Qualitative Analysis Questions in English

(B) The correct option is $(B)$.
In the Beilstein test,an organic compound is heated on a copper wire in a non-luminous flame.
The appearance of a green or bluish-green flame occurs due to the formation of volatile cupric halides $(CuX_2)$,which indicates the presence of halogens ($Cl$,$Br$,or $I$) in the organic compound.
This test is a qualitative test for halogens,although it does not specify which particular halogen is present.

(D) Lassaigne’s test (sodium fusion test) is a qualitative analysis method used to detect the presence of nitrogen,sulphur,and halogens in organic compounds.
In this test,the organic compound is fused with metallic sodium,which converts the elements present (nitrogen,sulphur,and halogens) into their corresponding water-soluble sodium salts:
$1. \text{Nitrogen} \rightarrow NaCN$
$2. \text{Sulphur} \rightarrow Na_2S$
$3. \text{Halogens} \rightarrow NaX$ (where $X = Cl, Br, I$)
These salts are then identified using specific chemical reagents.

(C) The $Beilstein$ test is a simple qualitative chemical test used to detect the presence of halogens in organic compounds. In this test,a copper wire is heated in a non-luminous flame until it no longer imparts any color to the flame. Then,the wire is dipped into the organic substance and heated again in the flame. The appearance of a green or blue-green flame indicates the presence of halogens due to the formation of volatile copper halides.

(C) The Lassaigne test (sodium fusion test) is a qualitative analysis used to detect the presence of elements like nitrogen $(N)$,sulfur $(S)$,and halogens ($X$,where $X = Cl, Br, I$) in organic compounds.
Since the test is used for all these elements,the correct answer is all of the given.

(B) When a sulfur-containing organic compound is fused with sodium metal,it forms sodium sulfide $(Na_2S)$.
$Na + S \rightarrow Na_2S$
When this $Na_2S$ is treated with sodium nitroprusside $(Na_2[Fe(CN)_5NO])$,it forms a violet-colored complex,sodium thio-nitroprusside $(Na_4[Fe(CN)_5NOS])$.
$Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]$
(Violet color)

(B) In the Lassaigne's test,when both nitrogen and sulfur are present in an organic compound,sodium thiocyanate $(NaSCN)$ is formed.
When this is treated with ferric chloride $(FeCl_3)$,it forms a blood-red colored complex,$[Fe(SCN)]^{2+}$.
However,if the question refers to the violet color specifically formed in the presence of sodium nitroprusside $(Na_2[Fe(CN)_5NO])$ during the test for sulfur,the reaction is:
$S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow [Fe(CN)_5NOS]^{4-}$
This complex,sodium thio-nitroprusside,is violet in color.

(C) In the qualitative analysis of phosphorus,the organic compound is heated with an oxidizing agent like sodium peroxide $(Na_2O_2)$.
This converts phosphorus into phosphate $(PO_4^{3-})$.
The solution is then boiled with concentrated nitric acid $(HNO_3)$ and treated with ammonium molybdate,$(NH_4)_2MoO_4$.
$A$ yellow precipitate of ammonium phosphomolybdate,$(NH_4)_3PO_4 \cdot 12MoO_3$,is formed,which confirms the presence of phosphorus.

(B) The detection of nitrogen in an organic compound via Lassaigne's test involves the following chemical reactions:
$Na + C + N \xrightarrow{\Delta} NaCN$
$2 NaCN + FeSO_4 \longrightarrow Fe(CN)_2 + Na_2SO_4$
$Fe(CN)_2 + 4 NaCN \longrightarrow Na_4[Fe(CN)_6]$
Finally,the sodium ferrocyanide reacts with ferric ions $(Fe^{3+})$ formed by the oxidation of some $Fe^{2+}$ to produce ferric ferrocyanide (Prussian blue):
$3 Na_4[Fe(CN)_6] + 4 Fe^{3+} \longrightarrow Fe_4[Fe(CN)_6]_3 + 12 Na^{+}$
The blue colour corresponds to the formula $Fe_4[Fe(CN)_6]_3$.

(A) Lassaigne's test is primarily used for the detection of nitrogen,sulphur,and halogens in organic compounds.
In this test,the organic compound is fused with metallic sodium to convert the elements into their corresponding water-soluble sodium salts:
$Na + C + N \rightarrow NaCN$
$2Na + S \rightarrow Na_2S$
$Na + X \rightarrow NaX$ (where $X = Cl, Br, I$)
Boron is not detected by Lassaigne's test because it does not form a stable ionic salt under these conditions that can be easily identified by standard qualitative reagents.

(D) Lassaigne's test is used to detect nitrogen $(N)$,sulfur $(S)$,and halogens $(Cl, Br, I)$ in organic compounds.
It involves the fusion of the organic compound with metallic sodium,which converts these elements into water-soluble sodium salts like $NaCN$,$Na_2S$,and $NaCl$.
Oxygen $(O)$ cannot be detected by Lassaigne's test because it does not form a stable,water-soluble sodium salt that can be identified through standard qualitative reagents used in this test.

(A) The process described is the Lassaigne's test modification or a standard qualitative analysis for phosphorus. $1$. The organic compound is fused with sodium peroxide $(Na_2O_2)$ to convert phosphorus into phosphate $(PO_4^{3-})$. $2$. Boiling with $HNO_3$ removes excess peroxide and ensures the phosphorus is in the phosphate form. $3$. Upon adding ammonium molybdate $(NH_4)_2MoO_4$ in the presence of $HNO_3$,a canary yellow precipitate of ammonium phosphomolybdate $(NH_4)_3PO_4 \cdot 12MoO_3$ is formed. $4$. Therefore,the element present is phosphorus.

(C) The presence of halogens in an organic compound is detected using the $Beilstein$ test or $Lassaigne's$ test.
In $Lassaigne's$ test,the sodium fusion extract is acidified with nitric acid $(HNO_3)$ and then treated with silver nitrate $(AgNO_3)$.
$A$ precipitate is formed,which is soluble in ammonium hydroxide $(NH_4OH)$.

(D) The qualitative analysis results are as follows:
$(i)$ Fehling's test is positive for aliphatic aldehydes. Aromatic aldehydes and heterocyclic aldehydes where the $-CHO$ group is directly attached to the ring (like in $A, B, C$) do not give a positive Fehling's test due to resonance stabilization. Only $D$ ($2$-(thiazol$-2-$yl)acetaldehyde) has the $-CHO$ group separated from the ring by a $-CH_2-CH_2-$ chain,making it behave like an aliphatic aldehyde,thus giving a positive Fehling's test.
$(ii)$ The formation of a blood-red colour with sodium nitroprusside in the $Na$ fusion extract indicates the presence of both Nitrogen $(N)$ and Sulfur $(S)$ in the compound. Among the options,only $C$ and $D$ contain both $N$ and $S$.
Combining both observations,only compound $D$ satisfies both conditions.

(D) Sulphanilic acid exists as a zwitterion $(H_3N^+-C_6H_4-SO_3^-)$. Due to the presence of the strong acidic $-SO_3H$ group and basic $-NH_2$ group,it does not behave as a typical carboxylic acid and does not undergo esterification.
Sulphanilic acid contains $C, H, N, O,$ and $S$. In Lassaigne's test,if both nitrogen and sulphur are present,they react to form sodium thiocyanate $(NaSCN)$. This compound reacts with $Fe^{3+}$ ions to form a blood-red colouration $([Fe(SCN)]^{2+})$,which confirms the presence of both $N$ and $S$.

(D) Lassaigne's test is a standard qualitative analysis method used to detect the presence of elements such as Nitrogen $(N)$,Sulphur $(S)$,Phosphorus $(P)$,and halogens ($X$,where $X = Cl, Br, I$) in organic compounds.

(C) In the qualitative analysis of phosphorus,the organic compound is heated with an oxidizing agent (like sodium peroxide) to convert phosphorus into phosphate ions $(PO_4^{3-})$.
This solution is then treated with concentrated nitric acid $(HNO_3)$ and ammonium molybdate $((NH_4)_2MoO_4)$.
The reaction produces a canary yellow precipitate of ammonium phosphomolybdate,which has the chemical formula $(NH_4)_3PO_4 \cdot 12MoO_3$.

(D) The presence of halogen in an organic compound can be detected by $Beilstein's$ test. In this test,a copper wire is heated in a flame until it shows no green color,then it is dipped in the organic compound and heated again. The appearance of a green flame indicates the presence of a halogen.

(D) In the qualitative elemental analysis of organic compounds (Lassaigne test),the sulphur atom of the organic molecule is converted into the water-soluble salt $Na_2S$.
This is tested by adding lead acetate $(CH_3COO)_2Pb$ solution,which results in the formation of a black precipitate of $PbS$.
The reaction is: $Na_2S_{(aq)} + (CH_3COO)_2Pb_{(aq)} \rightarrow PbS \downarrow (\text{Black}) + 2CH_3COONa_{(aq)}$.