JEE Practical Organic Chemistry Questions

The qualitative detection of extra-elements (nitrogen, sulphur, halogen and phosphorus) in an organic compound is carried out with the help of Lassaigne’s test, also called the sodium-fusion test.

Case 1 : Statement (I)
Statement (I) says that nitrogen, sulphur, halogen and phosphorus present in an organic compound are detected by Lassaigne’s test. During the test the compound is fused with metallic sodium. The fusion converts the covalently bound elements into ionic salts such as $$NaCN$$, $$Na_2S$$, $$NaX$$ (where $$X = Cl, Br, I$$) and $$Na_3PO_4$$, which can then be identified by suitable confirmatory tests. Hence Statement (I) is correct.

Case 2 : Statement (II)
Statement (II) claims that the elements are converted from covalent form into ionic form by fusing the compound with magnesium in Lassaigne’s test. However, the metal actually used in Lassaigne’s test is sodium, not magnesium. Therefore the statement describing fusion with magnesium is incorrect. Hence Statement (II) is false.

Thus Statement (I) is true while Statement (II) is false. The option that matches this combination is Option D.

Final answer : Option D

In qualitative inorganic analysis, sulphide ion $$S^{2-}$$ is identified by the sodium nitroprusside test.

First, the salt is fused with solid sodium carbonate and the fused mass is extracted with water to obtain the carbonate extract that contains all anions in their ionic forms.

When a few drops of freshly prepared aqueous sodium nitroprusside $$Na_2[Fe(CN)_5NO]$$ are added to this extract, the following reaction takes place if $$S^{2-}$$ is present:

$$Na_2[Fe(CN)_5NO]^{2-} + S^{2-} \;\longrightarrow\; Na_4[Fe(CN)_5NOS]^{4-}$$

The product $$Na_4[Fe(CN)_5NOS]$$ is a purple-coloured complex. The appearance of the purple colour is therefore a confirmatory test for the sulphide ion.

No such colour is produced with sulphate $$SO_4^{2-}$$, sulphite $$SO_3^{2-}$$, or simple metal cations such as $$Na^+$$, so these ions are ruled out.

Hence, the formation of the purple complex $$Na_4[Fe(CN)_5NOS]$$ indicates the presence of the sulphide ion.

Correct option: Option C (Sulphide ion).

Let us analyze both statements:

Statement I: In Lassaigne's test, the covalent organic molecules are transformed into ionic compounds.

In Lassaigne's test, the organic compound is fused with sodium metal. This converts covalent compounds containing N, S, or halogens into ionic compounds like NaCN, Na₂S, NaX (where X is halogen). This statement is true.

Statement II: The sodium fusion extract of an organic compound having N and S gives prussian blue colour with $$FeSO_4$$ and $$Na_4[Fe(CN)_6]$$.

When an organic compound contains both N and S, the fusion with sodium produces NaSCN (sodium thiocyanate) instead of NaCN and Na₂S separately. With $$FeSO_4$$, NaSCN would give a blood-red colour (ferric thiocyanate), not prussian blue.

For prussian blue test: $$Na_4[Fe(CN)_6]$$ reacts with $$Fe^{3+}$$ to give prussian blue $$Fe_4[Fe(CN)_6]_3$$. But in the presence of S, the CN⁻ ions are converted to SCN⁻, so prussian blue is not obtained directly. This statement is false.

Additionally, the reagent combination mentioned is incorrect. For the prussian blue test, we use $$FeSO_4$$ followed by $$FeCl_3$$, not $$Na_4[Fe(CN)_6]$$. Statement II is false.

The correct answer is Option 1: Statement I is true but Statement II is false.

The reagent $$K_4[Fe(CN)_6]$$ (potassium ferrocyanide) gives characteristic coloured precipitates with only certain metal ions. We shall examine tests (A) to (E) one by one.

$$Cu^{2+} + K_4[Fe(CN)_6] \xrightarrow{\text{acidic}} Cu_2[Fe(CN)_6] \downarrow$$
Copper(II) ferrocyanide is chocolate-brown in colour.
Observation stated in (A) - chocolate brown precipitate - is correct.

$$Fe^{3+} + K_4[Fe(CN)_6] \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow$$
Ferric ferrocyanide is the well-known Prussian blue.
Observation stated in (B) - Prussian blue precipitate - is correct.

$$Zn^{2+} + K_4[Fe(CN)_6] \xrightarrow{\text{neutral/alkaline}} Zn_2[Fe(CN)_6] \downarrow$$
Zinc ferrocyanide separates as a white or faint bluish-white precipitate; neutralisation with $$NH_4OH$$ furnishes the required medium.
Observation stated in (C) - white / bluish-white precipitate - is correct.

Magnesium ions do not give an insoluble ferrocyanide; $$Mg_2[Fe(CN)_6]$$ is soluble. Hence no blue precipitate is obtained.
Observation stated in (D) is incorrect.

Ferrocyanides of the alkaline-earth metals Ca, Sr and Ba are readily soluble in water; therefore $$Ba^{2+}$$ does not precipitate with $$K_4[Fe(CN)_6]$$ even after neutralisation with $$NaOH$$.
Observation stated in (E) is incorrect.

Thus only statements (A), (B) and (C) are correct.

The option containing A, B and C only is Option C.

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 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.

Let the major product $$P$$ obtained in the earlier sequence be stearic acid, $$C_{17}H_{35}COOH$$.
With an acid catalyst, glycerol esterifies completely with excess stearic acid to give the tri-ester (a triglyceride) $$Q$$, called tristearin.

Step-1 : Formation of $$Q$$
Glycerol possesses three -OH groups, so three molecules of stearic acid react with one molecule of glycerol and three molecules of water are eliminated:

$$C_3H_8O_3 + 3\,C_{17}H_{35}COOH \;\xrightarrow{\text{acid}}\; C_3H_5\big(OOC\,C_{17}H_{35}\big)_3 + 3\,H_2O$$

Molar mass of glycerol $$= 3\times12 + 8\times1 + 3\times16 = 92$$ g mol$$^{-1}$$
Molar mass of stearic acid $$= 18\times12 + 36\times1 + 2\times16 = 284$$ g mol$$^{-1}$$

Molar mass of $$Q$$ (tristearin):
$$M_Q = 92 + 3(284) - 3(18) = 92 + 852 - 54 = 890\;\text{g mol}^{-1}$$

Step-2 : Alkaline hydrolysis (saponification) of $$Q$$ with excess $$NaOH$$

$$C_3H_5\big(OOC\,C_{17}H_{35}\big)_3 + 3\,NaOH \;\rightarrow\; 3\,C_{17}H_{35}COO^-Na^+ + C_3H_8O_3$$

Thus 1 mol of $$Q$$ gives 3 mol of sodium stearate.

Molar mass of sodium stearate:
Carboxylate anion $$C_{17}H_{35}COO^-$$ has mass $$284 - 1 = 283$$ g.
Adding Na (23 g) gives $$M_{Na\text{-stearate}} = 283 + 23 = 306\;\text{g mol}^{-1}$$

Step-3 : Conversion of sodium stearate to calcium stearate (Ca-soap $$R$$) using excess $$CaCl_2$$

$$2\,C_{17}H_{35}COO^-Na^+ + CaCl_2 \;\rightarrow\; (C_{17}H_{35}COO^-)_2Ca^{2+} + 2\,NaCl$$

Stoichiometry: 2 mol Na-stearate → 1 mol Ca-stearate.
From Step-2 we have 3 mol Na-stearate, therefore

Number of moles of $$R$$ obtained
$$= \frac{3}{2} = 1.5\;\text{mol}$$

Molar mass of calcium stearate $$R$$:
$$M_R = 40 + 2(283) = 40 + 566 = 606\;\text{g mol}^{-1}$$

Mass of $$R$$ from 1 mol of $$Q$$:
$$m_R = 1.5 \times 606 = 909\;\text{g}$$

Therefore, starting with one mole of $$Q$$, the amount of calcium soap $$R$$ produced is 909 g.

We need to evaluate both statements about the confirmatory test for sulphide ions.

Statement-I Analysis:

When a sulphide salt is warmed with dilute $$H_2SO_4$$, hydrogen sulphide gas ($$H_2S$$) is liberated:

$$Na_2S + H_2SO_4 \rightarrow Na_2SO_4 + H_2S \uparrow$$

When this $$H_2S$$ gas comes in contact with paper dipped in lead acetate solution, it forms lead sulphide ($$PbS$$), which is black in colour:

$$(CH_3COO)_2Pb + H_2S \rightarrow PbS \downarrow (\text{black}) + 2CH_3COOH$$

This is indeed a standard confirmatory test for sulphide ions. So Statement-I is true.

Statement-II Analysis:

Statement-II claims the black colour is due to the formation of lead sulphite ($$PbSO_3$$). This is incorrect. Lead sulphite is white, not black. The black precipitate formed is lead sulphide ($$PbS$$), not lead sulphite.

So Statement-II is false.

Therefore, Statement-I is true but Statement-II is false.

The correct answer is Option 3.

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

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

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

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

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

Let us verify by checking the other options:

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

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

We need to identify which compound is white in color.

Analysis of each compound:

- Ammonium sulphide: $$(NH_4)_2S$$ is colorless in solution (may appear yellowish due to polysulfides). Not a white solid in typical form.

- Lead sulphate: $$PbSO_4$$ is a white crystalline solid. It is an insoluble white precipitate commonly encountered in qualitative analysis.

- Lead iodide: $$PbI_2$$ is yellow in color (bright yellow precipitate).

- Ammonium arsinomolybdate: This is yellow in color (canary yellow precipitate, used in testing for arsenic).

The correct answer is Option (2): lead sulphate.

In the Lassaigne's test for halogens, the sodium fusion extract may contain $$Na_2S$$ (if sulphur is present) and $$NaCN$$ (if nitrogen is present) along with $$NaX$$ (sodium halide).

When silver nitrate is added to test for halogens, both $$Na_2S$$ and $$NaCN$$ would also react with $$AgNO_3$$:

$$Na_2S + 2AgNO_3 \rightarrow Ag_2S \downarrow (\text{black}) + 2NaNO_3$$

$$NaCN + AgNO_3 \rightarrow AgCN \downarrow (\text{white}) + NaNO_3$$

These precipitates ($$Ag_2S$$ and $$AgCN$$) would interfere with the detection of silver halides.

To avoid this interference, the Lassaigne's extract is first boiled with dilute $$HNO_3$$. The dilute nitric acid decomposes $$Na_2S$$ and $$NaCN$$:

$$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$$

$$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$$

The gases $$H_2S$$ and $$HCN$$ escape on boiling, eliminating the interference. After this treatment, only the halide ions remain in solution and can be detected cleanly with $$AgNO_3$$.

The correct answer is Option 4: $$Na_2S$$ and $$NaCN$$ are decomposed by $$HNO_3$$.

Metal ions characterized by flame test:

$$Sr^{2+}$$: crimson red — YES

$$Ba^{2+}$$: apple green — YES

$$Ca^{2+}$$: brick red — YES

$$Cu^{2+}$$: blue-green — YES

$$Zn^{2+}$$: no characteristic flame color — NO

$$Co^{2+}$$: no characteristic flame color — NO

$$Fe^{2+}$$: no characteristic flame color — NO

Total = 4.

Therefore, the answer is $$\boxed{4}$$.

In qualitative analysis of organic compounds (Lassaigne's test):

A. Nitrogen detection: The sodium fusion extract containing NaCN is treated with FeSO₄ and then with dilute H₂SO₄. The formation of Prussian blue colour confirms nitrogen. The product is Fe₄[Fe(CN)₆]₃ → Match with III.

B. Sulphur detection: The sodium fusion extract containing Na₂S is treated with sodium nitroprusside Na₂[Fe(CN)₅NO]. A violet colour confirms sulphur → Match with I.

C. Phosphorus detection: The compound is heated with an oxidizing agent, dissolved in HNO₃, and treated with ammonium molybdate (NH₄)₂MoO₄. A yellow precipitate of ammonium phosphomolybdate confirms phosphorus → Match with IV.

D. Halogen detection: The sodium fusion extract is acidified with HNO₃ and treated with AgNO₃. The formation of a precipitate (AgCl, AgBr, or AgI) confirms halogens → Match with II.

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

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 paper chromatography:

• The cellulose paper contains water molecules trapped in its pores
• This water acts as the stationary phase
• The mobile phase is an organic solvent or a solvent mixture that moves up the paper by capillary action

The water present in the pores of the paper forms the stationary phase.

This matches option 1.

We need to identify organic compound X that, when treated with sodium peroxide in a Carius tube, forms a mineral acid Y, which on addition of $$BaCl_2$$ gives a precipitate Z used for quantitative estimation.

In the Carius method for elemental analysis, the organic compound is heated with a strong oxidizing agent (sodium peroxide, $$Na_2O_2$$) in a sealed tube. Different elements are converted to different products:

- Sulfur (S) is oxidized to sulfate ($$SO_4^{2-}$$)

- Halogens are converted to halide ions

- Phosphorus is converted to phosphate

The mineral acid Y is formed from the oxidation product. When $$BaCl_2$$ is added to Y and forms precipitate Z, the precipitate must be $$BaSO_4$$ (barium sulfate), because:

$$BaSO_4$$ is a white precipitate that is insoluble in water and dilute acids. This is the standard test for sulfate ions, and it is used for quantitative estimation of sulfur in organic compounds. The mineral acid Y is therefore $$H_2SO_4$$ (sulfuric acid).

Identify compound X (must contain sulfur).

Among the options:

(A) Cytosine - $$C_4H_5N_3O$$ - contains C, H, N, O but NO sulfur.

(B) A nucleotide - nucleotides contain C, H, N, O, P but typically NO sulfur.

(C) Methionine - $$C_5H_{11}NO_2S$$ - this is an amino acid that CONTAINS SULFUR.

(D) Chloroxylenol - $$C_8H_9ClO$$ - contains C, H, Cl, O but NO sulfur.

Only methionine contains sulfur. When methionine is treated with $$Na_2O_2$$ in a Carius tube, the sulfur is oxidized to $$Na_2SO_4$$, which on acidification gives $$H_2SO_4$$ (mineral acid Y). Adding $$BaCl_2$$ gives $$BaSO_4$$ precipitate (Z), which is weighed for quantitative estimation of sulfur.

The correct answer is Option 3: Methionine.

We need to identify the color change when SO$$_2$$ gas is passed through acidified K$$_2$$Cr$$_2$$O$$_7$$ solution.

Reaction:

SO$$_2$$ is a reducing agent. It reduces the orange dichromate ion (Cr$$_2$$O$$_7^{2-}$$, where Cr is in +6 state) to green chromic ion (Cr$$^{3+}$$):

$$\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + 3\text{SO}_2 \to \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$$

The Cr$$^{3+}$$ ions in solution give a green color.

This color change from orange to green is a confirmatory test for the presence of SO$$_3^{2-}$$ ions (which produce SO$$_2$$ on treatment with acid).

The correct answer is Option 3: Green.

During the preparation of ferrous ammonium sulphate (Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O), prolonged heating is avoided.

This is because Fe²⁺ ions are easily oxidized to Fe³⁺ ions when heated in the presence of air and water. Prolonged heating would cause oxidation of ferrous (Fe²⁺) to ferric (Fe³⁺) ions, which would contaminate the product.

The correct answer is Option 4: Prevent oxidation.

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$$_3^-$$.

We need to identify which transition element cation does NOT belong to Group IV in qualitative inorganic analysis.

Group IV cations are precipitated by H$$_2$$S in the presence of NH$$_4$$Cl and NH$$_4$$OH (basic medium). This group includes: Co$$^{2+}$$, Ni$$^{2+}$$, Mn$$^{2+}$$, and Zn$$^{2+}$$.

Fe$$^{3+}$$ belongs to Group III, where cations are precipitated by NH$$_4$$OH + NH$$_4$$Cl (as hydroxides). Fe$$^{3+}$$ precipitates as Fe(OH)$$_3$$ (brown precipitate) in Group III.

Among the given options:

- Fe$$^{3+}$$ → Group III (NOT Group IV) ✓

- Zn$$^{2+}$$ → Group IV

- Co$$^{2+}$$ → Group IV

- Ni$$^{2+}$$ → Group IV

The correct answer is Option 1: Fe$$^{3+}$$.

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.

We need to identify the anion in an inorganic salt based on the described test results. When BaCl₂ is added to the water extract of the salt, a white precipitate forms, indicating the presence of an anion that forms an insoluble barium salt. Possible candidates are SO₄²⁻ (BaSO₄ — white), SO₃²⁻ (BaSO₃ — white) and CO₃²⁻ (BaCO₃ — white). The white precipitate dissolves in dilute HCl with the release of a gas X having a characteristic odour. BaSO₄ does not dissolve in dilute HCl, so SO₄²⁻ is ruled out. BaSO₃ dissolves in dilute HCl releasing SO₂, which has a suffocating, pungent odour: $$\text{BaSO}_3 + 2\text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} + \text{SO}_2 \uparrow$$ BaCO₃ also dissolves in HCl but releases CO₂, which is odourless.

The evolution of a gas with a characteristic pungent odour thus confirms the presence of the SO₃²⁻ anion. The answer is Option B: SO₃²⁻.

We need to determine the number of chlorine atoms in bithionol. Bithionol is a well-known antiseptic and disinfectant used in medicated soaps. Its IUPAC name is 2,2'-thiobis(4,6-dichlorophenol), and its chemical formula is $$\text{C}_{12}\text{H}_{6}\text{Cl}_{4}\text{O}_{2}\text{S}$$.

We have two phenol rings connected by a sulfur bridge (thio linkage). Each phenol ring carries two chlorine atoms at the 4 and 6 positions. Since there are two such rings, the total number of chlorine atoms is $$2 \times 2 = 4$$.

Hence, the correct answer is $$\boxed{4}$$.

The reaction described here is the formation of a phenolphthalein-type dye. When a phenolic compound is heated with phthalic anhydride in the presence of concentrated $$H_2SO_4$$ (which acts as a dehydrating agent), a condensation reaction occurs. The resulting product, when dissolved in NaOH, gives a characteristic pink colour.

For this condensation to happen, the phenol must have a free $$-OH$$ group attached directly to the aromatic ring. The electrophilic substitution during the condensation takes place at the ortho or para positions relative to the $$-OH$$ group.

Now, 4-methylphenol (p-cresol) has the structure where the $$-OH$$ group is at position 1 and the $$-CH_3$$ group is at position 4. The para position (position 4) is occupied by $$-CH_3$$, but both ortho positions (positions 2 and 6) are free. The condensation with phthalic anhydride requires two available positions on the ring, and the ortho positions serve this purpose.

Among the other options, option 2 contains an alcoholic $$-OH$$ (not directly on the ring) which cannot undergo this phenol-specific condensation. Options 3 and 4 contain dihydroxy compounds (resorcinol-type), which would give fluorescein (a green fluorescent dye) rather than the pink phenolphthalein-type dye.

Hence, 4-methylphenol gives the pink colour with phthalic anhydride in concentrated $$H_2SO_4$$ followed by NaOH treatment.

Hence, the correct answer is Option 1.

In the Carius method for the estimation of halogens in an organic compound, the compound is heated with fuming nitric acid in a sealed tube (Carius tube). The organic matter is oxidised and the halogen is converted to the corresponding halide ion (X$$^-$$).

Silver nitrate (AgNO$$_3$$) is added to the resulting solution. The halide ions react with $$Ag^+$$ to form an insoluble silver halide precipitate (AgCl, AgBr, or AgI), which is then filtered, washed, dried and weighed to determine the amount of halogen present.

Therefore, the Carius method uses fuming nitric acid in the presence of $$AgNO_3$$.

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.

We match each test/reagent with the species it detects.

(a) Lassaigne's Test: In this test, the organic compound is fused with sodium metal to convert the elements present (nitrogen, sulphur, phosphorus, and halogens) into ionic compounds that can be detected in aqueous solution. Therefore, (a) matches with (iii) — N, S, P, and halogen.

(b) Cu(II) oxide: When an organic compound is heated with copper(II) oxide, the carbon in the compound is oxidised to $$CO_2$$, which can be detected by passing it through lime water (turning it milky). This test is used to detect the presence of carbon. Therefore, (b) matches with (i) — Carbon.

(c) Silver nitrate: The sodium fusion extract, when treated with silver nitrate solution, gives precipitates of silver halides ($$AgCl$$, $$AgBr$$, or $$AgI$$), which are used to specifically identify halogens. Therefore, (c) matches with (iv) — Halogen specifically.

(d) Black precipitate with acetic acid and lead acetate: When the sodium fusion extract containing sulphide ions is treated with acetic acid and lead acetate, a black precipitate of lead sulphide ($$PbS$$) is formed: $$Na_2S + Pb(CH_3COO)_2 \rightarrow PbS \downarrow + 2CH_3COONa$$. Therefore, (d) matches with (ii) — Sulphur.

The correct match is (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).

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 the Carius method for estimation of bromine, the organic compound is burnt in excess silver nitrate, and all the bromine is converted to silver bromide (AgBr), which is weighed.

The molar mass of AgBr = 108 + 80 = 188 g mol$$^{-1}$$.

Moles of AgBr formed: $$n_{\text{AgBr}} = \frac{0.2397}{188} = 1.275 \times 10^{-3}$$ mol

Since each mole of AgBr contains one mole of Br, the mass of bromine in the sample is:

$$m_{\text{Br}} = 1.275 \times 10^{-3} \times 80 = 0.1020 \text{ g}$$

The percentage of bromine in the organic compound is:

$$\% \text{Br} = \frac{m_{\text{Br}}}{m_{\text{compound}}} \times 100 = \frac{0.1020}{0.15} \times 100 = 68.0\%$$

The percentage of bromine in the organic compound is $$\mathbf{68}$$%.

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.

In column chromatography that uses silica gel as the stationary phase, the surface of the silica bears many $$\text{Si-OH}$$ (silanol) groups. These groups are polar and weakly acidic. Compounds reaching the column therefore compete with the solvent for hydrogen-bonding or acid-base interaction sites on the silica.

The governing principle is straightforward: the stronger a compound binds to the polar silica surface, the more slowly it travels down the column. Conversely, compounds that bind only weakly are swept out first by the mobile phase. Hence, we must compare the relative strength of interaction (and therefore the effective polarity on silica) of the three given compounds.

We list each structure and its key interaction sites:

• Compound (C) acetophenone, $$C_{6}H_{5}COCH_{3}$$, contains a single carbonyl oxygen. The carbonyl oxygen is only a hydrogen-bond acceptor and cannot donate a hydrogen bond. Moreover, there are no basic lone pairs that become protonated by the acidic silica surface. Thus, interaction with silica is limited to weak dipole attractions; overall, acetophenone is the least strongly adsorbed.

• Compound (A) benzanilide, $$C_{6}H_{5}CONHC_{6}H_{5}$$, is an amide. The carbonyl oxygen can accept a hydrogen bond and the $$N-H$$ can donate one. However, the lone pair on the nitrogen is delocalised into the carbonyl group, so basicity is strongly reduced; the molecule is also sterically bulky because of two phenyl rings, which lessens surface contact. Hence, benzanilide interacts with silica more than acetophenone but less than a free amine.

• Compound (B) aniline, $$C_{6}H_{5}NH_{2}$$, possesses a free amino group. On the acidic silica surface this amino group can become partially protonated $$\bigl(RNH_{2} + Si-OH -> RNH_{3}^{+ + Si-O^{-}}\bigr)$$, generating a strong ionic (acid-base) attraction in addition to hydrogen bonding. Therefore aniline is most strongly retained on silica.

Combining these observations, the adsorption (polarity) order on silica is

$$\text{aniline (B)} \;>\; \text{benzanilide (A)} \;>\; \text{acetophenone (C)}.$$

Because the stationary phase is polar and the mobile phase is the relatively polar mixture hexane : ethyl acetate (20 : 80), the compounds with weaker adsorption (lower effective polarity on silica) travel the fastest and elute first. Reversing the above order therefore gives the elution sequence:

$$\text{acetophenone (C)} \;\longrightarrow\; \text{benzanilide (A)} \;\longrightarrow\; \text{aniline (B)}.$$

Thus the column first furnishes (C), then (A), and finally (B).

Hence, the correct answer is Option C.

We have a mixture that contains three compounds dissolved in the organic solvent ethyl acetate:

$$$\text{m-chloroaniline (basic amine)}$$$ $$$\text{m-chlorophenol (weak acid, }pK_a\approx 10\text{)}$$$ $$$\text{m-chlorobenzoic acid (stronger acid, }pK_a\approx 4\text{)}$$$

The separation is carried out by successive extractions, using the difference in acid-base strength of the functional groups. A compound converts to its water-soluble ionic form only if the reagent present is strong enough to protonate or deprotonate it. We trace each step.

First extraction: a saturated solution of $$\text{NaHCO}_3$$ (a weak base) is shaken with the organic layer.

The reaction possible is:

$$$\text{m-chlorobenzoic acid (HA)} + \text{HCO}_3^- \;\longrightarrow\; \text{m-chlorobenzoate ion (A}^-)\;+\;\text{H}_2\text{CO}_3$$$

Because $$pK_a(\text{benzoic acid})\approx 4$$ is sufficiently low, $$\text{HCO}_3^-$$ can remove its proton. Phenol, with $$pK_a\approx 10$$, is not acidic enough to be deprotonated by $$\text{HCO}_3^-$$, and the basic amine is certainly not affected by a base. Thus only the salt of m-chlorobenzoic acid transfers to the aqueous layer. We label this aqueous layer as Fraction A.

So, after the first extraction:

$$$\text{Fraction A (aqueous)}:\; \text{m-chlorobenzoate ion }(\;\longrightarrow\; \text{recovers m-chlorobenzoic acid on acidification)}$$$ $$$\text{Organic phase remaining}:\; \text{m-chlorophenol + m-chloroaniline}$$$

Second extraction: the residual organic layer is now shaken with dilute $$\text{NaOH}$$ (a stronger base than $$\text{NaHCO}_3$$).

The reaction that occurs is:

$$$\text{m-chlorophenol (ArOH)} + \text{OH}^- \;\longrightarrow\; \text{m-chlorophenoxide ion (ArO}^-)\;+\;\text{H}_2\text{O}$$$

This time the base is strong enough to deprotonate the phenol. The resulting phenoxide ion is water-soluble and goes into the new aqueous layer, which we label as Fraction B. The amine does not react with base, so it stays in the organic phase.

Thus after the second extraction:

$$$\text{Fraction B (aqueous)}:\; \text{m-chlorophenoxide ion }(\;\longrightarrow\; \text{recovers m-chlorophenol on acidification)}$$$ $$$\text{Organic phase left}:\; \text{m-chloroaniline}$$$

The final organic layer is labelled Fraction C, and it contains only the unreacted base.

Summarising the contents in their neutral, isolated forms obtained after the usual acidification of the aqueous fractions:

$$$\text{Fraction A} \;:\; \text{m-chlorobenzoic acid}$$$ $$$\text{Fraction B} \;:\; \text{m-chlorophenol}$$$ $$$\text{Fraction C} \;:\; \text{m-chloroaniline}$$$

Matching this order with the given options, we see that Option B lists the compounds in exactly this sequence.

Hence, the correct answer is Option B.

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

We start by recalling a well-known qualitative test that analysts employ for detecting the presence of the element phosphorus in an organic compound. The sequence of reagents is distinctive: first the compound is fused or oxidised with an oxidising agent such as sodium peroxide, $$\text{Na}_2\text{O}_2$$, in order to convert any phosphorus present into the water-soluble phosphate ion $$\text{PO}_4^{3-}$$.

So, when the unknown organic compound “A” is treated with $$\text{Na}_2\text{O}_2$$, the following transformation takes place:

$$\text{(Phosphorus in A)} \; \overset{\text{Na}_2\text{O}_2}{\underset{\text{fusion}}{\rightarrow}} \; \text{Na}_3\text{PO}_4$$

Next, according to the stated procedure, the fusion mass is boiled with concentrated nitric acid, $$\text{HNO}_3$$. The purpose of this step is twofold: (i) to dissolve the sodium phosphate formed, producing orthophosphoric acid $$\text{H}_3\text{PO}_4$$ (or its nitrate solution), and (ii) to destroy any excess peroxide or other interfering species.

Symbolically we may write:

$$\text{Na}_3\text{PO}_4 + 3 \,\text{HNO}_3 \;\longrightarrow\; \text{H}_3\text{PO}_4 + 3 \,\text{NaNO}_3$$

Now we come to the characteristic identification step. The clear nitric-acid solution containing phosphate ions is treated with ammonium molybdate, $$(\text{NH}_4)_2\text{MoO}_4$$, in the presence of excess $$\text{HNO}_3$$ (the medium must be acidic). If phosphate ions are indeed present, a canary-yellow precipitate of ammonium phosphomolybdate is produced.

The balanced form of this precipitation reaction is generally represented as:

$$\text{H}_3\text{PO}_4 + 12 \,(\text{NH}_4)_2\text{MoO}_4 + 21 \,\text{HNO}_3 \;\longrightarrow\; (\text{NH}_4)_3\text{PO}_4 \cdot 12 \,\text{MoO}_3 \; \downarrow \; + 21 \,\text{NH}_4\text{NO}_3 + 12 \,\text{H}_2\text{O}$$

The appearance of the yellow precipitate is therefore an unmistakable confirmation test for phosphorus.

In the question, we are told exactly this sequence: oxidation with $$\text{Na}_2\text{O}_2$$, boiling with $$\text{HNO}_3$$, and finally the formation of a yellow precipitate upon addition of ammonium molybdate. Since this matches the classic test for phosphorus, we deduce that the element present in compound “A” must be phosphorus.

Fluorine, nitrogen and sulphur do not give such a yellow molybdate precipitate under these conditions; their identification relies on entirely different qualitative tests (e.g., the Lassaigne test for nitrogen and sulphur, or the precipitation of $$\text{BaSO}_4$$ for sulphate, etc.).

Hence, the correct answer is Option B.

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.

We are told that, in column chromatography, compound I is adsorbed more strongly than compound II, i.e. $$\text{adsorption of I} > \text{adsorption of II}.$$

In column chromatography the stationary phase is an adsorbent, often silica or alumina. A greater degree of adsorption means a stronger interaction with the stationary phase, so the compound travels down the column more slowly. Conversely, a compound that is adsorbed less strongly interacts weakly with the stationary phase and is carried downward more quickly by the mobile phase.

Therefore, because I is more strongly adsorbed, I moves slower through the column, while II, being less strongly adsorbed, moves faster.

Now let us connect this to the $$R_f$$ value, which is defined (for thin-layer chromatography, but the qualitative trend is the same) as

$$R_f = \dfrac{\text{distance travelled by the solute spot}}{\text{distance travelled by the solvent front}}.$$

If a compound travels a larger distance on the plate (or equivalently elutes faster in a column), the numerator in the above fraction is larger, giving a higher $$R_f$$. Thus, the compound that moves faster automatically acquires the higher $$R_f$$ value.

Since we have already established that II moves faster than I, it must also have the higher $$R_f$$ value.

Putting the two conclusions together:

• II moves faster than I, and
• II has the higher $$R_f$$ value.

This statement matches exactly with Option B.

Hence, the correct answer is Option B.

In Lassaigne’s test we first fuse the organic compound with metallic sodium. During this fusion the different hetero-elements present in the compound are converted into simple, water-soluble sodium salts:

$$\text{Halogen (X)} \; \rightarrow \; NaX$$ $$\text{Sulphur (S)} \; \rightarrow \; Na_2S$$ $$\text{Nitrogen (N)} \; \rightarrow \; NaCN$$

After the fusion, the water extract that contains these ions is called the sodium extract or Lassaigne filtrate. To detect halide ions we ordinarily add aqueous silver nitrate. The analytical reaction we depend upon is

$$Ag^+ + X^- \rightarrow AgX\downarrow \quad (\text{white, pale-yellow or yellow precipitate})$$

However, if the ions $$S^{2-}$$ or $$CN^-$$ are still present, they also give precipitates with silver ions:

$$2\,Ag^+ + S^{2-} \rightarrow Ag_2S\downarrow \;(\text{black})$$ $$Ag^+ + CN^- \rightarrow AgCN\downarrow \;(\text{white})$$

The black $$Ag_2S$$ or the white $$AgCN$$ would be mistaken for the halide precipitate and thus interfere with the test. Therefore, before we add silver nitrate we deliberately destroy $$S^{2-}$$ and $$CN^-$$ by boiling the sodium extract with concentrated nitric acid.

First we acidify:

$$Na_2S + 2\,HNO_3 \rightarrow 2\,NaNO_3 + H_2S$$

Then the strongly oxidising, hot $$HNO_3$$ converts sulphide further to sulphur or sulphate; a simplified oxidation step can be written as

$$3\,H_2S + 8\,HNO_3 \rightarrow 3\,S\downarrow + 8\,NO_2\uparrow + 4\,H_2O$$

In the same way, cyanide is oxidised and finally broken down into harmless gaseous products:

$$2\,NaCN + 4\,HNO_3 \rightarrow 2\,NaNO_3 + 2\,CO_2\uparrow + 2\,NO\uparrow + H_2O$$

Because $$S^{2-}$$ and $$CN^-$$ are completely destroyed, no $$Ag_2S$$ or $$AgCN$$ can form later. The halide ions, on the other hand, remain unaffected by concentrated nitric acid, so they alone precipitate with $$AgNO_3$$ and the test becomes reliable.

Thus the reason for heating the sodium extract with concentrated $$HNO_3$$ is that the possible interfering ions $$S^{2-}$$ and $$CN^-$$ are decomposed and cannot give misleading precipitates.

Hence, the correct answer is Option C.

In a Bunsen burner, the point of maximum temperature is always found just above the tip of the inner cone, where combustion is most efficient. According to the labels in the diagram, this corresponds to Region 2.

In qualitative inorganic analysis for group V cations, which include calcium ($$Ca^{2+}$$), barium ($$Ba^{2+}$$), and strontium ($$Sr^{2+}$$), we precipitate them as carbonates using ammonium carbonate ($$(NH_4)_2CO_3$$). The question asks why sodium carbonate ($$Na_2CO_3$$) cannot be used instead.

First, consider the precipitation reaction. The carbonate ion ($$CO_3^{2-}$$) reacts with these metal ions to form insoluble carbonates:

$$ Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3 \downarrow $$

$$ Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \downarrow $$

$$ Sr^{2+} + CO_3^{2-} \rightarrow SrCO_3 \downarrow $$

However, if we use sodium carbonate ($$Na_2CO_3$$), it dissociates completely in water:

$$ Na_2CO_3 \rightarrow 2Na^+ + CO_3^{2-} $$

This provides a high concentration of carbonate ions ($$CO_3^{2-}$$). Magnesium ion ($$Mg^{2+}$$), which belongs to group IV, is not supposed to precipitate in group V. But magnesium carbonate ($$MgCO_3$$) is sparingly soluble and can precipitate if the carbonate ion concentration is sufficiently high:

$$ Mg^{2+} + CO_3^{2-} \rightarrow MgCO_3 \downarrow $$

Ammonium carbonate ($$(NH_4)_2CO_3$$) is used instead because it contains ammonium ions ($$NH_4^+$$). These ions suppress the carbonate ion concentration by reacting with it:

$$ CO_3^{2-} + NH_4^+ \rightarrow HCO_3^- + NH_3 $$

This reaction reduces the concentration of $$CO_3^{2-}$$ ions, making it insufficient to precipitate $$MgCO_3$$ (since $$MgCO_3$$ has a higher solubility product compared to the carbonates of $$Ca^{2+}$$, $$Ba^{2+}$$, and $$Sr^{2+}$$). Thus, $$Mg^{2+}$$ remains in solution and is not precipitated in group V.

In contrast, sodium carbonate does not provide any suppressing ions. The sodium ions ($$Na^+$$) do not react with carbonate ions and do not reduce their concentration. Therefore, the high $$CO_3^{2-}$$ concentration from $$Na_2CO_3$$ causes $$Mg^{2+}$$ to precipitate as $$MgCO_3$$ along with the group V cations.

Now, evaluating the options:

Option A states that $$Mg^{2+}$$ ions will also be precipitated. This aligns with our reasoning, as the high carbonate concentration from sodium carbonate leads to unwanted precipitation of magnesium carbonate.

Option B claims the concentration of $$CO_3^{2-}$$ ions is very low. This is incorrect because sodium carbonate provides a high concentration of $$CO_3^{2-}$$ ions due to complete dissociation.

Option C suggests sodium ions react with acid radicals. Sodium ions ($$Na^+$$) are inert and do not react with acid radicals in a way that affects the precipitation of group V cations.

Option D says $$Na^+$$ ions interfere with the detection of $$Ca^{2+}$$, $$Ba^{2+}$$, $$Sr^{2+}$$. Sodium ions do not form insoluble compounds and do not interfere directly; the issue is the precipitation of $$Mg^{2+}$$.

Hence, the correct answer is Option A.

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.