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 $$N_{2}O$$

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.

When ethane-1,2-diamine (ethylenediamine, abbreviated as "en") is added progressively to an aqueous solution of nickel(II) chloride ($$NiCl_2$$), the following stepwise complexation occurs:

$$NiCl_2$$ dissolves in water to form $$[Ni(H_2O)_6]^{2+}$$, which is green in colour. This is an octahedral complex where the d-d transitions in $$Ni^{2+}$$ ($$d^8$$) with weak-field water ligands produce a green colour.

$$[Ni(H_2O)_6]^{2+} + en \rightarrow [Ni(en)(H_2O)_4]^{2+} + 2H_2O$$

As the stronger field ligand "en" replaces two water molecules, the crystal field splitting increases slightly, shifting the absorption and producing a pale blue colour.

$$[Ni(en)(H_2O)_4]^{2+} + en \rightarrow [Ni(en)_2(H_2O)_2]^{2+} + 2H_2O$$

With further increase in field strength, the complex turns blue.

$$[Ni(en)_2(H_2O)_2]^{2+} + en \rightarrow [Ni(en)_3]^{2+} + 2H_2O$$

The tris(ethylenediamine)nickel(II) complex has the highest crystal field splitting in this series, producing a violet colour.

The observed colour sequence is: Green $$\rightarrow$$ Pale Blue $$\rightarrow$$ Blue $$\rightarrow$$ Violet.

The correct answer is Option 3.

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.

Statement I: Dumas method is used for estimation of nitrogen in an organic compound.

This is true. The Dumas method involves heating the organic compound with copper oxide (CuO) in an atmosphere of $$CO_2$$, which converts nitrogen in the compound to $$N_2$$ gas. The volume of $$N_2$$ collected is measured to estimate the nitrogen content.

Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with concentrated $$H_2SO_4$$.

This is false. The description in Statement II corresponds to the Kjeldahl method, not the Dumas method. In the Kjeldahl method, the organic compound is heated with concentrated $$H_2SO_4$$, which converts nitrogen to ammonium sulphate. The Dumas method uses CuO, not $$H_2SO_4$$.

The answer is Option A: Statement I is true but Statement II is false.

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.

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.

We need to determine the molar mass of the final product (A) formed in the given reaction sequence starting from nitrobenzene.

Reaction Sequence Analysis:

• Step 1: Reduction ($$\text{Sn} + \text{HCl}$$)
  Nitrobenzene is reduced to aniline ($$\text{C}_6\text{H}_5\text{NH}_2$$).
• Step 2: Diazotization ($$\text{NaNO}_2 + \text{HCl, } 0^\circ\text{C}$$)
  Aniline reacts to form benzene diazonium chloride ($$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$$).
• Step 3: Sandmeyer Reaction ($$\text{Cu}_2\text{Cl}_2$$)
  The diazonium group is replaced by chlorine, yielding chlorobenzene ($$\text{C}_6\text{H}_5\text{Cl}$$).
• Step 4: Wurtz-Fittig / Fittig Reaction ($$\text{Na / dry ether}$$)
  Two molecules of chlorobenzene couple together in the presence of sodium metal to form biphenyl as the final product (A).

Molar Mass Calculation of Product (A):

• The molecular formula of biphenyl is $$\text{C}_{12}\text{H}_{10}$$.
• Atomic masses: $$\text{C} = 12\text{ g/mol}$$, $$\text{H} = 1\text{ g/mol}$$.
• $$\text{Molar Mass} = (12 \times 12) + (10 \times 1)$$
  $$\text{Molar Mass} = 144 + 10 = 154\text{ g/mol}$$

Answer: 154

The sequence of reagents supplied in the question carries out electrophilic bromination on a methyl-substituted benzene ring.
A methyl group is electron-releasing; therefore it directs the incoming electrophile ($$Br^+$$) preferentially to its ortho and para positions. Because of steric hindrance the para product is obtained in the largest amount.

Hence the major product $$\text{(A)}$$ is para-bromotoluene, having the molecular formula $$C_7H_7Br$$.

Molar-mass calculation (using the atomic masses given in the question):
$$\text{Molar mass of }C_7H_7Br = 7(12) + 7(1) + 80 = 84 + 7 + 80 = 171\ \text{g mol}^{-1}$$

Therefore, the molar mass of the major product $$\text{(A)}$$ is

$$\boxed{171}$$

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 identify the correct statements about p-block elements and their compounds from the given list.

Analysis of each statement:

(A) Non-metals have higher electronegativity than metals.

This is TRUE. Electronegativity is the tendency of an atom to attract shared electrons towards itself. Non-metals, being on the right side of the periodic table, have higher electronegativity because they need fewer electrons to complete their octet and have smaller atomic sizes.

(B) Non-metals have lower ionisation enthalpy than metals.

This is FALSE. Non-metals generally have higher ionisation enthalpy than metals. Non-metals hold their electrons more tightly due to their smaller atomic size and higher effective nuclear charge, making it harder to remove an electron.

(C) Compounds formed between highly reactive non-metals and highly reactive metals are generally ionic.

This is TRUE. Highly reactive metals (like Na, K) have very low ionisation enthalpies and readily lose electrons, while highly reactive non-metals (like F, Cl) have high electron affinity and readily gain electrons. The large electronegativity difference between them leads to the formation of ionic bonds through complete transfer of electrons.

(D) The non-metal oxides are generally basic in nature.

This is FALSE. Non-metal oxides are generally acidic in nature. For example, $$CO_2$$ forms carbonic acid, $$SO_3$$ forms sulphuric acid, $$P_4O_{10}$$ forms phosphoric acid when dissolved in water.

(E) The metal oxides are generally acidic or neutral in nature.

This is FALSE. Metal oxides are generally basic in nature. For example, $$Na_2O$$, $$CaO$$, $$MgO$$ are all basic oxides that form hydroxides when dissolved in water.

The correct statements are (A) and (C) only.

The correct answer is Option (2): (A) and (C) only.

Statement I: In group 13, the stability of +1 oxidation state increases down the group.

This is correct due to the inert pair effect. As we go down the group, the $$ns^2$$ electrons become less available for bonding, making the +1 state more stable. Tl predominantly shows +1 state. ✓

Statement II: The atomic size of gallium is greater than that of aluminium.

Due to d-block contraction (poor shielding by 3d electrons), Ga has a smaller atomic radius than Al. The actual order is Al (143 pm) > Ga (135 pm). Statement II is incorrect. ✗

The correct answer is Option (4): Statement I is correct but Statement II is incorrect.

We need to identify how many ions from $$Sn^{4+}, Sn^{2+}, Pb^{2+}, Tl^{3+}, Pb^{4+}, Tl^{+}$$ behave as oxidising agents.

Key Concept: Inert Pair Effect and Oxidising Behaviour

An oxidising agent is a species that readily accepts electrons (gets reduced). Due to the inert pair effect, the higher oxidation states of heavier p-block elements are less stable, making them strong oxidising agents as they tend to get reduced to the lower oxidation state.

Analysis of each ion:

- $$Sn^{4+}$$: Tin in +4 state. While $$Sn^{2+}$$ is more stable for heavier members, $$Sn^{4+}$$ is actually quite stable for tin (Period 5), so it is not a particularly strong oxidising agent.

- $$Sn^{2+}$$: This is a reducing agent (gets oxidised to $$Sn^{4+}$$), not an oxidising agent.

- $$Pb^{2+}$$: Lead in +2 state is the more stable oxidation state of lead. It is not an oxidising agent in this context.

- $$Tl^{3+}$$: Thallium in +3 state is unstable due to the strong inert pair effect in Period 6. $$Tl^{3+}$$ readily gets reduced to $$Tl^{+}$$, making it a strong oxidising agent.

- $$Pb^{4+}$$: Lead in +4 state is unstable due to the inert pair effect. $$Pb^{4+}$$ readily gets reduced to $$Pb^{2+}$$ (e.g., $$PbO_2$$ is a strong oxidising agent). Strong oxidising agent.

- $$Tl^{+}$$: This is the more stable state of thallium. It is not an oxidising agent.

The ions that act as oxidising agents are $$Tl^{3+}$$ and $$Pb^{4+}$$ -- a total of 2 ions.

The correct answer is Option 2: 2.

We evaluate statements about Group 14 elements (C, Si, Ge, Sn, Pb):

(A) Covalent radius decreases down the group from C to Pb in a regular manner.

This is incorrect. The covalent radius generally increases down the group, not decreases. Also, the increase from Si to Ge is less pronounced due to the d-orbital contraction (scandide contraction).

(B) Electronegativity decreases from C to Pb down the group gradually.

This is incorrect. While the general trend is a decrease in electronegativity going down the group, it does not decrease gradually/uniformly. There are irregularities due to d- and f-orbital effects.

(C) Maximum covalence of C is 4 whereas other elements can expand their covalence due to presence of d-orbitals.

This is correct. Carbon has no d-orbitals in its valence shell, so its maximum covalence is 4. Si, Ge, Sn, and Pb have vacant d-orbitals and can expand their covalence beyond 4 (e.g., $$SiF_6^{2-}$$).

(D) Heavier elements do not form $$p\pi - p\pi$$ bonds.

This is correct. Due to their large atomic sizes, Si, Ge, Sn, and Pb cannot form effective $$p\pi - p\pi$$ multiple bonds, unlike carbon which readily forms $$C=C$$ and $$C \equiv C$$ bonds.

(E) Carbon can exhibit negative oxidation states.

This is correct. Carbon can show negative oxidation states, such as $$-4$$ in $$CH_4$$, $$-1$$ in $$C_2H_6$$, etc.

The correct statements are (C), (D), and (E) only.

The answer is Option B: (C), (D) and (E) Only.

Find x (neutrons in more abundant isotope of boron).

Boron has two isotopes: $$^{10}B$$ (19.9%) and $$^{11}B$$ (80.1%). The more abundant isotope is $$^{11}B$$.

Number of neutrons in $$^{11}B = 11 - 5 = 6$$. So $$x = 6$$.

Find y (oxidation state of boron in the product with air).

When amorphous boron is heated with air, it forms boron trioxide ($$B_2O_3$$):

$$ 4B + 3O_2 \rightarrow 2B_2O_3 $$

In $$B_2O_3$$, the oxidation state of boron is $$+3$$. So $$y = +3$$.

Therefore, $$x + y = 6 + 3 = 9$$.

The correct answer is Option (2): 9.

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.

Assertion (A): Both rhombic and monoclinic sulphur exist as $$S_8$$ while oxygen exists as $$O_2$$.

This is correct. Both allotropes of sulphur (rhombic and monoclinic) contain $$S_8$$ molecules in a puckered ring structure, whereas oxygen naturally exists as $$O_2$$.

Reason (R): Oxygen forms $$p\pi - p\pi$$ multiple bonds with itself and other elements having small size and high electronegativity like C, N, which is not possible for sulphur.

The first part about oxygen forming $$p\pi - p\pi$$ bonds is correct. However, the statement "which is not possible for sulphur" is too absolute. Sulphur can form some degree of $$p\pi - p\pi$$ bonding in certain compounds (e.g., in SO$$_2$$, SO$$_3$$). The key distinction is that sulphur cannot form effective $$p\pi - p\pi$$ bonds with itself due to the larger size of 3p orbitals, which is why it forms $$S_8$$ rings with single bonds. Thus, the Reason as stated is not entirely correct.

Since (A) is correct but (R) is not correct, the answer is Option A.

Statement I: Gallium has a wide liquid range (m.p. 29.76°C, b.p. 2204°C), making it useful for high-temperature thermometers. True.

Statement II: Gallium melts at ~303 K (29.76°C), so it would be solid below this temperature. A Ga thermometer cannot measure 256 K since Ga is solid there. False.

The correct answer is Option (4): Statement I is true but Statement II is false.

We need to identify which product is NOT formed when lead sulphide (PbS) reacts with dilute nitric acid ($$HNO_3$$).

First, the balanced chemical equation for this reaction is:

$$ 3PbS + 8HNO_3(\text{dilute}) \rightarrow 3Pb(NO_3)_2 + 3S + 2NO\uparrow + 4H_2O $$

Next, we examine why these products form.

In this reaction:

- $$PbS$$ is oxidized: $$S^{2-}$$ in PbS is oxidized to elemental sulphur $$S^0$$ (oxidation state changes from $$-2$$ to $$0$$).

- $$HNO_3$$ acts as the oxidizing agent: $$N^{+5}$$ in $$HNO_3$$ is reduced to $$N^{+2}$$ in $$NO$$ (nitric oxide).

- $$Pb^{2+}$$ combines with $$NO_3^-$$ to form lead nitrate $$Pb(NO_3)_2$$.

Now, we identify which products are formed and which are not.

The products of this reaction are:

- Lead nitrate ($$Pb(NO_3)_2$$) — formed

- Sulphur ($$S$$) — formed

- Nitric oxide ($$NO$$) — formed

- Water ($$H_2O$$) — formed

Nitrous oxide ($$N_2O$$) is NOT a product of this reaction. Nitrous oxide would require nitrogen to be reduced to the $$+1$$ oxidation state, but with dilute $$HNO_3$$ and PbS, the reduction product is nitric oxide ($$NO$$, where nitrogen is in the $$+2$$ state).

The correct answer is Option (2): Nitrous oxide.

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)

Approach: Evaluate each statement about Group 15 elements and identify the incorrect ones.

Statement (A): Dinitrogen is a diatomic gas which acts like an inert gas at room temperature.

$$N_2$$ has a very strong triple bond (bond energy $$\approx 941$$ kJ/mol), making it unreactive at room temperature. It behaves like an inert gas under normal conditions. Statement A is correct.

Statement (B): The common oxidation states of these elements are $$-3, +3,$$ and $$+5$$.

Group 15 elements (N, P, As, Sb, Bi) commonly exhibit oxidation states of $$-3$$ (e.g., $$NH_3$$), $$+3$$ (e.g., $$PCl_3$$), and $$+5$$ (e.g., $$PCl_5$$). Statement B is correct.

Statement (C): Nitrogen has a unique ability to form $$p\pi - p\pi$$ multiple bonds.

Due to its small size, nitrogen can form effective lateral overlap of $$p$$-orbitals, enabling $$p\pi - p\pi$$ bonding (as in $$N_2$$, $$NO_2$$). Heavier elements of Group 15 cannot form such bonds effectively. Statement C is correct.

Statement (D): The stability of $$+5$$ oxidation state increases down the group.

Due to the inert pair effect, the stability of the $$+5$$ oxidation state actually decreases down the group (Bi prefers +3 over +5). Statement D is incorrect.

Statement (E): Nitrogen shows a maximum covalency of 6.

Nitrogen has no $$d$$-orbitals available for bonding (it is a second-period element). Its maximum covalency is 4 (e.g., $$NH_4^+$$). Statement E is incorrect.

The incorrect statements are (D) and (E) only, which corresponds to Option 3.

Hydroxyapatite, the enamel on teeth, has the formula $$[3(Ca_3(PO_4)_2) \cdot Ca(OH)_2]$$ or $$Ca_{10}(PO_4)_6(OH)_2$$.

When $$F^-$$ ions replace $$OH^-$$, the much harder fluoroapatite is formed according to the reaction:

$$Ca_{10}(PO_4)_6(OH)_2 + 2F^- \rightarrow Ca_{10}(PO_4)_6F_2 + 2OH^-$$

The formula of fluoroapatite is $$[3(Ca_3(PO_4)_2) \cdot CaF_2]$$ or equivalently $$Ca_{10}(PO_4)_6F_2$$.

The correct answer is Option 4: $$[3(Ca_3(PO_4)_2) \cdot CaF_2]$$.

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 — $$H_2N-(CH_2)_4-NH_2$$

A. Br2 water test  

This is a test for unsaturation (alkenes or alkynes). The reddish-orange colour of bromine water disappears (decolorizes) as the bromine reacts across the double or triple bond to form a colourless di-bromoalkane.

B. Ceric ammonium nitrate (CAN) 

This test is used to detect the Alcoholic group . When CAN is added to an alcohol, it forms a complex through an alkoxy exchange, resulting in a distinct pink or red colour.

C. Ferric chloride test 

This is a specific test for Phenols or enols. Neutral ferric chloride reacts with phenols to form complex coordination compounds that exhibit characteristic colours ranging from violet and blue to green or red, depending on the specific phenol present.

D. 2, 4 - DNP (Brady’s Reagent) test 

This test identifies the Carbonyl group (>C=O) found in aldehydes and ketones. The reagent reacts with the carbonyl carbon to form a hydrazone, which is a poorly soluble solid appearing as a yellow, orange, or red precipitate.

Correct Reaction: Option B

• Reaction: Addition of HI to 1-methylcyclohexene.
• Mechanism: This follows Markovnikov’s rule. The electrophile ($$H^+$$) adds to the double-bonded carbon with more hydrogens, and the nucleophile ($$I^-$$) adds to the more substituted tertiary ($$3^\circ$$) carbon.
• Product: 1-iodo-1-methylcyclohexane is correctly formed.

Incorrect Reactions

• Option A: Aliphatic primary amines ($$R-NH_2$$) react with $$HNO_2$$ to form alcohols ($$R-OH$$), but the carbon chain length must remain constant. The reactant has 3 carbons (propylamine), but the product shown (ethanol) only has 2.
• Option C: Reaction with $$Br_2$$ under UV light or Heat ($$\Delta$$) favors allylic substitution, not addition across the double bond.
• Option D: This is the Hoffmann Bromamide Degradation. It should result in an amine with one less carbon ($$C_2H_5NH_2$$) than the starting amide ($$C_2H_5CONH_2$$). The product shown incorrectly retains the original carbon count.

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. 

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.

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 $$\boxed{135}$$.

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 $$\boxed{95 \times 10^{-1}\text{ g}}$$.

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

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.

We need to evaluate two statements about sulphanilic acid ($$H_2N-C_6H_4-SO_3H$$, also known as 4-aminobenzenesulphonic acid).

The first statement claims that sulphanilic acid gives the esterification test for a carboxyl group. This is incorrect. Sulphanilic acid contains a sulphonic acid group ($$-SO_3H$$), not a carboxyl group ($$-COOH$$). The esterification test involves reacting a carboxyl group with an alcohol in the presence of an acid catalyst to form a fruity-smelling ester, a reaction that sulphonic acid groups do not undergo. Therefore, sulphanilic acid will not give a positive esterification test for a carboxyl group.

The second statement asserts that sulphanilic acid gives a red colour in Lassaigne’s test for extra element detection. This is correct. Sulphanilic acid contains both nitrogen (in the $$-NH_2$$ group) and sulphur (in the $$-SO_3H$$ group). When subjected to sodium fusion (Lassaigne’s) test, these elements form sodium thiocyanate according to:

$$ Na + C + N + S \to NaSCN $$

Upon adding ferric chloride (FeCl$$_3$$) solution to the extract, the reaction is:

$$ 3NaSCN + FeCl_3 \to Fe(SCN)_3 + 3NaCl $$

Ferric thiocyanate, Fe(SCN)$$_3$$, imparts a characteristic blood-red colour, confirming the presence of both nitrogen and sulphur.

Thus, the correct answer is Option 4: Statement I is incorrect but Statement II is correct.

Oxidation number (O.N.) is obtained by assigning the bonding electrons of every hetero-atomic bond to the more electronegative partner and adding the formal charge. Whenever an atom is bonded only to atoms of the same element, that bond contributes $$0$$ to its O.N.

Case 1: $$\mathbf{Br_3O_8}$$
Let the oxidation number of the three bromine atoms be $$x_1,x_2,x_3$$ (they need not be identical). Oxygen (except in peroxides and superoxides) has $$-2$$. Total charge is zero, so
$$x_1+x_2+x_3+8(-2)=0\;\;\Longrightarrow\;\;x_1+x_2+x_3=+16$$ Whatever individual values the three bromine atoms possess, none of them can be $$0$$ because their sum is $$+16$$. Number of atoms in O.N. $$0 = 0$$.

Case 2: $$\mathbf{F_2O}$$
Fluorine is the most electronegative element, so $$\text{O.N.(F)}=-1$$. Let O.N.(O) = $$x$$. $$x+2(-1)=0\;\;\Longrightarrow\;\;x=+2$$ No atom has O.N. $$0$$ here. Number of atoms in O.N. $$0 = 0$$.

Case 3: $$\mathbf{H_2S_4O_6}$$ (tetrathionic acid)
This molecule contains a chain $$HO_3S\!-\!S\!-\!S\!-\!SO_3H$$. • The two terminal sulphur atoms are each bonded to three oxygens and one sulphur. Such an environment is the same as in the dithionate ion, whose sulphur is known to be $$+5$$. Assign O.N.(terminal S) = $$+5$$.
• Let the O.N. of each internal sulphur be $$y$$.
For the whole neutral acid (with $$2$$ hydrogens at $$+1$$): $$2(+1)+2(+5)+2y+6(-2)=0$$ $$2+10+2y-12=0\;\;\Longrightarrow\;\;2y=0\;\;\Longrightarrow\;\;y=0$$ Thus both internal sulphur atoms are in the zero oxidation state. Number of atoms in O.N. $$0 = 2$$.

Case 4: $$\mathbf{H_2S_5O_6}$$ (pentathionic acid)
Structure: $$HO_3S\!-\!S\!-\!S\!-\!S\!-\!SO_3H$$ (a five-sulphur chain). • Terminal sulphur atoms again have O.N. $$+5$$.
• Let the three internal sulphur atoms have O.N.s $$y_1,y_2,y_3$$.
Overall charge $$0$$ gives $$2(+1)+2(+5)+(y_1+y_2+y_3)+6(-2)=0$$ $$2+10+(y_1+y_2+y_3)-12=0\;\;\Longrightarrow\;\;y_1+y_2+y_3=0$$ In a pure $$S\!-\!S$$ chain an individual sulphur is normally either $$0$$ or $$-1$$. A convenient integral solution is $$y_1=0,\;y_2=0,\;y_3=0$$ - fully consistent with the bonding scheme (each of those three S atoms is attached only to other sulphurs). Hence all three internal sulphur atoms are in O.N. $$0$$. Number of atoms in O.N. $$0 = 3$$.

Case 5: $$\mathbf{C_3O_2}$$ (carbon sub-oxide, $$O=C=C=C=O$$)
Let the oxidation numbers of the three carbons be $$x_1,x_2,x_3$$ (left to right). Each oxygen is $$-2$$, so total for both oxygens is $$-4$$: $$x_1+x_2+x_3-4=0\;\;\Longrightarrow\;\;x_1+x_2+x_3=+4$$ The two terminal carbonyl carbons (each double-bonded to an oxygen and singly to another carbon) are in O.N. $$+2$$. Therefore $$x_1=+2,\;x_3=+2 \;\;\Rightarrow\;\;x_2=0$$ for the central carbon. Number of atoms in O.N. $$0 = 1$$.

Total count of atoms in oxidation state 0:
$$0\;(Br_3O_8) + 0\;(F_2O) + 2\;(H_2S_4O_6) + 3\;(H_2S_5O_6) + 1\;(C_3O_2) = 6$$

Hence, the required sum is 6.

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.

The starting compound is p-nitrotoluene.

On bromination with $$Br_2$$, the $$-CH_3$$ group acts as an ortho/para director and the $$-NO_2$$ group acts as a meta director. Both groups direct the incoming bromine to the same position, giving product A as 2-bromo-4-nitrotoluene.

Reduction with $$Sn/HCl$$ converts the nitro group into an amino group, forming product B (2-bromo-4-methylaniline).

Treatment with $$NaNO_2/HCl$$ at $$273-278\text{ K}$$ converts the amino group into the corresponding diazonium salt, giving product C.

Reaction with $$H_3PO_2/H_2O$$ replaces the diazonium group by hydrogen (deamination), yielding product D as o-bromotoluene.

Oxidation with $$KMnO_4/KOH$$ followed by acidic workup $$H_3O^+$$ converts the benzylic $$-CH_3$$ group into $$-COOH$$ without affecting the bromine substituent.

Thus, product E is o-bromobenzoic acid (2-bromobenzoic acid).

Therefore,

• Product A = 2-bromo-4-nitrotoluene

• Product E = 2-bromobenzoic acid

Hence, the correct option is $$\boxed{\text{Option B}}$$.

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 ($$\text{p-CH}_3\text{--C}_6\text{H}_4\text{--CONH--C}_6\text{H}_5$$).

The required transformations are:

- Bromination at position 2

- Conversion of $$-NO_2$$ to $$-Cl$$

- Oxidation of $$-CH_3$$ to $$-COOH$$

Bromination must be carried out first. In 4-nitrotoluene, the $$-CH_3$$ group is ortho/para-directing and the $$-NO_2$$ group is meta-directing; both direct bromination to position 2.

$$\text{4-Nitrotoluene} \xrightarrow{Br_2/Fe} \text{2-Bromo-4-nitrotoluene}$$

The nitro group is then reduced to an amino group.

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

The amino group is converted into a diazonium salt and subsequently replaced by chlorine via the Sandmeyer reaction.

$$-NH_2 \xrightarrow{HONO} -N_2^+Cl^- \xrightarrow{CuCl} -Cl$$

Finally, the methyl group is oxidized to a carboxylic acid.

$$-CH_3 \xrightarrow{KMnO_4} -COOH$$

Thus, the correct sequence is

$$Br_2/Fe \;\rightarrow\; Fe/H^+ \;\rightarrow\; HONO \;\rightarrow\; CuCl \;\rightarrow\; KMnO_4$$

Hence, the correct answer is

$$\boxed{\text{Option D}}$$

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,

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

Firstly CN- nucleophile gets attached to the carbonyl reducing it to alcohol. In the next step EtOH in acidic medium on reaction with a Cyanide group results in the formation of Ethyl Ester. At last the 2 CH3- from Grignard Reagent get attached to the ester, firstly removing -OEt since it is a good leaving group then reducing the carbonyl carbon to -OH.

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.

Explanation: Benzene first undergoes Friedel-Crafts alkylation with isopropyl chloride in the presence of $$AlCl_3$$ to form cumene.

Cumene is oxidized by air to form cumene hydroperoxide, which on acidic hydrolysis gives phenol.

Therefore, intermediate $$P$$ is phenol.

Phenol reacts with:

$$Na_2Cr_2O_7/H_2SO_4$$

to undergo oxidation and form $$p$$-benzoquinone.

Phenol also reacts with bromine in non-polar solvent $$CS_2$$ to undergo monobromination.

Since the $$-OH$$ group is ortho/para-directing and steric hindrance favors substitution at the para position, the major product formed is $$p$$-bromophenol.

Thus, $$A = p\text{-benzoquinone}$$, $$B = p\text{-bromophenol}$$. The right option is B.

Hence, Option C is correct.