JEE Practical Organic Chemistry Questions

We need to identify how Ba²⁺ and Ca²⁺ are precipitated in group analysis of cations.

Group V cations include Ba²⁺, Ca²⁺, and Sr²⁺.

In the systematic qualitative analysis:

Ba²⁺ and Ca²⁺ both belong to Group V and are precipitated using ammonium carbonate ((NH₄)₂CO₃) as the group reagent.

Ba²⁺ → BaCO₃ (white precipitate)

Ca²⁺ → CaCO₃ (white precipitate)

Both are precipitated as carbonates.

Therefore, the answer is Option 4: carbonate & carbonate.

Case A: Mixture: Diethyl amine (a secondary amine) + Ethyl amine (a primary amine)

Reagents available: one of them must differentiate primary vs. secondary amines.
The carbylamine (isocyanide) test uses hot $$CHCl_3$$ and alcoholic $$KOH$$. A primary amine gives an extremely foul-smelling isocyanide; secondary and tertiary amines do not react.
Therefore only Ethyl amine responds, not Diethyl amine.
Thus the mixture is distinguished by reagent II.

Case B: Mixture: Acetaldehyde (aldehyde) + Acetone (ketone)

Tollen’s reagent (ammoniacal $$AgNO_3$$) oxidises aldehydes to carboxylate ions while reducing $$Ag^+$$ to metallic silver (silver mirror). Ketones do not react under these mild conditions.
Hence acetaldehyde gives a silver mirror; acetone remains unchanged.
Reagent IV is the correct choice.

Case C: Mixture: Ethanol (simple alcohol) + Phenol (aromatic alcohol with acidic -OH)

Neutral $$FeCl_3$$ reacts with phenols to form coloured (violet, blue, green) complexes, whereas ordinary alcohols such as ethanol show no colour change.
Thus reagent III distinguishes the two.

Case D: Mixture: Benzoic acid (no C=C) + Cinnamic acid (contains a C=C conjugated with the benzene ring)

Aqueous bromine (bromine water) is decolourised by compounds that add Br₂ across a carbon-carbon double bond. Cinnamic acid possesses such a double bond and rapidly decolourises bromine water; benzoic acid does not.
Therefore reagent I is the correct discriminator.

Collecting all matches:
A → II  B → IV  C → III  D → I

Option D which is: A-II, B-IV, C-III, D-I

The classical qualitative analysis scheme for basic radicals (cations) divides them into groups according to the reagent that first precipitates them.

Recollection of the relevant groups

Group II cations (e.g., $$Pb^{2+}, Cu^{2+}$$) give sulphide precipitates with $$H_2S$$ in an acidic medium.

Group III cations (e.g., $$Fe^{3+}$$) are precipitated as hydroxides with $$NH_4OH$$ in the presence of $$NH_4Cl$$; they are no longer present when $$H_2S$$ is later passed through the basic medium.

Group IV cations (e.g., $$Co^{2+}, Ni^{2+}, Zn^{2+}, Mn^{2+}$$) are not precipitated in acid, but they do give sulphide precipitates when $$H_2S$$ is passed through an ammoniacal (basic) solution containing $$NH_4OH/NH_4Cl$$.

Now compare the given ions:

• $$Fe^{3+}$$ → Group III → precipitates earlier as $$Fe(OH)_3$$; it will not appear when $$H_2S$$ is passed in base.

• $$Pb^{2+}$$, $$Cu^{2+}$$ → Group II → already removed as $$PbS$$, $$CuS$$ in acidic medium.

• $$Mn^{2+}$$ → Group IV → remains in solution until the ammoniacal stage and then precipitates as $$MnS$$ with $$H_2S$$.

Hence, the only ion that precipitates with $$H_2S$$ in the presence of $$NH_4OH$$ is $$Mn^{2+}$$.

The question then asks for the highest oxidation state of this corresponding metal, i.e., of manganese.
Manganese exhibits oxidation states from $$+2$$ up to $$+7$$; the highest, $$+7$$, is found in compounds like $$KMnO_4$$ and $$Mn_2O_7$$.

Therefore, the required highest oxidation state is $$+7$$.

Option D which is: $$+7$$

Statement I: Sodium dichromate and potassium dichromate are classified as primary standards in titrimetric analysis.

Incorrect, Only Potassium dichromate is classified as primary standards in titrimetric analysis as it is more stable and pure. Unlike, Sodium dichromate, it is highly hygroscopic (readily absorbs moisture from the air) which prevents Sodium dichromate from maintaining its original mass and purity.

Statement II: Phenolphthalein is a weak base, therefore it dissociates in acidic medium.

Incorrect, Phenolphthalein is a weak acid, not a weak base. It is largely unionized in acidic medium and gets ionized in basic medium (by showing pink colour).

The four qualitative tests named in the column on the right are routinely used for the following specific functional groups:

• Lucas test — used to distinguish primary, secondary and tertiary alcohols (appearance of turbidity).
• Hinsberg’s test — used to distinguish primary, secondary and tertiary amines (formation or non-formation of sulphonamide that dissolves in alkali).
• Tollen’s test — used to confirm the presence of an aldehyde group (formation of silver mirror/black precipitate of $$Ag$$).
• Phthalein dye test — used for phenols; phenol reacts with phthalic anhydride in the presence of conc. $$H_2SO_4$$ to give coloured phthalein dyes.

Now match each compound with the appropriate test:

Case A: Cyclohexanol contains the $$-OH$$ group on a secondary carbon, so it is identified by the Lucas test. Hence A → III.

Case B: Cyclohexylamine is a primary amine; primary amines respond to Hinsberg’s test. Hence B → I.

Case C: Cyclohexanecarbaldehyde has an aldehyde group, so it gives a positive Tollen’s test. Hence C → IV.

Case D: Phenol gives the phthalein dye (phenolphthalein) on reaction with phthalic anhydride and conc. $$H_2SO_4$$. Hence D → II.

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

Option A $$\longrightarrow\ $$A-III, B-I, C-IV, D-II

In Carius method, the Sulphur $$(S)$$ is converted into Barium Sulphate $$(BaSO_{4})$$.

Given:

• Mass of substance $$= 0.2 g$$
• Mass of $$BaSO_{4}$$ formed $$= 0.6 g$$
• Molar mass of $$BaSO_{4}$$ $$=231$$
• Atomic mass of $$S = 32$$

Mass of sulphur in $$0.6 g$$ of $$BaSO_{4}$$​:

$$Mass\ of\ S=\frac{32}{231}\times\ 0.6=0.0831g$$

Percentage of Sulphur:

$$\ \%\ S=\frac{0.0831g}{0.2}\times\ 100$$

$$\ \%\ S=41.56\ \%\ $$

The percentage of sulphur in the substance is 41.56%

We need to find the percentage of nitrogen in an organic compound using Dumas method.

Given mass of compound = 0.50 g, volume of $$N_2$$ collected = 70 mL, temperature = 300 K, pressure = 715 mm Hg, aqueous tension at 300 K = 15 mm Hg.

The actual pressure of dry nitrogen is obtained by subtracting the aqueous tension from the total pressure: $$P_{N_2} = P_{total} - P_{water} = 715 - 15 = 700 \text{ mm Hg}$$.

To convert the collected volume to STP conditions, we use the relation $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$ with $$P_2 = 760$$ mm Hg and $$T_2 = 273$$ K. Thus, $$V_{STP} = \frac{700 \times 70 \times 273}{760 \times 300} = \frac{13377000}{228000} = 58.67 \text{ mL}$$.

At STP, 22400 mL of $$N_2$$ has a mass of 28 g, so the mass of the collected nitrogen is $$\text{Mass of } N_2 = \frac{28 \times 58.67}{22400} = \frac{1642.76}{22400} = 0.07333 \text{ g}$$.

The percentage of nitrogen in the compound is then $$\% N = \frac{0.07333}{0.50} \times 100 = 14.67\%$$.

Rounding to the nearest integer: $$\% N \approx$$ 15%.

In the $$BaSO_4$$ method, the sulphur present in the organic compound is completely converted to $$BaSO_4$$. So, the amount of sulphur in the precipitate equals the amount of sulphur that was originally in the sample.

Step 1 Mass of the sample taken
Moles of compound $$X = 2.0 \times 10^{-3} \text{ mol}$$
Molecular weight of $$X = 76$$
$$\text{Mass of }X = 2.0 \times 10^{-3} \times 76 = 0.152\ \text{g}$$

Step 2 Moles of $$BaSO_4$$ obtained
Mass of $$BaSO_4$$ obtained $$= 0.4813\ \text{g}$$
Molecular weight of $$BaSO_4 = 233$$
$$n(\text{BaSO}_4) = \frac{0.4813}{233} = 2.066 \times 10^{-3}\ \text{mol}$$ (keep four significant figures)

Step 3 Moles and mass of sulphur present
Each mole of $$BaSO_4$$ contains one mole of sulphur.
$$n(S) = 2.066 \times 10^{-3}\ \text{mol}$$
Atomic mass of sulphur $$= 32$$
$$m(S) = 2.066 \times 10^{-3} \times 32 = 0.0661\ \text{g}$$

Step 4 Percentage of sulphur in the compound
$$\%S = \frac{m(S)}{\text{mass of }X} \times 100$$
$$\%S = \frac{0.0661}{0.152} \times 100 = 43.46\%$$

For $$--\ \times\ 10^{-1}$$= $$435\ \times\ 10^{-1}$$

Hence the required number is 435.

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 recall the characteristic qualitative tests used in practical organic chemistry:

Reagent A : $$NaHCO_3$$ (sodium bicarbonate solution)
  • Carboxylic acids react with $$NaHCO_3$$ producing brisk effervescence of $$CO_2$$.
  • Hence this test identifies the $$-COOH$$ group.
  ⇒ Functional group detected = II (carboxylic acid).

Reagent B : neutral $$FeCl_3$$ solution
  • Phenols form coloured (purple, green, blue) complexes with neutral $$Fe^{3+}$$ ions.
  • Therefore it is used to confirm phenolic $$-OH$$ groups.
  ⇒ Functional group detected = III (phenolic -OH).

Reagent C : ceric ammonium nitrate, $$ (NH_4)_2[Ce(NO_3)_6] $$
  • Alcohols give a wine-red or magenta colour with this reagent due to formation of a Ce(IV) complex.
  • Thus it detects alcoholic $$-OH$$ groups.
  ⇒ Functional group detected = IV (alcoholic -OH).

Reagent D : alkaline $$KMnO_4$$ (Baeyer’s test)
  • Alkaline $$KMnO_4$$ is decolourised by compounds containing $$C=C$$ or $$C \equiv C$$ bonds (unsaturation).
  • Hence it is the standard test for double or triple bonds.
  ⇒ Functional group detected = I (double bond / unsaturation).

Collecting the matches:
  • A → II
  • B → III
  • C → IV
  • D → I

This corresponds to Option A: A-II, B-III, C-IV, D-I.

Therefore, the correct answer is Option A.

We need to find the percentage of bromine in an organic compound using the Carius method.

In the Carius method for halogen estimation, the organic compound is heated with fuming nitric acid in the presence of silver nitrate, converting halogens into silver halides. For bromine this reaction can be represented as $$\text{Organic-Br} \rightarrow AgBr$$.

We know the following data: mass of the organic compound is 0.25 g, mass of AgBr formed is 0.15 g, molar mass of Ag is 108 g/mol, molar mass of Br is 80 g/mol, and hence the molar mass of AgBr is 108 + 80 = 188 g/mol.

Therefore the moles of AgBr formed are calculated by $$\text{Moles of AgBr} = \frac{0.15}{188}$$. Since one mole of AgBr contains one mole of Br, the moles of bromine are also $$\text{Moles of Br} = \frac{0.15}{188}$$.

Substituting into the mass calculation gives $$\text{Mass of Br} = \frac{0.15}{188} \times 80 = \frac{12}{188} = \frac{12}{188} \approx 0.06383 \, \text{g}$$.

Next, the percentage of bromine in the compound is given by $$\% \text{Br} = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 = \frac{0.06383}{0.25} \times 100$$, which yields $$\% \text{Br} = 25.53\%$$.

Expressing this in the required format $$\_ \times 10^{-1}$$ gives $$25.53 = 255.3 \times 10^{-1}$$. Rounding to the nearest integer results in $$255 \times 10^{-1}$$.

The answer is 255.

We need to determine how many of the given cations give a characteristic precipitate with potassium ferrocyanide $$K_4[Fe(CN)_6]$$.

Key Concept: Potassium ferrocyanide $$K_4[Fe(CN)_6]$$ is used as a group reagent and gives characteristic coloured precipitates with certain metal ions by forming insoluble double salts or complex compounds.

Testing each cation:

1. $$Cu^{2+}$$: Gives a chocolate brown (reddish-brown) precipitate of copper(II) ferrocyanide: $$Cu_2[Fe(CN)_6]$$. This is a characteristic test for $$Cu^{2+}$$. ✓

2. $$Fe^{3+}$$: Gives a deep Prussian blue precipitate of iron(III) ferrocyanide: $$Fe_4[Fe(CN)_6]_3$$. This is a very well-known confirmatory test for $$Fe^{3+}$$. ✓

3. $$Ba^{2+}$$: Does not give a characteristic precipitate with $$K_4[Fe(CN)_6]$$. Barium is typically identified using sulphate or chromate reagents. ✗

4. $$Ca^{2+}$$: Does not give a characteristic precipitate. Calcium is identified by flame test (brick red) or oxalate precipitation. ✗

5. $$NH_4^+$$: Does not give a precipitate with $$K_4[Fe(CN)_6]$$. $$NH_4^+$$ is detected by Nessler's reagent or by heating with NaOH. ✗

6. $$Mg^{2+}$$: Does not give a characteristic precipitate with $$K_4[Fe(CN)_6]$$. ✗

7. $$Zn^{2+}$$: Gives a white (bluish-white) precipitate of zinc ferrocyanide: $$Zn_2[Fe(CN)_6]$$. This is a standard test for $$Zn^{2+}$$. ✓

Count: $$Cu^{2+}$$, $$Fe^{3+}$$, and $$Zn^{2+}$$ give characteristic precipitates. That is 3 cations.

The answer is 3.

We need to evaluate two statements about the Kjeldahl method and its applicability to pyridine.

Analysis of Statement (I): "Kjeldahl method is applicable to estimate nitrogen in pyridine."

This statement is FALSE. The Kjeldahl method is NOT applicable to compounds where nitrogen is present in a ring structure (as in pyridine, quinoline, and similar heterocyclic compounds), or in nitro groups ($$-NO_2$$) and azo groups ($$-N=N-$$). This is because the nitrogen in these structures is particularly resistant to being converted to ammonium sulphate by digestion with concentrated sulphuric acid.

Analysis of Statement (II): "The nitrogen present in pyridine can easily be converted into ammonium sulphate in Kjeldahl method."

This statement is also FALSE. The nitrogen in pyridine's ring structure cannot be easily liberated and converted to ammonium sulphate $$(NH_4)_2SO_4$$ during the digestion step of the Kjeldahl method. In the Kjeldahl method, the organic compound is digested with concentrated $$H_2SO_4$$ in the presence of a catalyst (such as $$K_2SO_4$$ and $$CuSO_4$$), which converts organic nitrogen to $$(NH_4)_2SO_4$$. However, the nitrogen locked in the aromatic ring of pyridine resists this conversion because of the extra stability of the aromatic ring.

Since both statements are false, the correct answer is Option (2): Both Statement I and Statement II are false.

We need to determine which element(s) are indicated by a blood red colour when the sodium fusion extract is treated with $$FeSO_4$$ and concentrated $$H_2SO_4$$.

Understand the Lassaigne (sodium fusion) test.

In the sodium fusion test, the organic compound is fused with sodium metal. During fusion:

- Nitrogen in the compound forms $$NaCN$$ (sodium cyanide)

- Sulphur forms $$Na_2S$$ (sodium sulphide)

- If both N and S are present, $$NaSCN$$ (sodium thiocyanate) is formed

Describe the reaction that produces blood red colour.

When $$FeSO_4$$ is added to the sodium fusion extract containing $$NaSCN$$:

$$FeSO_4 + 2NaSCN \rightarrow Fe(SCN)_2 + Na_2SO_4$$

In the presence of concentrated $$H_2SO_4$$ (which oxidises $$Fe^{2+}$$ to $$Fe^{3+}$$):

$$Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+}$$

The ferric thiocyanate complex $$[Fe(SCN)]^{2+}$$ has a characteristic blood red colour.

Identify the elements.

The formation of thiocyanate ($$SCN^-$$) requires BOTH nitrogen (for the CN part) and sulphur (for the S part). Therefore, the blood red colour indicates the presence of both nitrogen and sulphur in the organic compound.

Note: If only nitrogen were present (without sulphur), the test with $$FeSO_4$$ would give Prussian blue colour (from ferric ferrocyanide), not blood red.

The correct answer is Option (3): N and S.

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.

In the borax bead test, borax (Na$$_2$$B$$_4$$O$$_7$$ · 10H$$_2$$O) is heated and decomposes as follows:

$$\text{Na}_2\text{B}_4\text{O}_7 \xrightarrow{\Delta} 2\text{NaBO}_2 + \text{B}_2\text{O}_3$$

When CuSO$$_4$$ is used, in the oxidising flame, copper remains as Cu$$^{2+}$$ (in the form of CuO).

The boric anhydride (B$$_2$$O$$_3$$) reacts with the metal oxide:

$$\text{B}_2\text{O}_3 + \text{CuO} \to \text{Cu(BO}_2\text{)}_2$$

Copper metaborate, Cu(BO$$_2$$)$$_2$$, gives the characteristic blue-green colour to the bead in the oxidising flame.

The answer is Option C: Cu(BO$$_2$$)$$_2$$.

In the Dumas method for estimation of nitrogen, an organic compound is heated with excess copper oxide (CuO) in an atmosphere of CO$$_2$$. During this process the nitrogen in the compound is converted to N$$_2$$ gas, while some oxides of nitrogen (like NO, N$$_2$$O) can also form.

To ensure collection of pure N$$_2$$, the gas mixture is passed over heated copper gauze, which reduces any oxides of nitrogen back to N$$_2$$:
$$2Cu + 2NO \rightarrow 2CuO + N_2$$ The pure nitrogen gas is then collected over a KOH solution, which absorbs CO$$_2$$.

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.

We need to identify the two anions based on the given tests:

Test 1: Brown ring test

When treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed. This is the characteristic brown ring test for nitrate (NO$$_3^-$$) ions.

$$\text{NO}_3^- + 3\text{Fe}^{2+} + 4\text{H}^+ \rightarrow 3\text{Fe}^{3+} + \text{NO} + 2\text{H}_2\text{O}$$

$$[\text{Fe(H}_2\text{O)}_5\text{NO}]^{2+}$$ gives the brown ring.

Test 2: Neutral FeCl$$_3$$ test

Treatment with neutral FeCl$$_3$$ gave a deep red colour which disappeared on boiling and a brown-red precipitate was formed. This is the characteristic test for acetate (CH$$_3$$COO$$^-$$) ions.

$$3\text{CH}_3\text{COO}^- + \text{Fe}^{3+} \rightarrow (\text{CH}_3\text{COO})_3\text{Fe}$$

The deep red colour is due to ferric acetate. On boiling, it hydrolyses to form a brown-red precipitate of basic ferric acetate.

Therefore, the mixture contains CH$$_3$$COO$$^-$$ and NO$$_3^-$$.

1. Fehling's Test (Positive):

   A positive Fehling's test indicates the presence of an aliphatic aldehyde group ($$\text{--CHO}$$). This eliminates any option that contains a ketone, ester, or other non-reducing functional groups, focusing our attention on compounds with an active formyl functionality.




2. Sodium Fusion Extract (Lassaigne's Test) with Sodium Nitroprusside:

   When an organic compound containing both Nitrogen ($$\text{N}$$) and Sulfur ($$\text{S}$$) is fused with sodium metal, they react together with carbon to form sodium thiocyanate ($$\text{NaSCN}$$):

   $$\text{Na} + \text{C} + \text{N} + \text{S} \rightarrow \text{NaSCN}$$
   • Blood-Red Coloration: When ferric ions ($$\text{Fe}^{3+}$$) or specific complexing reagents like sodium nitroprusside are introduced into an acidified thiocyanate extract solution, it yields a characteristic blood-red color due to the formation of ferric thiocyanate complexes:
   $$\text{Fe}^{3+} + \text{SCN}^\ominus \rightarrow [\text{Fe}(\text{SCN})]^{2+} \quad \text{(Blood-Red Solution)}$$
   • Absence of Prussian Blue: Because both nitrogen and sulfur are fixed together as thiocyanate ($$\text{SCN}^\ominus$$) rather than isolated cyanide ($$\text{CN}^\ominus$$) ions, there are no free cyanide ions available to form the classic Prussian blue complex ($$\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$$).

   Therefore, the compound must contain both nitrogen and sulfur atoms alongside its aldehyde group.

Evaluating the Structural Candidates:

• Oxazole derivatives: Contain Nitrogen ($$\text{N}$$) and Oxygen ($$\text{O}$$), but lack Sulfur ($$\text{S}$$). They do not give a blood-red color during Lassaigne's element detection.
• Pyrrole derivatives: Contain Nitrogen ($$\text{N}$$), but lack Sulfur ($$\text{S}$$), giving a Prussian blue test rather than blood-red thiocyanate formations.
• 2-Thiazolecarboxaldehyde ($$\text{C}_4\text{H}_3\text{NOS}$$): The thiazole ring features both an aromatic nitrogen atom and a sulfur heteroatom in its cyclic core, along with a directly attached formyl group ($$\text{--CHO}$$). This perfectly satisfies both the structural and chemical criteria.

Conclusion:

The simultaneous presence of the aldehyde moiety and both nitrogen and sulfur within the thiazole ring framework satisfies all requirements for a positive Fehling's reaction and a blood-red Lassaigne's outcome.

Answer: Option C (2-Thiazolecarboxaldehyde)

We need to match each test with what it detects.

Molisch's Test (A) is used to detect carbohydrates (II). In this test, concentrated H$$_2$$SO$$_4$$ dehydrates the carbohydrate to furfural, which reacts with $$\alpha$$-naphthol to form a purple ring at the junction of two layers.

Biuret Test (B) is used to detect peptides/proteins (I). When a compound containing two or more peptide bonds ($$-CONH-$$) is treated with dilute NaOH and CuSO$$_4$$ solution, a violet colour is produced due to the formation of a copper complex.

Carbylamine Test (C) is specific for primary amines (III). When a primary amine is heated with chloroform and alcoholic KOH, an isocyanide with a foul smell is formed:

$$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$$

Schiff's Test (D) is used to detect aldehydes (IV). Aldehydes restore the pink colour of Schiff's reagent (decolorized fuchsin dye).

Hence, the correct answer is (A)-II, (B)-I, (C)-III, (D)-IV.

The iodoform test is positive for compounds containing either a $$CH_3CO-$$ group (methyl ketone) or a $$CH_3CH(OH)-$$ group (secondary alcohol with methyl group adjacent to OH).

Analyzing each compound:

(a) 1-Phenylbutan-2-one (C₆H₅CH₂COCH₂CH₃): The carbonyl has CH₂ groups on both sides, not a methyl group. Negative ✗

(b) 2-Methylbutan-2-ol ((CH₃)₂C(OH)CH₂CH₃): This is a tertiary alcohol with no CH₃CHOH pattern. Negative ✗

(c) 3-Methylbutan-2-ol (CH₃CH(OH)CH(CH₃)₂): Has a CH₃CHOH group. Positive ✓

(d) 1-Phenylethanol (C₆H₅CH(OH)CH₃): Has a CH₃CHOH group. Positive ✓

(e) 3,3-Dimethylbutan-2-one (CH₃COC(CH₃)₃): Has a CH₃CO group. Positive ✓

(f) 1-Phenylpropan-2-ol (C₆H₅CH₂CH(OH)CH₃): Has a CH₃CHOH group. Positive ✓

Number of compounds giving positive iodoform test = 4 (c, d, e, f).

We need to identify which substance enhances the lathering property of soap.

Lathering refers to the formation of foam when soap is mixed with water. To improve lathering, substances called builders or lather boosters are added to soap.

Considering the given options:

- Sodium stearate: This is itself a soap (sodium salt of stearic acid). It does not specifically enhance lathering.

- Sodium carbonate: This is a water softener (washing soda). It helps soap work in hard water but does not directly enhance lathering.

- Sodium rosinate: This is the sodium salt of rosin (abietic acid), obtained from pine trees. It is specifically added to soap to increase lather formation. Rosin soap produces rich, stable foam.

- Trisodium phosphate: This is a cleaning agent and water softener but does not enhance lathering.

Sodium rosinate is added to soaps as a lather-enhancing agent. It improves the foaming ability of the soap. Hence, the correct answer is Option C.

We need to identify which statement is NOT true with respect to the brown ring test for nitrate ions.

Recall the Brown Ring Test procedure

In the brown ring test:

1. The suspected nitrate salt is dissolved in water.

2. Freshly prepared ferrous sulfate ($$FeSO_4$$) solution is added.

3. Concentrated $$H_2SO_4$$ is carefully added along the side of the test tube so that it forms a separate layer below the aqueous solution.

Analyze each statement

Option A: "A dark brown ring is formed at the junction of two solutions."

This is TRUE. The brown ring appears at the interface of the $$H_2SO_4$$ layer and the aqueous solution.

Option B: "Ring is formed due to nitroferrous sulphate complex."

The brown ring is due to the formation of $$[Fe(H_2O)_5(NO)]^{2+}$$ complex, which is called nitrosonium ferrous sulphate (or nitroso ferrous sulphate), NOT "nitroferrous sulphate." The correct name involves "nitroso" prefix. However, the key issue is whether the name used is acceptable.

Option C: "The brown complex is $$[Fe(H_2O)_5(NO)]SO_4$$"

This is TRUE. The brown ring is formed by the complex $$[Fe(H_2O)_5(NO)]SO_4$$, where NO is a neutral ligand coordinated to $$Fe^{2+}$$ (though the actual oxidation involves $$Fe^+ $$ with $$NO^+$$).

Option D: "Heating the nitrate salt with conc. $$H_2SO_4$$, light brown fumes are evolved."

When a nitrate salt is heated with concentrated $$H_2SO_4$$, brown fumes of $$NO_2$$ are evolved. This is TRUE and is a confirmatory test for nitrate ions.

Identify the incorrect statement

Option B states the ring is due to "nitroferrous sulphate complex." The correct name is nitrosoferrous sulphate (or nitrosonium pentaaquairon(II) sulphate), not "nitro" ferrous sulphate. "Nitro" would imply $$NO_2$$ as the ligand, but the actual ligand is $$NO$$ (nitrosyl/nitroso). This makes Option B the incorrect statement.

The correct answer is Option B: Ring is formed due to nitroferrous sulphate complex.

First recall the characteristic qualitative tests involved:

(i) Lassaigne (sodium-fusion) test for nitrogen - the fusion extract, on boiling with $$FeSO_4$$ and then acidifying with conc. $$H_2SO_4$$, gives a Prussian-blue colour due to formation of $$Fe_4[Fe(CN)_6]_3$$. (Observation P)

(ii) Sodium-fusion extract + sodium nitroprusside $$\rightarrow$$ blood-red colour when sulphide ion ($$S^{2-}$$) is present. (Observation Q)

(iii) Addition of the organic compound to a saturated $$NaHCO_3$$ solution - brisk effervescence (release of $$CO_2$$) is obtained only when the compound possesses an acidic $$-COOH$$ group. (Observation R)

(iv) Bromine water test - phenols and aromatic amines are rapidly brominated to give a white precipitate of the tribromo derivative. (Observation S)

(v) Neutral $$FeCl_3$$ test - phenolic $$-OH$$ groups form coloured complexes (violet, blue, green etc.). (Observation T)

Now examine each compound in LIST-I.

Case I: Aniline, $$C_6H_5NH_2$$
• Contains nitrogen ⇒ positive Lassaigne test ⇒ Observation P.
• Aromatic amine reacts instantaneously with bromine water to give 2,4,6-tribromoaniline (white ppt.) ⇒ Observation S.
Hence I $$\to$$ P, S.

Case II: o-Cresol (2-methylphenol)
• It is a phenol; neutral $$FeCl_3$$ gives a violet complex ⇒ Observation T.
• Although phenols also give a bromine-water test, the option set that satisfies all compounds uniquely associates o-cresol only with Observation T.
Hence II $$\to$$ T.

Case III: Cysteine, $$HSCH_2CH(NH_2)COOH$$
• Contains sulphur ⇒ sodium-fusion extract with sodium nitroprusside gives blood-red colour ⇒ Observation Q.
• Contains a carboxylic acid group; addition to $$NaHCO_3$$ liberates $$CO_2$$ gas ⇒ effervescence ⇒ Observation R.
Hence III $$\to$$ Q, R.

Case IV: Caprolactam (the lactam of 6-aminocaproic acid)
• Contains nitrogen but no sulphur or acidic $$-COOH$$ group. Therefore only the Prussian-blue Lassaigne test is positive ⇒ Observation P.
Hence IV $$\to$$ P.

Collecting all the matches:
I $$\to$$ P, S ; II $$\to$$ T ; III $$\to$$ Q, R ; IV $$\to$$ P

This correspondence is exactly the one given in Option D.

Therefore, the correct choice is:
Option D which is: I $$\to$$ P, S; II $$\to$$ T; III $$\to$$ Q, R; IV $$\to$$ P.

In ion-exchange chromatography, a cation exchange resin must have acidic functional groups that can release $$\text{H}^+$$ ions and bind cations from the solution. The sulfonic acid group $$-\text{SO}_3\text{H}$$ is the standard functional group used in strongly acidic cation exchange resins.

An anion exchange resin must have basic functional groups that can release anions (such as $$\text{OH}^-$$) and bind anions from solution. The amino group $$-\text{NH}_2$$ (or quaternary ammonium groups) serves this purpose in anion exchange resins.

Therefore, the correct pair of functional groups responsible for the ion-exchange property of cation and anion exchange resins is $$-\text{SO}_3\text{H}$$ and $$-\text{NH}_2$$, respectively.

The Kjeldahl method fails and gives inaccurate results for the following types of compounds:

1. Nitrogen present in aromatic rings

Examples:

$$\mathrm{Pyridine,\ Quinoline}$$

2. Nitrogen directly bonded to oxygen

Examples:

$$\mathrm{Nitro\ compounds}$$

3. Nitrogen-nitrogen bonded compounds

Examples:

$$\mathrm{Azo\ compounds}$$

However, benzamide can be estimated by the Kjeldahl method because the nitrogen present in the amide group is converted into ammonia during digestion.

$$\mathrm{C_6H_5CONH_2}$$

Therefore, benzamide gives a positive Kjeldahl test. Thus, the correct option is B.

Ceric ammonium nitrate (CAN) test is used to identify alcohols. When an alcohol is added to a solution of ceric ammonium nitrate, the yellow solution turns red due to the formation of a cerium(IV) alkoxide complex. This confirms the presence of the $$-OH$$ group in alcohols.

The $$CHCl_3$$ / alcoholic KOH test is known as the carbylamine (isocyanide) test. When a primary amine is heated with chloroform and alcoholic KOH, it produces an isocyanide (carbylamine) with an extremely foul smell. The reaction is:

$$R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O$$

This test is specific to primary amines only.

Therefore, ceric ammonium nitrate is used for alcohol identification and $$CHCl_3$$ / alc. KOH is used for amine identification. The correct answer is alcohol, amine.

First, let us recall the main reagents that appear in each of the four classical qualitative tests listed in the question.

We know that the presence or absence of copper(II) ions, written chemically as $$Cu^{2+}$$, is decisive here, because the wording asks which test “does not use copper reagent.” So we will look at every test one by one and explicitly state whether copper ions participate in the procedure.

1. Seliwanoff’s test: This test is designed to distinguish ketoses from aldoses. The reagent is an ethanolic solution of resorcinol mixed with concentrated $$HCl$$. The ketose, if present, undergoes rapid dehydration in the strongly acidic medium, forming hydroxymethylfurfural, which then condenses with resorcinol to give a cherry-red complex. There is no mention, requirement, or role of copper ions in this entire sequence.

2. Biuret test: The Biuret reagent contains $$CuSO_{4}$$ in a strongly alkaline solution (usually $$NaOH$$ or $$KOH$$). The peptide bonds in proteins form a violet-coloured coordination complex with $$Cu^{2+}$$. Hence this test unequivocally employs copper ions.

3. Barfoed’s test: Barfoed’s reagent is prepared from $$Cu(CH_{3}COO)2$$ (copper(II) acetate) dissolved in dilute acetic acid. It is used to detect monosaccharides specifically, and the Cu(II) is reduced to Cu(I) oxide, giving a brick-red precipitate. Thus copper ions are essential here as well.

4. Benedict’s test: Benedict’s qualitative reagent contains $$CuSO_{4}$$, sodium citrate, and sodium carbonate. Reducing sugars reduce the blue $$Cu^{2+}$$ to insoluble red $$Cu_{2}O$$. Therefore, copper ions again play the central role.

Looking at all four results together, only Seliwanoff’s test operates without any copper reagent; the other three rely directly on $$Cu^{2+}$$ chemistry.

Hence, the correct answer is Option A.

In Duma's method, nitrogen in the organic compound is converted to $$N_2$$ gas, which is collected over water. We need to correct for the aqueous tension (water vapour pressure).

The pressure of dry nitrogen gas is: $$P_{N_2} = P_{total} - P_{water} = 758 - 14 = 744$$ mm of Hg.

Using the ideal gas law to find moles of $$N_2$$: $$n = \frac{PV}{RT}$$, where $$P = \frac{744}{760}$$ atm $$= 0.9789$$ atm, $$V = \frac{30}{1000}$$ L $$= 0.030$$ L, $$R = 0.0821$$ L atm mol$$^{-1}$$ K$$^{-1}$$, and $$T = 287$$ K.

$$n = \frac{0.9789 \times 0.030}{0.0821 \times 287} = \frac{0.029368}{23.5627} = 1.2463 \times 10^{-3}$$ mol.

Mass of nitrogen = $$1.2463 \times 10^{-3} \times 28 = 0.03490$$ g.

Percentage of nitrogen = $$\frac{0.03490}{0.1840} \times 100 = 18.97\%$$.

Rounding off to the nearest integer, the percentage composition of nitrogen is $$19$$.

We begin by recalling the characteristic reagents that are classically associated with each of the given organic tests or nitrogen-estimation methods.

For the Lucas Test we know that the reagent is called Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and anhydrous zinc chloride. The purpose of this reagent is to convert alcohols to their corresponding chlorides so that the relative reactivity of 1°, 2° and 3° alcohols can be judged by the turbidity time. Hence for Lucas Test the correct reagent from the list must be $$\text{Conc. HCl + ZnCl}_2$$, which is option (d).

Next, we think of the Dumas method of nitrogen estimation. In the Dumas method the organic compound is strongly heated with excess cupric oxide to bring about complete combustion. The nitrogen present is converted to molecular nitrogen, while carbon is oxidised to $$\text{CO}_2$$. Thus the reagent invariably mentioned in text books is cupric oxide, written as $$\text{CuO}$$ together with an atmosphere that sweeps away the produced gases (often quoted as $$\text{CO}_2$$). Therefore the pair that best represents the Dumas method is (c) $$\text{CuO / CO}_2$$.

Now we recall Kjeldahl’s method. In Kjeldahl’s digestion the organic compound is heated with concentrated sulphuric acid so that nitrogen is ultimately converted to ammonium sulphate. Hence concentrated $$\text{H}_2\text{SO}_4$$ is the indispensable reagent. Looking at the reagent list, (e) is $$\text{H}_2\text{SO}_4$$, which fits perfectly.

The Hinsberg Test is a classic qualitative test used to distinguish primary, secondary and tertiary amines. The reagent employed is benzene­sulphonyl chloride, written chemically as $$\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$$, used in the presence of aqueous potassium hydroxide. This corresponds exactly to reagent (a).

Collecting all the correct pairings we have

$$\begin{aligned} \text{(i) Lucas Test} &\ \longrightarrow\ (d)\; \text{Conc.\ HCl + ZnCl}_2,\\ \text{(ii) Dumas method} &\ \longrightarrow\ (c)\; \text{CuO / CO}_2,\\ \text{(iii) Kjeldahl's method} &\ \longrightarrow\ (e)\; \text{H}_2\text{SO}_4,\\ \text{(iv) Hinsberg Test} &\ \longrightarrow\ (a)\; \text{C}_6\text{H}_5\text{SO}_2\text{Cl / aq.\ KOH}. \end{aligned}$$

Comparing these matches with the given options, we see that Option C lists the exact same correspondence:

(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a).

Hence, the correct answer is Option C.

The Kjeldahl method involves converting organic nitrogen into ammonium sulfate by heating with concentrated sulfuric acid. However, this method fails for compounds containing nitrogen in:

1. Nitro groups (-NO2)
2. Azo groups (-N=N-)
3. Nitrogen present in a ring structure (like Pyridine)

This is because the nitrogen in these functional groups cannot be quantitatively converted to ammonium sulfate under standard Kjeldahl digestion conditions.

Note: Amines (-NH2), amides, and nitriles (-CN) generally give a positive Kjeldahl test because they can be successfully reduced/hydrolyzed to ammonia.

Based on the provided qualitative analysis tests, the unknown organic compound can be identified by evaluating each step logically. First, the compound's insolubility in dilute HCl indicates that it lacks strongly basic functional groups, such as aliphatic or aromatic amines, which would otherwise form soluble hydrochloride salts. Second, its solubility in an NaOH solution reveals that the compound possesses an acidic functional group capable of reacting with a strong base to form a water-soluble sodium salt. Since it does not specify solubility in weaker bases like sodium bicarbonate, this acidic nature points heavily toward a phenolic compound rather than a carboxylic acid. Finally, the Bromine Water test resulting in decolorization confirms the presence of highly electron-rich aromatic rings or carbon-carbon multiple bonds. Phenols have highly activated benzene rings due to the strong resonance electron-donation of the hydroxyl group, which allows them to undergo rapid electrophilic aromatic substitution with bromine water, decolorizing the orange-brown solution and typically forming a halogenated precipitate. Therefore, the qualitative analysis strongly demonstrates that the organic compound is a phenol

 2,4 DNP Test

• What it specifies: The presence of a carbonyl group, meaning the compound is either an aldehyde or a ketone.

 Iodoform Test

• What it specifies: The presence of a methyl ketone group (CH3-C=O) or a terminal methyl alcohol group (CH3-CH-OH).

Azo-Dye Test

• What it specifies: The presence of an aromatic primary amine (an -NH2 group attached directly to a benzene ring, like aniline).

1. Step 1: Acid-Catalyzed Hydrolysis (Cleavage)

   Treating the substrate with dilute $$\text{HCl}$$ and heat ($$\Delta$$) selectively hydrolyzes both the methoxy ether ($$-\text{OCH}_3$$) and the ester group ($$-\text{O--C}(=\text{O})\text{CH}_3$$). This unmasks two phenolic hydroxyl groups ($$-\text{OH}$$) on the aromatic ring, yielding a catechol-like intermediate with a net total of two reactive phenolic hydroxyl groups.




2. Step 2: Condensation Polymerization

   The resulting dihydroxy aromatic intermediate reacts with oxalic acid ($$(\text{COOH})_2$$), which acts as a dicarboxylic acid monomer. A step-growth condensation polymerization takes place, eliminating water ($$\text{H}_2\text{O}$$) molecules to form alternating ester linkages (a polyester backbone).

First, we recall the characteristic qualitative tests that are commonly used in organic chemistry:

$$\text{Ferric chloride test}$$ is given by compounds containing the phenolic $$-OH$$ group, producing a coloured complex.

$$\text{Carbylamine test}$$ is positive for aliphatic or aromatic primary amines. When the compound is warmed with chloroform and alcoholic $$KOH$$, the foul-smelling isocyanide (carbylamine) is formed.

$$\text{Sodium hydrogen carbonate (NaHCO}_3\text{) test}$$ detects carboxylic acids. The acid reacts with $$NaHCO_3$$ to liberate $$CO_2$$ gas, visible as brisk effervescence.

$$\text{Baeyer’s test}$$ employs dilute alkaline $$KMnO_4$$ solution. Compounds containing unsaturation (double or triple bonds) decolourise the purple $$KMnO_4$$ solution while producing a brown precipitate of $$MnO_2$$.

Now we examine each drug in Column-I one by one and match it with the appropriate test from Column-II.

We have Chloroxylenol. Its structure possesses a phenolic $$-OH$$ group attached to an aromatic ring. A phenolic group responds positively to the ferric chloride test, forming a violet to blue-green complex. So, for Chloroxylenol:

$$\text{Chloroxylenol} \longrightarrow \text{Ferric chloride test}$$

Hence, $$A \rightarrow R$$.

Next, consider Norethindrone. This steroidal contraceptive molecule contains an ethynyl $$(-C \equiv CH)$$ group, i.e., a carbon-carbon triple bond. According to Baeyer’s test, any compound with such unsaturation reacts with dilute alkaline $$KMnO_4$$ and discharges its purple colour. Therefore, for Norethindrone:

$$\text{Norethindrone} \longrightarrow \text{Baeyer’s test}$$

Thus, $$B \rightarrow S$$.

Now, we take Sulphapyridine. The ring nitrogen is bonded to a primary $$-NH_2$$ group on an aromatic ring. A primary amine gives the carbylamine reaction:

$$\text{ArNH}_2 + CHCl_3 + 3\,KOH \;\longrightarrow\; \text{ArNC} + 3\,KCl + 3\,H_2O$$

The unpleasant odour of isocyanide confirms the presence of a primary amine. Hence, Sulphapyridine matches the carbylamine test:

$$\text{Sulphapyridine} \longrightarrow \text{Carbylamine test}$$

Therefore, $$C \rightarrow P$$.

Finally, we analyse Penicillin. Every penicillin molecule carries a free carboxylic acid $$-COOH$$ group. When this acidic hydrogen reacts with sodium hydrogen carbonate, carbon dioxide gas is evolved:

$$RCOOH + NaHCO_{3} \rightarrow RCOONa + CO_{2} \uparrow + H_{2}O$$

The effervescence confirms a positive NaHCO3 test. So, Penicillin corresponds to the sodium hydrogen carbonate test:

$$\text{Penicillin} \longrightarrow \text{Sodium hydrogen carbonate test}$$

Consequently, $$D \rightarrow Q$$.

Collecting all the results, we obtain the complete matching:

$$A \rightarrow R,\; B \rightarrow S,\; C \rightarrow P,\; D \rightarrow Q$$

This sequence is exactly the same as Option D.

Hence, the correct answer is Option D.

First, we carefully examine the four compounds in Item I and recall the most characteristic qualitative reagent for each.

We have $$\text{Lysine}$$, an α-amino acid that contains two amino ($$-\mathrm{NH_2}$$) groups. A very common laboratory test for amino acids is the ninhydrin test; the reagent $$\text{ninhydrin}$$ reacts with free $$-\mathrm{NH_2}$$ groups to give the deep purple-blue Ruhemann’s purple complex. Hence, Lysine will definitely match with $$\text{ninhydrin} \;(q).$$

Now, consider $$\text{Furfural}$$ (2-furancarboxaldehyde). It is an aldehyde derived from a furan ring, and aldehydes that arise from carbohydrates give the well-known Molisch reaction. The Molisch reagent is alcohol-soluble $$\alpha$$- or $$1$$-naphthol. When furfural is mixed with $$1$$-naphthol and concentrated $$\mathrm{H_2SO_4}$$, a violet ring appears, confirming the aldehydic derivative of a carbohydrate. Therefore, Furfural must be paired with $$1\text{-naphthol} \;(p).$$

Next, look at $$\text{Benzyl alcohol} \;( \mathrm{C_6H_5CH_2OH} ).$$ Alcohols in general give an immediate colour change (usually yellow-orange to red) with the ceric ammonium nitrate test. The reagent $$\displaystyle \mathrm{(NH_4)_2[Ce(NO_3)_6]}$$, abbreviated as CAN, is specific for alcoholic $$-\mathrm{OH}$$ groups. Thus Benzyl alcohol will respond to $$\text{ceric ammonium nitrate} \;(s).$$

Lastly, $$\text{Styrene} \;(\mathrm{C_6H_5CH=CH_2})$$ possesses a carbon-carbon double bond. Alkenes readily decolourise cold, dilute, alkaline or neutral $$\mathrm{KMnO_4}$$ (Baeyer’s test) by reducing the purple $$\mathrm{MnO_4^-}$$ ion to brown $$\mathrm{MnO_2}$$. Consequently, Styrene is best matched with $$\mathrm{KMnO_4} \;(r).$$

Collecting all our pairings, we have

$$ \begin{aligned} a &\;\;\text{(Lysine)} &\longrightarrow&\; q \;(\text{ninhydrin}), \\ b &\;\;\text{(Furfural)} &\longrightarrow&\; p \;(1\text{-naphthol}), \\ c &\;\;\text{(Benzyl alcohol)} &\longrightarrow&\; s \;(\text{ceric ammonium nitrate}), \\ d &\;\;\text{(Styrene)} &\longrightarrow&\; r \;(\mathrm{KMnO_4}). \\ \end{aligned} $$

This exact correspondence is listed in Option B.

Hence, the correct answer is Option B.

In Item I the three organic compounds are:

 $$\text{(A) Phenol, } \quad \text{(B) Aniline, } \quad \text{(C) Benzoic acid}$$

Item II lists four characteristic qualitative tests:

 $$\text{(P) Brisk effervescence with } NaHCO_3$$  (carboxylic ‑COOH group)
 $$\text{(Q) Violet colour with } FeCl_3$$  (phenolic -OH group)
 $$\text{(R) White ppt. of } 2,4,6\text{-tribromophenol on } Br_2/H_2O$$  (highly activated phenolic ring)
 $$\text{(S) Foul-smelling isothiocyanate in the carbylamine test}$$  (primary aromatic amine (-NH2))

We now match each compound from Item I with the test(s) it answers in Item II.

Collecting all the matches:

 $$\boxed{\text{(A) → (Q, R); (B) → (S); (C) → (P)}}$$

This corresponds to Option A.

Therefore, the correct answer is:
Option A which is: (A) → (Q,R); (B) → (S); (C) → (P)

Kjeldahl’s method works only when the nitrogen present in an organic compound is first converted to $$\left(NH_4\right)_2SO_4$$ by boiling the substance with excess conc. $$H_2SO_4$$. For this to happen, the nitrogen atom must be present as an -NH, -NH2, -CONH-, -CNH- etc., i.e. it must be bonded only to carbon and/or hydrogen (and, after protonation, be able to form $$NH_4^+$$).
The method fails whenever nitrogen is present in an oxidation state or environment that prevents the formation of ammonium ion. Thus compounds containing

• a nitro group $$(-NO_2)$$,
• azo / diazo linkages $$(-N=N-,\; -N_2^+Cl^-)$$,
• nitrile $$(-C\equiv N)$$, isonitrile $$(-NC)$$,
• nitroso $$(-NO)$$, or
• ring nitrogen as in pyridine, quinoline etc.

are not estimated correctly by the Kjeldahl procedure.

Keeping this rule in mind, inspect the four choices (structures as given in the paper):

Therefore, the only compound that fulfils the requirement is the amide in Option C.

Final answer: Option C which is: acetamide $$\left(CH_3CONH_2\right)$$.

First we recall the principle of qualitative analysis: cations of the fourth analytical group are those whose sulphides are precipitated when colourless hydrogen sulphide gas is passed through the original solution kept slightly alkaline with ammonium chloride and ammonium hydroxide. The common period-IV cations that behave in this way are $$Mn^{2+},\; Fe^{2+},\; Co^{2+},\; Ni^{2+},\; Zn^{2+}.$$ In the present question the list of possibilities has already been narrowed down to four cations of this group.

We are told that on passing $$H_2S$$ a precipitate appears. Let us represent the general reaction that takes place when a divalent metal ion $$M^{2+}$$ of this group encounters hydrogen sulphide in alkaline medium:

$$M^{2+}+ S^{2-}\;\rightarrow\; MS\downarrow$$

Here $$MS$$ is the metal sulphide. Among the four given cations the colours of these sulphides are well known:

$$ZnS$$ is white, $$MnS$$ is flesh-coloured (light buff), $$NiS$$ is black and $$CoS$$ is black.

The statement of the problem merely says “a precipitate” without giving its colour, so at this stage all four contenders are still formally possible.

The precipitate is next treated with dilute hydrochloric acid and it dissolves. We write the generic dissolution reaction with zinc as an example:

$$ZnS + 2\,HCl \;\rightarrow\; ZnCl_2 + H_2S\uparrow$$

$$ZnCl_2$$ thus formed is a colourless solution containing $$Zn^{2+}$$ ions. Importantly, both $$NiS$$ and $$CoS$$ are insoluble in dilute hydrochloric acid, whereas $$ZnS$$ and $$MnS$$ dissolve. Hence, the very fact that the sulphide dissolves rules out $$Ni^{2+}$$ and $$Co^{2+}$$ from further consideration.

Now we have to distinguish between the two remaining possibilities, $$Mn^{2+}$$ and $$Zn^{2+}.$$ For that we look at the next operation: the acid solution gives a white precipitate with sodium hydroxide solution. Let us write the hydroxide formation reactions for the two ions still in the race:

For zinc:

$$ZnCl_2 + 2\,NaOH \;\rightarrow\; Zn(OH)_2\downarrow \;+\; 2\,NaCl$$

For manganese:

$$MnCl_2 + 2\,NaOH \;\rightarrow\; Mn(OH)_2\downarrow \;+\; 2\,NaCl$$

Although freshly precipitated $$Mn(OH)_2$$ is almost white, it very rapidly absorbs oxygen from the air and turns brown due to the formation of hydrated $$MnO_2$$. In contrast, $$Zn(OH)_2$$ is a persistent white gelatinous precipitate. The question explicitly says “white precipitate”, which is fully consistent with $$Zn(OH)_2$$ and not with the easily darkening $$Mn(OH)_2$$. This observation strongly favours $$Zn^{2+}$$.

The decisive confirmation comes from the final test: the solution gives a bluish-white precipitate with basic potassium ferrocyanide, $$K_4[Fe(CN)_6]$$. We first state the qualitative formula that governs such tests:

$$n\,M^{2+} + [Fe(CN)_6]^{4-} \;\rightarrow\; M_n[Fe(CN)_6]\downarrow$$

For zinc the reaction is written explicitly as

$$2\,Zn^{2+} + [Fe(CN)_6]^{4-} \;\rightarrow\; Zn_2[Fe(CN)_6]\downarrow$$

The precipitate $$Zn_2[Fe(CN)_6]$$ is described in the literature as bluish white. In contrast, $$Mn^{2+}$$ with the same reagent gives a light pink or flesh-coloured precipitate, quite distinct from bluish white. Therefore the observation provided in the problem matches $$Zn^{2+}$$ and no other.

Let us collect the evidence:

1. Sulphide precipitated by $$H_2S$$ dissolves in dilute $$HCl$$   →   eliminates $$Ni^{2+}$$ and $$Co^{2+}$$.
2. Hydroxide obtained with $$NaOH$$ remains white   →   favours $$Zn^{2+}$$ over $$Mn^{2+}$$.
3. Basic ferrocyanide gives bluish-white precipitate   →   characteristic of $$Zn^{2+}$$.

All three experimental facts converge on the same conclusion:

$$\boxed{Zn^{2+}}$$

Hence, the correct answer is Option B.

Kjeldahl’s method is not applicable for compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring because nitrogen of these compounds does not change to ammonium sulphate under these conditions. Hence, only Alanine can be used to determine percentage of nitrogen.

Lassaignes' test for nitrogen involves fusing the organic compound with sodium metal. This fusion converts any nitrogen present into sodium cyanide (NaCN). The fused mass is then extracted with water, and the extract is tested by adding ferrous sulfate, ferric chloride, and acidifying with dilute sulfuric acid. If nitrogen is present, a Prussian blue precipitate forms. For nitrogen to be detected, it must be bonded to carbon in the compound so that during fusion, it can form cyanide ion (CN⁻).

Now, let's evaluate each option:

Option A: Propanenitrile (CH₃CH₂CN)

Propanenitrile contains a nitrile group (-C≡N), where nitrogen is bonded to carbon. During sodium fusion, the nitrile group is reduced to cyanide ion. Thus, it will give a positive Lassaignes' test for nitrogen.

Option B: Hydroxylamine hydrochloride (NH₂OH·HCl)

Hydroxylamine hydrochloride has the formula NH₂OH·HCl. The nitrogen atom is bonded to hydrogen and oxygen but not to carbon. During sodium fusion, nitrogen cannot form cyanide ion because cyanide requires nitrogen to be bonded to carbon. Instead, it may decompose to form ammonia or other nitrogen compounds that do not yield cyanide. Therefore, it will not give a positive Lassaignes' test for nitrogen.

Option C: Nitromethane (CH₃NO₂)

Nitromethane contains a nitro group (-NO₂), where nitrogen is bonded to carbon and oxygen atoms. During sodium fusion, the nitro group is reduced to cyanide ion. Thus, it will give a positive Lassaignes' test for nitrogen.

Option D: Ethanamine (CH₃CH₂NH₂)

Ethanamine is an amine where nitrogen is bonded to carbon atoms. During sodium fusion, amines are converted to cyanide ion. Thus, it will give a positive Lassaignes' test for nitrogen.

Hence, hydroxylamine hydrochloride (Option B) is not expected to show Lassaignes' test for nitrogen because its nitrogen is not bonded to carbon, preventing cyanide formation.

So, the answer is Option B.

• Ethyl acetate ($$\text{CH}_3\text{COOC}_2\text{H}_5$$): An organic ester. Esters undergo chemical reactions like hydrolysis under strongly acidic or strongly basic conditions, but are quite stable in neutral environments.
• Sodium chloride solution ($$\text{NaCl}_{(aq)}$$): A salt of a strong acid ($$\text{HCl}$$) and a strong base ($$\text{NaOH}$$). Dissolving it in water yields a completely neutral solution ($$\text{pH} = 7$$) containing stable, unreactive $$\text{Na}^+$$ and $$\text{Cl}^-$$ ions.

When ethyl acetate is mixed with aqueous sodium chloride, no chemical transformation takes place. This is due to several key factors:

• The solution is neutral, so there are no catalytic $$\text{H}^+$$ or $$\text{OH}^-$$ ions available to drive ester hydrolysis or saponification.
• The chloride ion ($$\text{Cl}^-$$) is an extremely weak nucleophile in an aqueous environment because it is heavily stabilized by hydration shells. It cannot attack the carbonyl carbon of the ester to displace the ethoxide leaving group.

Because there is no thermodynamic driving force or kinetic pathway for a reaction, the components simply coexist as a physical mixture without changing their chemical identities.

Since no chemical reaction occurs, the final composition of the solution remains exactly the same as the initial reactants mixed together:

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaCl}$$

Answer: Option A — $$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaCl}$$

We need to find the percentage of nitrogen using Kjeldahl's method. Given 1 g of compound, 15 mL of 1 M $$H_2SO_4$$ is neutralised by the evolved $$NH_3$$.

In Kjeldahl's method, nitrogen in the compound is converted to ammonium sulphate, then ammonia is liberated and absorbed in excess $$H_2SO_4$$: $$2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$$

Each mole of $$H_2SO_4$$ reacts with 2 moles of $$NH_3$$ (since $$H_2SO_4$$ is diprotic). Moles of $$H_2SO_4$$ = $$15 \times 10^{-3} \times 1 = 0.015$$ mol, which corresponds to Moles of $$NH_3$$ = $$2 \times 0.015 = 0.03$$ mol. Since each $$NH_3$$ contains one nitrogen atom, moles of N = 0.03 mol.

Substituting into the formula for mass gives $$\text{Mass of N} = 0.03 \times 14 = 0.42 \text{ g}$$

Therefore, the percentage of nitrogen is $$\% N = \frac{0.42}{1} \times 100 = 42\%$$

The correct answer is Option 1: 42.

Chemistry Solution

We need to match the chemical formulas given in List-I with their characteristic colors given in List-II.

The compounds and their corresponding colors are based on standard qualitative analysis tests:

1. A. $$Fe_4[Fe(CN)_6]_3 \cdot xH_2O$$ is known as Prussian Blue, which is formed during the Lassaigne's test for nitrogen detection. Therefore, A matches with III.
2. B. $$[Fe(CN)_5NOS]^{4-}$$ is a violet-colored iron nitroprusside complex formed during the test for sulfide ions (sulfur detection). Therefore, B matches with I.
3. C. $$[Fe(SCN)]^{2+}$$ is a blood-red complex formed when ferric ions react with thiocyanate ions. Therefore, C matches with II.
4. D. $$(NH_4)_3PO_4 \cdot 12MoO_3$$ is an ammonium phosphomolybdate precipitate that has a characteristic yellow color, used to detect phosphorus. Therefore, D matches with IV.

Combining these matches, we get: $$A-III, B-I, C-II, D-IV$$.

Therefore, the correct option is $$A-III, B-I, C-II, D-IV$$.

In the estimation of sulphur, the organic compound is converted to $$BaSO_4$$.

The formula for percentage of sulphur is $$\% S = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of compound}} \times 100$$.

Here the molar mass of $$BaSO_4$$ is $$137 + 32 + 4(16) = 233$$ g/mol.

For a compound mass of 160 mg and a $$BaSO_4$$ precipitate of 466 mg, substitution into the formula gives $$\% S = \frac{32}{233} \times \frac{466}{160} \times 100$$.

This can be rewritten as $$= \frac{32 \times 466}{233 \times 160} \times 100$$ and then as $$= \frac{14912}{37280} \times 100$$.

Since $$\frac{466}{233} = 2$$, the calculation simplifies to $$\% S = \frac{32 \times 2}{160} \times 100 = \frac{64}{160} \times 100 = 40\%$$.

The answer is 40%.

We need to identify the yellow coloured precipitate formed in the qualitative test for phosphorus.

Describe the test procedure

In the qualitative detection of phosphorus in organic compounds:

1. The organic compound is first heated with an oxidising agent (such as sodium peroxide or concentrated nitric acid) to convert all the phosphorus into phosphate ions ($$PO_4^{3-}$$).

2. The resulting solution is treated with concentrated nitric acid ($$HNO_3$$) to ensure an acidic medium.

3. Ammonium molybdate $$(NH_4)_2MoO_4$$ solution is then added.

Identify the product

The phosphate ions react with ammonium molybdate in the presence of nitric acid to form a characteristic canary yellow precipitate of ammonium phosphomolybdate:

$$PO_4^{3-} + 12(NH_4)_2MoO_4 + 21HNO_3 \rightarrow (NH_4)_3PO_4 \cdot 12MoO_3 \downarrow + 21NH_4NO_3 + 12H_2O$$

The yellow precipitate has the formula $$(NH_4)_3PO_4 \cdot 12MoO_3$$ (ammonium phosphomolybdate). This is a standard confirmatory test for the presence of phosphate ions.

The correct answer is Option (2): $$(NH_4)_3PO_4 \cdot 12MoO_3$$.

We need to identify the acidic radical that gives a pale yellow precipitate soluble with difficulty in NH$$_4$$OH when tested with AgNO$$_3$$. When the sodium carbonate extract of a salt is acidified with dilute HNO$$_3$$ and treated with AgNO$$_3$$, the halide ions form silver halide precipitates with characteristic colours and solubilities: AgCl (from Cl$$^-$$) gives a white curdy precipitate easily soluble in dilute NH$$_4$$OH (forms soluble [Ag(NH$$_3$$)$$_2$$]Cl), AgBr (from Br$$^-$$) gives a pale yellow precipitate soluble with difficulty in NH$$_4$$OH (partially soluble, requires concentrated NH$$_4$$OH), and AgI (from I$$^-$$) gives a yellow precipitate insoluble in NH$$_4$$OH.

Since the observed precipitate is pale yellow, Cl$$^-$$ is ruled out and Br$$^-$$ is consistent; moreover, its solubility with difficulty in NH$$_4$$OH matches AgBr, whereas AgI would be completely insoluble and AgCl would dissolve easily.

Furthermore, CO$$_3^{2-}$$ would not give a precipitate with AgNO$$_3$$ in acidic medium, as it reacts with acid to form CO$$_2$$, eliminating Option (4).

The correct answer is Option (1): Br$$^-$$.

A. Unsaturation (Baeyer's Test) $$\rightarrow$$ IV. Discharge of pink colour

Baeyer's reagent is a cold, alkaline solution of potassium permanganate ($$\text{KMnO}_4$$), which is deep pink/purple. When it reacts with an alkene or alkyne, the $$\text{MnO}_4^-$$ ion is reduced to $$\text{MnO}_2$$ (a brown precipitate), causing the pink colour to disappear (decolourise).

B. Alcoholic group (Ceric ammonium nitrate test) $$\rightarrow$$ I. Red colour appears

Ceric ammonium nitrate ($$\text{CAN}$$) is an orange-yellow reagent. When added to an alcohol, it forms a alkoxy cerium(IV) complex, which results in a characteristic wine-red colouration.

C. Aldehyde group (Tollen's reagent) $$\rightarrow$$ II. Silver mirror appears

Tollen's reagent contains the silver-ammonia complex $$[\text{Ag}(\text{NH}_3)_2]^+$$. Aldehydes reduce these silver ions to metallic silver, which deposits on the inner walls of the test tube, creating a shining silver mirror.

D. Phenolic group ($$\text{FeCl}_3$$ test) $$\rightarrow$$ III. Violet colour appears

Phenols react with neutral ferric chloride to form a coordination complex with the iron. For simple phenol, this complex is typically violet or purple, though other phenols can produce green or blue colours.