Chemical analysis of organic compounds Questions in English

(A) When an organic compound is heated with $CuO$,carbon is oxidized to $CO_{2}$ and hydrogen is oxidized to $H_{2}O$.
Since the compound produces $CO_{2}$,it must contain carbon.
Since it does not produce water,it does not contain hydrogen.
Among the given options,$CCl_{4}$ (carbon tetrachloride) is the only compound that contains carbon but no hydrogen.
Therefore,the correct option is $A$.

(B) In Lassaigne's test,the organic compound is fused with sodium metal to convert nitrogen into sodium cyanide $(NaCN)$.
$Na + C + N \rightarrow NaCN$
When the extract is treated with ferrous sulphate $(FeSO_4)$,sodium ferrocyanide is formed:
$6NaCN + FeSO_4 \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4$
Some $Fe^{2+}$ ions are oxidized to $Fe^{3+}$ ions by the atmosphere or concentrated $H_2SO_4$ added during acidification.
These $Fe^{3+}$ ions react with the ferrocyanide ions to form ferric ferrocyanide,which is Prussian blue in colour:
$4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3$
Thus,the complex responsible for the Prussian blue colour is $Fe_4[Fe(CN)_6]_3$.

(D) In Lassaigne's test,if nitrogen and sulphur are present in the organic compound,they form $NaCN$ and $Na_2S$ respectively during sodium fusion.
These ions interfere with the test for halogens (like $Cl^-$,$Br^-$,$I^-$) by forming precipitates with $AgNO_3$.
To remove these,the sodium fusion extract is boiled with concentrated $HNO_3$.
This decomposes $NaCN$ and $Na_2S$ into volatile $HCN$ and $H_2S$ gases,respectively.
$NaCN (X) + HNO_3 (Z) \rightarrow NaNO_3 + HCN \uparrow$
$Na_2S (Y) + 2HNO_3 (Z) \rightarrow 2NaNO_3 + H_2S \uparrow$
Therefore,$X = NaCN$,$Y = Na_2S$,and $Z = \text{conc. } HNO_3$.

(D) In the Lassaigne sodium fusion test,the organic compound is fused with metallic sodium.
If both nitrogen and sulphur are present in the organic compound,they react with sodium to form sodium thiocyanate,represented as $NaCNS$.
If only nitrogen is present,it forms $NaCN$.
If only sulphur is present,it forms $Na_2S$.
Therefore,$N$ and $S$ are converted into $NaCN$,$Na_2S$,and $NaCNS$ depending on the elements present.

(A) Ammonium phosphomolybdate $(X)$ is formed during the qualitative test for phosphorus,and its chemical formula is $(NH_4)_3PO_4 \cdot 12MoO_3$.
Prussian blue $(Y)$ is formed during the Lassaigne's test for nitrogen,and its chemical formula is $Fe_4[Fe(CN)_6]_3 \cdot xH_2O$.

(A) Kjeldahl's method is used for the estimation of nitrogen in organic compounds.
However,it is not applicable to compounds containing nitrogen in nitro $(-NO_2)$,azo $(-N=N-)$ groups,or nitrogen present in the ring (like in pyridine).
In the given compounds:
$I$ $(C_6H_5NH_2)$: Aniline,contains nitrogen in the amino group. Suitable.
$II$ $(C_6H_5N_2^+Cl^-)$: Benzenediazonium chloride,contains nitrogen in the azo group. Not suitable.
$III$ $(C_6H_5NO_2)$: Nitrobenzene,contains nitrogen in the nitro group. Not suitable.
$IV$ $(C_5H_5N)$: Pyridine,nitrogen is in the ring. Not suitable.
$V$ $(C_6H_5CH_2NH_2)$: Benzylamine,contains nitrogen in the amino group. Suitable.
Therefore,compounds $I$ and $V$ are suitable for Kjeldahl's method.

(D) The test described is the Lassaigne's test for nitrogen.
$1$. The sodium fusion extract contains sodium cyanide $(NaCN)$ if nitrogen is present in the organic compound.
$2$. When $NaCN$ reacts with iron$(II)$ sulphate $(FeSO_4)$,it forms sodium hexacyanoferrate$(II)$.
$3$. Upon adding concentrated $H_2SO_4$,some $Fe^{2+}$ ions are oxidized to $Fe^{3+}$ ions.
$4$. These $Fe^{3+}$ ions react with the hexacyanoferrate$(II)$ ions to form ferric ferrocyanide,$Fe_4[Fe(CN)_6]_3$,which is known as Prussian blue.
$5$. Thus,the appearance of Prussian blue colour confirms the presence of nitrogen.

(B) The carbon and nitrogen present in the organic compound,upon fusion with sodium metal,form sodium cyanide $(NaCN)$,which is soluble in water.
This is converted into sodium ferrocyanide by the addition of a sufficient quantity of ferrous sulphate.
Ferric ions generated during the process react with ferrocyanide to form the Prussian blue precipitate of ferric ferrocyanide.
The Prussian blue colour is due to the formation of ferric ferrocyanide,$Fe_4\left[Fe(CN)_6\right]_3 \cdot xH_2O$.
Reaction:
$(i) Fe^{2+} + 6CN^{-} \longrightarrow \left[Fe(CN)_6\right]^{4-}$ (Hexacyanoferrate$(II)$)
$(ii) Fe^{2+} \xrightarrow[Conc. H_2SO_4]{} Fe^{3+} + e^{-}$
$(iii) 3\left[Fe(CN)_6\right]^{4-} + 4Fe^{3+} \longrightarrow Fe_4\left[Fe(CN)_6\right]_3 \cdot xH_2O$

(B) $1$. When an organic compound containing phosphorus is fused with sodium peroxide $(Na_2O_2)$,the phosphorus is oxidized to phosphate ions,forming sodium phosphate $(Na_3PO_4)$ as compound '$X$'.
$2$. When '$X$' $(Na_3PO_4)$ is boiled with concentrated nitric acid $(HNO_3)$ and then treated with ammonium molybdate reagent,it forms a yellow precipitate of ammonium phosphomolybdate,which is $(NH_4)_3PO_4 \cdot 12MoO_3$ (or $(NH_4)_3[P(Mo_3O_{10})_4]$),identified as '$Y$'.
$3$. Therefore,'$X$' is $Na_3PO_4$ and '$Y$' is $(NH_4)_3PO_4 \cdot 12MoO_3$.

(B) Lassaigne's test is used to detect nitrogen,sulfur,and halogens in organic compounds. The test involves fusing the organic compound with metallic sodium to convert the elements into water-soluble sodium salts like $NaCN$,$Na_2S$,and $NaCl$.
In the case of benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$,the diazonium group is unstable. Upon heating with sodium metal,the nitrogen atoms are released as $N_2$ gas before they can react with sodium to form $NaCN$. Consequently,the test for nitrogen becomes negative.
Other compounds like $o$-nitrophenol,sulfanilic acid,and $m$-chloroaniline contain nitrogen in more stable forms that successfully convert to $NaCN$ during the fusion process,thus giving a positive test.

(C) In the Lassaigne's test,the sodium extract containing $NaCN$ is treated with $FeSO_4$ followed by $FeCl_3$ and acidified with $H_2SO_4$.
The chemical reactions are as follows:
$Na + C + N \rightarrow NaCN$
$FeSO_4 + 2NaCN \rightarrow Fe(CN)_2 + Na_2SO_4$
$Fe(CN)_2 + 4NaCN \rightarrow Na_4[Fe(CN)_6]$ (Sodium ferrocyanide)
$3Na_4[Fe(CN)_6] + 4FeCl_3 \rightarrow Fe_4[Fe(CN)_6]_3 + 12NaCl$
The compound $Fe_4[Fe(CN)_6]_3$ is known as ferric ferrocyanide,which gives the characteristic Prussian blue colour.

(A) Lassaigne's test is used to detect elements like $N, S,$ and halogens in organic compounds.
$1$. $A$ blue colouration or precipitate (Prussian blue,$Fe_4[Fe(CN)_6]_3$) is formed when the compound contains both $C$ and $N$,which react with molten $Na$ to form $NaCN$.
$2$. $A$ white precipitate $(AgCl)$ is formed when the compound contains chlorine,which reacts with molten $Na$ to form $NaCl$,which then reacts with $AgNO_3$.
$3$. $N$-methylaniline $(C_6H_5NHCH_3)$ contains $C$ and $N$,thus it gives a blue colouration.
$4$. $2$-chlorobenzoic acid contains $Cl$,thus it gives a white precipitate with $AgNO_3$.
Therefore,the correct pair is $N$-methylaniline and $2$-chlorobenzoic acid.

(C) The correct matches are as follows:
$A$. Unsaturation (Baeyer's test): The pink colour of alkaline $KMnO_4$ is discharged $(IV)$.
$B$. Alcoholic group (Ceric ammonium nitrate test): $A$ red or violet colour complex is formed $(III)$.
$C$. Aldehyde group (Tollen's reagent): $A$ silver mirror is formed due to the reduction of $Ag^+$ to $Ag$ $(II)$.
$D$. Phenolic group ($FeCl_3$ test): $A$ violet or red colour complex is formed $(I)$.
Therefore,the correct sequence is $A-IV, B-III, C-II, D-I$.

(B) In Lassaigne's test,the organic compound is fused with metallic sodium.
This process converts the covalently bonded elements (like $N$,$S$,and Halogens) present in the organic compound into their corresponding water-soluble sodium salts,which are ionic in nature (e.g.,$NaCN$,$Na_2S$,$NaX$).