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 $$N_{2}O$$

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.

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).

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.

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:

Hence, the correct answer 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: $$[3(Ca_3(PO_4)_2) \cdot CaF_2]$$.

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.

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

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

$$\mathrm{ArN_2^+Cl^- + H_3PO_2 + H_2O \rightarrow ArH + N_2 + H_3PO_3 + HCl}$$ (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.

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.

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.

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 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.

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.

Hoffmann Bromamide Degradation: This reaction converts a primary amide $$(-(RCONH_{2}))$$ to a primary amine $$(-(RNH_{2}))$$.

The reduction of ketones using lithium aluminum hydride ($$LiAlH_4$$) produces secondary alcohols. It is a powerful, irreversible nucleophilic addition reaction.

The reaction of an amide with hypobromite is the Hofmann bromamide degradation (Hofmann rearrangement). In this reaction, an amide $$\text{RCONH}_2$$ is treated with $$\text{Br}_2$$ in aqueous $$\text{NaOH}$$ (which generates $$\text{OBr}^-$$, the hypobromite ion).

The mechanism proceeds through formation of an N-bromoamide, which undergoes base-induced rearrangement to an isocyanate $$\text{R-N=C=O}$$. In the aqueous alkaline medium, the isocyanate is hydrolysed to a carbamic acid intermediate that decomposes to give the primary amine $$\text{RNH}_2$$ and $$\text{CO}_2$$. Since the reaction is carried out in strongly basic (NaOH) solution, the $$\text{CO}_2$$ is immediately absorbed to form the carbonate ion $$\text{CO}_3^{2-}$$.

Thus the carbonyl carbon of the original amide is lost as $$\text{CO}_3^{2-}$$, which is option (1).

The Hinsberg test, which uses para-toluenesulphonyl chloride ($$\text{TsCl}$$, also called tosyl chloride), is the standard method to distinguish and separate primary, secondary, and tertiary amines.

When a primary amine reacts with $$\text{TsCl}$$, it forms a sulphonamide $$\text{R-NH-Ts}$$ that still has an acidic N-H hydrogen. This product dissolves in aqueous NaOH to form a soluble sodium salt, and can be separated from the mixture.

A secondary amine also reacts with $$\text{TsCl}$$ to form a sulphonamide $$\text{R}_2\text{N-Ts}$$, but this product has no N-H hydrogen, so it does not dissolve in NaOH. It remains as an insoluble solid, which can be filtered off.

A tertiary amine does not react with $$\text{TsCl}$$ at all (it has no N-H bond), so it remains unchanged in solution.

Since all three classes of amines give different, physically separable products, para-toluenesulphonyl chloride can be used to separate primary, secondary, and tertiary amines. The correct answer is Option A.

To form a colored azo dye with $$\beta$$-naphthol in NaOH, the compound must be a primary aromatic amine. This means the amine group ($$-NH_2$$) must be attached directly to a benzene ring (for example, aniline or a substituted aniline). Thus the correct option is B.

We know that sodium hydrogencarbonate, written as $$\text{NaHCO}_3$$, is a salt of a weak acid $$\text{H}_2\text{CO}_3$$ (carbonic acid) and a strong base $$\text{NaOH}$$. Whenever an acid that is stronger than $$\text{H}_2\text{CO}_3$$ comes in contact with $$\text{NaHCO}_3$$, an acid-base reaction occurs in which $$\text{HCO}_3^-$$ is converted to $$\text{H}_2\text{CO}_3$$, and the unstable $$\text{H}_2\text{CO}_3$$ immediately decomposes to give carbon dioxide gas:

First write the general sequence of steps:

$$\text{H}^+ + \text{HCO}_3^- \;\longrightarrow\; \text{H}_2\text{CO}_3$$

$$\text{H}_2\text{CO}_3 \;\longrightarrow\; \text{H}_2\text{O} + \text{CO}_2\uparrow$$

So, out of the given options we have to identify the compound that can really supply the necessary $$\text{H}^+$$ (i.e. behaves as an acid in water) more strongly than carbonic acid.

Let us analyse each option in turn.

Option A: $$\text{CH}_3\text{NH}_3^+ \text{Cl}^-$$   (methyl­ammonium chloride)
This is a salt formed from a weak base $$\text{CH}_3\text{NH}_2$$ (methylamine) and a strong acid $$\text{HCl}$$. In aqueous solution the cation $$\text{CH}_3\text{NH}_3^+$$ behaves as a conjugate acid and releases a proton:

$$\text{CH}_3\text{NH}_3^+ \;\rightleftharpoons\; \text{CH}_3\text{NH}_2 + \text{H}^+$$

The liberated $$\text{H}^+$$ can react with $$\text{HCO}_3^-$$ exactly as outlined above. Writing the full ionic equation with $$\text{NaHCO}_3$$ we obtain

$$\text{CH}_3\text{NH}_3^+\,\text{Cl}^- + \text{Na}^+\,\text{HCO}_3^- \;\longrightarrow\; \text{CH}_3\text{NH}_2 + \text{Na}^+\,\text{Cl}^- + \text{H}_2\text{CO}_3$$

and then

$$\text{H}_2\text{CO}_3 \;\longrightarrow\; \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Because a gas is evolved, brisk effervescence of $$\text{CO}_2$$ will be observed.

Option B: $$\text{CH}_3\text{NHO}^-$$ (methyloxime anion) is actually the conjugate base of hydroxylamine derivative; it does not donate a proton, hence it cannot supply $$\text{H}^+$$ to drive the above reaction with $$\text{NaHCO}_3$$.

Option C: $$\text{CH}_3\text{-CO-NH}_2$$ (acetamide) is a very weak acid (essentially neutral) because the -NH2 group is not able to donate a proton readily; therefore it fails to react with $$\text{NaHCO}_3$$.

Option D: $$\text{CH}_3\text{NH}_2$$ (methylamine) is itself a base, so instead of giving $$\text{H}^+$$ it would accept protons; there is no possibility of evolving $$\text{CO}_2$$ from $$\text{NaHCO}_3$$.

Among all choices, only the salt $$\text{CH}_3\text{NH}_3^+ \text{Cl}^-$$ is sufficiently acidic to protonate $$\text{HCO}_3^-$$ and liberate $$\text{CO}_2$$.

Hence, the correct answer is Option A.

Let us first identify each amine:

A. Phenyl methanamine is $$C_6H_5CH_2NH_2$$ — benzylamine, an aliphatic primary amine (the $$-NH_2$$ group is on the $$-CH_2-$$ carbon, not directly on the ring).

B. N,N-Dimethylaniline is $$C_6H_5N(CH_3)_2$$ — a tertiary aromatic amine.

C. N-Methylaniline is $$C_6H_5NHCH_3$$ — a secondary aromatic amine.

D. Benzenamine is $$C_6H_5NH_2$$ — aniline, a primary aromatic amine.

Benzylamine (A) is an aliphatic amine because the nitrogen lone pair is not involved in resonance with the benzene ring. In aromatic amines (B, C, D), the lone pair on nitrogen delocalizes into the ring, reducing basicity. Therefore, benzylamine is the most basic of all four.

Among the aromatic amines, methyl groups on nitrogen donate electron density through their inductive effect (+I effect), which partially compensates for the resonance withdrawal. N,N-Dimethylaniline (B) has two methyl groups on nitrogen, making it more basic than N-Methylaniline (C), which has one methyl group. Aniline (D) has no methyl groups on nitrogen and is the least basic.

Therefore, the order of basic nature is: $$A > B > C > D$$.

This problem uses the Hinsberg test, where an amine reacts with benzene sulfonyl chloride ($$\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$$). The key observation is that the product B is soluble in dilute NaOH, which tells us that B contains an acidic $$\text{N-H}$$ proton, meaning the original amine A must be a primary amine ($$\text{R-NH}_2$$).

A primary amine reacts with benzenesulfonyl chloride to give $$\text{R-NHSO}_2\text{C}_6\text{H}_5$$, a sulfonamide that still has one N-H bond. This proton is acidic enough (due to the electron-withdrawing sulfonyl group) to be removed by dilute NaOH, making the product soluble in base. A secondary amine would give a sulfonamide with no N-H bond, so it would be insoluble in NaOH. A tertiary amine does not react with benzenesulfonyl chloride at all.

Among the options, $$\text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2$$ (option 1) is a tertiary amine. $$\text{C}_6\text{H}_5\text{NHCH}_2\text{CH}_3$$ (option 2) is a secondary amine. $$\text{C}_6\text{H}_5\text{CH}_2\text{NHCH}_3$$ (option 3) is also a secondary amine. Only $$\text{C}_6\text{H}_5\text{CH}(\text{NH}_2)(\text{CH}_3)$$ (option 4) is a primary amine with a free $$-\text{NH}_2$$ group.

Therefore compound A is $$\text{C}_6\text{H}_5\text{CH}(-\text{NH}_2)(-\text{CH}_3)$$, option (4).

Step 1: Formation of Product A

• The Reaction: Aniline reacts with acetic anhydride in an acylation reaction. The nitrogen's lone pair attacks the carbonyl carbon of the anhydride.
• Product A: This reaction converts the primary amine into an amide, specifically forming acetanilide ($N$-phenylacetamide).
• The Purpose: This step "protects" the amine. The lone pair on the nitrogen is now involved in resonance with the adjacent carbonyl oxygen of the acetyl group. This pulls electron density away from the benzene ring, significantly reducing the ring's reactivity. It changes the group from strongly activating to only moderately activating, allowing for controlled, single substitutions in the next step.

Step 2: Formation of Product B

• The Reaction: Acetanilide (Product A) is treated with bromine ($Br_2$) in acetic acid ($CH_3COOH$) at room temperature. This is an electrophilic aromatic substitution.
• Product B: The acetamido group ($-NHCOCH_3$) is an ortho/para-directing group. Because this group is physically quite bulky, it creates steric hindrance, blocking incoming electrophiles from easily attacking the adjacent ortho positions. Therefore, the bromine attaches almost exclusively at the opposite end of the ring, making the major product p-bromoacetanilide (para-bromoacetanilide).

We begin by recalling some well-established facts about aliphatic primary amines.

(i) Gabriel\,phthalimide\,synthesis: This method converts potassium phthalimide to a primary amine after two simple steps (alkylation and hydrolysis). Only the first hydrogen on nitrogen is replaced, so the product obtained is always a primary amine and never a secondary or tertiary one. Hence primary aliphatic amines can be prepared by this synthesis.

(ii) Basicity order in aqueous medium: The general order for aliphatic amines is secondary $$\gt \text{primary}\gt \text{tertiary}.$$ This comes from a compromise between the $$+I$$ effect (which favours basicity) and solvation of the protonated base (which disfavours steric crowding). Therefore a primary amine is indeed less basic than the corresponding secondary amine.

(iii) Hydrogen bonding (intermolecular association): A primary amine has two N-H bonds whereas a secondary amine possesses only one. The more N-H bonds present, the more hydrogen bonds can form between two molecules: No.\ of hydrogen bonds $$\propto\text{No.\ of N-H bonds}.$$ Consequently the intermolecular association in primary amines is actually greater than in secondary amines, not less.

(iv) Reaction with nitrous acid, HONO: For an aliphatic primary amine RNH$$_2$$, diazotisation gives the diazonium ion R-N$$_2^+$$, which immediately loses N$$_2$$ to generate the carbocation R$$^+$$. Water present in the reaction mixture captures this carbocation and furnishes the corresponding alcohol ROH. The overall transformation is sketched below:

$$\mathrm{RNH_2+HONO\;\longrightarrow\; R^+ + N_2 + 2H_2O\; ;\qquad R^+ + H_2O \longrightarrow ROH + H^+}.$$

For higher homologues (ethylamine, propylamine, etc.) the reaction proceeds cleanly to the alcohol. In the special case of methylamine, the methyl diazonium ion is so unstable that several side reactions compete; the major gaseous product is N$$_2$$, but the amount of methanol obtained is negligible. Hence in laboratory practice we say that “all primary aliphatic amines give the corresponding alcohol on treatment with nitrous acid except methylamine.”

Now we examine each option in the light of the above discussion:

Option A is consistent with fact (i) and is therefore a correct statement.

Option B matches fact (ii) and is likewise correct.

Option C claims that primary amines associate less than secondary amines, whereas fact (iii) tells us the reverse. So Option C is incorrect.

Option D agrees with fact (iv); methylamine is indeed treated as an exception in standard texts, so the statement is regarded as correct.

Only Option C fails to conform to the established chemistry.

Hence, the correct answer is Option C.

Ammonolysis is the reaction in which ammonia ($$NH_3$$) acts as the nucleophile and replaces a leaving group (such as a halide) in a substrate, forming a new C-N bond. The term specifically refers to cleavage by ammonia, analogous to how hydrolysis is cleavage by water.

Looking at the options: option (1) is an acylation reaction of an amine with an acid chloride (Schotten-Baumann reaction), option (2) is a reduction of a nitrile to a primary amine, and option (3) is a simple acid-base reaction forming an ammonium salt.

Option (4), $$C_6H_5CH_2Cl + NH_3 \to C_6H_5CH_2NH_2$$, is the reaction of benzyl chloride with ammonia where $$NH_3$$ displaces the chloride via nucleophilic substitution, which is the textbook definition of ammonolysis. The correct answer is option (4).

To determine the percentage yield of the amide product, we first calculate the molar masses of the compounds involved using the given atomic masses, which are

$$140.5 \text{ g/mol}$$for benzoyl chloride ($$C_7H_5OCl$$), $$169 \text{ g/mol}$$ for diphenylamine ($$C_{12}H_{11}N$$), and $$273 \text{ g/mol}$$

for the resulting amide product ($$C_{19}H_{15}NO$$). Next, we find the number of moles for both reactants to identify the limiting reagent. The moles of benzoyl chloride are calculated as

$$\frac{0.140 \text{ g}}{140.5 \text{ g/mol}} \approx 0.000996 \text{ mol}$$, while the moles of diphenylamine are $$\frac{0.388 \text{ g}}{169 \text{ g/mol}} \approx 0.002296 \text{ mol}$$. Since $$0.001 < 0.002296$$, benzoyl chloride is the limiting reagent and dictates the maximum amount of product that can be formed. Using the 1:1 stoichiometric ratio, the theoretical yield of the amide product is $$0.001 \text{ mol} \times 273 \text{ g/mol} \approx 0.272 \text{ g}$$.

Finally, we calculate the percentage yield by dividing the actual given yield ($$0.210 \text{ g}$$) by the theoretical yield ($$0.272 \text{ g}$$) and multiplying by 100, which gives$$\left( \frac{0.210}{0.272} \right) \times 100 \approx 76.9\%$$. Rounding off to the nearest integer, the final percentage yield is 77.

Aniline reacts with acetic anhydride (or acetyl chloride) to form acetanilide in an acetylation reaction:

$$C_6H_5NH_2 + CH_3COCl \rightarrow C_6H_5NHCOCH_3 + HCl$$

The molar mass of aniline ($$C_6H_5NH_2$$) is $$93$$ g/mol, and the molar mass of acetanilide ($$C_6H_5NHCOCH_3$$) is $$135$$ g/mol.

Moles of aniline = $$\frac{1.86}{93} = 0.02$$ mol.

Since 1 mole of aniline produces 1 mole of acetanilide, the theoretical yield of acetanilide = $$0.02 \times 135 = 2.70$$ g.

However, 10% of the product is lost during purification. So the actual amount obtained = $$2.70 \times 0.9 = 2.43$$ g = $$243 \times 10^{-2}$$ g.

The value of the answer is $$\textbf{243}$$.

• (A) Isobutylamine: This is a primary aliphatic amine. Its corresponding alkyl halide (isobutyl halide) can successfully undergo the required $$S_N2$$ reaction.
• (B) Ethylamine: This is a simple primary aliphatic amine. Ethyl halides are excellent substrates for $$S_N2$$ reactions, making this highly suitable.
• (C) Benzylamine: While it contains a benzene ring, the amine group is attached to a $$CH_2$$ carbon, making it a primary aliphatic amine. The corresponding benzyl halide is actually highly reactive towards $$S_N2$$ nucleophilic attack, so this works perfectly.

Amines that CANNOT be synthesized:

• (D) Aniline: This is a primary aromatic amine, meaning the nitrogen is attached directly to the benzene ring. It cannot be synthesized using this method.

Reasoning: The Gabriel synthesis is a classic and highly specific method used exclusively for preparing primary aliphatic amines. Thus ,it cannot be used to prepare aniline.

We begin with the basic fact that reduction of the nitro group $$\left( -\,\mathrm{NO_2}\right)$$ to the amino group $$\left( -\,\mathrm{NH_2}\right)$$ requires a source of nascent hydrogen or a catalytic hydrogenation system. A reagent or a reagent‐combination capable of supplying such hydrogen will convert nitrobenzene $$\left(\mathrm{C_6H_5NO_2}\right)$$ into aniline $$\left(\mathrm{C_6H_5NH_2}\right).$$

Now we examine each option one by one.

I. $$\mathrm{Sn + HCl}$$    In the acidic medium, metallic tin reacts to give $$\mathrm{Sn^{2+}}$$ ions and nascent hydrogen: $$\mathrm{Sn + 2\,HCl \;\rightarrow\; SnCl_2 + 2\,[H]}.$$ The nascent hydrogen reduces the nitro group stepwise $$\bigl(-\mathrm{NO_2}\;\rightarrow\;-\mathrm{NHOH}\;\rightarrow\;-\mathrm{NH_2}\bigr)$$ so that $$\mathrm{C_6H_5NO_2 \xrightarrow[HCl]{Sn} C_6H_5NH_2}.$$ Hence this reagent works.

II. $$\mathrm{Sn + NH_4OH}$$    Tin needs an acidic medium to generate nascent hydrogen. In the alkaline medium furnished by $$\mathrm{NH_4OH},$$ tin does not produce the necessary hydrogen; therefore the nitro group is not reduced all the way to aniline. So this combination fails for the required conversion.

III. $$\mathrm{Fe + HCl}$$    Exactly analogous to Sn/HCl, iron in acid gives nascent hydrogen: $$\mathrm{Fe + 2\,HCl \;\rightarrow\; FeCl_2 + 2\,[H]}.$$ So we get $$\mathrm{C_6H_5NO_2 \xrightarrow[HCl]{Fe} C_6H_5NH_2}.$$ Thus this reagent is effective.

IV. $$\mathrm{Zn + HCl}$$    Zinc in hydrochloric acid behaves similarly: $$\mathrm{Zn + 2\,HCl \;\rightarrow\; ZnCl_2 + 2\,[H]},$$ followed by $$\mathrm{C_6H_5NO_2 \xrightarrow[HCl]{Zn} C_6H_5NH_2}.$$ Hence this reagent also works.

V. $$\mathrm{H_2}\;-\;\mathrm{Pd}$$    This is catalytic hydrogenation. The general statement is: “Hydrogen gas in the presence of a transition-metal catalyst (Pd, Pt, Ni, Raney Ni) reduces $$\mathrm{-NO_2}$$ to $$\mathrm{-NH_2}$$.” Therefore $$\mathrm{C_6H_5NO_2 + 3\,H_2 \xrightarrow{Pd} C_6H_5NH_2 + 2\,H_2O}.$$ So it is a valid reagent.

VI. $$\mathrm{H_2}\;-\;\text{Raney Ni}$$    Raney nickel is another well-known hydrogenation catalyst, hence $$\mathrm{C_6H_5NO_2 + 3\,H_2 \xrightarrow{Raney\;Ni} C_6H_5NH_2 + 2\,H_2O}.$$ Thus this reagent also accomplishes the conversion.

Counting all the successful cases we have:

Sn/HCl},\;$$\text{Fe/HCl}$$,\;$$\text{Zn/HCl}$$,\;$$\text{H}_2\text{/Pd}$$,\;$$\text{H}_2\text{/Raney$$ Ni

Total $$= 5.$$

Hence, the correct answer is Option 5.

First, recall the principle of Kjeldahl’s method. In this method, the compound is digested with concentrated $$\text{H}_2\text{SO}_4$$, which converts all nitrogen present in the compound into ammonium sulphate $$\left((\text{NH}_4)_2\text{SO}_4\right)$$. The general reaction written symbolically is

$$\text{Organic nitrogen} \;+\; \text{H}_2\text{SO}_4 \;\xrightarrow{\text{heat}}\; (\text{NH}_4)_2\text{SO}_4$$

After digestion, the reaction mixture is made alkaline with excess $$\text{NaOH}$$. This liberates ammonia gas:

$$(\text{NH}_4)_2\text{SO}_4 + 2\,\text{NaOH} \;\longrightarrow\; 2\,\text{NH}_3 \uparrow + \text{Na}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$$

The evolved $$\text{NH}_3$$ is then absorbed in a known excess of standard acid and finally estimated by back-titration. Thus, the success of Kjeldahl’s method depends on the ability of hot concentrated $$\text{H}_2\text{SO}_4$$ to convert every form of nitrogen in the sample into $$\text{NH}_4^+$$.

However, if nitrogen is present in such a way that it cannot be reduced to $$\text{NH}_4^+$$ under these conditions, the method fails. Specifically, nitrogen present in its oxidised form (such as the nitro group $$\left(-\text{NO}_2\right)$$ or the nitroso group $$\left(-\text{NO}\right)$$) resists conversion to ammonium ions during sulphuric-acid digestion. Because of this resistance, the total nitrogen is not quantitatively transformed into $$\text{NH}_3$$ and the estimation becomes inaccurate.

Now we examine each option to see how the nitrogen is bonded:

Option A: $$\text{C}_6\text{H}_5\text{NH}_2$$ is an aniline molecule where nitrogen is present as a primary amine $$(-\text{NH}_2)$$. In concentrated $$\text{H}_2\text{SO}_4$$, an amine nitrogen is easily protonated and ultimately converted to ammonium sulphate. So, Kjeldahl’s method works here.

Option B: $$\text{CH}_3\text{CH}_2{-}\text{C}\equiv\text{N}$$ is propionitrile, containing a cyano group $$(-\text{C}\equiv\text{N})$$. Nitrile nitrogen is in a reduced state and, on digestion with hot $$\text{H}_2\text{SO}_4$$, it is hydrolysed to give ammonium sulphate. Thus, Kjeldahl’s method also works for nitriles.

Option C: $$\text{C}_6\text{H}_5\text{NO}_2$$ is nitrobenzene. Here nitrogen is present in the nitro group $$(-\text{NO}_2)$$ where its oxidation state is $$+5$$. Because this nitrogen is already highly oxidised, concentrated $$\text{H}_2\text{SO}_4$$ cannot further oxidise it to $$\text{NH}_4^+$$. Instead, the nitro group remains largely unaffected or sometimes even forms $$\text{NO}_2$$ or $$\text{NO}$$ gases that escape. Therefore, nitro-nitrogen is not estimated by Kjeldahl’s method.

Option D: $$\text{NH}_2{-}\text{CO}{-}\text{NH}_2$$ (urea) contains two amide nitrogens $$(-\text{NH}_2)$$. Amide nitrogens are also reduced forms of nitrogen and get converted to ammonium sulphate during digestion, so Kjeldahl’s method works here as well.

From the above discussion, only the compound with a nitro group, namely $$\text{C}_6\text{H}_5\text{NO}_2$$ (Option C), cannot be analysed by the Kjeldahl procedure.

Hence, the correct answer is Option C.

We have the starting compound $$\text{C}_2\text{H}_5\text{CN}$$, which is also written as $$\text{CH}_3\text{CH}_2\text{CN}$$. The functional group present is a nitrile $$(-\,\text{C}≡\text{N})$$. Our target molecule is $$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$, that is, a primary amine containing three carbon atoms.

To go from a nitrile $$\text{R-C}≡\text{N}$$ to a primary amine $$\text{R-CH}_2\text{NH}_2$$, the standard method is complete reduction of the triple bond. The commonly used laboratory reagent for this purpose is lithium aluminium hydride, $$\text{LiAlH}_4$$.

First, recall the general reduction formula:

$$\text{R-C}≡\text{N} \overset{\text{LiAlH}_4}{\underset{\text{ether}}{\rightarrow}} \text{R-CH}_2\text{NH}_2$$

This tells us that every nitrile carbon receives two hydride ions from $$\text{LiAlH}_4$$, converting the $$C≡N$$ triple bond first to an imine intermediate and finally to the saturated $$\text{CH}_2\text{NH}_2$$ group. No carbon atoms are lost or gained in the process.

Applying the same transformation to our substrate:

$$\text{CH}_3\text{CH}_2\text{C}≡\text{N} \overset{\text{LiAlH}_4}{\underset{\text{ether}}{\rightarrow}} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$

Thus the reagent that achieves the desired conversion is clearly $$\text{LiAlH}_4$$.

Why the other options fail:

• $$\text{NaBH}_4$$ is a milder hydride donor; it reduces aldehydes and ketones but does not normally touch nitriles.
• $$\text{CaH}_2$$ is not a typical hydride reagent for organic reductions; it is mainly a drying agent.
• $$\text{Na(CN)BH}_3$$ (sodium cyanoborohydride) is even milder than $$\text{NaBH}_4$$ and is used for reductive amination, not for reducing nitriles.

So, only $$\text{LiAlH}_4$$ can reduce $$\text{C}_2\text{H}_5\text{CN}$$ all the way to $$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$$.

Hence, the correct answer is Option C.

We have the reagent $$CHCl_3$$ in the presence of alcoholic $$KOH$$. In organic chemistry this specific combination is called the carbylamine test. The underlying reaction is written in general form as

$$R{-}NH_2 \;+\; CHCl_3 \;+\; 3KOH \;\longrightarrow\; R{-}NC \;+\; 3KCl \;+\; 3H_2O$$

It is important to note that the reaction proceeds only when the substrate contains a primary amine group, i.e., a nitrogen atom bonded to exactly one carbon atom and two hydrogen atoms. Secondary and tertiary amines, amides, imides and other nitrogen-containing functional groups do not undergo this reaction.

Now we examine each substance mentioned in the options in terms of the presence or absence of a primary $$NH_2$$ group.

Adenine: In the purine ring of adenine there is an exocyclic $$NH_2$$ group attached to carbon-6. Since this nitrogen is bonded to two hydrogens and one carbon, it is a primary amine and therefore adenine will give the carbylamine reaction.

Thymine: Thymine is a pyrimidine derivative that contains two carbonyl groups and an $$NH$$ in the ring, but no exocyclic $$NH_2$$ group. Hence it lacks a primary amine and will not respond to the carbylamine reagent.

Proline: Proline is a cyclic amino acid in which the nitrogen is incorporated into a five-membered ring (pyrrolidine). That nitrogen is bonded to two carbons and one hydrogen, making it a secondary amine, so the carbylamine reaction is negative for proline.

Lysine: Lysine possesses two separate $$NH_2$$ groups: one attached to the α-carbon (the normal amino acid amino group) and another on the ε-carbon in its side chain. Both are primary amines, so lysine readily undergoes the carbylamine reaction.

Putting these observations together, we see that the compounds giving a positive test are adenine and lysine.

Option C lists exactly these two compounds. None of the other options contains the complete and correct set.

Hence, the correct answer is Option C.