Chemical analysis of organic compounds Questions in English

(B) In the Lassaigne's test for nitrogen in an organic compound,the sodium fusion extract is boiled with iron $(II)$ sulphate and then acidified with sulphuric acid.
In the process,sodium cyanide first reacts with iron $(II)$ sulphate to form sodium hexacyanoferrate $(II)$.
Then,on heating with sulphuric acid,some iron $(II)$ gets oxidised to iron $(III)$,which reacts with hexacyanoferrate $(II)$ to form iron $(III)$ hexacyanoferrate $(II)$,which is Prussian blue in colour.
The chemical equations involved in the reaction are:
$6CN^{-} + Fe^{2+} \longrightarrow [Fe(CN)_{6}]^{4-}$
$3[Fe(CN)_{6}]^{4-} + 4Fe^{3+} \xrightarrow{xH_{2}O} Fe_{4}[Fe(CN)_{6}]_{3} \cdot xH_{2}O$ (Prussian blue)
Hence,the Prussian blue colour is due to the formation of $Fe_{4}[Fe(CN)_{6}]_{3}$.

(B) $(i)$ The purity of a compound is ascertained by determining its melting or boiling point. Most pure compounds exhibit sharp melting and boiling points.
$(ii)$ Modern methods for checking the purity of an organic compound are based on various chromatographic and spectroscopic techniques.

(N/A) In an organic compound,carbon and hydrogen are detected by heating the compound with copper$(II)$ oxide $(CuO)$.
$i$. Detection of carbon: Carbon present in the compound is oxidized to carbon dioxide $(CO_2)$,which turns lime water milky due to the formation of calcium carbonate.
Reaction: $C + 2CuO \rightarrow 2Cu + CO_{2(g)}$
$CO_{2(g)} + Ca(OH)_{2(aq)} \rightarrow CaCO_{3(s)} + H_2O_{(l)}$
$ii$. Detection of hydrogen: Hydrogen present in the compound is oxidized to water $(H_2O)$. This water is tested using anhydrous copper$(II)$ sulphate $(CuSO_4)$,which turns blue due to the formation of hydrated copper sulphate.
Reaction: $2H + CuO \xrightarrow{\Delta} Cu + H_2O$
$5H_2O + CuSO_{4(white)} \rightarrow CuSO_4 \cdot 5H_2O_{(blue)}$

(N/A) Nitrogen $(N)$,Sulphur $(S)$,Halogens $(Cl, Br, I)$,and Phosphorus $(P)$ present in an organic compound are detected by "Lassaigne's test".
$(a)$ The elements $(N, X, S)$ present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. The following reactions take place:
$(i)$ $Na + C + N \xrightarrow{\Delta} NaCN$
$(ii)$ $2Na + S \xrightarrow{\Delta} Na_{2}S$
$(iii)$ $Na + X \xrightarrow{\Delta} NaX$ (where $X = Cl, Br, \text{ or } I$)
Cyanide,sulphide,and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
$(b)$ Test for Nitrogen:
Procedure: The sodium fusion extract is boiled with iron $(II)$ sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron $(II)$ sulphate and forms sodium hexacyanidoferrate $(II)$. On heating with concentrated sulphuric acid,some iron $(II)$ ions are oxidised to iron $(III)$ ions which react with sodium hexacyanidoferrate $(II)$ to produce iron $(III)$ hexacyanidoferrate $(II)$ (ferriferrocyanide),which is Prussian blue in colour.
$6CN_{(aq)}^{-} + Fe_{(aq)}^{2+} \rightarrow [Fe(CN)_{6}]_{(aq)}^{4-}$
$3[Fe(CN)_{6}]_{(aq)}^{4-} + 4Fe^{3+} \xrightarrow{xH_{2}O} Fe_{4}[Fe(CN)_{6}]_{3} \cdot xH_{2}O$ (Prussian blue)
$(c)$ Test for Sulphur: Following two tests occur for detection of sulphur:
$(i)$ The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. $A$ black precipitate of lead sulphide indicates the presence of sulphur.
$S_{(aq)}^{2-} + Pb_{(aq)}^{2+} \rightarrow PbS_{(s)}$ (Black precipitate)
$(ii)$ On treating sodium fusion extract with sodium nitroprusside,the appearance of a violet colour further indicates the presence of sulphur.

(B) The sodium fusion extract often contains sodium cyanide $(NaCN)$ and sodium sulphide $(Na_2S)$ if nitrogen and sulphur are present in the organic compound.
If $AgNO_3$ is added directly,these ions would form precipitates like $AgCN$ and $Ag_2S$,which interfere with the halide test.
Adding $HNO_3$ decomposes these species:
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$
This ensures that only the halide ions ($Cl^-$,$Br^-$,$I^-$) react with $Ag^+$ to form the characteristic silver halide precipitate.

(N/A) Organic compounds contain elements like nitrogen,sulphur,and halogens in a covalent form.
These elements cannot be detected directly by standard chemical tests.
Fusion with metallic sodium converts these elements into their corresponding water-soluble ionic salts.
The reactions are as follows:
$Na + C + N \rightarrow NaCN$
$2Na + S \rightarrow Na_{2}S$
$Na + X \rightarrow NaX$ (where $X = Cl, Br, I$).
These ionic salts can then be easily detected using specific reagents.

(B) When an organic compound is fused with sodium metal,the elements present in the compound (such as $N$,$S$,and halogens $X$) are converted from their covalent form into their corresponding ionic forms (such as $NaCN$,$Na_2S$,and $NaX$). This is the principle behind the Lassaigne's test.

(N/A) The detection of nitrogen involves the following steps:
$1$. Fusion with sodium metal: $N + C + Na \longrightarrow NaCN$
$2$. Formation of ferrocyanide: $6 CN^{-} + Fe^{2+} \longrightarrow [Fe(CN)_{6}]^{4-}$
$3$. Formation of ferric ferrocyanide (Prussian blue): $4 Fe^{3+} + 3 [Fe(CN)_{6}]^{4-} \cdot x H_{2}O \longrightarrow Fe_{4}[Fe(CN)_{6}]_{3} \cdot x H_{2}O$

(A) In the Lassaigne's test for nitrogen,the sodium fusion extract containing sodium cyanide $(NaCN)$ is treated with ferrous sulphate $(FeSO_{4})$.
This leads to the formation of sodium ferrocyanide,$Na_{4}[Fe(CN)_{6}]$.
On further reaction with ferric chloride $(FeCl_{3})$,it forms ferric ferrocyanide,which is known as Prussian blue.
The chemical formula is $Fe_{4}[Fe(CN)_{6}]_{3} \cdot x H_{2}O$.
This compound is chemically known as Iron$(III)$ hexacyanoferrate$(II)$.

(B) In organic compounds,elements like nitrogen,sulphur,and halogens are covalently bonded to the carbon skeleton.
Because they are not present as free ions,they do not react directly with reagents to give characteristic ionic tests.
Therefore,the organic compound is fused with sodium metal to convert these covalent elements into their corresponding water-soluble ionic salts (e.g.,$NaCN$,$Na_2S$,$NaCl$),which can then be detected using standard reagents.

(C) When $AgNO_3$ is added to the Lassaigne solution acidified with $HNO_3$,the halide ions $(X^-)$ react with silver ions $(Ag^+)$ to form a precipitate of silver halide $(AgX)$.
$Ag^+ (aq) + X^- (aq) \longrightarrow AgX (s) \downarrow$

(N/A) $i$. White precipitate: Chlorine is present,and these precipitates are more soluble in $NH_4OH$.
$ii$. Light yellowish precipitate: Bromine is present. The precipitate is sparingly soluble in ammonium hydroxide and becomes soluble upon heating.
$iii$. Yellow precipitate: It is insoluble in $NH_4OH$,indicating the presence of iodine.

(B) The sodium extract often contains sodium cyanide $(NaCN)$ and sodium sulphide $(Na_2S)$ along with sodium halides $(NaX)$.
If these are not removed,they would react with silver nitrate $(AgNO_3)$ to form silver cyanide $(AgCN)$ and silver sulphide $(Ag_2S)$,which are also precipitates and would interfere with the test for halogens.
Nitric acid $(HNO_3)$ is added to decompose these interfering ions into volatile gases like hydrogen cyanide $(HCN)$ and hydrogen sulphide $(H_2S)$,which are then boiled off.
Therefore,the correct reason is to decompose sodium cyanide and sodium sulphide if present.

(N/A) $(i)$ $A$ blood-red colour is obtained due to the formation of $[Fe(SCN)]^{2+}$.
$(ii)$ By adding $AgNO_3$,a pale yellow precipitate of $AgBr$ is formed.
$(iii)$ It indicates that sulphur is absent in the organic compound.
$(iv)$ $FeSO_4$ is added to react with $CN^-$ ions to form ferrocyanide ions,which then react with $Fe^{3+}$ ions.
$(v)$ It is due to the formation of ferric ferrocyanide,$Fe_4[Fe(CN)_6]_3$.
$(vi)$ It indicates that sulphur is present in the compound,forming $Na_4[Fe(CN)_5NOS]$.

(N/A) When $AgNO_3$ is added to the sodium fusion extract (acidified with $HNO_3$),a precipitate of silver halide $(AgX)$ is formed.
$(i)$ $AgCl$ forms a white precipitate.
(ii) $AgBr$ forms a light yellow precipitate.
(iii) $AgI$ forms a yellow precipitate.

(B) In the Dumas method,the organic compound is heated with copper oxide $(CuO)$ in an atmosphere of carbon dioxide. The nitrogen gas evolved is collected over a potassium hydroxide $(KOH)$ solution. The volume of nitrogen gas is measured using an apparatus called a $Nitrometer$.

(C) The correct matches are as follows:
$(i).$ Prussian blue is formed in the test for the presence of nitrogen $(Fe_4[Fe(CN)_6]_3)$.
$(ii).$ Yellow precipitate is formed in the test for the presence of phosphorus (ammonium phosphomolybdate).
$(iii).$ Purple solution is formed in the test for the presence of sulphur (sodium nitroprusside test).
$(iv).$ Black precipitate is formed in the test for the presence of sulphur (lead sulphide).
Therefore,the correct matching is $(i-c, ii-a, iii-b, iv-b)$.

(D) The correct matches are as follows:
$(i).$ Lassaigne's test involves $(d).$ Fusion with sodium.
$(ii).$ Test for nitrogen involves $(c).$ Acidify with conc. sulphuric acid (to form Prussian blue).
$(iii).$ Test for sulphur involves $(b).$ Acidify with acetic acid (to detect sulphide ions).
$(iv).$ Test for phosphorus involves $(a).$ By sodium peroxide (oxidative fusion).
Therefore,the correct sequence is $(i-d), (ii-c), (iii-b), (iv-a)$.

(B) In the Lassaigne's test,the sodium extract is acidified with dilute $HNO_{3}$ to decompose any $NaCN$ or $Na_{2}S$ present,which would otherwise interfere with the test.
If dilute $H_{2}SO_{4}$ is used instead of $HNO_{3}$,$Ag^{+}$ ions from $AgNO_{3}$ will react with $SO_{4}^{2-}$ ions to form $Ag_{2}SO_{4}$.
$2AgNO_{3} + H_{2}SO_{4} \rightarrow Ag_{2}SO_{4} + 2HNO_{3}$
Since $Ag_{2}SO_{4}$ has low solubility,it forms a white precipitate that can be incorrectly identified as $AgCl$,leading to a false positive result.

(A) If the organic compound contains both $N$ and $S$,then during fusion it may form either sodium thiocyanate $(NaSCN)$ or a mixture of sodium cyanide $(NaCN)$ and sodium sulphide $(Na_2S)$ depending upon the amount of sodium metal used. If the sodium metal used is less,only $NaSCN$ is produced.
This then reacts with $Fe^{3+}$ ions (produced by oxidation of $Fe^{2+}$ ions during the preparation of Lassaigne's extract) to give red colouration due to the formation of ferric thiocyanate.
$Fe^{2+} \xrightarrow{\text{Aerial oxidation}} Fe^{3+}$; $Fe^{3+} + 3SCN^- \longrightarrow [Fe(SCN)_3]$ (Blood red colour)
If excess sodium is used,$NaCN$ and $Na_2S$ are formed. $NaCN$ reacts with $FeSO_4$ to give Prussian blue colour due to the formation of ferric ferrocyanide,$Fe_4[Fe(CN)_6]_3$.
$6NaCN + FeSO_4 \longrightarrow Na_4[Fe(CN)_6] + Na_2SO_4$
$3Na_4[Fe(CN)_6] + 4Fe^{3+} \longrightarrow Fe_4[Fe(CN)_6]_3 + 12Na^+$
From the above discussion,it follows that Manish and Rajni used excess sodium,forming $NaCN$ which gave Prussian blue colour,while Ramesh used less sodium,forming $NaSCN$ which gave red colouration.

(A-3, B-5, C-1, D-2, E-4) The correct matching is: $A-3, B-5, C-1, D-2, E-4$.
$(A).$ Dumas method: In this method,nitrogen is estimated as nitrogen gas $(N_2)$.
$(B).$ Kjeldahl's method: In this method,nitrogen is converted into ammonium sulphate $((NH_4)_2SO_4)$.
$(C).$ Carius method: This method is used for the estimation of halogens,where the compound is heated with $AgNO_3$ in the presence of $HNO_3$.
$(D).$ Chromatography: In thin layer chromatography,a thin layer $(0.2 \ mm)$ of an adsorbent such as silica gel is spread over a glass plate,which acts as a stationary phase.
$(E).$ Homolytic fission: When the homolytic fission of a covalent bond occurs,free radicals are formed.

(C) The correct matches are as follows:
$(a)$ Lassaigne's test is used for the detection of nitrogen,sulphur,phosphorus,and halogens in organic compounds. Thus,$(a)-(iii)$.
$(b)$ $Cu(II)$ oxide is used in the detection of carbon in organic compounds (via the formation of $CO_2$). Thus,$(b)-(i)$.
$(c)$ Silver nitrate $(AgNO_3)$ is used specifically for the detection of halogens. Thus,$(c)-(iv)$.
$(d)$ The sodium fusion extract gives a black precipitate with acetic acid and lead acetate,which confirms the presence of sulphur $(PbS)$. Thus,$(d)-(ii)$.
Therefore,the correct sequence is $(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)$.

(A) For the testing of halogens,$HNO_3$ (Nitric acid) is added to the sodium extract.
This is done because if $CN^-$ or $S^{2-}$ ions are present in the extract,they would interfere with the test by forming $AgCN$ or $Ag_2S$ precipitates.
$HNO_3$ decomposes these ions into $HCN$ and $H_2S$ gases,respectively,which are then removed by boiling,ensuring that only halides react with $AgNO_3$.

(B) The sodium fusion extract (Lassaigne's extract) is used to detect elements like nitrogen,sulfur,halogens,and phosphorous in organic compounds.
These elements are converted into their respective water-soluble ionic forms by fusing the organic compound with sodium metal.
For example,nitrogen is converted to $NaCN$,sulfur to $Na_2S$,halogens to $NaX$ (where $X = Cl, Br, I$),and phosphorous to $Na_3PO_4$.

(B) The Kjeldahl's method is not applicable to compounds containing nitrogen in nitro $(-NO_2)$ or azo $(-N=N-)$ groups,or nitrogen present in a ring (like pyridine),because the nitrogen in these compounds is not quantitatively converted to ammonium sulphate under the conditions of the Kjeldahl's method.
Among the given options:
$A$. Nitrobenzene contains a nitro group.
$B$. Aniline $(C_6H_5NH_2)$ contains an amino group,which is easily converted to ammonium sulphate.
$C$. Azobenzene contains an azo group.
$D$. Pyridine contains nitrogen in the ring.
Therefore,the Kjeldahl's method can be used for Aniline.

(B) The sodium fusion extract is boiled with concentrated $HNO_3$ to decompose sodium cyanide $(NaCN)$ and sodium sulphide $(Na_2S)$ present in the extract.
This is necessary because these ions interfere with the silver nitrate $(AgNO_3)$ test for halogens by forming precipitates like $AgCN$ or $Ag_2S$.

(A) In Lassaigne's test,if an organic compound contains both nitrogen and sulphur,they react with fused sodium to form sodium thiocyanate $(NaSCN)$.
The chemical reaction is: $Na + C + N + S \rightarrow NaSCN$.
However,if an excess of sodium is used,the sodium thiocyanate is decomposed into sodium cyanide $(NaCN)$ and sodium sulphide $(Na_{2}S)$.
The chemical reaction is: $NaSCN + 2Na \rightarrow NaCN + Na_{2}S$.
Therefore,both Statement $I$ and Statement $II$ are correct.

(A) In $Lassaigne's$ test for sulphur,the sodium fusion extract is treated with sodium nitroprusside.
The reaction is: $Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5(NOS)]$.
The complex $Na_4[Fe(CN)_5(NOS)]$ is responsible for the deep violet or purple colouration.

(B) Lassaigne's test is used to detect elements like nitrogen,sulfur,and halogens in organic compounds.
For a positive test,the compound must contain carbon along with nitrogen and halogen to form sodium cyanide $(NaCN)$ and sodium halide $(NaX)$ upon fusion with sodium metal.
$CH_3NH_2 \cdot HCl$ is an organic compound containing carbon,nitrogen,and chlorine.
Upon fusion with sodium $(Na)$,it forms $NaCN$ (which gives a positive test for nitrogen) and $NaCl$ (which gives a positive test for halogen).
$N_2H_4 \cdot HCl$,$NH_4Cl$,and $NH_2OH \cdot HCl$ are inorganic compounds that do not contain carbon,hence they do not form $NaCN$ during the fusion process.

(D) The correct matches are:
$A$. Nitrogen: Detected by forming Prussian blue,$Fe_4[Fe(CN)_6]_3$ $(III)$.
$B$. Sulphur: Detected by Sodium nitroprusside,$Na_2[Fe(CN)_5NO]$ $(I)$.
$C$. Phosphorus: Detected by ammonium molybdate,$(NH_4)_2MoO_4$ $(IV)$.
$D$. Halogen: Detected by $AgNO_3$ $(II)$.
Therefore,the correct sequence is $A-III, B-I, C-IV, D-II$.

(A) When both nitrogen and sulphur are present in an organic compound,sodium thiocyanate $(NaSCN)$ is formed during the preparation of Lassaigne's extract.
$Na + C + N + S \rightarrow NaSCN$
This $NaSCN$ reacts with $Fe^{3+}$ ions to form a blood-red coloured complex,$[Fe(SCN)]^{2+}$.
$Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$
Since all nitrogen is consumed in the formation of $SCN^-$,no free cyanide ions $(CN^-)$ are available to form the prussian blue precipitate.

(C) The sodium fusion extract contains $NaCN$ and $Na_2S$ if both nitrogen and sulphur are present in the organic compound.
These react to form sodium thiocyanate: $Na^{+} + CN^{-} + S^{2-} \rightarrow NaSCN$.
In the presence of $Fe^{3+}$ ions (formed by the oxidation of $Fe^{2+}$ by concentrated $H_2SO_4$),the thiocyanate ion forms a blood-red complex: $Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$.
Thus,the appearance of a blood-red colour confirms the presence of both $N$ and $S$.

(D) In the Lassaigne's test,if the organic compound contains nitrogen or sulphur,the sodium fusion extract will contain $NaCN$ or $Na_2S$ respectively.
These ions interfere with the silver nitrate test for halogens by forming precipitates like $AgCN$ or $Ag_2S$.
Boiling the extract with dilute $HNO_3$ decomposes these species into volatile gases ($HCN$ and $H_2S$),thereby removing them from the solution before the addition of $AgNO_3$.

(D) Lassaigne's test is used to detect nitrogen,sulfur,and halogens in organic compounds.
It involves the fusion of the organic compound with sodium metal,which converts the elements present into their corresponding sodium salts.
For nitrogen,it forms sodium cyanide $(NaCN)$.
Since $NaCN$ requires both carbon and nitrogen,the organic compound must contain both elements.
Hydrazine $(NH_2-NH_2)$ contains nitrogen but lacks carbon,therefore it cannot form $NaCN$ and does not give a positive Lassaigne's test.

(A) $A. H_2N-NH_3^+Cl^-$ contains $Cl^-$ (gives white precipitate with $AgNO_3$,$r$) and contains $N$ (gives Prussian blue with $Na$ fusion,$p$). It also reacts with aldehydes to form hydrazones $(s)$.
$B. HO-C_6H_4-CH(NH_3^+)COOH$ (with $I^-$) contains $I^-$ (gives yellow precipitate with $AgNO_3$,not white,so $r$ is excluded). It has a phenol group (gives $FeCl_3$ test,$q$) and contains $N$ (gives $p$).
$C. HO-C_6H_4-NH_3^+Cl^-$ contains $Cl^-$ (gives $r$),phenol group (gives $q$),and $N$ (gives $p$).
$D. (NO_2)_2C_6H_3-NH-NH_3^+Br^-$ contains $Br^-$ (gives cream/white precipitate with $AgNO_3$,$r$),$N$ (gives $p$),and is a hydrazine derivative (reacts with aldehydes,$s$).
Matching the options: $A-p, r, s; B-p, q; C-p, q, r; D-p, r, s$. Based on the provided options,$A-r, s; B-p, q; C-p, q, r; D-p, s$ is the most accurate match.

(A) The scheme shows that one component of the mixture reacts with both $NaOH(aq)$ and $NaHCO_3(aq)$ to become soluble,while the other remains insoluble.
$1.$ $C_6H_5COOH$ (benzoic acid) is a strong acid $(pK_a \approx 4.2)$ and reacts with both $NaOH$ and $NaHCO_3$ to form soluble sodium benzoate.
$2.$ $C_6H_5CH_2COOH$ (phenylacetic acid) is also a strong acid and reacts with both $NaOH$ and $NaHCO_3$ to form soluble sodium phenylacetate.
$3.$ $C_6H_5OH$ (phenol) is a weak acid $(pK_a \approx 10)$ and reacts with $NaOH$ to form soluble sodium phenoxide,but it does not react with $NaHCO_3$.
$4.$ $C_6H_5CH_2OH$ (benzyl alcohol) is neutral and does not react with either $NaOH$ or $NaHCO_3$.
Analyzing the options:
- $(B)$ Mixture of $C_6H_5COOH$ (acid) and $C_6H_5CH_2OH$ (neutral): $C_6H_5COOH$ reacts with both $NaOH$ and $NaHCO_3$ (soluble),while $C_6H_5CH_2OH$ does not (insoluble). This fits the scheme.
- $(D)$ Mixture of $C_6H_5CH_2COOH$ (acid) and $C_6H_5CH_2OH$ (neutral): $C_6H_5CH_2COOH$ reacts with both $NaOH$ and $NaHCO_3$ (soluble),while $C_6H_5CH_2OH$ does not (insoluble). This fits the scheme.
Therefore,both $(B)$ and $(D)$ follow the given separation scheme.

(D) $1$. $I$. Aniline $(C_6H_5NH_2)$: Contains $N$,so it gives a positive Lassaigne's test for nitrogen (Prussian blue color with $FeSO_4/H_2SO_4$),matching $P$. It also reacts with bromine water to form $2,4,6$-tribromoaniline (white precipitate),matching $S$.
$2$. $II$. $o$-Cresol: It is a phenol,so it gives a violet color with neutral $FeCl_3$,matching $T$.
$3$. $III$. Cysteine: Contains $N$ and $S$. Sodium fusion extract with sodium nitroprusside gives a violet/blood red color due to $S^{2-}$ ions,matching $Q$. It also contains a carboxylic acid group $(-COOH)$,which reacts with $NaHCO_3$ to release $CO_2$ gas (effervescence),matching $R$.
$4$. $IV$. Caprolactam: Contains $N$,so it gives a positive Lassaigne's test for nitrogen,matching $P$.
Matching: $I \rightarrow P, S$; $II \rightarrow T$; $III \rightarrow Q, R$; $IV \rightarrow P$. The correct option is $D$.

(C) Statement $I$ is true because Lassaigne's test is used to detect nitrogen,sulphur,halogens,and phosphorus in organic compounds.
Statement $II$ is false because in Lassaigne's test,the organic compound is fused with metallic sodium,not magnesium,to convert the elements from covalent form into ionic form.

(D) Statement $I$ is true because in Lassaigne's test,covalent organic compounds are fused with sodium metal to form ionic salts like $NaCN$,$Na_2S$,and $NaSCN$.
Statement $II$ is false because an organic compound containing both $N$ and $S$ forms sodium thiocyanate $(NaSCN)$. When this extract is treated with $Fe^{3+}$ ions (usually from $FeCl_3$),it forms ferric thiocyanate,which gives a blood-red colour,not Prussian blue. Prussian blue is obtained only when $N$ is present without $S$.

(A) $(1)$ Carboxylic acid reacts with sodium bicarbonate solution to release $CO_2$ gas,causing effervescence.
$(2)$ Phenolic $-OH$ group reacts with neutral $FeCl_3$ to form a violet-coloured complex.
$(3)$ Alcoholic $-OH$ group reacts with ceric ammonium nitrate to produce a red-coloured complex.
$(4)$ Alkaline $KMnO_4$ (Baeyer's reagent) reacts with unsaturated compounds (alkenes or alkynes) to cause the disappearance of the purple colour,indicating unsaturation.

(D) $2 CuO + C \xrightarrow[\Delta]{ } 2 Cu + CO_2$ is not part of "Lassaigne's test".
"Lassaigne's test" involves the fusion of an organic compound with metallic sodium to convert covalently bonded elements like $N$,$S$,and $X$ (halogens) into their ionic forms ($NaCN$,$Na_2S$,and $NaX$).
The reaction $2 CuO + C \xrightarrow[\Delta]{ } 2 Cu + CO_2$ is used for the detection of carbon in organic compounds (Liebig's method),not for the elements detected by "Lassaigne's test".

(D) Kjeldahl's method is used for the estimation of nitrogen in organic compounds.
It is not applicable to compounds containing nitrogen in nitro $(-NO_2)$ groups,azo $(-N=N-)$ groups,or nitrogen present in a ring (cyclic) structure (e.g.,pyridine,quinoline).
This is because the nitrogen in these functional groups or structures is not quantitatively converted to ammonium sulfate under the conditions of the Kjeldahl method.
Therefore,the correct answer is $D$.

(D) Lassaigne's test is used to detect the presence of nitrogen $(N)$,sulfur $(S)$,and halogens $(X)$ in organic compounds.
When both nitrogen $(N)$ and sulfur $(S)$ are present in an organic compound,they react during the fusion with sodium metal to form sodium thiocyanate $(NaSCN)$.
$Na + C + N + S \rightarrow NaSCN$
This sodium thiocyanate reacts with ferric ions $(Fe^{3+})$ to form a blood-red coloured complex,ferric thiocyanate,$[Fe(SCN)]^{2+}$.
$Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+} \text{ (Blood Red Colour)}$
Therefore,the compound must contain both nitrogen $(N)$ and sulfur $(S)$.
Option $D$ contains both an amino group ($-NH_2$,source of $N$) and a methylthio group ($-SCH_3$,source of $S$). Thus,it will give a blood-red colour.

(A) The correct matches are as follows:
$A$. Chromatography is the most modern and versatile technique for the separation and purification of organic compounds.
$B$. In Lassaigne's test for $N$,sodium fusion extract reacts with ferrous sulphate to form Prussian blue coloured complex,$Fe_4[Fe(CN)_6]_3$.
$C$. In Lassaigne's test for $S$,the sodium fusion extract reacts with sodium nitroprusside to form a violet coloured complex,$Na_4[Fe(CN)_5NOS]$.
$D$. Aniline is separated from water by steam distillation as it is steam volatile.
$E$. Glycerol is purified from spent lye by vacuum distillation because it decomposes at its boiling point.
Thus,the correct sequence is $A-i, B-iv, C-iii, D-v, E-ii$.

(B) The purpose of the sodium fusion test (Lassaigne's test) is to convert the elements present in the organic compound (like $N$,$S$,and halogens) from their covalent form into their corresponding water-soluble ionic forms (like $NaCN$,$Na_2S$,and $NaX$).
This allows for the subsequent qualitative detection of these elements using standard chemical reagents.

(D) The sodium fusion extract is boiled with concentrated nitric acid before adding silver nitrate $(AgNO_3)$ to test for halogens.
This step is essential to decompose any sodium sulfide $(Na_2S)$ or sodium cyanide $(NaCN)$ present in the extract,which would otherwise interfere with the test by forming precipitates like $Ag_2S$ or $AgCN$.

(C) During the Lassaigne's test,when an organic compound containing nitrogen is fused with sodium metal,the nitrogen present in the compound reacts with carbon and sodium to form sodium cyanide.
The chemical reaction involved is:
$Na + C + N \xrightarrow{\Delta} NaCN$

(B) When both nitrogen $(N)$ and sulfur $(S)$ are present in an organic compound,sodium fusion produces sodium thiocyanate $(NaCNS)$.
Upon adding acidified ferric chloride $(FeCl_3)$ solution to the extract,the ferric ions react with thiocyanate ions to form ferric thiocyanide,which exhibits a blood-red colouration.
The chemical reaction is: $3 \ NaCNS + FeCl_3 \longrightarrow Fe(CNS)_3 + 3 \ NaCl$ (blood-red colour).