$NO$$_{3}$$ at 200-250 $$^\circ$$C
The anion is nitrate, $$NO_{3}^{-}$$. At about 200-250 $$^\circ$$C, ammonium nitrate similarly decomposes, but because the nitrate ion is a stronger oxidising agent, the oxidation state change stops at +1, producing nitrous oxide (laughing gas) instead of $$N_{2}$$:

$$\mathrm{NH_{4}NO_{3} \;\xrightarrow[\;200\;-\;250^\circ C\;]\; N_{2}O + 2\,H_{2}O}$$

Therefore compound Y is $$\mathbf{N_{2}O}$$.

Comparing with the given options, X = $$N_{2}$$ and Y = $$N_{2}O$$ correspond to Option A.

Answer : Option A which is: $$N_{2}$$ and $$N_{2}O$$

Nylon 6,6 is prepared by condensation of hexamethylenediamine and adipic acid.
Only primary amines give the carbylamine test, therefore the monomer $$X$$ must be the diamine:

$$X \;=\; NH_2\!-\!(CH_2)_6\!-\!NH_2 \;$$ (hexamethylenediamine)

Each molecule of $$X$$ contains two nitrogen atoms.

Number of moles of $$X$$ taken  =  10

Moles of nitrogen atoms present
$$= 10 \times 2 = 20 \text{ mol of N-atoms}$$

In the Dumas method all the nitrogen is converted to molecular nitrogen: $$2\,\text{N-atoms} \rightarrow N_2$$.
Hence, moles of $$N_2$$ formed
$$= \frac{20}{2} = 10 \text{ mol}$$

Mass of $$N_2$$ evolved
$$= 10 \text{ mol} \times 28 \text{ g mol}^{-1} = 280 \text{ g}$$

Therefore, the amount of nitrogen gas obtained is 280 g.

The reaction takes place in two steps involving diazotization followed by an azo coupling reaction.

In the first step, sulphanilic acid reacts with nitrous acid at

$$273\ \text{K}$$

to form the corresponding diazonium salt.

$$\text{Sulphanilic Acid}\xrightarrow{HNO_2,\ 273,\text{K}}\text{Benzene diazonium-4-sulfonate}$$

In the second step, the diazonium salt undergoes coupling with 1-naphthylamine in a weakly acidic medium to produce the azo dye.

$$\text{Benzene diazonium-4-sulfonate}+\text{1-Naphthylamine}\xrightarrow{H^+}\text{4-(4-Sulphophenylazo)-1-naphthylamine}$$

To determine the required mass, we first calculate the molecular formula of the product.

The sulphanilic acid fragment contributes

$$C_6H_5O_3S.$$

The azo linkage contributes

$$N_2.$$

The 1-naphthylamine fragment contributes

$$C_{10}H_8N,$$

since one hydrogen atom of the aromatic ring is replaced during coupling.

Hence, the molecular formula of the final product is

$$C_{16}H_{13}N_3O_3S.$$

Using the given atomic masses,

• Carbon:

$$16\times12=192\ \text{g mol}^{-1}$$

• Hydrogen:

$$13\times1=13\ \text{g mol}^{-1}$$

• Nitrogen:

$$3\times14=42\ \text{g mol}^{-1}$$

• Oxygen:

$$3\times16=48\ \text{g mol}^{-1}$$

• Sulfur:

$$1\times32=32\ \text{g mol}^{-1}$$

Therefore, the molar mass of the product is

$$192+13+42+48+32=327\ \text{g mol}^{-1}.$$

For

$$0.1\ \text{mol},$$

the required mass is

$$\text{Mass}=\text{Moles}\times\text{Molar Mass}$$

$$=0.1\times327$$

$$=32.7\ \text{g}.$$

Approximating to the nearest whole number,

$$32.7\ \text{g}\approx33\ \text{g}.$$

Therefore, the correct answer is

$$\boxed{\text{Option (C) }33.}$$

The carbylamine test (also known as Hofmann's carbylamine test) is a specific chemical test used exclusively for the detection of primary ($$1^\circ$$) amines.

When a primary amine is heated with chloroform ($$\text{CHCl}_3$$) and alcoholic potassium hydroxide ($$\text{KOH}$$), it forms an isocyanide (carbylamine), which has an incredibly foul, offensive smell. Secondary ($$2^\circ$$) and tertiary ($$3^\circ$$) amines do not show this reaction.

$$\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-NC} \ (\text{Isocyanide}) + 3\text{KCl} + 3\text{H}_2\text{O}$$

Because only A and C are primary amines, they are the only ones that show a positive carbylamine test.

We examine each statement in turn.

Statement (A): “Primary amines do not give diazonium salts when treated with $$NaNO_{2}$$ in acidic condition.”
An aromatic primary amine undergoes diazotization as follows: $$ArNH_{2} + NaNO_{2} + 2HCl \xrightarrow{0-5^\circ C} ArN_{2}^{+}Cl^{-} + 2H_{2}O \quad-(1)$$ Aliphatic primary amines give unstable diazonium salts that decompose immediately, but aromatic primary amines do give stable diazonium salts under cold acidic conditions. Therefore the blanket statement “primary amines do not give diazonium salts” is false.

Statement (B): “Aliphatic and aromatic primary amines on heating with $$CHCl_{3}$$ and ethanolic KOH form carbylamines.”
The carbylamine (isocyanide) test is: $$RNH_{2} + CHCl_{3} + 3KOH \xrightarrow{heat} RNC + 3KCl + 3H_{2}O \quad-(2)$$ Both aliphatic and aromatic primary amines give the foul-smelling isocyanide, so statement (B) is correct.

Statement (C): “Secondary and tertiary amines also give carbylamine test.”
Only primary amines have an N-H proton to form the isocyanide. Secondary and tertiary amines do not give this test. Thus statement (C) is false.

Statement (D): “Benzenesulfonyl chloride is known as Hinsberg’s reagent.”
By definition, benzenesulfonyl chloride (PhSO₂Cl) is the Hinsberg reagent used to distinguish amine classes. Hence statement (D) is correct.

Statement (E): “Tertiary amines react with benzenesulfonyl chloride very easily.”
Tertiary amines lack an N-H proton and therefore do not form sulfonamides with PhSO₂Cl. They only form ammonium salts with HCl byproduct: $$R_{3}N + HCl \rightarrow R_{3}NH^{+}Cl^{-} \quad-(3)$$ Thus statement (E) is false.

Therefore the only correct statements are (B) and (D).
The correct answer is Option C: (B) and (D) only.

Hinsberg’s reagent is $$C_6H_5SO_2Cl$$ (benzenesulphonyl chloride). It undergoes nucleophilic substitution with species that possess at least one N-H bond. Hence:

(i) Primary amines (one N-H bond) give $$C_6H_5SO_2\,NH\,R$$, whose remaining hydrogen is acidic, so the product dissolves in aqueous alkali.
(ii) Secondary amines (no acidic hydrogen after substitution but still have one N-H bond initially) give $$C_6H_5SO_2\,NRR'$$, an insoluble sulphonamide.
(iii) Ammonia itself behaves like a primary amine and forms $$C_6H_5SO_2NH_2$$.
(iv) Tertiary amines, nitro compounds, nitriles, etc. lack an N-H bond and therefore do not react.

Among the given compounds:
A - primary amine ⇒ reacts.
B - tertiary amine ⇒ does not react.
C - secondary amine ⇒ reacts.
D - nitro (or any compound without N-H) ⇒ does not react.
E - ammonia ⇒ reacts.

Thus the compounds that react with Hinsberg’s reagent are A, C and E.

Option B which is: A, C and E Only

An azo dye contains the chromophore $$\,\text{-N}=\text{N-}\,$$ that is created by coupling an aromatic diazonium ion with an electron-rich aromatic ring (phenol, aniline, $$\beta$$-naphthol, etc.).
Hence two essential stages are required:

1. Diazo­tisation: convert a primary aromatic amine to a diazonium salt with $$NaNO_2 + HCl$$ at $$0{-}5^{\circ}C$$.
2. Coupling: react that diazonium salt, in weakly alkaline (or mildly acidic) medium, with a strongly activated aromatic ring to give $$Ar{-}N=N{-}Ar'$$, which is coloured.

Option A:
$$C_6H_5NH_2 \xrightarrow[\;0-5^{\circ}C\;]{NaNO_2/HCl} C_6H_5N_2^+Cl^-$$ (benzenediazonium chloride)
$$C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow[\;278{-}283\;K\;]{NaOH} C_6H_5N=N{-}C_6H_4OH\;(\text{p-hydroxy­azobenzene,\;Orange I})$$

The product contains the $$\text{-N}=\text{N-}$$ linkage and is an intensely coloured azo dye. Therefore Option A satisfies both necessary steps and definitely yields an azo dye.

Why the other options fail (briefly):
• They either never generate an aromatic diazonium salt, or
• the intermediate produced is not allowed to couple with an activated ring, or
• further reactions (Sandmeyer, reduction, hydrolysis etc.) destroy the diazonium ion before coupling can occur.
Thus no $$\text{-N}=\text{N-}$$ chromophore, hence no azo dye.

Option A which is: diazotisation of aniline followed by alkaline coupling with phenol, is the correct choice.

To determine the major product $$A$$ of the given reaction sequence, we analyze the transformation occurring at each stage. This sequence is a standard method for synthesizing para-substituted aniline derivatives while preventing unwanted polysubstitution.

The starting compound is nitrobenzene. Treatment with $$Sn/HCl$$ reduces the nitro group $$(-NO_2)$$ to an amino group $$(-NH_2)$$, producing aniline.

The aniline formed is then treated with acetic anhydride $$\left(Ac_2O\right)$$ in the presence of pyridine. This reaction acetylates the amino group to form acetanilide. The acetyl group acts as a protecting group, reducing the strong activating effect of the amino group and thereby preventing tribromination during the subsequent bromination step.

When acetanilide is treated with $$Br_2$$ in acetic acid, electrophilic aromatic substitution takes place. The acetamido group $$(-NHCOCH_3)$$ is an ortho/para-directing group. However, because of steric hindrance near the ortho positions, bromination occurs predominantly at the para position, yielding para-bromoacetanilide as the major product.

Finally, hydrolysis with aqueous $$NaOH$$ removes the acetyl protecting group, regenerating the amino functionality. This converts para-bromoacetanilide into para-bromoaniline.

Thus, the overall sequence is

$$\text{Nitrobenzene} \xrightarrow{Sn/HCl} \text{Aniline} \xrightarrow{Ac_2O,;Pyridine} \text{Acetanilide} \xrightarrow{Br_2,;AcOH} p\text{-Bromoacetanilide} \xrightarrow{NaOH(aq)} p\text{-Bromoaniline}.$$

Therefore, the major product $$A$$ is para-bromoaniline, in which the $$-NH_2$$ and $$-Br$$ groups are positioned para to each other on the benzene ring.

Hence, the correct answer is $$p$$-bromoaniline, corresponding to the first option.

The basicity of an amine depends on how easily the lone pair on the nitrogen can accept a proton. Greater electron density on N and better stabilization of the resulting ammonium ion both increase basicity.

Important effects that decide electron density on nitrogen are:

(i) Inductive effect ($$+I$$ or $$-I$$) from alkyl/aryl substituents.
(ii) Resonance (mesomeric) effect ($$+M$$ or $$-M$$) that may delocalize or donate electron density.
(iii) Hybridisation of the nitrogen (sp$$^3$$ in aliphatic > sp$$^2$$ in aromatic for basicity).

Case 1: Aliphatic amines

$$CH_3NH_2$$ (D) is a primary aliphatic amine.
$$(CH_3)_2NH$$ (E) is a secondary aliphatic amine.

Alkyl groups exhibit a $$+I$$ effect. Two alkyl groups in E push more electron density toward N than one alkyl group in D. In aqueous medium, secondary > primary for basicity (tertiary suffers from poor solvation).
Hence $$\text{basicity: } E \gt D$$.

Case 2: Aromatic amines (anilines)

In aniline, the lone pair on N is partly delocalised into the benzene ring by resonance, making it less available for protonation; therefore aniline is less basic than aliphatic amines.

(A) Aniline - no extra substituent.
(B) p-MeO-Aniline - $$\text{MeO}$$ group shows a strong $$+M$$ (electron-donating) effect, increasing electron density on the ring and on N through resonance, so it is more basic than aniline.
(C) p-NO$$_2$$-Aniline - $$NO_2$$ group shows a powerful $$-M$$ (electron-withdrawing) effect, withdrawing electron density from the ring and N, thus making the amine much less basic.

Therefore, among the three anilines: $$\text{basicity: } B \gt A \gt C$$.

Combining both sets

Overall descending basicity order becomes:
$$(CH_3)_2NH \;(E) \gt CH_3NH_2 \;(D) \gt p\text{-}MeO\text{-Aniline} \;(B) \gt \text{Aniline} \;(A) \gt p\text{-}NO_2\text{-Aniline} \;(C)$$

This matches Option B.

Answer: Option B  (E > D > B > A > C).

To identify the compounds $$[A]$$, $$[B]$$, and $$[C]$$, we analyze each step of the given reaction sequence involving the transformation of an aromatic amine.

The first step involves the reaction of compound $$[A]$$ with $$NaNO_2/HCl$$ at $$273{-}278\text{ K}$$. These conditions correspond to the diazotization reaction, which is characteristic of primary aromatic amines. Therefore, compound $$[A]$$ must be aniline ($$C_6H_5NH_2$$), which is converted into benzene diazonium chloride.

The benzene diazonium chloride then reacts with $$KI$$. In this reaction, the diazonium group is replaced by an iodine atom with the liberation of nitrogen gas, producing iodobenzene. Hence,

$$[B] = C_6H_5I.$$

In the final step, iodobenzene is treated with sodium metal in dry ether. This is the Wurtz-Fittig reaction, in which two phenyl groups couple together to form biphenyl.

Thus,

$$[C] = C_6H_5{-}C_6H_5.$$

Therefore, the identities of the compounds are:

• $$[A] = \text{Aniline } (C_6H_5NH_2)$$
• $$[B] = \text{Iodobenzene } (C_6H_5I)$$
• $$[C] = \text{Biphenyl } (C_6H_5{-}C_6H_5)$$

Hence, the correct sequence corresponds to the green-checked option shown in the given question.

Option A (Incorrect Reaction): When benzene diazonium chloride ($$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$$) is treated with ethanol ($$\text{EtOH}$$), ethanol acts as a mild reducing agent. Instead of undergoing nucleophilic substitution to form phenetole ($$\text{C}_6\text{H}_5\text{OEt}$$), it reduces the diazonium salt to benzene, while ethanol itself gets oxidized to ethanal (acetaldehyde).

The correct chemical equation for this reaction is:

$$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 \uparrow + \text{CH}_3\text{CHO} + \text{HCl}$$

Therefore, Option A is correct.

We need to arrange the following bases in increasing order of basic nature in aqueous solution: $$NH_3$$, $$H_2N-NH_2$$, $$CH_3CH_2NH_2$$, $$(CH_3CH_2)_2NH$$, and $$(CH_3CH_2)_3N$$.

The basicity of amines in aqueous solution depends on two main factors:

(i) The inductive effect (+I effect) of alkyl groups, which increases electron density on nitrogen and increases basicity.

(ii) The solvation of the conjugate acid (ammonium ion) by water molecules through hydrogen bonding, which stabilises the conjugate acid and increases basicity.

Step 1: Compare $$H_2N-NH_2$$ (hydrazine) and $$NH_3$$

In hydrazine, the lone pair on one nitrogen is partially withdrawn by the adjacent electronegative $$NH_2$$ group (due to the -I effect of the $$-NH_2$$ group). This makes hydrazine a weaker base than ammonia.

So: $$H_2N-NH_2 \lt NH_3$$

Step 2: Compare alkylamines with $$NH_3$$

All ethylamines (primary, secondary, and tertiary) are more basic than $$NH_3$$ because the ethyl group donates electron density to nitrogen via the +I effect.

So: $$NH_3 \lt CH_3CH_2NH_2$$, $$(CH_3CH_2)_2NH$$, $$(CH_3CH_2)_3N$$

Step 3: Order among the ethylamines in aqueous solution

In aqueous solution, the observed basicity order among ethylamines is:

$$(CH_3CH_2)_2NH \gt CH_3CH_2NH_2 \gt (CH_3CH_2)_3N$$

This is because:

- Secondary amines like $$(CH_3CH_2)_2NH$$ have a good balance of +I effect (two alkyl groups) and solvation ability (one N-H bond available for hydrogen bonding).

- Primary amines like $$CH_3CH_2NH_2$$ have fewer alkyl groups but better solvation (two N-H bonds).

- Tertiary amines like $$(CH_3CH_2)_3N$$ have the strongest +I effect (three alkyl groups) but poor solvation of the conjugate acid due to steric hindrance and only one site for H-bonding, which reduces their basicity in water.

Step 4: Final order

Combining all comparisons, the increasing order of basic nature in aqueous solution is:

$$$H_2N-NH_2 \lt NH_3 \lt (CH_3CH_2)_3N \lt CH_3CH_2NH_2 \lt (CH_3CH_2)_2NH$$$

Hence, the correct answer is Option C.

We need to determine the final products formed in the given reaction sequence starting from nitrobenzene.

Reaction Sequence Step-by-Step:

• Step 1: Bromination ($$\text{Br}_2 / \text{AcOH}$$)
  Nitrobenzene undergoes electrophilic aromatic substitution. Since the nitro group ($$-\text{NO}_2$$) is strongly deactivating and meta-directing, bromination occurs at the meta-position to form 1-bromo-3-nitrobenzene.
• Step 2: Reduction ($$\text{Sn} / \text{HCl}$$)
  The nitro group ($$-\text{NO}_2$$) is selectively reduced to a primary amino group ($$-\text{NH}_2$$), yielding 3-bromoaniline.
• Step 3: Diazotization ($$\text{NaNO}_2 + \text{HCl}$$)
  The amino group reacts with nitrous acid at low temperature to form a diazonium salt intermediate: 3-bromobenzenediazonium chloride.
• Step 4: Reductive Deamination ($$\text{EtOH}$$)
  Treating the diazonium salt with ethanol ($$\text{CH}_3\text{CH}_2\text{OH}$$) reduces the diazonium group ($$-\text{N}_2^+\text{Cl}^-$$) to a hydrogen atom ($$-\text{H}$$), releasing nitrogen gas ($$\text{N}_2$$). Concurrently, ethanol is oxidized to acetaldehyde ($$\text{CH}_3\text{CHO}$$).

Conclusion:

The final organic products formed at the end of the sequence are bromobenzene and acetaldehyde ($$\text{CH}_3\text{CH}=\text{O}$$).

Answer: Option C

The starting compound is aniline ($$C_6H_5NH_2$$) and the target is a mono-brominated aniline. Because the -$$NH_2$$ group is a very strong ortho/para-directing activator, direct treatment of aniline with $$Br_2$$ would give mainly 2,4,6-tribromo­aniline. Hence we must first protect the amino group, then carry out controlled bromination, and finally remove the protecting group.

Step 1 : Protonation
Concentrated $$H_2SO_4$$ converts aniline into its anilinium hydrogen-sulphate salt $$C_6H_5NH_3^+HSO_4^-$$. This salt is less reactive toward electrophilic substitution and also dissolves readily so that acetylation takes place smoothly in the next step.

Step 2 : Acetylation (protection of $$NH_2$$)
With acetic anhydride ($$Ac_2O$$) the anilinium ion is converted to acetanilide: $$C_6H_5NHCOCH_3$$. The -$$NHCOCH_3$$ group is only moderately activating and mainly directs incoming electrophiles to the para position because the ortho positions are sterically hindered by the bulky $$-NHCOCH_3$$.

Step 3 : Controlled bromination
Bromine ($$Br_2$$) now reacts with acetanilide to give almost exclusively p-bromoacetanilide: $$C_6H_4Br(NHCOCH_3)$$.

Step 4 : Hydrolysis of the protecting group
Heating with water ($$H_2O,\,\Delta$$) hydrolyses p-bromoacetanilide back to p-bromoanilinium ion, releasing acetic acid: $$C_6H_4BrNH_3^+$$.

Step 5 : Liberation of free amine
Finally, aqueous $$NaOH$$ neutralises the anilinium ion and furnishes the free base, p-bromoaniline ($$C_6H_4BrNH_2$$).

Thus the only sequence that carries out protection → bromination → deprotection in the correct order is:

$$H_2SO_4,\; Ac_2O,\; Br_2,\; H_2O(\Delta),\; NaOH$$.

Option B is therefore the correct answer.

The correctness of the two statements has to be checked only on the basis of the availability of the lone-pair on the nitrogen atom. Whenever that lone pair can participate in resonance (delocalisation) its availability for donation to an external proton decreases, hence basicity decreases. Conversely, if the lone pair remains localised on nitrogen it is freely available and the compound is more basic.

Case 1: Lone pair involved in resonance with a carbonyl group (amide).
In an amide, the nitrogen is directly attached to the $$C=O$$ group. The lone pair on nitrogen overlaps with the vacant $$\pi^{*}$$ orbital of the carbonyl carbon giving the resonating structures
$$R-C(=O)-NH_2 \rightleftharpoons R-C(O^{-})=NH_2^{+}$$
Because of this conjugation, the lone pair is not freely available for protonation, so amides are very weak bases. Statement I—which attributes the low basicity of amides to delocalisation of the lone pair—is therefore correct.

Case 2: Lone pair involved in resonance with an aromatic ring (aniline).
In aniline, the nitrogen lone pair can conjugate with the benzene ring to give structures such as
$$\mathrm{Ph-NH_2 \;\rightleftharpoons\; Ph^{-}=NH_2^{+}}$$
Again the lone pair is partially withdrawn from nitrogen, so aniline is less basic than aliphatic amines. Statement II—which explains reduced basicity of aniline (or its acylated derivative) through similar resonance—is also correct.

Since both statements correctly explain the lowered basicity by the same underlying factor—resonance delocalisation of the nitrogen lone pair—both are true.

Option C which is: Both Statement I and Statement II are correct

Direct bromination of aniline does not produce 3,4,5-tribromoaniline because the strongly activating $$-NH_2$$ group directs substitution to the ortho and para positions, giving 2,4,6-tribromoaniline.

Therefore, an indirect synthetic route must be followed.

The correct sequence is option (C).

In the first step, p-nitroaniline is treated with excess bromine in acetic acid.

The amino group ($-NH_2$) is a strong ortho/para director and dominates the directing influence of the nitro group.

Since the para position is already occupied by $$-NO_2$$, bromination occurs at both ortho positions.

The product formed is 2,6-dibromo-4-nitroaniline.

In the second step, the amino group is converted into a diazonium salt using

$$NaNO_2/HCl.$$

The diazonium group is then replaced by bromine through the Sandmeyer reaction using

$$CuBr.$$

This transformation replaces the original $$-NH_2$$ group with a bromine atom, producing 3,4,5-tribromonitrobenzene after appropriate numbering.

In the final step, the nitro group is reduced using

$$Sn/HCl,$$

converting

$$-NO_2 \longrightarrow -NH_2.$$

The final product obtained is 3,4,5-tribromoaniline.

The remaining options do not produce the desired compound:

• Option (A): Bromination of nitrobenzene followed by reduction gives only 3-bromoaniline.
• Option (B): Bromination of benzene followed by treatment with ammonia does not lead to 3,4,5-tribromoaniline.
• Option (D): Direct bromination of aniline gives 2,4,6-tribromoaniline because of the strong ortho/para directing effect of the amino group.

Hence, the correct sequence is option (C).

(A) Amide $$\rightarrow$$ (III) $$LiAlH_4$$

• Why this reagent: $$LiAlH_4$$ is a powerful nucleophilic reducing agent capable of reducing the highly stable $$C=O$$ bond of an amide all the way to a $$CH_2$$ group to yield a primary amine.
• Why others fail: $$H_2/Ni$$ is usually not strong enough to reduce amides efficiently under standard conditions. $$Sn/HCl$$ is specifically used for nitro groups and won't touch the amide carbonyl. $$NaOH$$ would simply cause hydrolysis to a carboxylate salt and ammonia, not reduction.

(B) Nitrobenzene $$\rightarrow$$ (IV) $$Sn, HCl$$

• Why this reagent: This is a classic laboratory method for preparing aniline. The metal/acid combination ($$Sn/HCl$$ or $$Fe/HCl$$) provides the protons and electrons necessary to reduce $$-NO_2$$ to $$-NH_2$$.
• Why others fail: While $$H_2/Ni$$ could technically work, $$Sn/HCl$$ is the preferred "textbook" match for aromatic nitro compounds in this list. $$LiAlH_4$$ is avoided for aromatic nitro compounds because it can lead to azo compounds (coupling) rather than pure amines.

(C) Nitrile $$\rightarrow$$ (II) $$H_2/Ni$$

• Why this reagent: Catalytic hydrogenation is the most efficient way to add four hydrogen atoms across the $$C\equiv N$$ triple bond to produce a primary amine ($$R-CH_2NH_2$$).
• Why others fail: $$NaOH$$ would hydrolyze the nitrile to an amide and then a carboxylic acid. $$Sn/HCl$$ is not the standard reagent for full nitrile reduction (it is used in the Stephen reduction to stop at the aldehyde, but only with $$SnCl_2$$).

(D) Phthalimide $$\rightarrow$$ (I) $$NaOH$$

• Why this reagent: This refers to the final step of the Gabriel Phthalimide Synthesis. Once an alkyl group is attached to the nitrogen, base-catalyzed hydrolysis ($$NaOH$$) is used to break the cyclic imide bonds and release the primary amine.
• Why others fail: The goal here is "cleavage" to release the amine, not "reduction" of the aromatic ring or carbonyls. Reducing agents like $$LiAlH_4$$ or $$H_2/Ni$$ would attack the carbonyl groups of the phthalimide ring itself, destroying the starting material instead of releasing the desired amine.

The given sequence involves two standard transformations:
(i) $$\text{alkyl halide} \xrightarrow{KCN,\;\text{alc.},\;\Delta} \text{nitrile}$$ (nucleophilic substitution)
(ii) $$\text{nitrile} \xrightarrow{LiAlH_4/\;\text{ether}} \text{primary amine}$$ (complete reduction)

Step-1 (Substitution):
Starting alkyl halide $$A$$ is $$\left(CH_3\right)_2CHBr$$ (2-bromopropane).
Treatment with alcoholic $$KCN$$ gives the nitrile $$B$$ by $$\mathrm{S_N2}$$ substitution:

$$\left(CH_3\right)_2CHBr \;+\; KCN \;\longrightarrow\; \left(CH_3\right)_2CHCN \;+\; KBr$$

Step-2 (Reduction):
The nitrile $$B$$ is reduced with $$LiAlH_4$$ to a primary amine $$C$$:

$$\left(CH_3\right)_2CHCN \;\xrightarrow{LiAlH_4/\text{ether}}\; \left(CH_3\right)_2CHCH_2NH_2$$

Thus the major product $$C$$ is isobutylamine.

Counting $$sp^3$$ hybridised carbons in $$C$$:
• Two methyl carbons: $$2$$
• One secondary carbon (attached to two methyls and $$CH_2$$): $$1$$
• One methylene carbon attached to $$NH_2$$: $$1$$

Total $$sp^{3}$$ hybridised carbon atoms $$= 2 + 1 + 1 = 4$$

Therefore, the required number is 4.

We need to determine whether aminobenzene and aniline are the same compound.

What is aniline?

Aniline is an aromatic amine with the molecular formula $$C_6H_5NH_2$$. It consists of a benzene ring ($$C_6H_5-$$) directly bonded to an amino group ($$-NH_2$$).

What is aminobenzene?

The name "aminobenzene" literally means a benzene ring with an amino ($$-NH_2$$) group attached. Breaking down the name: "amino" = $$-NH_2$$ group, "benzene" = $$C_6H_6$$ ring. The structure is $$C_6H_5NH_2$$.

Comparison:

Both names refer to the same compound with structure $$C_6H_5NH_2$$:

- "Aniline" is the common (trivial) name.

- "Aminobenzene" is the substitutive IUPAC-style name (also called benzenamine in strict IUPAC nomenclature).

Therefore, Statement I is correct (they are the same compound) and Statement II is incorrect (they are NOT different compounds).

The correct answer is Option B: Statement I is correct but Statement II is incorrect.

During electrophilic bromination, aniline first donates its lone-pair electrons to the aromatic ring through resonance. This makes the ring highly activated toward electrophiles and directs the attack exclusively to the ortho and para positions.

The key mechanistic intermediate of any aromatic electrophilic substitution is the σ-complex (also called the arenium ion). For aniline these σ-complexes must satisfy two conditions:
1. The electrophile attaches at an ortho or para carbon (because $$-NH_2$$ is an ortho/para director).
2. At least one resonance form of the σ-complex must place the positive charge on the nitrogen atom. Such resonance delocalisation greatly stabilises the intermediate.

If the electrophile were to attack at the meta position, none of the resonance structures would be able to shift the positive charge onto the nitrogen atom; the charge would remain on ring carbons only. Consequently, the meta σ-complex is far less stabilised and is not formed under ordinary conditions.

Among the four structures given in the options, the three that correspond to ortho or para attack all allow resonance forms in which the positive charge resides on nitrogen, so they are genuine intermediates. The remaining structure (Option C) shows the electrophile attached at the meta position; its resonance forms keep the positive charge on ring carbons and never on nitrogen. Therefore it cannot arise during bromination of aniline.

Hence, the arenium ion that is not involved in the bromination mechanism of aniline is the meta σ-complex shown in Option C.

Option C which is: (meta σ-complex)

We need to identify the compound $$B$$ formed in the given two-step reaction sequence starting from 1,4-dichlorobutane.

Key Concept: When an alkyl halide reacts with excess ammonia ($$NH_3$$), it undergoes nucleophilic substitution ($$S_N2$$) to form an amine salt. Treating this salt with a strong base like sodium hydroxide ($$NaOH$$) liberates the free amine.

Step 1: Reaction of 1,4-dichlorobutane with excess ammonia to form intermediate A

In this step, both terminal chlorine atoms of $$Cl-(CH_2)_4-Cl$$ are substituted by ammonia molecules. Because ammonia is used in excess, it prevents further alkylation and yields the corresponding dicationic ammonium salt as intermediate $$A$$:

$$Cl-(CH_2)_4-Cl + 2NH_3 \rightarrow [^{-}ClH_3N^{+}-(CH_2)_4-^{+}NH_3Cl^{-}]$$

Thus, intermediate $$A$$ is butane-1,4-diaminium chloride.

Step 2: Reaction of intermediate A with sodium hydroxide to form B

When the ammonium salt intermediate $$A$$ is treated with $$NaOH$$, a neutralization reaction takes place. The hydroxide ions ($$OH^-$$) deprotonate the ammonium groups, releasing the free diamine, water, and sodium chloride:

$$[^{-}ClH_3N^{+}-(CH_2)_4-^{+}NH_3Cl^{-}] + 2NaOH \rightarrow H_2N-(CH_2)_4-NH_2 + 2H_2O + 2NaCl$$

The free amine $$B$$ formed is butane-1,4-diamine (putrescine), which has the structural formula $$H_2N-(CH_2)_4-NH_2$$.

Therefore, compound $$B$$ is $$H_2N-(CH_2)_4-NH_2$$, which corresponds to Option B.

Answer: Option B — $$H_2N-(CH_2)_4-NH_2$$

We need to identify gas $$X$$ that gives a brown precipitate with Nessler's reagent.

Nessler's reagent is an alkaline solution of potassium mercuric iodide ($$K_2[HgI_4]$$). It is a specific test reagent for ammonia ($$NH_3$$) and ammonium ions ($$NH_4^+$$).

When ammonia gas is passed through Nessler's reagent, the following reaction occurs:

$$NH_3 + 2K_2[HgI_4] + 3KOH \rightarrow \underset{\text{(brown ppt)}}{Hg_2ONH_2I} + 7KI + 2H_2O$$

The brown precipitate formed is known as the iodide of Millon's base ($$Hg_2ONH_2I$$).

Let us verify by checking the other options:

• $$H_2S$$ gives a black precipitate ($$HgS$$) with Nessler's reagent, not brown.
• $$CO_2$$ does not react with Nessler's reagent to give a precipitate.
• $$Cl_2$$ does not give a brown precipitate with Nessler's reagent.

Therefore, the correct answer is Option 3: NH$$_3$$.

Statement I: The $$-NH_2$$ group in aniline is ortho and para directing and a powerful activating group.

Analysis: The $$-NH_2$$ group is indeed an ortho/para director due to the lone pair on nitrogen which donates electron density to the ring via resonance. It is a powerful activating group — it strongly activates the benzene ring toward electrophilic aromatic substitution.

Statement I is correct.

Statement II: Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

Analysis: Aniline does NOT undergo Friedel-Craft's reaction because the lone pair on the $$-NH_2$$ group coordinates with the Lewis acid catalyst ($$AlCl_3$$), forming a salt. This effectively deactivates the catalyst and also makes the ring electron-poor (as the nitrogen lone pair is no longer available for resonance with the ring).

Statement II is correct.

Conclusion: Both statements are correct.

The answer is Option A: Both Statement I and Statement II are correct.

In excess CH3-MgBr, firstly the ester group gets reduced to substituted alcohol then the Grignard's reagent attacks on the CN group first converting it to imine and then upon addition of water the result is formation of ketone. Since, the aforementioned imine salt is stable it does not undergo further Grignard reagent attacks and upon hydrolysis Oxygen being more electronegative takes up the double bond resulting in removal of -NH2 group in form of Ammonia. 

We need to evaluate both statements about the Hinsberg test (reaction with p-toluenesulfonyl chloride, TsCl).

Statement (I): All the following compounds react with TsCl: $$C_6H_5NH_2$$ (aniline, primary amine), $$(C_6H_5)_2NH$$ (diphenylamine, secondary amine), $$(C_6H_5)_3N$$ (triphenylamine, tertiary amine).

In the Hinsberg test:

- Primary amines react with TsCl to form sulfonamides (N-monosubstituted).

- Secondary amines react with TsCl to form sulfonamides (N,N-disubstituted).

- Tertiary amines do not react with TsCl because they have no N-H bond to undergo substitution.

Since $$(C_6H_5)_3N$$ (a tertiary amine) does not react with TsCl, Statement I is false.

Statement (II): Their products in the above reaction are soluble in aqueous NaOH.

For primary amines, the product (N-monosubstituted sulfonamide) has an acidic N-H and is soluble in NaOH. For secondary amines, the product (N,N-disubstituted sulfonamide) has no acidic N-H and is insoluble in NaOH. Since not all products are soluble in NaOH, Statement II is also false.

The correct answer is Option 4: Both Statement I and Statement II are false.

Statement I deals with the sulfonation of aniline and the Lassaigne’s test for sulfur. Sulfonation of aniline with conc. $$H_2SO_4$$ at 453-473 K proceeds as follows:
$$C_6H_5NH_2 + H_2SO_4 \xrightarrow{453-473\,K} p\text{-}H_2NC_6H_4SO_3H$$ The product is p-aminobenzenesulfonic acid (sulfanilic acid).

In the Lassaigne’s test for sulfur, the sodium fusion extract is treated with sodium nitroprusside reagent. The presence of sulfur gives a blood-red coloration. Since p-aminobenzenesulfonic acid contains a sulfonic acid group (-SO₃H), it yields a blood-red color in this test. Therefore, Statement I is true.

Statement II concerns Friedel-Crafts reactions of aniline. Aniline’s -lone pair forms a complex with the Lewis acid catalyst $$AlCl_3$$:
$$C_6H_5NH_2 + AlCl_3 \longrightarrow [C_6H_5NH_2\cdot AlCl_3] \longrightarrow [C_6H_5NH_3]^+ + AlCl_4^-$$ In this complex, the nitrogen carries a positive charge, withdrawing electron density from the benzene ring and making it less reactive toward electrophilic substitution. Hence aniline acts as a deactivating group under Friedel-Crafts conditions, and no alkylation or acylation occurs. Therefore, Statement II is true.

Both Statement I and Statement II are true. The correct answer is Option D.

1.Formation of Product A

• Reaction: Benzene undergoes nitration using $$\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4$$ to form nitrobenzene.
• Reduction: Nitrobenzene is then reduced by $$\text{Sn/HCl}$$ to convert the nitro group ($$-\text{NO}_2$$) into an amino group ($$-\text{NH}_2$$).
• Product A: Aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$).

2. Formation of Product B

• Diazotization: Aniline reacts with $$\text{NaNO}_2 + \text{HCl}$$ at a low temperature ($$0\text{-}5^\circ\text{C}$$ / $$273\text{-}278\text{K}$$) to form the stable benzene diazonium chloride salt ($$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$$).
• Azo Coupling: The diazonium ion acts as an electrophile and couples with phenol at its highly reactive, less sterically hindered para-position.
• Product B: $$p$$-Hydroxyazobenzene (an orange azo dye).

Aniline reacts with bromine water to give 2,4,6-tribromoaniline (white precipitate).

$$ C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr $$

Moles of aniline.

Molar mass of aniline = 93 g/mol. Moles = $$\frac{9.3}{93} = 0.1$$ mol.

Theoretical yield of 2,4,6-tribromoaniline.

Molar mass of $$C_6H_2Br_3NH_2$$ = 72 + 4 + 14 + 240 = 330 g/mol.

Theoretical yield = $$0.1 \times 330 = 33$$ g.

Percentage yield.

$$ \text{\% yield} = \frac{26.4}{33} \times 100 = 80\% $$

The answer is 80.

The reaction of aniline with acetic anhydride:

$$C_6H_5NH_2 + (CH_3CO)_2O \to C_6H_5NHCOCH_3 + CH_3COOH$$

Molar mass of aniline = $$6(12) + 7(1) + 14 = 93$$ g/mol.

Molar mass of acetanilide = $$8(12) + 9(1) + 14 + 16 = 135$$ g/mol.

Moles of aniline = $$\frac{9.3}{93} = 0.1$$ mol.

Moles of acetanilide = 0.1 mol (1:1 ratio).

Mass of acetanilide = $$0.1 \times 135 = 13.5$$ g = $$135 \times 10^{-1}$$ g.

The answer is $$\boxed{135}$$.

We need to find the molar mass of amine X, prepared by ammonolysis of benzyl chloride, that gives a clear solution with the Hinsberg reagent (p-toluenesulphonyl chloride).

The Hinsberg test distinguishes between primary, secondary, and tertiary amines: (i) Primary amines react with p-toluenesulphonyl chloride to form a sulphonamide that is soluble in NaOH (acidic NH, dissolves in alkali). (ii) Secondary amines form a sulphonamide that is insoluble in NaOH (precipitate). (iii) Tertiary amines do not react — the solution remains clear.

Since the solution remains clear with p-toluenesulphonyl chloride, the amine X is a tertiary amine.

In ammonolysis of benzyl chloride (C$$_6$$H$$_5$$CH$$_2$$Cl) with excess benzyl chloride, the following sequence occurs: $$C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2$$ (primary), then $$C_6H_5CH_2Cl + C_6H_5CH_2NH_2 \rightarrow (C_6H_5CH_2)_2NH$$ (secondary), and finally $$C_6H_5CH_2Cl + (C_6H_5CH_2)_2NH \rightarrow (C_6H_5CH_2)_3N$$ (tertiary).

Therefore, the tertiary amine is tribenzylamine: $$(C_6H_5CH_2)_3N$$.

The molecular formula of tribenzylamine is $$C_{21}H_{21}N$$, and substituting into the molar mass calculation gives $$M = 21(12) + 21(1) + 14 = 252 + 21 + 14 = 287 \, \text{g/mol}$$.

Option A: 287 g mol$$^{-1}$$.

In this reaction, an H atom (mass 1) on the $$-\text{NH}_2$$ group is replaced by an acetyl group $$-\text{COCH}_3$$ (mass 43). The net increase in molar mass per amino group is $$43 - 1 = 42$$ g/mol.

Since the product’s molar mass increases from 108 to 192, the total increase in molar mass is $$\Delta M = 192 - 108 = 84 \, \text{g/mol}$$.

Substituting into the relation for the number of amino groups gives $$n = \frac{\Delta M}{42} = \frac{84}{42} = 2$$.

Therefore, the compound $$X$$ has 2 amino groups, corresponding to Option 2.

Aniline is first converted to acetanilide by acetylation. The balanced equation is

$$C_6H_5NH_2 + CH_3COOH \;(\text{or } (CH_3CO)_2O)\;\rightarrow\; C_6H_5NHCOCH_3 + H_2O$$

One mole of aniline gives one mole of acetanilide, so the reaction follows a 1 : 1 molar ratio.

Step 1 - Molar mass of aniline
$$M_{\text{aniline}} = 6(12) + 7(1) + 14 = 72 + 7 + 14 = 93 \text{ g mol}^{-1}$$

Step 2 - Moles of aniline taken
Given mass of aniline = $$6.55\text{ g}$$
$$n_{\text{aniline}} = \frac{6.55}{93} = 0.07043 \text{ mol}$$ (keep four significant figures for accuracy)

Step 3 - Molar mass of acetanilide
Acetanilide formula: $$C_8H_9NO$$
$$M_{\text{acetanilide}} = 8(12) + 9(1) + 14 + 16 = 96 + 9 + 14 + 16 = 135 \text{ g mol}^{-1}$$

Step 4 - Maximum (theoretical) mass of acetanilide
Moles of acetanilide produced = moles of aniline (1 : 1 ratio): $$0.07043 \text{ mol}$$
$$m_{\text{acetanilide}} = 0.07043 \times 135 = 9.517 \text{ g}$$

Step 5 - Expressing the answer as “×10−1 g”
Write 9.517 g as a multiple of $$10^{-1}$$ g:
$$9.517 \text{ g} = 95.17 \times 10^{-1} \text{ g}$$

Rounded to the nearest whole number (as required in JEE integer-type answers), the coefficient is $$95$$.

Therefore, the maximum amount of acetanilide obtainable is $$\boxed{95 \times 10^{-1}\text{ g}}$$.

Find the mass of aniline yellow formed from 279 g of aniline with one equivalent of benzenediazonium chloride.

Since aniline reacts with benzenediazonium chloride to give p-aminoazobenzene (aniline yellow) and HCl, the reaction can be written as $$C_6H_5NH_2 + C_6H_5N_2^+Cl^- \rightarrow C_6H_5N=NC_6H_4NH_2 + HCl.$$

The molar mass of aniline ($$C_6H_5NH_2$$) is 93 g/mol, so the number of moles in 279 g is $$\mathrm{Moles} = \frac{279}{93} = 3\ \text{mol}.$$

One equivalent of benzenediazonium chloride reacts with each mole of aniline in a 1:1 ratio, so 3 mol of aniline yields 3 mol of the azo product.

The molar mass of aniline yellow ($$C_{12}H_{11}N_3$$) is $$12\times12 + 11\times1 + 3\times14 = 144 + 11 + 42 = 197\ \text{g/mol},$$ and hence the mass of product is $$3 \times 197 = 591\ \text{g}.$$

The correct answer is 591.

Hinsberg's reagent reacts with primary ($$1^{\circ }$$) and secondary ($$2^{\circ }$$ amines). It does not react with tertiary ($$3^{\circ }$$ amines), amides, or diazonium salts under standard conditions.Analysis of the Compounds:

1. Benzene diazonium chloride: Does not react.
2. Benzimide (Amide): Does not react.
3. Aniline ($$1^{\circ }$$ amine): Reacts.
4. $$(N)$$-phenylpiperidine ($$3^{\circ }$$ amine): Does not react.
5. Benzyl methyl amine ($$2^{\circ }$$ amine): Reacts.
6. Ethane-1,2-diamine ($$1^{\circ }$$ amine): Reacts.
7. Piperidine ($$2^{\circ }$$ amine): Reacts.
8. Pyridine ($$3^{\circ }$$ amine): Does not react.
9. Propan-1-amine ($$1^{\circ }$$ amine): Reacts.
10. Urea (Amide): Does not react

We have 9.3 g of pure aniline which undergoes diazotisation followed by coupling with phenol to give an orange dye.

In the diazotisation step aniline reacts with nitrous acid (generated in situ from $$NaNO_2 + HCl$$) at 0-5 °C to form benzenediazonium chloride:

$$C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$

In the coupling step the diazonium salt couples with phenol in a weakly basic medium to produce p-hydroxyazobenzene (an orange azo dye):

$$C_6H_5N_2^+Cl^- + C_6H_5OH \rightarrow C_6H_5-N=N-C_6H_4-OH + HCl$$

To calculate the moles of aniline, we use its molar mass: $$6(12) + 5(1) + 14 + 2(1) = 93\text{ g/mol}$$, which gives $$\text{Moles of aniline} = \frac{9.3}{93} = 0.1\text{ mol}$$.

Since the stoichiometry is 1:1, one mole of aniline yields one mole of dye. The molar mass of p-hydroxyazobenzene ($$C_{12}H_{10}N_2O$$) is $$12(12) + 10(1) + 2(14) + 16 = 198\text{ g/mol}$$. Thus, the mass of dye produced is $$0.1 \times 198 = 19.8 \approx 20\text{ g}$$. Therefore, 20 g of the orange dye is obtained (to the nearest integer).

Ethylamine + NaNO₂/HCl → Diazonium salt → N₂ + Ethanol

C₂H₅NH₂ → N₂

At STP: Volume of N₂ = 2.24 L, so moles of N₂ = 2.24/22.4 = 0.1 mol

Moles of ethylamine = 0.1 mol (1:1 ratio)

Mass = 0.1 × 45 = 4.5 g = 45 × 10⁻¹ g

The answer is 45.

Explanation

The starting compound is $$N,N$$-dimethylaniline, which is a tertiary aromatic amine.

In the presence of $$NaNO_2$$ and mineral acid $$HX$$, nitrous acid is generated in situ.

$$NaNO_2 + HX \rightarrow HNO_2$$

In acidic medium, nitrous acid generates the nitrosonium ion:

$$HNO_2 + H^+ \rightarrow NO^+ + H_2O$$

The electrophile in this reaction is:

$$NO^+$$

Since the $$-N(CH_3)_2$$ group is strongly electron-donating, it activates the benzene ring towards electrophilic substitution.

Due to steric hindrance, substitution occurs predominantly at the para position to give $$p$$-nitroso-$$N,N$$-dimethylaniline.

Statement A is correct because the product formed is the para-nitroso derivative of $$N,N$$-dimethylaniline.

Statement C is correct because nitrous acid is unstable at room temperature. Therefore, the reaction is carried out at low temperature, typically $$0^\circ\text{C}$$ to $$5^\circ\text{C}$$.

Statement D is correct because the electrophile responsible for the reaction is the nitrosonium ion:

$$NO^+$$

Statement B is incorrect.

Secondary amines undergo $$N$$-nitrosation to form $$N$$-nitrosoamines.

Tertiary aromatic amines do not undergo $$N$$-nitrosation because the nitrogen atom has no replaceable hydrogen.

Instead, they undergo $$C$$-nitrosation on the aromatic ring to form para-nitroso substituted products.

Hence, the stable product is a $$C$$-nitroso compound and not an $$N$$-nitroso ammonium compound.

Therefore, the incorrect statement is Option B.

Benzyl bromide reacts with silver cyanide ((AgCN)) to form benzyl isocyanide because (AgCN) is predominantly covalent and the nucleophilic attack occurs through the nitrogen atom.

The reaction is

$$C_6H_5CH_2Br+AgCN\rightarrow C_6H_5CH_2NC+AgBr.$$

Benzylamine reacts with chloroform and aqueous potassium hydroxide in the carbylamine reaction. Since benzylamine is a primary amine, it gives the corresponding isocyanide.

The reaction is

$$C_6H_5CH_2NH_2+CHCl_3+3KOH\rightarrow C_6H_5CH_2NC+3KCl+3H_2O.$$

(N)-Methylbenzylamine is a secondary amine and therefore does not undergo the carbylamine reaction, so it does not produce benzyl isocyanide.

Benzyl tosylate reacts with potassium cyanide ((KCN)) to give benzyl cyanide because (KCN) is ionic and the nucleophilic attack occurs through the carbon atom of the cyanide ion.

The reaction is

$$C_6H_5CH_2OTs+KCN\rightarrow C_6H_5CH_2CN+KOTs.$$

Hence, benzyl isocyanide is formed in

$$\boxed{\text{Option A and Option B}}.$$

Statement I: Pure aniline and other arylamines are usually colourless.

Pure aniline and other aromatic amines are colourless in their pure state. Therefore, Statement I is correct.

Statement II: Arylamines get coloured on storage due to atmospheric reduction.

Arylamines become coloured on prolonged storage because they undergo slow oxidation in the presence of atmospheric oxygen and light. The colour formation is due to atmospheric oxidation, not reduction.

Hence, Statement II is incorrect.

Therefore:

• Statement I is correct.

• Statement II is incorrect.

Hence, the correct answer is

$$\boxed{\text{Option C: Statement I is correct but Statement II is incorrect.}}$$

Order of basic strength of methyl substituted amines in aqueous medium.

Basicity in aqueous solution depends on both the inductive effect of methyl groups, which increases basicity, and steric or solvation effects, since bulky groups hinder solvation of the conjugate acid and thus lower basicity.

The experimentally observed order in aqueous medium is:

$$ \text{Me}_2\text{NH} > \text{MeNH}_2 > \text{Me}_3\text{N} > \text{NH}_3 $$

Me$$_2$$NH has a strong inductive effect and its conjugate acid is well solvated through two N-H bonds, MeNH$$_2$$ has a moderate inductive effect with good solvation, and although Me$$_3$$N has the strongest inductive effect, its conjugate acid is poorly solvated due to having only one N-H bond.

The correct answer is Option A: $$\boxed{\text{Me}_2\text{NH} > \text{MeNH}_2 > \text{Me}_3\text{N} > \text{NH}_3}$$.

1. Formation of Product A (Bromination)

• Reagents: $$\text{Br}_2\text{ / AcOH}$$ (Bromine in acetic acid)
• Reaction Type: Electrophilic Aromatic Substitution (EAS)
• Mechanism: The $$-\text{NHCOCH}_3$$ (acetamido) group attached to the benzene ring has a lone pair of electrons on the nitrogen atom. This makes it an ortho/para-directing activator.

  However, due to the steric bulk of the acetamido group, substitution at the ortho-position is heavily hindered. Therefore, electrophilic substitution predominantly occurs at the less hindered para-position.

• Product A: 4-bromoacetanilide (or p-bromoacetanilide)

2. Formation of Product B (Reduction)

• Reagent: $$\text{LiAlH}_4$$ (Lithium Aluminum Hydride)
• Reaction Type: Amide Reduction
• Mechanism:

  $$\text{LiAlH}_4$$ is a powerful reducing agent that selectively reduces the carbonyl carbon ($$\text{-C=O}$$) of amides entirely into a methylene group ($$\text{-CH}_2\text{-}$$). Unlike ester or ketone reductions that yield alcohols, amides are converted directly into amines.

  Thus, the $$-\text{NH-CO-CH}_3$$ amide group is reduced to a $$-\text{NH-CH}_2\text{-CH}_3$$ secondary amine group.

• Product B: $$N$$-ethylaniline

Compound A has molecular formula C$$_4$$H$$_{11}$$N. The given clues are:

1. Optical activity: A has a chiral center.

2. Gives N$$_2$$ with HNO$$_2$$: A is a primary amine ($$-NH_2$$).

3. Reacts with PhSO$$_2$$Cl to give a product soluble in KOH: This is the Hinsberg test result for a primary amine (the sulfonamide product has an acidic N-H that dissolves in KOH).

So A is an optically active primary amine with formula C$$_4$$H$$_{11}$$N, which is C$$_4$$H$$_9$$NH$$_2$$.

Among the isomers of C$$_4$$H$$_9$$NH$$_2$$:

- n-Butylamine: no chiral center

- Isobutylamine: no chiral center

- sec-Butylamine (2-aminobutane): CH$$_3$$CH(NH$$_2$$)CH$$_2$$CH$$_3$$ — has a chiral center at C-2 ✔

- tert-Butylamine: no chiral center

Therefore, A is sec-Butylamine (2-aminobutane).

The reaction is an azo coupling reaction.

The diazonium salt of sulfanilic acid acts as the electrophile, while 1-naphthylamine acts as the activated aromatic nucleophile.

Coupling occurs predominantly at the para position with respect to the $$-NH_2$$ group of 1-naphthylamine, forming an azo compound containing the linkage $$-N=N-$$.

The product possesses an extended conjugated system involving the benzene ring, the azo bridge, and the naphthalene ring.

This extensive $$\pi$$-conjugation lowers the energy gap between molecular orbitals, causing absorption in the visible region and giving rise to a strongly coloured azo dye.

The azo dye formed from diazotized sulfanilic acid and 1-naphthylamine is the dye produced in the Griess reaction and is characteristically red in colour.

Hence, the colour of the product is

$$\boxed{\text{Red}}$$

Therefore, the correct answer is

$$\boxed{\text{Option C}}$$

Step 1: Aniline reacts with NaNO₂ at 0-5°C (diazotization):

$$C_6H_5NH_2 + NaNO_2 + HCl \xrightarrow{0-5°C} C_6H_5N_2^+Cl^-$$

Product 'A' is benzenediazonium chloride.

Step 2: Benzenediazonium salt couples with N,N-dimethylaniline (azo coupling):

The diazonium salt attacks the para position of N,N-dimethylaniline (which is a strongly activated ring).

Product 'B' is an azo compound: p-(N,N-dimethylamino)azobenzene (also known as butter yellow or methyl yellow).

Structure: C₆H₅-N=N-C₆H₄-N(CH₃)₂ (with the -N=N- azo bridge connecting the two aromatic rings, NMe₂ at para position).

This matches option 3.

1. Step 1: Intermolecular Nucleophilic Acyl Substitution (Carbamate Formation)

   The primary amino group ($$\text{--NH}_2$$) of the starting material is a stronger nucleophile than the primary hydroxyl group ($$\text{--CH}_2\text{OH}$$). It attacks the carbonyl carbon of diethyl carbonate ($$\text{EtO--CO--OEt}$$), displacing one ethoxide ($$\text{--OEt}$$) leaving group to form an ethyl carbamate intermediate:

   $$\text{R--NH}_2 + \text{EtO--CO--OEt} \rightarrow \text{R--NH--CO--OEt} + \text{EtOH}$$



2. Step 2: Intramolecular Cyclization (Oxazolidin-2-one Ring Closure)

   The adjacent hydroxyl group ($$\text{--CH}_2\text{OH}$$) now acts as an internal nucleophile. It attacks the carbonyl group of the newly formed carbamate line, displacing the second ethoxide group ($$\text{--OEt}$$) through an intramolecular nucleophilic acyl substitution:

   $$\text{R--NH--CO--OEt} \xrightarrow{\text{Intramolecular}} \text{5-membered cyclic carbamate (oxazolidin-2-one)} + \text{EtOH}$$

Stereochemistry Retention:

Crucially, throughout both nucleophilic substitution steps, neither bond connected to the chiral center is ever broken or altered. Because the chiral center remains completely undisturbed, its original configuration (wedge/dash orientation of the tert-butyl group) is fully retained in the final product.

The reaction of an amide with bromine in the presence of aqueous potassium hydroxide is known as the Hofmann bromamide degradation (Hofmann rearrangement). In this reaction, an amide is converted into a primary amine containing one carbon atom fewer than the original amide.

The general reaction is

$$RCONH_2+Br_2+4KOH\rightarrow RNH_2+K_2CO_3+2KBr+2H_2O.$$

For propanamide ((CH_3CH_2CONH_2)), the carbonyl carbon is removed during the rearrangement, producing a primary amine with a two-carbon chain.

Thus,

$$CH_3CH_2CONH_2\xrightarrow{Br_2/KOH}CH_3CH_2NH_2.$$

Hence, the product formed is ethylamine.

Therefore, the correct answer is

$$\boxed{\text{Ethylamine }(CH_3CH_2NH_2)}.$$

The reaction proceeds in two distinct steps.

In the first step, $$Br_2$$ and $$NaOH$$ (on heating) bring about the Hoffmann bromamide degradation.

This reaction converts a primary amide ($-CONH_2$) into a primary amine ($-NH_2$) with the loss of one carbon atom.

For the given substrate,

$$-CH_2CONH_2 \xrightarrow{Br_2/NaOH,\ \Delta} -CH_2NH_2.$$

After this transformation, the intermediate contains an ortho-substituted benzene ring with the following groups:

• Position 1: $$-CH_2NH_2$$
• Position 2: $$-COOCH_3$$

In the second step, the amino group acts as a nucleophile.

The lone pair on the nitrogen atom attacks the carbonyl carbon of the neighboring ester group, resulting in an intramolecular nucleophilic acyl substitution.

During this process, the methoxy group ($-OCH_3$) leaves, and a new bond is formed between the nitrogen atom and the carbonyl carbon, producing a cyclic amide (lactam).

To determine the ring size, count the atoms involved in ring formation:

1. Benzene carbon attached to the carbonyl group
2. Carbonyl carbon
3. Nitrogen atom
4. $$CH_2$$ carbon
5. Benzene carbon attached to the $$CH_2$$ group

Thus, a five-membered fused lactam ring is formed.

Examining the given options:

• Option (A): An anhydride derivative and therefore incorrect.
• Option (B): Represents the five-membered fused lactam formed through intramolecular cyclization and is correct.
• Option (C): Does not contain a nitrogen atom and is therefore incorrect.
• Option (D): Represents a six-membered ring and does not account for the loss of one carbon atom during Hoffmann degradation.

Hence, the major product formed in the reaction is option (B).

Step 1: Calculating the Double Bond Equivalent (DBE)

The Double Bond Equivalent (also known as the Index of Hydrogen Deficiency) tells us the total number of rings and/or $$\pi$$-bonds present in the molecule. The general formula for a compound with the molecular formula $$\text{C}_\text{c}\text{H}_\text{h}\text{O}_\text{o}\text{N}_\text{n}$$ is:

$$\text{DBE} = \text{c} + 1 - \frac{\text{h} - \text{n}}{2}$$

Substituting the values from the given molecular formula ($$\text{C}_{14}\text{H}_{13}\text{ON}$$):

$$\text{DBE} = 14 + 1 - \frac{13 - 1}{2}$$

$$\text{DBE} = 15 - 6 = 9$$

A DBE value of 9 strongly indicates the presence of:

• Two benzene rings: Each benzene ring accounts for 4 structural units (1 ring + 3 double bonds), totaling $$2 \times 4 = 8$$ units of unsaturation.
• One carbonyl or amide group: The remaining 1 unit of unsaturation ($$9 - 8 = 1$$) corresponds to a double bond, specifically a carbonyl/amide carbon-oxygen double bond ($$\text{C=O}$$).

Step 2: Deducing the Functional Groups from Chemical Tests

1. Neutral Nature of Compound P:

   Compound P is described as neutral. Since it contains a nitrogen atom and an oxygen atom but lacks acidic or basic traits, it points directly toward a substituted amide group ($$\text{--CONH--}$$) rather than a free amine or carboxylic acid.




2. Acidic Hydrolysis of Amides:

   When an amide is heated with dilute hydrochloric acid ($$\text{HCl}$$), it undergoes cleavage into a carboxylic acid and an amine salt:

   $$\text{R--CONH--R'} \xrightarrow{\text{HCl, }\Delta} \text{R--COOH (Residue Q)} + \text{R'--NH}_3^\oplus\text{Cl}^\ominus \xrightarrow{\text{NaOH}} \text{R'--NH}_2\text{ (Oily Liquid R)}$$



3. Analysis of Residue Q:

   Residue Q gives brisk effervescence when treated with sodium bicarbonate ($$\text{NaHCO}_3$$), liberating carbon dioxide ($$\text{CO}_2$$) gas. This test is a hallmark confirmation for the presence of a carboxylic acid group ($$\text{--COOH}$$).




4. Analysis of Oily Liquid R (Hinsberg Test):

   Oily liquid R reacts with Hinsberg's reagent (benzenesulfonyl chloride, $$\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$$) to produce a solid sulfonamide compound that dissolves readily in aqueous sodium hydroxide ($$\text{NaOH}$$). This specific solubility profile proves that R must be a primary amine ($1^\circ\text{ amine}$ marketing as $$\text{R'--NH}_2$$), as its sulfonamide derivative still contains an acidic hydrogen on the nitrogen atom.

Step 3: Reconstructing Compound P

Since the hydrolysis fragments are a carboxylic acid and a primary amine, the parent structure P is a secondary amide linking two aromatic rings:

$$\text{4-methyl-N-phenylbenzamide} \quad (\text{p-CH}_3\text{--C}_6\text{H}_4\text{--CONH--C}_6\text{H}_5)$$

Let's verify the atom count for this structure:

• Carbons: 7 from the 4-methylbenzoyl part + 6 from the aniline ring = 14 carbons.
• Hydrogens: 7 from the tolyl fragment + 1 from the amide nitrogen + 5 from the phenyl ring = 13 hydrogens.
• Heteroatoms: 1 Oxygen and 1 Nitrogen.

The total formula is precisely $$\text{C}_{14}\text{H}_{13}\text{ON}$$, fitting every physical parameter and chemical constraint perfectly.

Conclusion:

The combination of a calculated DBE of 9, its neutral nature, and the formation of a carboxylic acid along with a primary amine upon acid hydrolysis identifies compound P as N-(4-methylphenyl)benzamide / 4-methyl-N-phenylbenzamide.

Answer: The option displaying the structure with a 4-methylphenyl ring on one side and a phenyl group attached to the amide nitrogen ($$\text{p-CH}_3\text{--C}_6\text{H}_4\text{--CONH--C}_6\text{H}_5$$).

Gabriel phthalimide synthesis:Look for a primary amine where the -NH2 group is attached to an aliphatic carbon chain instead of the benzene ring directly.

Hinsberg's reagent:Secondary amines react with Hinsberg's reagent to give a solid that is insoluble in NAOH.

primary aromatic amines give red dye.

hence answer would be D

Reactions of diazonium salt with different reagents:

A. With ArNH₂ (amine): Diazonium coupling reaction with amine gives an azo dye (N=N linkage) → Match with III

B. With HBF₄, Δ: Balz-Schiemann reaction. ArN₂⁺ → ArF (Fluorobenzene) → Match with I

C. With Cu/HCl: Sandmeyer reaction. ArN₂⁺ → ArCl (Chlorobenzene) → Match with IV

D. With CuCN/KCN: Sandmeyer reaction (with cyanide). ArN₂⁺ → ArCN (Cyanobenzene) → Match with II

The correct matching is: A-III, B-I, C-IV, D-II.

We need to match each amine with its correct $$pK_b$$ value. From NCERT data the $$pK_b$$ values of common amines are: aniline ($$C_6H_5NH_2$$, primary aromatic) $$pK_b = 9.38$$; ethanamine ($$C_2H_5NH_2$$, primary aliphatic) $$pK_b = 3.29$$; N-ethylethanamine ($$(C_2H_5)_2NH$$, secondary aliphatic) $$pK_b = 3.00$$; and N,N-diethylethanamine ($$(C_2H_5)_3N$$, tertiary aliphatic) $$pK_b = 3.25$$.

List II gives I = 3.25, II = 3.00, III = 9.38, and IV = 3.29. Aniline $$\rightarrow$$ III ($$pK_b = 9.38$$) because resonance with the benzene ring weakens its basicity. Ethanamine $$\rightarrow$$ IV ($$pK_b = 3.29$$) as a primary aliphatic amine. N-ethylethanamine $$\rightarrow$$ II ($$pK_b = 3.00$$) since the +I effect of two alkyl groups and minimal steric hindrance makes it the strongest base here. N,N-diethylethanamine $$\rightarrow$$ I ($$pK_b = 3.25$$) because steric effects in the tertiary amine reduce its basicity compared to the secondary amine.

The complete matching is A-III, B-IV, C-II, D-I, so the correct answer is Option D.

Nitrobenzene undergoes catalytic hydrogenation in the presence of (H_2/Pd) to reduce the nitro group (({-}NO_2)) into a primary amino group (({-}NH_2)).

Thus,

$$C_6H_5NO_2\xrightarrow{H_2/Pd,\ C_2H_5OH}C_6H_5NH_2.$$

Hence,

$$\boxed{[A]=\text{Aniline }(C_6H_5NH_2).}$$

Aniline then reacts with acetic anhydride in the presence of pyridine to undergo acetylation, forming the corresponding amide, acetanilide.

The reaction is

$$C_6H_5NH_2+(CH_3CO)_2O\xrightarrow{\text{Pyridine}}C_6H_5NHCOCH_3+CH_3COOH.$$

Hence,

$$\boxed{[B]=\text{Acetanilide }(C_6H_5NHCOCH_3).}$$

Therefore,

$$\boxed{[A]=\text{Aniline},\qquad [B]=\text{Acetanilide}.}$$

Lactams are cyclic amides. Under strong aqueous base and heat they undergo alkaline hydrolysis (amide cleavage). The hydroxide ion attacks the carbonyl carbon, opening the ring and producing the salt of an ω-amino-acid. Subsequent acidification (step ii : $$H^+$$) protonates both the carboxylate and the amide-derived anion, giving the neutral amino-acid.

For the lactam given in the paper (the monomer of Nylon-6, $$\varepsilon$$-caprolactam), the ring contains six methylene units between the nitrogen and the carbonyl carbon. Ring opening therefore produces a straight chain of six carbon atoms that carries a terminal $$COOH$$ group at one end and an $$NH_2$$ group at the other:

$$\varepsilon\text{-caprolactam} \xrightarrow[\Delta]{NaOH} \;^-OOC-(CH_2)_5-NH^- \;\xrightarrow{H^+}\; HOOC-(CH_2)_5-NH_2$$

Thus ‘A’ is 6-aminocaproic acid (also called 6-aminohexanoic acid), whose structure is

$$NH_2-(CH_2)_5-COOH$$

That corresponds to Option C.

Answer: Option C which is: 6-aminocaproic acid, $$NH_2-(CH_2)_5-COOH$$.

Step 1: Nitrosation ($$\text{NaNO}_2 / \text{HCl}$$)

The starting material is a substituted phenol (thymol derivative). When treated with nitrous acid ($$\text{NaNO}_2 + \text{HCl}4$$ forming $$\text{HNO}_2$$) at a low temperature, it undergoes electrophilic aromatic substitution (nitrosation).

• The hydroxyl group ($$\text{-OH}$$) is a strongly activating, ortho/para-directing group.
• The ortho positions relative to the $$\text{-OH}$$ group are hindered or occupied.
• The carbon position directly para to the $$\text{-OH}$$ group (the top-left vertex of the ring) is completely vacant and sterically accessible.

the nitroso electrophile ($$\text{NO}^+$$) attacks at this para-position to form Compound A (a para-nitrosophenol derivative).

Step 2: Selective Reduction ($$\text{NH}_4\text{SH}$$)

Compound A is then treated with ammonium hydrosulfide ($$\text{NH}_4\text{SH}$$).

• $$\text{NH}_4\text{SH}$$ is a well-known mild and selective reducing agent.
• It reduces the nitroso group ($$\text{-NO}$$) directly to a primary amino group ($$\text{-NH}_2$$) without affecting the other alkyl or phenolic substituents on the ring.

Option A

The reagents used in the four successive steps correspond to four very standard transformations in the chemistry of aromatic nitrogen-containing compounds. We identify each step first and then follow the fate of the functional group round by round.

Step 1 : Reduction of an aromatic nitro group
If the starting compound is $$C_6H_5NO_2$$ (nitrobenzene) and it is treated with a reducing system such as $$Sn/HCl$$, $$Fe/HCl$$ or $$H_2/Pd$$, the nitro group is converted into an amino group:$$C_6H_5NO_2 \;\xrightarrow[\;HCl\;]{\;Sn\;}\; C_6H_5NH_2$$.
Thus, after step 1 we obtain aniline.

Step 2 : Diazotisation of the amine
Primary aromatic amines react with nitrous acid (generated in situ from $$NaNO_2/HCl$$ at $$0{-}5^{\circ}\text{C}$$) to give diazonium salts:$$C_6H_5NH_2 \;\xrightarrow[\;0{-}5^{\circ}C\;]{\;NaNO_2/HCl\;}\; C_6H_5N_2^+Cl^-$$.

Step 3 : Sandmeyer cyanation
Aryl diazonium chloride reacts with cuprous cyanide to introduce the cyano group, giving an aromatic nitrile (Sandmeyer reaction):$$C_6H_5N_2^+Cl^- \;\xrightarrow{\;CuCN\;}\; C_6H_5CN$$.

Step 4 : Reduction of the nitrile to a primary amine
Lithium aluminium hydride $$\big(LiAlH_4\big)$$ is a powerful reducing agent that converts nitriles to primary amines by adding two hydrogen atoms to the carbon of the -CN group and two to the nitrogen:$$C_6H_5CN \;\xrightarrow{\;LiAlH_4\;}\; C_6H_5CH_2NH_2$$ (benzylamine).

Hence, after the entire sequence the major product $$P$$ is benzylamine, $$C_6H_5CH_2NH_2$$.

Among the given structures, benzylamine corresponds to Option B.

Therefore, the correct answer is:
Option B which is: benzylamine ($$C_6H_5CH_2NH_2$$).

Reaction Mechanism:

1. Diazotization: When an aromatic primary amine reacts with HNO2 it undergoes diazotization to form a highly reactive diazonium ion.
2. Intramolecular Cyclization:

   Since o-phenylenediamine has two amino groups ortho to each other, one of the $$-\text{NH}_2$$ groups is converted into a diazonium group ($$-\text{N}_2^+$$).

3. Ring Closure:

   The adjacent, unreacted primary amino group ($$-\text{NH}_2$$) acts as an internal nucleophile. The lone pair on its nitrogen atom attacks the neighboring diazonium group, causing a rapid intramolecular cyclization to close a stable 5-membered heterocyclic ring.

4. Deprotonation:

   After losing a proton ($$\text{H}^+$$), the stable heterocyclic compound Benzotriazole is formed.

Structure of the Product:

The resulting compound contains a benzene ring fused to a triazole ring (a 5-membered ring containing three nitrogen atoms), with the structure:

• A benzene core.
• Fused to a ring containing $$-\text{N}=\text{N}-\text{NH}-$$.

We need to evaluate the Assertion and Reason about amino acid synthesis.

Assertion: $$\alpha$$-halocarboxylic acid on reaction with dilute NH$$_3$$ gives good yield of $$\alpha$$-amino carboxylic acid, whereas the yield of amines is very low when prepared from alkyl halides.

This is correct. When alkyl halides react with NH$$_3$$, the product amine is more nucleophilic than NH$$_3$$ itself, leading to over-alkylation (polyalkylation) and low yield of primary amine. However, $$\alpha$$-halocarboxylic acids give good yield of amino acids because the product amino acid exists as a zwitter ion, which is much less nucleophilic than NH$$_3$$, preventing further reaction.

Reason: Amino acids exist in zwitter ion form in aqueous medium.

This is correct. Amino acids exist as $$^+\text{NH}_3\text{-CHR-COO}^-$$ in aqueous medium.

The reason (R) does explain the assertion (A). The zwitter ion form makes the amino group in the product positively charged ($$-\text{NH}_3^+$$), so it cannot act as a nucleophile to undergo further alkylation. This is why the yield is good for amino acid synthesis but poor for amine synthesis from alkyl halides.

Therefore, both (A) and (R) are correct, and (R) is the correct explanation of (A).

The correct answer is Option A: Both (A) and (R) are correct and (R) is the correct explanation of (A).

Step (i): Bromination ($$\text{Br}_2$$)

When 4-nitrotoluene reacts with $$\text{Br}_2$$, electrophilic aromatic substitution occurs. We must look at the directing influence of both groups to determine where the bromine atom ($$\text{-Br}$$) will attach:

• The $$\text{-CH}_3$$ group directs incoming electrophiles to its ortho positions (positions 2 and 6 relative to itself).
• The $$\text{-NO}_2$$ group directs incoming electrophiles to its meta positions (which also happen to be positions 2 and 6 relative to the $$\text{-CH}_3$$ group).

Because both groups reinforce each other, the bromine atom is introduced ortho to the methyl group and meta to the nitro group.

$$\text{Intermediate 1} = \text{2-bromo-4-nitrotoluene}$$

Step (ii): Reduction ($$\text{H}_2 / \text{Pd}$$)

Catalytic hydrogenation using hydrogen gas over a palladium catalyst ($$\text{H}_2/\text{Pd}$$) selectively reduces the strongly electrophilic nitro group ($$\text{-NO}_2$$) into an amino group ($$\text{-NH}_2$$). The ring-bound bromine and methyl groups remain unaffected under these typical conditions.

$$\text{Intermediate 2} = \text{3-bromo-4-methylaniline}$$

Step (iii): Diazotization ($$\text{NaNO}_2, \text{HCl}, 0^\circ\text{C}$$)

Reacting a primary aromatic amine ($$\text{-NH}_2$$) with nitrous acid (generated in situ from $$\text{NaNO}_2$$ and $$\text{HCl}$$) at a low temperature ($$0^\circ\text{C} - 5^\circ\text{C}$$) converts the amino group into a highly reactive diazonium salt group ($$\text{-N}_2^+\text{Cl}^-$$).

$$\text{Intermediate 3} = \text{3-bromo-4-methylbenzenediazonium chloride}$$

Step (iv): Deamination ($$\text{H}_3\text{PO}_2$$)

Hypophosphorous acid ($$\text{H}_3\text{PO}_2$$) acts as a mild reducing agent that replaces the diazonium group ($$\text{-N}_2^+$$) with a hydrogen atom ($$\text{-H}$$), releasing nitrogen gas ($$\text{N}_2$$). This effectively removes the original functional group at that position entirely.

Therefore, the final product (P) is 2-bromotoluene (also called o-bromotoluene).

Transformation (1):

Benzamide reacts with $$Br_2/NaOH$$ via Hoffmann bromamide degradation.

A primary amide is converted into a primary amine containing one less carbon atom.

$$C_6H_5CONH_2\xrightarrow{Br_2/NaOH}C_6H_5NH_2$$

Hence, transformation (1) forms an aromatic amine.

Transformation (2):

Gabriel phthalimide synthesis proceeds through an $$S_N2$$ mechanism.

Chlorobenzene is an aryl halide and does not undergo $$S_N2$$ substitution because the $$C-Cl$$ bond has partial double-bond character due to resonance.

Hence, transformation (2) does not form an aromatic amine.

Transformation (3):

Nitrobenzene undergoes catalytic reduction with $$H_2/Pd$$.

$$C_6H_5NO_2\xrightarrow{H_2/Pd}C_6H_5NH_2$$

Hence, transformation (3) forms an aromatic amine.

Transformation (4):

Acetanilide undergoes acidic hydrolysis.

$$C_6H_5NHCOCH_3\xrightarrow{\Delta,\ \mathrm{dil.}\ H_2SO_4}C_6H_5NH_2+CH_3COOH$$

Hence, transformation (4) forms an aromatic amine.

Therefore, transformations $$(1)$$, $$(3)$$, and $$(4)$$ produce an aromatic amine.

$$\boxed{3}$$

Gabriel phthalimide synthesis can only be used to prepare primary amines. It involves treating potassium phthalimide with an alkyl halide, followed by hydrolysis.

Molecular formula: C₈H₁₁N. Degrees of unsaturation = $$\frac{2(8)+2-11+1}{2} = 4$$ (indicating a benzene ring).

The aromatic primary amines with this formula that can be made by Gabriel synthesis (excluding arylamines where NH₂ is directly on the ring, as those cannot be made by Gabriel synthesis) are:

These are the methylbenzylamine isomers (where -CH₂NH₂ is the reactive part):

1. 2-Methylbenzylamine (o-CH₃C₆H₄CH₂NH₂)

2. 3-Methylbenzylamine (m-CH₃C₆H₄CH₂NH₂)

3. 4-Methylbenzylamine (p-CH₃C₆H₄CH₂NH₂)

The number of isomeric aromatic amines is 3.

To determine the decreasing order of stability for the given benzenediazonium salts, we must evaluate the electronic effects of the substituents on the benzene ring. The stability of a diazonium ion depends heavily on how well the positive charge on the nitrogen atoms can be dispersed.

Electron-donating groups (+R effect) push electron density into the ring, which helps neutralize and delocalize the positive charge on the diazonium group, thereby increasing stability. Conversely, electron-withdrawing groups (-R effect) pull electron density away from the ring, which intensifies the positive charge and decreases stability.

In compound (A), the methoxy group (-OCH₃) exhibits a strong electron-donating +R effect, making the diazonium ion the most stable. Compound (C) has no substituent and serves as the neutral baseline. In compounds (D) and (B), the cyano (-CN) and nitro (-NO₂) groups act as powerful electron-withdrawing groups via the -R effect, highly destabilizing the ion. Because the nitro group exerts a stronger electron-withdrawing effect than the cyano group, compound (B) is the least stable of all.

Therefore, the decreasing order of stability is: (A) > (C) > (D) > (B).

We evaluate the assertion that the reaction of $$CH_3Cl$$ with aniline and anhydrous $$AlCl_3$$ does not give ortho- and para-methylaniline. This is true because when aniline is treated with $$CH_3Cl$$ in the presence of anhydrous $$AlCl_3$$ (Friedel-Crafts alkylation), the expected ortho- and para-methylanilines are not obtained.

The $$-NH_2$$ group reacts with $$AlCl_3$$ to form a salt:

$$C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5\overset{+}{N}H_2\text{-}AlCl_3^-$$

This salt formation converts the activating, ortho/para-directing $$-NH_2$$ group into the deactivating, meta-directing $$-\overset{+}{N}H_2AlCl_3^-$$ group, so the Friedel-Crafts reaction either does not proceed or gives poor yields.

The stated reason is that the $$-NH_2$$ group becomes deactivating because of salt formation with anhydrous $$AlCl_3$$ and hence yields meta-methylaniline. While the first part is true, the claim that it yields meta-methylaniline is false: the ring is so deactivated that the Friedel-Crafts reaction essentially does not occur, rather than giving the meta product, since Friedel-Crafts reactions do not work on strongly deactivated rings.

The assertion is true and the reason is false.

Hence, the correct answer is Option C: A is true, but R is false.

Thus, the right option is C.

We need to determine how many of the six diaminobenzoic acid isomers give each diaminobenzene product upon decarboxylation, and find the melting point of product C (obtained from only one acid).

Identify all six isomers of diaminobenzoic acid.

Diaminobenzoic acid has the formula $$C_6H_3(NH_2)_2COOH$$. On a benzene ring with a COOH group at position 1, the two $$NH_2$$ groups can be at:

1. Positions 2,3

2. Positions 2,4

3. Positions 2,5

4. Positions 2,6

5. Positions 3,4

6. Positions 3,5

Determine the decarboxylation product for each isomer.

Decarboxylation removes the COOH group. The relative positions of the two $$NH_2$$ groups determine the product:

1. (2,3)-diaminobenzoic acid $$\rightarrow$$ 1,2-diaminobenzene (o-phenylenediamine)

2. (2,4)-diaminobenzoic acid $$\rightarrow$$ 1,3-diaminobenzene (m-phenylenediamine)

3. (2,5)-diaminobenzoic acid $$\rightarrow$$ 1,4-diaminobenzene (p-phenylenediamine)

4. (2,6)-diaminobenzoic acid $$\rightarrow$$ 1,3-diaminobenzene (m-phenylenediamine) [positions 2,6 relative to removed COOH become 1,3]

5. (3,4)-diaminobenzoic acid $$\rightarrow$$ 1,2-diaminobenzene (o-phenylenediamine)

6. (3,5)-diaminobenzoic acid $$\rightarrow$$ 1,3-diaminobenzene (m-phenylenediamine)

Count the products.

Product A (from 3 acids): 1,3-diaminobenzene (m-phenylenediamine) — from isomers 2, 4, and 6. Melting point = 63°C.

Product B (from 2 acids): 1,2-diaminobenzene (o-phenylenediamine) — from isomers 1 and 5. Melting point = 104°C.

Product C (from 1 acid): 1,4-diaminobenzene (p-phenylenediamine) — from isomer 3 only. Melting point = 142°C.

The melting point of product C is 142°C.

The correct answer is Option D.

We need to identify what happens during Friedel-Crafts alkylation of aniline. Aniline (C₆H₅NH₂) has a lone pair on nitrogen, and in Friedel-Crafts alkylation, the Lewis acid catalyst (like AlCl₃) is used to generate a carbocation. The lone pair on the nitrogen atom of aniline reacts with the Lewis acid catalyst AlCl₃, forming a coordination complex where nitrogen donates its lone pair to the Lewis acid:

$$\text{C}_6\text{H}_5\text{NH}_2 + \text{AlCl}_3 \rightarrow \text{C}_6\text{H}_5\text{NH}_2^{+}-\text{AlCl}_3^{-}$$

This effectively puts a positive charge on the nitrogen atom, and the positively charged nitrogen acts as a strong deactivating group (meta-directing). The nitrogen withdraws electron density from the ring through its positive charge, making the ring highly electron-deficient. This deactivates the benzene ring so much that Friedel-Crafts reaction does not proceed on the ring. In fact, Friedel-Crafts reactions generally fail with aniline and other amines because the amine group gets protonated/complexed, deactivating the ring.

The answer is Option D: positively charged nitrogen at benzene ring.

We need to identify the statement that is NOT correct about p-toluenesulphonyl chloride (Hinsberg's reagent).

p-Toluenesulphonyl chloride ($$CH_3C_6H_4SO_2Cl$$) is known as Hinsberg's reagent. It is used to distinguish between primary, secondary, and tertiary amines based on their different reactions with this reagent.

Option A: "It is Hinsberg's reagent" — This is correct. p-Toluenesulphonyl chloride is indeed Hinsberg's reagent.

Option B: "It is used to distinguish primary and secondary amines" — This is correct. Primary amines form N-alkyl sulphonamides that are soluble in alkali (due to the acidic N-H), while secondary amines form N,N-dialkyl sulphonamides that are insoluble in alkali (no N-H for salt formation).

Option C: "On treatment with secondary amine, it leads to a product that is soluble in alkali" — This is NOT correct. When a secondary amine reacts with Hinsberg's reagent, the product is $$CH_3C_6H_4SO_2NR_2$$, which has no acidic hydrogen on nitrogen. Therefore, it is insoluble in alkali. Only the product from primary amines ($$CH_3C_6H_4SO_2NHR$$) is soluble in alkali because the N-H hydrogen is acidic enough to form a sodium salt.

Option D: "It does not react with tertiary amine" — This is correct. Tertiary amines have no N-H bond, so they cannot undergo the sulphonamide reaction with Hinsberg's reagent.

Therefore, the correct answer is Option C.

For any carbonyl compound the sequence for enamine formation is : (1) nucleophilic addition of a secondary amine on the $$C=O$$ carbon, (2) proton transfers, and (3) loss of water accompanied by removal of an $$\alpha$$-hydrogen to give the C=C-N system. The position of equilibrium

$$\text{carbonyl compound}\;+\;\text{(sec. amine)} \rightleftharpoons \text{enamine}+H_2O$$

depends on the stability of the enamine that is produced. A more stable enamine pulls the equilibrium to the right and therefore “forms most readily”. The stability of an enamine is governed by the same factors that stabilise ordinary alkenes:

• greater alkyl substitution on the C=C bond (hyperconjugation and +I effect)
• conjugation of the C=C bond with another $$\pi$$-system (aromatic ring, C=C, C≡C, etc.)

All alkyl groups are simple methyl/ethyl groups, so the resulting C=C bond is only modestly substituted and not conjugated with any other $$\pi$$-system.

The situation is similar to Case A; no extra substitution and no conjugation.

The carbonyl carbon is directly attached to a phenyl ring. When the $$\alpha$$-proton is removed, the resulting C=C bond becomes

$$\text{Ph}-C(=C)-N$$

This double bond is in conjugation with the benzene ring. Resonance delocalises the enamine’s double bond into the aromatic system, giving several resonance contributors such as

$$\text{Ph}^{\ominus}\!-\!C-C^{\oplus}\!=\!N \longleftrightarrow \text{Ph}-C^{\oplus}\!=\!C-N^{\ominus}$$

Conjugation with the aromatic ring therefore provides a large extra stabilisation on top of the normal alkyl-substitution stabilisation.

Here again the C=C bond produced in the enamine is only alkyl-substituted and not conjugated with any extended $$\pi$$-system.

Comparing all four cases, the enamine derived from Case C is the only one that enjoys resonance conjugation with an aromatic ring in addition to normal alkyl substitution. That additional delocalisation lowers its energy the most, so Case C gives the most stable enamine and therefore shows the easiest / most favourable enamine formation.

Hence the correct choice is:
Option C which is: the carbonyl compound having the phenyl ring directly attached to the carbonyl carbon.

Explanation: The formation of an orange azo dye with nitrous acid and phenol confirms the presence of a primary aromatic amine.

Only primary aromatic amines form stable diazonium salts, which couple with phenol to give azo dyes.

Thus, the separated liquid is aniline $$C_6H_5NH_2$$.

Treatment of Compound $$A$$ with aqueous $$NaOH$$ liberates free aniline, indicating that Compound $$A$$ is the acidic salt of aniline.

Therefore, Compound $$A$$ is anilinium chloride:

$$C_6H_5NH_3^+Cl^-$$

Anilinium chloride is acidic in water because it is formed from a weak base and a strong acid, explaining the red litmus result.

Its molecular weight is:

$$6(12) + 8(1) + 14 + 35.5 = 129.5\ g/mol$$

which matches the given range $$131 \pm 2$$.

Therefore, Compound $$A$$ is anilinium chloride, and the right option is D.

The reaction sequence described is characteristic of the Gabriel Phthalimide Synthesis used for preparing primary amines.

In this synthesis, the key intermediate is phthalimide.

Compound $$C$$ has the molecular formula:

$$C_8H_5NO_2$$

which corresponds to phthalimide.

Phthalimide reacts with ethanolic $$KOH$$ to form potassium phthalimide, which then reacts with an alkyl halide $$R-Cl$$.

Subsequent alkaline hydrolysis yields a primary amine.

Therefore, Compound $$C$$ must be phthalimide.

Phthalimide is prepared from phthalic acid through the following sequence.

Phthalic acid reacts with ammonia to form ammonium phthalate.

On heating, ammonium phthalate loses water to form phthalamide.

Further heating causes cyclization and loss of ammonia to produce phthalimide.

Thus, Compound $$A$$ is phthalic acid.

The starting compound is aniline, $$C_6H_5NH_2$$.

Step-1 : Diazotisation (formation of A)
When aniline is treated with sodium nitrite in the presence of dilute hydrochloric acid at $$0\!-\!5^{\circ}\text{C}$$, it undergoes diazotisation. The nitrous acid generated in situ converts the -NH2 group into a diazonium group.

$$C_6H_5NH_2 + NaNO_2 + 2\,HCl \;\xrightarrow[0-5^{\circ}\!C]{}\; C_6H_5N_2^+Cl^- + NaCl + 2\,H_2O$$

Thus, product $$A$$ is benzene diazonium chloride, $$C_6H_5N_2^+Cl^-$$.

Step-2 : Sandmeyer reaction (formation of B)
A diazonium salt reacts with cuprous cyanide (generated from $$CuCN/KCN$$) to give the corresponding aryl nitrile. This transformation is called the Sandmeyer cyanation.

$$C_6H_5N_2^+Cl^- + CuCN \;\xrightarrow{\,50-60^{\circ}\!C\,}\; C_6H_5CN + CuCl + N_2\uparrow$$

Therefore, product $$B$$ is benzonitrile, $$C_6H_5CN$$.

Hence, the pair of products is:
$$A : C_6H_5N_2^+Cl^- \quad\text{and}\quad B : C_6H_5CN$$

These correspond to Option A.

Explanation: Aniline contains an amino group $$\mathrm{(-NH_2)}$$ directly attached to the benzene ring. The $$\mathrm{-NH_2}$$ group is normally an activating and ortho/para-directing group.

However, direct nitration of aniline gives a mixture containing significant amounts of meta product:

$$\mathrm{para\text{-}nitroaniline \approx 51%}$$

$$\mathrm{meta\text{-}nitroaniline \approx 47%}$$

$$\mathrm{ortho\text{-}nitroaniline \approx 2%}$$

Hence, the Assertion is correct.

The nitrating mixture consists of concentrated $$\mathrm{HNO_3}$$ and concentrated $$\mathrm{H_2SO_4}$$, making the medium strongly acidic.

Aniline behaves as a base and gets protonated to form the anilinium ion:

$$\mathrm{C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+}$$

The $$\mathrm{-NH_3^+}$$ group is strongly deactivating and meta-directing.

Therefore, a large fraction of the reaction occurs through the meta-directing anilinium ion, producing substantial amounts of meta-nitroaniline.

Hence, both Assertion and Reason are correct, and the Reason correctly explains the Assertion.

The Hofmann degradation (Hofmann rearrangement) converts a primary amide ($$RCONH_2$$) to a primary amine ($$RNH_2$$) with one fewer carbon atom, using $$Br_2$$ and a strong base such as NaOH or KOH. The mechanism begins with the amide reacting with $$Br_2$$ in the presence of NaOH, which brominates the nitrogen to form an N-bromoamide. The base then abstracts the N-H proton, and loss of bromide generates a nitrene intermediate in which the nitrogen atom is electron deficient, possessing only a sextet of electrons. Next, the R group migrates from the carbonyl carbon to the electron-deficient nitrogen in a 1,2-shift, yielding an isocyanate intermediate. Finally, hydrolysis of the isocyanate with water produces the primary amine and $$CO_2$$.

In analyzing Statement I, the claim is that in the Hofmann degradation only an alkyl group migrates from the carbonyl carbon to the nitrogen atom. In fact, the R group attached to the carbonyl carbon—whether it is alkyl or aryl—migrates to nitrogen, and no other group undergoes migration. Within the typical context of JEE-level examples involving alkyl amides, this statement is considered correct.

Analysis of Statement II reveals that the migrating group indeed moves to an electron-deficient atom. During the rearrangement, nitrogen becomes electron deficient in the nitrene intermediate (having only six electrons), and the R group shifts to this electron-poor nitrogen. This feature is characteristic of 1,2-rearrangement reactions.

Since both Statement I and Statement II are correct, the answer is Option A.

Hence, Option C is correct.

This is an azo coupling reaction. As per NCERT, the right answer is C.

We need to match List-I (chemical reactions/reagents) with List-II (their descriptions). To do this, each item is examined in turn.

(A) Benzenesulphonyl chloride: Benzenesulphonyl chloride ($$C_6H_5SO_2Cl$$) is known as the Hinsberg reagent. It is used in the Hinsberg test to distinguish between primary, secondary, and tertiary amines, so A matches with (III) Hinsberg reagent.

(B) Hoffmann bromamide reaction: In the Hoffmann bromamide degradation, a primary amide is treated with $$Br_2$$ and NaOH to form an amine with one fewer carbon atom. The intermediate in this reaction is an isocyanate (R-N=C=O), and thus B matches with (IV) Known reaction of Isocyanates.

(C) Carbylamine reaction: The carbylamine reaction (also called isocyanide test) involves heating a primary amine with chloroform and alcoholic KOH to form an isocyanide with a foul smell. Being a test for primary amines, C matches with (I) Test for primary amines.

(D) Hoffmann orientation: Hoffmann elimination (or Hoffmann orientation) gives the less substituted alkene as the major product when a quaternary ammonium salt undergoes elimination. This is opposite to Saytzeff’s rule, hence it is called Anti-Saytzeff elimination, and D matches with (II) Anti Saytzeff.

Accordingly, the final matching is A-III, B-IV, C-I, D-II, and the correct answer is Option C: A-III, B-IV, C-I, D-II.

Explanation: Propan-1-ol $$\mathrm{CH_3-CH_2-CH_2-OH}$$ contains a 3-carbon chain, while n-butylamine $$\mathrm{CH_3-CH_2-CH_2-CH_2-NH_2}$$ contains a 4-carbon chain.

Therefore, a one-carbon chain extension is required.

Step 1: Treatment with $$\mathrm{SOCl_2}$$ converts the alcohol into an alkyl chloride:

$$\mathrm{CH_3-CH_2-CH_2-OH + SOCl_2 \rightarrow CH_3-CH_2-CH_2-Cl + SO_2 + HCl}$$

This forms $$\mathrm{1\text{-}chloropropane}$$.

Step 2: Reaction with $$\mathrm{KCN}$$ gives nucleophilic substitution through an $$\mathrm{S_N2}$$ mechanism:

$$\mathrm{CH_3-CH_2-CH_2-Cl + KCN \rightarrow CH_3-CH_2-CH_2-C \equiv N + KCl}$$

This forms $$\mathrm{butanenitrile}$$.

Step 3: Reduction of the nitrile group using $$\mathrm{H_2/Ni}$$ converts it into a primary amine:

$$\mathrm{CH_3-CH_2-CH_2-C \equiv N + 4[H] \rightarrow CH_3-CH_2-CH_2-CH_2-NH_2}$$

Thus, the required sequence is $$\mathrm{(i)\ SOCl_2 \quad (ii)\ KCN \quad (iii)\ H_2/Ni}$$

Brønsted basic strength depends on the availability of the nitrogen lone pair for accepting: $$\mathrm{H^+}$$ 

Basicity is mainly influenced by: $$\mathrm{+I\ effect}$$ and  $$\mathrm{Steric\ Hindrance}$$ 

In: $$\mathrm{N(CH_2CH_3)_3}$$ (triethylamine), three ethyl groups increase electron density on nitrogen through: $$\mathrm{+I\ effect}$$

However, the bulky ethyl groups create significant steric hindrance around nitrogen, reducing protonation and solvation.

Hence, its basic strength decreases.

In:$$\mathrm{HN(CH_2CH_3)_2}$$ (diethylamine), two ethyl groups provide strong: $$\mathrm{+I\ effect}$$ with much lower steric hindrance.

Thus, secondary amines are generally stronger bases than ordinary tertiary amines in aqueous medium.

In: $$\mathrm{1\text{-}Methyl\text{-}3\text{-}pyrroline}$$ the adjacent: $$\mathrm{sp^2}$$ hybridised carbon atoms exert a slight: $$\mathrm{-I\ effect}$$

which decreases electron density on nitrogen. Hence, it is less basic than saturated aliphatic amines.

In: $$\mathrm{Quinuclidine}$$ the nitrogen lone pair is highly available. The bicyclic cage structure provides strong $$\mathrm{+I\ effect}$$ while preventing steric crowding around the lone pair.

Thus, protonation occurs very easily and the conjugate acid formed is highly stable.

Therefore, the strongest Brønsted base is:

$$\mathrm{Quinuclidine}$$

Correct option:

$$\mathrm{D}$$

Given reaction:

$$C_6H_5NH_2\xrightarrow[\;288\,K\;]{HNO_3/H_2SO_4}\text{Nitroanilines}$$

Step 1: Generation of the electrophile

$$HNO_3 + H_2SO_4 \rightarrow H_2NO_3^+ + HSO_4^-$$

$$H_2NO_3^+ \rightarrow NO_2^+ + H_2O$$

Hence,

$$H_2SO_4$$acts as an acid, while $$HNO_3$$ acts as a base.

Step 2: Protonation of aniline

$$C_6H_5NH_2 + H^+$$ $$\rightleftharpoons C_6H_5NH_3^+$$

The anilinium ion $$-NH_3^+$$ is a strong meta-directing group.

Therefore, direct nitration of aniline gives predominantly:

$$m\text{-nitroaniline}$$ along with a significant amount of $$p\text{-nitroaniline}$$

Evaluation of Statements:

For Statement (A):

$$\text{o-Nitroaniline and p-Nitroaniline are the predominant products}$$

$${\text{False}}$$

For Statement (B):

$$\text{p-Nitroaniline and m-Nitroaniline are the predominant products}$$

$${\text{True}}$$

For Statement (C):

$$HNO_3 \text{ acts as an acid}$$

$${\text{False}}$$

For Statement (D):

$$H_2SO_4 \text{ acts as an acid}$$

$${\text{True}}$$

Therefore,

$${\text{Statements (B) and (D) are correct}}$$

When a primary aliphatic amine reacts with nitrous acid ($$HNO_2$$) in cold conditions (273 K) and the temperature is then raised to room temperature (298 K), the initial reaction at low temperature involves a primary aliphatic amine ($$R-NH_2$$) reacting with nitrous acid (prepared in situ from $$NaNO_2 + HCl$$) to form an aliphatic diazonium salt:

$$R-NH_2 + HNO_2 \rightarrow R-N_2^+Cl^- + H_2O$$

Aliphatic diazonium salts are extremely unstable and decompose immediately even at low temperatures. Upon warming to room temperature (298 K), the diazonium salt loses nitrogen:

$$R-N_2^+Cl^- \xrightarrow{298 K} R^+ + N_2 \uparrow$$

The resulting carbocation reacts with water to give an alcohol:

$$R^+ + H_2O \rightarrow R-OH + H^+$$

Overall, the primary aliphatic amine is converted into an alcohol with the evolution of nitrogen gas:

$$R-NH_2 \xrightarrow{NaNO_2/HCl, 273K \rightarrow 298K} R-OH + N_2 + HCl$$

This behavior contrasts with that of aromatic primary amines, which form stable diazonium salts at 273 K that can undergo coupling reactions. The correct answer is Option A: Alcohol.

The final step uses $$\mathrm{Br_2/NaOH}$$, which are the reagents for the Hoffmann bromamide degradation.

This reaction converts a primary amide $$\mathrm{(-CONH_2)}$$ into a primary amine $$\mathrm{(-NH_2)}$$ with loss of the carbonyl carbon.

Since the final product is $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylaniline}$$, the intermediate must be $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylbenzamide}$$.

The first step involves reaction with ammonia $$\mathrm{(NH_3)}$$, which converts an acid chloride $$\mathrm{(-COCl)}$$ into an amide $$\mathrm{(-CONH_2)}$$.

Therefore, compound $$\mathrm{A}$$ must be $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylbenzoyl\ chloride}$$.

Its molecular formula $$\mathrm{C_8H_6Cl_2O}$$ matches the given formula.

Correct Answer: $$\mathrm{4\text{-}chloro\text{-}3\text{-}methylbenzoyl\ chloride}$$

This question asks about the correct sequential order of reagents for a given reaction. Since the question references a reaction diagram that is not available, we will reason based on the given answer.

The correct answer is Option B: $$HNO_2$$, KI, $$Fe/H^+$$, $$HNO_2$$, $$H_2O$$/warm

Since the starting amine group ($$-NH_2$$) is converted to a diazonium salt ($$-N_2^+$$), nitrous acid at 0-5 °C (from $$NaNO_2 + HCl$$) is used for diazotization:

$$Ar-NH_2 \xrightarrow{HNO_2} Ar-N_2^+Cl^-$$

After diazotization, the diazonium group is replaced by iodine using potassium iodide in a Sandmeyer-type reaction:

$$Ar-N_2^+Cl^- \xrightarrow{KI} Ar-I + N_2 + KCl$$

Next, a nitro group ($$-NO_2$$) present elsewhere on the ring is reduced to an amine group ($$-NH_2$$) with iron in acidic medium:

$$Ar-NO_2 \xrightarrow{Fe/H^+} Ar-NH_2$$

Since the newly formed $$-NH_2$$ group must be diazotized again, nitrous acid is once more employed:

$$Ar-NH_2 \xrightarrow{HNO_2} Ar-N_2^+Cl^-$$

Finally, hydrolysis by warming with water converts the diazonium salt into a phenol ($$-OH$$) group:

$$Ar-N_2^+Cl^- \xrightarrow{H_2O/\text{warm}} Ar-OH + N_2 + HCl$$

Therefore, the correct sequence is Option B: $$HNO_2$$, KI, $$Fe/H^+$$, $$HNO_2$$, $$H_2O$$/warm.

Benzenesulphonyl chloride: $$\mathrm{C_6H_5SO_2Cl}$$ is also known as  $$\mathrm{Hinsberg's\ Reagent}$$ It reacts with primary and secondary amines to form: $$\mathrm{Sulphonamides}$$

Explanation: The reaction converts a primary amide $$\mathrm{(R-CONH_2)}$$ into a primary amine $$\mathrm{(R-NH_2)}$$ with one less carbon atom using $$\mathrm{Br_2/KOH}$$.

An N-bromoamide intermediate is first formed: $$\mathrm{R-CONHBr}$$

Rearrangement then occurs with migration of the $$\mathrm{R}$$ group from the carbonyl carbon to nitrogen and simultaneous loss of $$\mathrm{Br^-}$$.

This produces the key intermediate: $$\mathrm{R-N=C=O}$$

The intermediate $$\mathrm{R-N=C=O}$$ is an alkyl isocyanate.

Hydrolysis of the isocyanate finally gives the primary amine.

Thus, the characteristic intermediate in Hoffmann bromamide degradation is: $$\mathrm{R-N=C=O}$$

Correct Option: $$\mathrm{(C)}$$

We begin by recalling the definition of basicity in organic molecules. A base is a species which can donate a lone pair of electrons to accept a proton ($$H^{+}$$). Hence, the more freely available the lone pair on nitrogen, the stronger is the basic character of the compound.

Now we consider the two compounds in the question:

(i) Acetamide : $$CH_{3}CONH_{2}$$

(ii) Aniline  : $$C_{6}H_{5}NH_{2}$$

For acetamide, the nitrogen lone pair interacts with the adjacent carbonyl group. We write the resonance structures first, stating the resonance principle: “If a lone pair on an atom is adjacent to a multiple bond, $$\pi$$-electron delocalisation can occur.” Using this rule we obtain

$$CH_{3}CONH_{2}\; \longleftrightarrow \; CH_{3}C^{(-)}=O^{(+)}NH_{2}$$

Because of this $$n \rightarrow \pi^{*}$$ conjugation, the lone pair is pulled toward the more electronegative oxygen. Thus it is very poorly available for protonation. Quantitatively, the conjugate acid $$(CH_{3}CONH_{3}^{+})$$ has a $$pK_{a}$$ around 0, showing extreme weakness of the base.

For aniline, the nitrogen lone pair is adjacent to an aromatic ring. Again applying the same resonance principle, we write the resonance forms:

$$C_{6}H_{5}NH_{2}\; \longleftrightarrow \; C_{6}H_{5}^{(-)} \!-\! NH_{2}^{(+)}$$

symbolically extended over the ring as

$$\begin{aligned} &\; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; N \!H_{2}\\ C_{6}H_{5}NH_{2}&\; \longleftrightarrow \; C_{6}H_{5}^{(-)}\!=\!NH_{2}^{(+)} \end{aligned}$$

Thus the nitrogen lone pair is delocalised into the benzene ring, reducing its electron density and consequently its tendency to accept a proton. Nevertheless, the withdrawal in aniline is less severe than in acetamide, because in acetamide the electron-withdrawing carbonyl oxygen is far more electronegative than any carbon in the aromatic ring.

Therefore the experimentally observed order of basicity is

$$\text{acetamide} \; < \; \text{aniline}$$

or, equivalently,

$$\text{aniline is more basic than acetamide.}$$

This conclusion directly contradicts Statement I, which claims “Aniline is less basic than acetamide.” Hence Statement I is false.

Statement II says, “In aniline, the lone pair of electrons on nitrogen atom is delocalised over benzene ring due to resonance and hence less available to a proton.” The resonance structures shown above confirm exactly this reasoning, so Statement II is true.

We have now established:

$$\text{Statement I : false}, \qquad \text{Statement II : true}$$

Among the given options, the description “Statement I is false but statement II is true” corresponds to Option B.

Hence, the correct answer is Option B.

Correct Option (First Choice):

“Reaction is possible and compound (A) will be major product.”

This statement is correct. Although a large amount of meta product is formed, the para product $$\mathrm{(A)}$$ is obtained slightly more, with approximately $$51\%$$ yield, making it the major product.

Option B is incorrect:

While the $$\mathrm{-NH_2}$$ group is normally an ortho/para-directing group, compound $$\mathrm{(B)}$$ is still formed in significant amount because, in the acidic nitrating medium, aniline gets protonated to form the anilinium ion.

The anilinium ion is meta-directing.

Option C is incorrect:

Compound $$\mathrm{(B)}$$ is formed in considerable quantity (approximately $$47\%$$), but it is not the major product since compound $$\mathrm{(A)}$$ is formed slightly more.

Option D is incorrect:

Sulphonation of aniline is possible under different reaction conditions such as heating with concentrated sulfuric acid. However, under standard nitration conditions at $$288\ \mathrm{K}$$, nitration is the dominant reaction pathway.

The formation of such a large amount of the meta-nitro product is due to the strongly acidic conditions used during the nitration process. Under normal circumstances, the amine group attached to the benzene ring is a powerful activating group. Because the nitrogen atom has a lone pair of electrons, it pushes those electrons into the ring through resonance, which naturally directs any incoming electrophiles, like the nitro group, to the ortho and para positions. This is why you still see a significant amount of the para product forming in the reaction mixture, as a portion of the molecules react in their standard state.

However, the reagents used for this nitration, concentrated nitric acid and concentrated sulfuric acid, create an intensely acidic environment. Because the amine group on the aniline molecule is basic, it rapidly reacts with the abundant protons in this acidic mixture. This acid-base reaction protonates the amine group, converting it into an anilinium ion. This transformation is the crucial key to the puzzle because the nitrogen atom now holds a positive charge and no longer has a free lone pair of electrons to donate to the benzene ring.

Once converted into the anilinium ion, the functional group entirely changes its chemical personality. Instead of pushing electrons into the ring, the positively charged nitrogen acts as a strong electron-withdrawing group, pulling electron density away from the benzene ring through the inductive effect. Electron-withdrawing groups inherently deactivate the ortho and para positions much more heavily than the meta position. Consequently, the anilinium ions in the mixture direct the incoming nitro groups almost exclusively to the meta position, which perfectly explains why you end up with a massive forty-seven percent yield of the meta-nitroaniline product alongside the expected para product.

Thus the right answer is the formation of anilinium ion, which is option A.

The reaction sequence can be understood by analyzing each step separately.

In the first step, 2-methylbutanal reacts with hydrogen cyanide ($HCN$) to form a cyanohydrin.

The reaction proceeds through nucleophilic addition, where the cyanide ion ($CN^-$) attacks the electrophilic carbonyl carbon.

Pure $$HCN$$ is a weak acid and ionizes only slightly, so it does not provide a sufficient concentration of $$CN^-$$ ions. Therefore, a base catalyst is required to generate the active nucleophile.

Hence, reagent $$X$$ must be $$NaOH$$, which converts $$HCN$$ into $$CN^-$$. Using an acid such as $$HNO_3$$ would suppress the ionization of $$HCN$$ and inhibit the reaction.

The cyanohydrin formed in the first step contains both a hydroxyl group ($-OH$) and a nitrile group ($-C \equiv N$).

In the second step, treatment with $$LiAlH_4$$ reduces the nitrile group to a primary amine.

The reduction follows the transformation

$$-C \equiv N \longrightarrow -CH_2NH_2.$$

The hydroxyl group remains unchanged during this process.

Therefore, the carbon atom of the nitrile group is retained in the final product, and the product must contain a $$-CH_2NH_2$$ substituent attached to the carbon bearing the hydroxyl group.

Any structure showing the amino group attached directly as $$-NH_2$$ to that carbon is incorrect because it omits the carbon originally present in the cyanide group.

Hence,

• $$X = NaOH$$
• $$Y$$ is the structure containing the $$-CH_2NH_2$$ group.

Therefore, the correct answer is the first option.

$$\mathrm{ArN_2^+Cl^- + H_3PO_2 + H_2O \rightarrow ArH + N_2 + H_3PO_3 + HCl}$$ (From Amines chapter in NCERT).

$$\mathrm{C_6H_6 + CH_3Cl \xrightarrow[\ ]{Anhyd.\ AlCl_3} C_6H_5CH_3 + HCl}$$ (From Hydrocarbons chapter in NCERT, Friedel-Crafts Alkylation Reaction)
Similarly if we use $$ CH_3CH_2Cl$$

$$\mathrm{C_6H_6 + CH_3CH_2Cl \xrightarrow[\ ]{Anhyd.\ AlCl_3} C_6H_5CH_2CH_3 + HCl}$$

Thus the right option is A.

The starting material is nitrobenzene and the final product is 3-chlorophenol ($$m$$-chlorophenol). Since the chlorine atom and the hydroxyl group are in a $$meta$$ relationship, the chlorination must be carried out while the $$-NO_2$$ group is still present because it is a strong $$meta$$-directing group.

The $$-NO_2$$ group is strongly deactivating and $$meta$$ directing, whereas both the $$-NH_2$$ and $$-OH$$ groups are strongly activating and $$ortho/para$$ directing. Therefore, reducing the nitro group before chlorination would lead predominantly to $$ortho$$ and $$para$$ substituted products rather than the desired $$meta$$ isomer.

The correct sequence is:

• Step 1: $$Cl_2/FeCl_3$$
  Nitrobenzene undergoes electrophilic aromatic substitution, and the $$-NO_2$$ group directs chlorine to the $$meta$$ position, giving $$m$$-chloronitrobenzene.
• Step 2: $$Fe/HCl$$
  The nitro group is reduced to an amino group, producing $$m$$-chloroaniline.
• Step 3: $$NaNO_2,\ HCl,\ 0^\circ\mathrm{C}$$
  The amino group undergoes diazotization to form $$m$$-chlorobenzenediazonium chloride.
• Step 4: $$H_2O/H^+$$ (warm)
  Hydrolysis of the diazonium salt replaces the diazonium group with a hydroxyl group, yielding 3-chlorophenol.

The other sequences are incorrect because:

• Option (A): Reduction occurs before chlorination. The resulting $$-NH_2$$ group is $$ortho/para$$ directing, leading to the wrong products.
• Option (B): Conversion to phenol occurs before chlorination. The $$-OH$$ group is also $$ortho/para$$ directing, so chlorination does not give the required $$meta$$ isomer.
• Option (D): Diazotization is attempted directly on a nitro compound. Only primary aromatic amines undergo diazotization, so the nitro group must first be reduced.

Hence, the correct sequence is Option (C).