Chemical analysis of organic compounds Questions in English

(B) Alkaline pyrogallol is a well-known absorbent for $O_{2}$ gas in gas analysis.
Oil of cinnamon (which contains cinnamaldehyde) is used to absorb $O_{3}$ gas.
Therefore,the correct pair is $O_{2}$ and $O_{3}$.

(D) In $Kjeldahl’s$ method,the organic compound containing nitrogen is heated with concentrated $H_2SO_4$ to convert nitrogen into $(NH_4)_2SO_4$.
$CuSO_4$ is added to the reaction mixture to act as a catalyst,which speeds up the decomposition process.

(C) In $Lassaigne's$ test,the organic compound is fused with sodium metal to convert covalently bonded elements $(N, S, X)$ into their corresponding water-soluble ionic salts.
$1$. Nitrogen $(N)$ is converted to sodium cyanide $(NaCN)$.
$2$. Sulfur $(S)$ is converted to sodium sulfide $(Na_2S)$.
$3$. Halogens $(X)$ are converted to sodium halides $(NaX)$.
$NaNC$ (sodium isocyanide) is not a product of this fusion reaction,as the cyanide ion formed is $CN^-$,not $NC^-$.
Therefore,the correct option is $C$.

(C) Halogens are detected by Beilstein’s test.
In this test,a copper wire is dipped in the original solution and heated in a Bunsen burner flame.
$A$ green colour is imparted to the flame due to the formation of a volatile copper halide.
This confirms the presence of a halogen.

(C) Kjeldahl's method is based on the principle that organic compounds containing nitrogen are quantitatively decomposed to form $(NH_4)_2SO_4$ when heated with concentrated $H_2SO_4$.
In this process,$CuSO_4$ is added to act as a catalyst to speed up the reaction.

(C) In the Lassaigne test for nitrogen,the organic compound is fused with sodium metal to form sodium cyanide $(NaCN)$.
This $NaCN$ is then treated with ferrous sulfate $(FeSO_4)$ and ferric chloride $(FeCl_3)$.
The final product formed is ferric ferrocyanide,which is $Fe_4[Fe(CN)_6]_3$,also known as Prussian blue.

(B) Lassaigne's test requires the presence of both carbon and nitrogen in an organic compound to form sodium cyanide $(NaCN)$ upon fusion with sodium metal.
Compounds that contain nitrogen but lack carbon cannot form $NaCN$.
Hydrazine hydrochloride $(NH_2NH_2 \cdot HCl)$ contains nitrogen but no carbon,therefore it fails the test.

(A) When an organic compound containing both carbon and nitrogen is fused with sodium metal,the carbon and nitrogen react to form sodium cyanide $(NaCN)$.
This compound is soluble in water and is detected during the Lassaigne's test for nitrogen.

(B) In Lassaigne's test,the organic compound is fused with metallic sodium. If the compound contains nitrogen,it reacts with sodium to form sodium cyanide $(NaCN)$.
The chemical reaction is: $C + N + Na \rightarrow NaCN$.
This $NaCN$ is then extracted with water and tested for the presence of cyanide ions.

(B) When both nitrogen $(N)$ and sulphur $(S)$ are present in an organic compound,they react with fused sodium $(Na)$ during the Lassaigne test to form sodium thiocyanate $(NaSCN)$.
$Na + C + N + S \to NaSCN$
This compound gives a blood-red coloration with ferric ions $(Fe^{3+})$ due to the formation of ferric thiocyanate: $Fe^{3+} + 3SCN^- \to [Fe(SCN)_3]$.

(B) The Lassaigne test is used to detect nitrogen,sulfur,and halogens in organic compounds.
For the detection of nitrogen,the compound must contain both carbon and nitrogen atoms to form sodium cyanide $(NaCN)$ upon fusion with metallic sodium.
In the compound $NH_2NH_2 \cdot HCl$ (hydrazine hydrochloride),there is no carbon atom present.
Therefore,it cannot form $NaCN$ during the fusion process,and thus,it fails the Lassaigne test for nitrogen.

(C) The Middleton test is a chemical test used for the detection of halogens $(X)$ in organic compounds.
In this test,the organic compound is heated with copper oxide $(CuO)$ on a charcoal block,which produces copper halides.
These copper halides impart a green or blue-green color to the flame,confirming the presence of halogens.

(D) Lassaigne's test (sodium fusion test) is a qualitative analysis used to detect the presence of elements like nitrogen,sulfur,and halogens $(X)$ in organic compounds.
It converts these elements into their corresponding water-soluble sodium salts,such as sodium cyanide $(NaCN)$,sodium sulfide $(Na_2S)$,and sodium halides $(NaX)$.

(B) In the Lassaigne's test for nitrogen,the sodium fusion extract contains sodium cyanide $(NaCN)$.
When this extract is treated with ferrous sulfate $(FeSO_4)$,it forms sodium ferrocyanide: $6NaCN + FeSO_4 \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4$.
Upon further reaction with ferric ions $(Fe^{3+})$ formed by the oxidation of some $Fe^{2+}$ ions,ferric ferrocyanide is produced: $3Na_4[Fe(CN)_6] + 4Fe^{3+} \rightarrow Fe_4[Fe(CN)_6]_3 \cdot xH_2O$.
This compound,$Fe_4[Fe(CN)_6]_3$,is known as Prussian blue,which gives the characteristic blue color.

(A) Lassaigne's test is used to detect nitrogen,sulfur,and halogens in organic compounds. For a positive nitrogen test,the compound must contain both carbon and nitrogen atoms to form sodium cyanide $(NaCN)$ during fusion with sodium metal. $Hydrazine$ $(NH_2NH_2)$ does not contain carbon,therefore it cannot form $NaCN$ and will not give a positive Lassaigne's test for nitrogen.

(C) The purity of an organic compound can be verified using both chromatography and spectroscopy.
Chromatography (such as thin-layer chromatography or gas chromatography) is used to separate components and check for the presence of impurities.
Spectroscopy (such as $IR$,$NMR$,or $UV-Vis$) is used to confirm the identity and purity of the compound by analyzing its structural features.
Therefore,both methods are essential for characterization and purity testing.

(A) To detect carbon and hydrogen in an organic compound,the compound is heated with copper$(II)$ oxide $(CuO)$.
Carbon is oxidized to carbon dioxide $(CO_2)$,which turns lime water milky.
Hydrogen is oxidized to water $(H_2O)$,which turns anhydrous copper sulfate blue.
The reaction is: $C + 2CuO \rightarrow CO_2 + 2Cu$ and $2H + CuO \rightarrow H_2O + Cu$.

(A) In the Lassaigne's test for sulfur,the sodium fusion extract is treated with sodium nitroprusside,$Na_2[Fe(CN)_5NO]$.
When sulfur is present,it reacts with the nitroprusside ion to form a violet-colored complex,sodium thio-nitroprusside,which is $Na_4[Fe(CN)_5NOS]$.
The reaction is: $[Fe(CN)_5NO]^{2-} + S^{2-} \rightarrow [Fe(CN)_5NOS]^{4-}$.

(A) The Lassaigne test is used for the detection of nitrogen,sulfur,and halogens in organic compounds.
Since these elements are covalently bonded in organic compounds,they cannot be detected directly by standard precipitation reactions.
Therefore,the fundamental principle of the Lassaigne test is to fuse the organic compound with metallic sodium $(Na)$.
This process converts the covalently bonded elements into their corresponding water-soluble ionic salts (e.g.,$NaCN$,$Na_2S$,$NaCl$),which can then be easily identified using standard chemical reagents.

(A) The Lassaigne's test is used to detect nitrogen,sulfur,and halogens in organic compounds.
In this test,the organic compound is fused with metallic sodium $(Na)$ to convert the elements present into their corresponding water-soluble sodium salts.
For nitrogen,the reaction is: $Na + C + N \rightarrow NaCN$.
$NH_2NH_2$ (hydrazine) is an organic compound containing nitrogen,which upon fusion with sodium forms $NaCN$,thus giving a positive Lassaigne's test.
$NaNO_3$ and $NH_4Cl$ are inorganic compounds and do not contain carbon,which is required to form $NaCN$.
$NaCN$ is already a salt and does not represent the organic compound being tested.

(C) In the Lassaigne's test for sulfur,the sodium fusion extract contains sodium sulfide $(Na_2S)$.
When this extract is treated with sodium nitroprusside,a violet color is formed.
However,for the detection of sulfur using lead acetate,the $Na_2S$ reacts with lead acetate $((CH_3COO)_2Pb)$ to form lead sulfide $(PbS)$,which appears as a black precipitate.
The chemical reaction is: $Na_2S + (CH_3COO)_2Pb \rightarrow PbS \downarrow (\text{black}) + 2CH_3COONa$.

(A) The $Lassaigne's$ test is used to detect nitrogen,sulfur,and halogens in organic compounds.
When an organic compound contains both nitrogen $(N)$ and sulfur $(S)$,the fusion with sodium metal produces sodium thiocyanate $(NaSCN)$.
$Na + C + N + S \rightarrow NaSCN$
When this $Lassaigne's$ extract is treated with ferric chloride $(FeCl_3)$,the ferric ions react with thiocyanate ions to form ferric thiocyanate,which has a blood-red color.
$Fe^{3+} + 3SCN^- \rightarrow Fe(SCN)_3$ (Blood red color)
Among the given options,$Thiourea$ $(NH_2CSNH_2)$ contains both nitrogen and sulfur,thus it will give a positive test for both $N$ and $S$ leading to the formation of $NaSCN$.

(D) The molecular mass of organic compounds can be determined by various methods:
$1$. $Silver \ salt \ method$: Used for determining the equivalent mass of organic acids.
$2$. $Chloroplatinate \ salt \ method$: Used for determining the molecular mass of organic bases.
$3$. $Mass \ spectrometry$: $A$ powerful analytical technique used to measure the mass-to-charge ratio of ions to determine the molecular mass of a compound.
Therefore,all the given methods are used for determining the molecular mass of organic compounds.

(B) When ferric ions $(Fe^{3+})$ react with potassium ferrocyanide $(K_4[Fe(CN)_6])$,they form a deep blue precipitate known as Prussian blue.
The chemical reaction is:
$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$
This compound is ferric ferrocyanide,which is responsible for the blue color.

(A) In the Lassaigne's test,when an organic compound containing nitrogen is fused with $Na$ metal,it forms $NaCN$.

(A) If $Na_2S$ and $NaCN$ are present in the extract,they will interfere with the test for halogens by forming precipitates with $AgNO_3$.
Boiling the extract with conc. $HNO_3$ decomposes these compounds into volatile $H_2S$ and $HCN$ gases:
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$
These gases escape from the solution,ensuring they do not interfere with the halogen test.

(C) The formation of Prussian blue colour in the Lassaigne's test (sodium fusion test) indicates the presence of both nitrogen and carbon in the organic compound,which react to form sodium cyanide $(NaCN)$.
$Na + C + N \rightarrow NaCN$
This $NaCN$ then reacts with ferrous sulphate $(FeSO_4)$ to form sodium ferrocyanide,which further reacts with ferric ions to produce ferric ferrocyanide,known as Prussian blue.
Among the given options,only $CH_3-CH=NH$ contains both carbon and nitrogen atoms.
Therefore,it will give a positive test for nitrogen.

(C) When an organic compound contains both nitrogen $(N)$ and sulfur $(S)$,sodium fusion produces sodium thiocyanate $(NaSCN)$.
$Na + C + N + S \rightarrow NaSCN$
On adding $FeCl_3$ to the acidified extract,the ferric ions react with thiocyanate ions to form a blood-red complex:
$Fe^{3+} + SCN^{-} \rightarrow [Fe(SCN)]^{2+}$ (Blood red colouration).
Therefore,the presence of both $N$ and $S$ is indicated.

(A) In Lassaigne's test,the organic compound is fused with metallic sodium.
When nitrogen $(N)$ is present,it reacts with sodium and carbon to form sodium cyanide: $Na + C + N \xrightarrow{\Delta} NaCN$.
When sulphur $(S)$ is present,it reacts with sodium to form sodium sulphide: $2Na + S \xrightarrow{\Delta} Na_2S$.
If both nitrogen and sulphur are present together,they form sodium thiocyanate: $Na + C + N + S \xrightarrow{\Delta} NaSCN$.

(A) Lassaigne's test is used to detect nitrogen,sulfur,and halogens in organic compounds.
In this test,the organic compound is fused with metallic sodium $(Na)$.
If nitrogen is present,it reacts with sodium to form sodium cyanide $(NaCN)$.
However,compounds that do not contain carbon along with nitrogen,or compounds where nitrogen is present in a form that does not easily form $NaCN$ upon fusion,may give a negative result.
$NH_2OH$ (Hydroxylamine) does not contain carbon.
Since Lassaigne's test requires the formation of $NaCN$ (which requires both $Na$ and $C$),compounds lacking carbon cannot form cyanide ions.
Therefore,$NH_2OH$ gives a negative Lassaigne's test for nitrogen.

(C) Sodium extract is heated with conc. $HNO_3$ to convert any $NaCN$ formed during fusion into $HCN$ and sulphide into $H_2S$.
This is done when the compound contains $N$ or $S$ along with halogens.
$HCN$ and $H_2S$ formed escape as they are volatile.
Otherwise,they interfere with the test for halogens as they also form a precipitate with $AgNO_3$.
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$

(A) In Lassaigne's test,fusion with sodium takes place and the following species are formed:
$A$. Aniline $(C_6H_5NH_2)$ contains $N$,which forms $CN^-$ ions. $CN^-$ reacts with hot and acidic solution of $FeSO_4$ to form $Fe_4[Fe(CN)_6]_3$ (Prussian blue),which is blue in colour. Thus,$A-iii$.
$B$. Benzene sulfonic acid $(C_6H_5SO_3H)$ contains $S$,which forms $S^{2-}$ ions. However,the question implies a phenolic derivative or specific test context. Given the options,if we consider the formation of a phenoxide-like species or specific test,$FeCl_3$ is used for phenol tests (red/violet colour). Thus,$B-i$.
$C$. Thiourea $(NH_2CSNH_2)$ contains both $N$ and $S$. The $S^{2-}$ ions formed react with sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ to give a violet colour. Thus,$C-ii$.
Therefore,the correct match is $A-iii, B-i, C-ii$.

(B) Lassaigne's test is used for the detection of nitrogen in organic compounds.
This test involves the fusion of the organic compound with metallic sodium to convert nitrogen into sodium cyanide $(NaCN)$.
Therefore,the compound must contain both carbon $(C)$ and nitrogen $(N)$ atoms to form $NaCN$.
Hydroxylamine hydrochloride $(H_2NOH \cdot HCl)$ contains nitrogen but lacks carbon,so it cannot form $NaCN$ and will not show a positive result for Lassaigne's test.

(B) Beilstein's test: Organic compounds containing halogens,when heated over a $Cu$ wire loop,produce a blue or green colored flame due to the formation of volatile copper halides.

(B) The Beilstein test is a simple chemical test used for the detection of halogens ($Cl$,$Br$,$I$) in organic compounds.
In this test,a copper wire is heated in a flame until it no longer imparts any color to the flame.
The wire is then dipped into the organic compound and heated again in the flame.
$A$ green or blue-green flame indicates the presence of a halogen due to the formation of volatile copper halides.

(B) The given qualitative analysis indicates the following properties of the compound:
$1$. Insoluble in dilute $HCl$: This suggests the compound is not basic (like an amine).
$2$. Soluble in $NaOH$ solution: This indicates the compound is acidic (like a phenol or carboxylic acid).
$3$. Decolourization of $Br_2/\text{water}$: This indicates the presence of an unsaturated system or a highly activated aromatic ring (like phenol,which forms $2,4,6-\text{tribromophenol}$).
Comparing these properties:
- Cyclohexanol is neutral and does not dissolve in $NaOH$.
- Phenol is acidic,dissolves in $NaOH$ to form sodium phenoxide,and reacts with bromine water to decolourise it.
- Cyclohexylamine is basic and dissolves in $HCl$.
- Aniline is basic and dissolves in $HCl$.
Therefore,the compound is Phenol.

(A) In the Lassaigne's test,sulphur present in the organic compound reacts with fused sodium to form sodium sulphide $(Na_2S)$.
When this extract is treated with sodium nitroprusside,$Na_2[Fe(CN)_5NO]$,it reacts with the sulphide ion $(S^{2-})$ to form a complex known as sodium thionitroprusside,which is $Na_4[Fe(CN)_5NOS]$.
The purple colour is due to the formation of the complex ion $[Fe(CN)_5NOS]^{4-}$.

(D) Lassaigne's test is used to detect nitrogen,sulfur,and halogens in organic compounds.
In this test,the organic compound is fused with metallic sodium,which converts nitrogen into sodium cyanide $(NaCN)$.
However,compounds that contain nitrogen but do not contain carbon,such as hydrazine $(NH_2NH_2)$,do not form $NaCN$ upon fusion with sodium.
Therefore,hydrazine gives a negative result for Lassaigne's test for nitrogen.

(C) In Lassaigne's test,an organic compound is fused with metallic sodium $(Na)$ to convert covalently bonded elements like nitrogen,sulfur,and halogens into water-soluble ionic salts.
$1$. Nitrogen is converted to sodium cyanide $(NaCN)$.
$2$. Sulfur is converted to sodium sulfide $(Na_2S)$.
$3$. Halogens $(X)$ are converted to sodium halides $(NaX)$.
$NaNC$ (sodium isocyanide) is not a product of this fusion reaction,as the cyanide ion formed is $CN^-$,not $NC^-$. Therefore,$NaNC$ is the correct answer.

(C) The organic compound is fused with sodium peroxide $(Na_2O_2)$. The fused mass is then extracted with water.
The aqueous solution obtained is boiled with concentrated nitric acid $(HNO_3)$,and ammonium molybdate solution is added to it.
$A$ yellow precipitate indicates the presence of phosphorus in the organic compound.
The chemical reaction is:
$Na_3PO_4 + 12(NH_4)_2MoO_4 + 21HNO_3 \rightarrow (NH_4)_3PO_4 \cdot 12MoO_3 + 21NH_4NO_3 + 12H_2O$
The yellow precipitate is ammonium phosphomolybdate,which is $(NH_4)_3PO_4 \cdot 12MoO_3$.

(D) In Lassaigne's test for nitrogen,the sodium fusion extract contains sodium cyanide $(NaCN)$.
When this extract is treated with ferrous sulfate $(FeSO_4)$,sodium ferrocyanide is formed: $6NaCN + FeSO_4 \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4$.
On adding ferric chloride $(FeCl_3)$,the ferric ions react with the ferrocyanide ions to form ferric ferrocyanide,which is Prussian blue in color: $3Na_4[Fe(CN)_6] + 4FeCl_3 \rightarrow Fe_4[Fe(CN)_6]_3 + 12NaCl$.
Thus,the blue color is due to the formation of $Fe_4[Fe(CN)_6]_3$.

(B) If the organic compound contains nitrogen or sulphur,the Lassaigne's extract is boiled with $HNO_3$ to decompose sodium cyanide $(NaCN)$ or sodium sulphide $(Na_2S)$ formed during fusion.
$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow $
$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow $
If cyanide and sulphide ions are not removed,they react with silver nitrate $(AgNO_3)$ and interfere with the halogen test by forming precipitates:
$NaCN + AgNO_3 \rightarrow AgCN \text{ (White ppt.)} + NaNO_3$
$Na_2S + 2AgNO_3 \rightarrow Ag_2S \text{ (Black ppt.)} + 2NaNO_3$

(C) The Lassaigne test involves the fusion of an organic compound with sodium metal to convert elements like nitrogen,sulfur,and halogens into their water-soluble ionic salts ($NaCN$,$Na_2S$,$NaCl$,etc.).
Sodium is preferred for this test because it is sufficiently reactive to fuse with the organic compound but not so reactive that it becomes hazardous or uncontrollable.
Potassium is not used in the Lassaigne test because it is significantly more reactive than sodium,making the reaction dangerously vigorous and difficult to handle.
Therefore,the Assertion is incorrect,and the Reason is correct.

(A) When both nitrogen and sulphur are present in an organic compound,sodium thiocyanate $(NaSCN)$ is formed during the preparation of Lassaigne's extract.
$Na + C + N + S \xrightarrow{\Delta} NaSCN$
On adding $FeCl_3$ to this extract,the $Fe^{3+}$ ions react with $SCN^-$ ions to form ferric thiocyanate,which gives a blood-red coloration.
$Fe^{3+} + 3SCN^- \longrightarrow [Fe(SCN)_3]$ (blood-red color)
Thus,both the Assertion and the Reason are correct,and the Reason explains the Assertion.

(C) $1$. $m-$chlorobenzoic acid is a strong acid and reacts with $NaHCO_3$ to form a water-soluble salt,which is extracted into fraction $A$.
$2$. $m-$chlorophenol is a weaker acid than $m-$chlorobenzoic acid but stronger than water; it does not react with $NaHCO_3$ but reacts with $NaOH$ to form a water-soluble salt,which is extracted into fraction $B$.
$3$. $m-$chloroaniline is a base and does not react with $NaHCO_3$ or $NaOH$. It remains in the organic layer,which is fraction $C$.
Therefore,fraction $A$ contains $m-$chlorobenzoic acid,fraction $B$ contains $m-$chlorophenol,and fraction $C$ contains $m-$chloroaniline.

(N/A) Lassaigne's test is employed to detect the presence of nitrogen,sulphur,halogens,and phosphorus in an organic compound. These elements are present in the covalent form in an organic compound and are converted into the ionic form by fusing the compound with sodium metal.
$Na + C + N \xrightarrow{\Delta} NaCN$
$2Na + S \xrightarrow{\Delta} Na_2S$
$Na + X \xrightarrow{\Delta} NaX$ (where $X = Cl, Br, I$)
The cyanide,sulphide,and halide of sodium formed are extracted by boiling the fused mass in distilled water. This is called Lassaigne's extract.
$(a)$ Test for nitrogen:
Lassaigne's extract is boiled with iron $(II)$ sulphate and then acidified with concentrated sulphuric acid. Sodium cyanide reacts with iron $(II)$ sulphate to form sodium hexacyanoferrate $(II)$. Upon heating with sulphuric acid,some iron $(II)$ is oxidised to iron $(III)$,which reacts to form iron $(III)$ hexacyanoferrate $(II)$,known as Prussian blue.
$6 CN^{-} + Fe^{2+} \longrightarrow [Fe(CN)_6]^{4-}$
$3[Fe(CN)_6]^{4-} + 4Fe^{3+} \xrightarrow{xH_2O} Fe_4[Fe(CN)_6]_3 \cdot xH_2O$ (Prussian blue)
$(b)$ Test for sulphur:
$(i)$ Lassaigne's extract is acidified with acetic acid and treated with lead acetate. $A$ black precipitate of lead sulphide indicates sulphur.
$S^{2-} + Pb^{2+} \to PbS$ (Black)
$(ii)$ Treatment with sodium nitroprusside gives a violet colour.
$S^{2-} + [Fe(CN)_5NO]^{2-} \to [Fe(CN)_5NOS]^{4-}$ (Violet)
If both nitrogen and sulphur are present,$NaSCN$ is formed,which gives a blood-red colour with $Fe^{3+}$.
$Fe^{3+} + SCN^{-} \to [Fe(SCN)]^{2+}$ (Blood red)
$(c)$ Test for halogens:
The extract is acidified with nitric acid and treated with silver nitrate. The resulting precipitate ($AgCl$ white,$AgBr$ pale yellow,$AgI$ yellow) confirms the halogen.

(N/A) While testing the Lassaigne's extract for the presence of halogens,it is first boiled with dilute nitric acid.
This is done to decompose $NaCN$ to $HCN$ and $Na_{2}S$ to $H_{2}S$ and to expel these gases.
If nitrogen and sulphur are present in the form of $NaCN$ and $Na_{2}S$,they would interfere with the test for halogens by forming $AgCN$ and $Ag_{2}S$ precipitates,which are also insoluble in water.
The chemical equations involved in the reaction are:
$NaCN + HNO_{3} \longrightarrow NaNO_{3} + HCN \uparrow$
$Na_{2}S + 2HNO_{3} \longrightarrow 2NaNO_{3} + H_{2}S \uparrow$

(N/A) The $Na_2S$ present in the sodium extract reacts with lead acetate to form lead sulphide $(PbS)$ precipitate.
If $H_2SO_4$ is used for acidification,it reacts with lead acetate to form lead sulphate $(PbSO_4)$ precipitate,which interferes with the test.
Acetic acid is a weak acid and does not form a precipitate with lead acetate,thus allowing the specific detection of sulphur as $PbS$.