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

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

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.

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.

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 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.

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.

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:

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

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.

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

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).

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.

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.

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.

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

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

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

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

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

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

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

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

Therefore, a one-carbon chain extension is required.

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence, its basic strength decreases.

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

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

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

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

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

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

Therefore, the strongest Brønsted base is:

$$\mathrm{Quinuclidine}$$

Correct option:

$$\mathrm{D}$$

Given reaction:

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

Step 1: Generation of the electrophile

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

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

Hence,

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

Step 2: Protonation of aniline

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

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

Therefore, direct nitration of aniline gives predominantly:

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

Evaluation of Statements:

For Statement (A):

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

$${\text{False}}$$

For Statement (B):

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

$${\text{True}}$$

For Statement (C):

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

$${\text{False}}$$

For Statement (D):

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

$${\text{True}}$$

Therefore,

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

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

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

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

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

The resulting carbocation reacts with water to give an alcohol:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hydrolysis of the isocyanate finally gives the primary amine.

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

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

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

Now we consider the two compounds in the question:

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

(ii) Aniline  : $$C_{6}H_{5}NH_{2}$$

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

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

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

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

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

symbolically extended over the ring as

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

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

Therefore the experimentally observed order of basicity is

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

or, equivalently,

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

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

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

We have now established:

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

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

Hence, the correct answer is Option B.

Correct Option (First Choice):

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

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

Option B is incorrect:

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

The anilinium ion is meta-directing.

Option C is incorrect:

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

Option D is incorrect:

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

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

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

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

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

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

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

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

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

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

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

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

The reduction follows the transformation

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

The hydroxyl group remains unchanged during this process.

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

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

Hence,

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

Therefore, the correct answer is the first option.

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

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

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

Thus the right option is A.

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

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

The correct sequence is:

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

The other sequences are incorrect because:

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

Hence, the correct sequence is Option (C).

We need to determine which reaction(s) will not give $$p$$-aminoazobenzene.

Reaction A: Nitrobenzene is first reduced with Sn/HCl to give aniline. Then treatment with $$HNO_2$$ (i.e., $$NaNO_2 + HCl$$) converts aniline to benzenediazonium chloride. Finally, coupling with aniline (a weakly activating group) gives $$p$$-aminoazobenzene. This reaction works correctly.

Reaction B: Nitrobenzene is treated with $$NaBH_4$$. However, $$NaBH_4$$ is a mild reducing agent that cannot reduce nitrobenzene to aniline. $$NaBH_4$$ is typically used for reducing aldehydes, ketones, and acid chlorides, not nitro groups. Reduction of nitrobenzene requires stronger reducing agents like Sn/HCl, Fe/HCl, or catalytic hydrogenation. Since aniline is not formed, the subsequent steps to make $$p$$-aminoazobenzene will not proceed. This reaction does not give the desired product.

Reaction C: Aniline is treated with $$HNO_2$$ (diazotization) to form benzenediazonium chloride, which then undergoes coupling with aniline in the presence of HCl to give $$p$$-aminoazobenzene. This is the standard method and works correctly.

Therefore, only reaction B will not give $$p$$-aminoazobenzene, which corresponds to option (1).

In the ammonolysis of alkyl halides, the alkyl halide reacts with ammonia to form an alkylammonium halide salt. For example, $$R-X + NH_3 \rightarrow R-NH_3^+X^-$$.

The product is an ammonium salt, not the free amine. The ammonium salt is formed because $$HX$$ (the hydrogen halide produced during the reaction) protonates the amine. To liberate the free amine from this salt, a strong base such as $$NaOH$$ is added.

The $$NaOH$$ neutralises the acid ($$HX$$) present in the salt: $$R-NH_3^+X^- + NaOH \rightarrow R-NH_2 + NaX + H_2O$$. This effectively removes the acidic impurity (hydrogen halide) and releases the free amine.

Therefore, the purpose of $$NaOH$$ in this reaction is to remove acidic impurities.

In the Gabriel phthalimide synthesis we first convert phthalimide into its potassium salt, represented as $$$\mathrm{C_{6}H_{4}(CO)_{2}NH + KOH \;\longrightarrow\; C_{6}H_{4}(CO)_{2}NK + H_{2}O}.$$$

This potassium phthalimide then reacts with an alkyl (or aryl) halide through a nucleophilic substitution (specifically an $$\mathrm{S_{N}2}$$ mechanism) to give an N-substituted phthalimide: $$$\mathrm{C_{6}H_{4}(CO)_{2}NK + R{-}X \;\longrightarrow\; C_{6}H_{4}(CO)_{2}N{-}R + KX}.$$$

Finally, basic or acidic hydrolysis of the N-substituted phthalimide liberates the corresponding primary amine: $$$\mathrm{C_{6}H_{4}(CO)_{2}N{-}R + 2\,H_{2}O \;\longrightarrow\; C_{6}H_{4}(COOH)_{2} + RNH_{2}.}$$$

For an aromatic primary amine we would need $$\mathrm{R = Ar},$$ i.e. the halide must be an aryl halide such as $$\mathrm{Ar{-}Cl}$$ or $$\mathrm{Ar{-}Br}.$$ Therefore the key reaction step would have to be

$$$\mathrm{C_{6}H_{4}(CO)_{2}NK + Ar{-}X \;\longrightarrow\; C_{6}H_{4}(CO)_{2}N{-}Ar + KX}.$$$

This step can occur only if the aryl halide undergoes nucleophilic substitution. However, aryl halides possess a carbon-halogen bond that is strengthened by resonance with the benzene ring, giving it partial double-bond character. They also lack the steric and electronic conditions required for the $$\mathrm{S_{N}1}$$ or $$\mathrm{S_{N}2}$$ mechanisms. Hence, aryl halides do not undergo nucleophilic substitution reactions under normal Gabriel-synthesis conditions.

Because this nucleophilic substitution does not take place, the intermediate $$\mathrm{C_{6}H_{4}(CO)_{2}N{-}Ar}$$ cannot be formed, and consequently aromatic primary amines cannot be prepared by the Gabriel method. Therefore the assertion that “Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines” is true, and the stated reason “Aryl halides do not undergo nucleophilic substitution reaction” is also true.

Moreover, the inability of aryl halides to participate in nucleophilic substitution is precisely the reason why the Gabriel synthesis fails for aromatic substrates. Thus the reason correctly explains the assertion.

Hence, the correct answer is Option 3.

This is Gabriel Phthalimide synthesis. The end product is:

In the general aliphatic reaction, when a primary aliphatic amine ($$R-NH_2$$) is treated with nitrous acid ($$HNO_2$$)—which is generated in situ by mixing sodium nitrite ($$NaNO_2$$) and hydrochloric acid ($$HCl$$)—it rapidly undergoes diazotization to form an aliphatic diazonium salt intermediate ($$[R-N_2^+Cl^-]$$). Unlike aromatic diazonium salts, these aliphatic diazonium salts are highly unstable even at very low temperatures. Nitrogen gas is an excellent leaving group and immediately departs, allowing the resulting carbocation to be instantly attacked by water ($$H_2O$$) from the aqueous solvent. This transformation is represented by the general equation:

$$R-NH_2 \xrightarrow{NaNO_2 + HCl} [R-N_2^+Cl^-] \xrightarrow{H_2O} R-OH + N_2 \uparrow + HCl$$

Consequently, as shown in the final step of the reaction sequence, the major compound formed is an aliphatic alcohol ($$R-OH$$), accompanied by the brisk effervescence of nitrogen gas ($$N_2$$). Therefore, this corresponds to Option D.

Explanation: The evolution of nitrogen gas $$N_2$$ indicates that the reaction involves an arenediazonium salt.

Ethanol acts as a mild reducing agent and replaces the diazonium group $$-N_2^+Cl^-$$ with hydrogen.

Since the final product is anisole $$C_6H_5OCH_3$$, the starting compound $$A$$ must be methoxybenzenediazonium chloride.

The molecular formula $$C_7H_7N_2OCl$$ matches methoxybenzenediazonium chloride.

During the reaction, ethanol is oxidized to acetaldehyde $$CH_3CHO$$, and the chloride ion combines with the proton released from ethanol to form $$HCl$$.

Thus:

$$A = p\text{-methoxybenzenediazonium chloride}$$

$$X = CH_3CHO$$

$$Y = HCl$$

The correct answer is p-aminophenol for A and N-(4-hydroxyphenyl)acetamide for B.

According to NCERT Class 12 Chemistry, Chapter 13: Amines:

"Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum." "Primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is known as acylation."

The given reaction sequence proceeds through the following steps:

First, the starting material, p-nitrophenol, undergoes catalytic hydrogenation using H₂ and a palladium catalyst in ethanol. This reducing agent selectively reduces the nitro group (-NO₂) to an amino group (-NH₂). The phenolic hydroxyl group (-OH) is not reduced under these conditions. Thus, product A is p-aminophenol, a benzene ring with an -OH group and an -NH₂ group at para positions.

Next, intermediate A is treated with 1.0 equivalent of acetic anhydride. p-Aminophenol contains two nucleophilic functional groups: the amino group (-NH₂) and the phenolic hydroxyl group (-OH). Because nitrogen is less electronegative than oxygen, it holds its lone pair of electrons less tightly, making the -NH₂ group a stronger nucleophile than the -OH group.

Since only one equivalent of the acylating agent (acetic anhydride) is provided, selective acetylation occurs at the more reactive site. The stronger amine nucleophile attacks the acetic anhydride, converting the -NH₂ group into an acetamido group (-NHCOCH₃). The less reactive -OH group remains unaffected.

Therefore, the major product B is N-(4-hydroxyphenyl)acetamide, where the original nitro group has been fully transformed into an amide, while the starting hydroxyl group remains intact.

The reaction sequence can be understood by following the transformation of the functional group at each stage.

In the first step, 4-methylpentanoyl chloride reacts with alcoholic ammonia.

This is a nucleophilic acyl substitution reaction in which ammonia attacks the carbonyl carbon and replaces the chloride ion, producing the corresponding primary amide.

Thus, the first intermediate formed is

$$\boxed{\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CONH_2},}$$

which is 4-methylpentanamide.

In the second step, the amide is treated with

$$\mathrm{Br_2/NaOH}.$$

These reagents carry out the Hoffmann bromamide degradation, converting a primary amide into a primary amine containing one carbon atom less than the original amide.

The carbonyl carbon is eliminated during the reaction.

Hence, the second intermediate is

$$\boxed{\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-NH_2},}$$

which is 3-methylbutan-1-amine.

In the third step, the primary aliphatic amine is treated with

$$\mathrm{NaNO_2/HCl},$$

which generates nitrous acid in situ.

The amine undergoes diazotization to form an unstable aliphatic diazonium salt. This intermediate immediately loses nitrogen gas to produce a carbocation.

In the presence of water, the carbocation is attacked by a water molecule, resulting in hydrolysis to the corresponding alcohol.

Therefore, the final product obtained is

$$\boxed{\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-OH},}$$

which is 3-methylbutan-1-ol.

Hence, the correct answer is

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

The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom. When the lone pair is delocalised (especially into electron-withdrawing groups like carbonyl groups), the basicity decreases significantly.

Consider each compound: $$(C_2H_5)_3N$$ and $$(C_2H_5)_2NH$$ are simple aliphatic amines with alkyl groups that donate electron density to nitrogen through the inductive effect, making them reasonably basic.

$$(CH_3CO)NHC_2H_5$$ is an amide (N-ethylacetamide) where the nitrogen lone pair is delocalised into one carbonyl group through resonance: $$-N-C=O \leftrightarrow -N^+=C-O^-$$. This significantly reduces the basicity compared to simple amines.

$$(CH_3CO)_2NH$$ is a diacylamide (diacetamide) where the nitrogen lone pair is delocalised into two carbonyl groups. The presence of two electron-withdrawing acetyl groups causes even greater delocalisation of the lone pair compared to a mono-acylamide, making this compound the least basic of all the given options.

Therefore, $$(CH_3CO)_2NH$$ is the least basic compound among the given choices.

Hinsberg reagent is:

$$\mathrm{C_6H_5SO_2Cl}$$

(benzenesulfonyl chloride)

It reacts with:

$$\mathrm{1^\circ\ and\ 2^\circ\ amines}$$

but does not react with:

$$\mathrm{3^\circ\ amines,\ aldehydes,\ ketones}$$

(A)

$$\mathrm{p\text{-}NO_2C_6H_4CN \xrightarrow{Na/Hg,\ C_2H_5OH}}$$

Both:

$$\mathrm{-NO_2 \rightarrow -NH_2}$$

and

$$\mathrm{-CN \rightarrow -CH_2NH_2}$$

Product contains primary amine groups.

Hence, it reacts with Hinsberg reagent.

(B)

$$\mathrm{C_6H_5CN \xrightarrow[\ H_2O\ ]{SnCl_2/HCl} C_6H_5CHO}$$

This is the Stephen reaction.

Product formed is benzaldehyde:

$$\mathrm{C_6H_5CHO}$$

which is an aldehyde.

Hence, it does not react with Hinsberg reagent.

(C)

$$\mathrm{p\text{-}CNC_6H_4CHO \xrightarrow[\ H_3O^+\ ]{LiAlH_4}}$$

$$\mathrm{-CN \rightarrow -CH_2NH_2}$$

$$\mathrm{-CHO \rightarrow -CH_2OH}$$

Product contains a primary amine.

Hence, it reacts with Hinsberg reagent.

(D)

$$\mathrm{m\text{-}CH_3C_6H_4CN \xrightarrow{H_2/Ni}}$$

$$\mathrm{-CN \rightarrow -CH_2NH_2}$$

Product formed is a primary amine.

Hence, it reacts with Hinsberg reagent.

Therefore, the only product that does not react with Hinsberg reagent is:

$$\boxed{\mathrm{Option\ B}}$$

The Reaction Name: The transformation of Product A into Product B using

$$SnCl_2$$, $$HCl$$, and $$H_3O^+$$ is a specific named reaction known as the Stephen reduction (or Stephen aldehyde synthesis).

• The Starting Material: Product A, formed from the first step (diazotization and reaction with $$KCN$$), is benzonitrile.
• Step 1 - Partial Reduction: Stannous chloride ($$SnCl_2$$) and hydrochloric acid ($$HCl$$) act as a reducing agent. They supply hydrogen to the carbon-nitrogen triple bond of benzonitrile. However, they only reduce it partially to a double bond, forming a stannic chloride byproduct and an intermediate salt called an imine hydrochloride.
• Equation for Reduction: $$C_6H_5-C\equiv N + 2[H] + HCl \xrightarrow{SnCl_2} C_6H_5-CH=NH \cdot HCl$$
• Step 2 - Acidic Hydrolysis: The imine hydrochloride is highly susceptible to hydrolysis. When the acidic water ($$H_3O^+$$) is added, it cleaves the carbon-nitrogen double bond. The nitrogen atom is removed (forming ammonium chloride) and is replaced by an oxygen atom.
• Equation for Hydrolysis: $$C_6H_5-CH=NH \cdot HCl + H_2O \xrightarrow{H_3O^+} C_6H_5-CHO + NH_4Cl$$
• The Final Product: Because the nitrile group ($$-C\equiv N$$) has been formally replaced by an aldehyde group ($$-CHO$$), your final Product B is benzaldehyde.

The reaction described is the carbylamine reaction (isocyanide test): when a primary amine is treated with CHCl₃ and KOH (alcoholic), it gives an isocyanide (isonitrile), which has a characteristic foul smell and is highly toxic. The reaction is: R-NH₂ + CHCl₃ + 3KOH → R-N≡C: + 3KCl + 3H₂O.

This reaction is specific to primary amines — secondary and tertiary amines do not give this positive test. Therefore, compound A must be a primary amine.

Compound B is the isonitrile (isocyanide) formed, which is indeed toxic. This is R-N≡C (isocyanide / isonitrile), not a nitrile (which would be R-C≡N).

The isonitrile B can be destroyed (decomposed / hydrolyzed) by heating with concentrated HCl: R-N≡C + HCl + H₂O → R-NH₂ + HCO₂H (or HCOOH / formic acid). This converts the toxic isonitrile back to a primary amine and formic acid.

Thus: A is a primary amine, B is an isonitrile compound, and C is concentrated HCl. This corresponds to Option 3.

Nitrobenzene undergoes reduction with:

$$\mathrm{Sn/HCl}$$

The nitro group is completely reduced to a primary amine.

$$\mathrm{C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2}$$

Thus, intermediate $$\mathrm{A}$$ is:

$$\mathrm{Aniline\ (C_6H_5NH_2)}$$

In the next step, aniline reacts with benzenediazonium chloride:

$$\mathrm{C_6H_5N_2^+Cl^-}$$ through azo coupling.

The diazonium ion acts as a weak electrophile, while aniline activates the benzene ring through the:

$$\mathrm{-NH_2}$$ group. Since: $$\mathrm{-NH_2}$$ is ortho/para directing, coupling occurs mainly at the para position due to lower steric hindrance.

The reaction is:

$$\mathrm{C_6H_5NH_2 + C_6H_5N_2^+Cl^- \longrightarrow p\text{-}H_2NC_6H_4-N=N-C_6H_5 + HCl}$$

The final product $$\mathrm{P}$$ is:

$$\mathrm{p\text{-}aminoazobenzene}$$

which is a yellow azo dye.