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(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^-

Vedclass · Accepted Answer

$A-iii, B-i, C-ii$

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(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 $A-iii, B-i, C-ii$.

Column $I$	Column $II$
$A$. Aniline	$i$. Red colour with $FeCl_3$
$B$. Benzene sulfonic acid	$ii$. Violet colour with sodium nitroprusside
$C$. Thiourea	$iii$. Blue colour with hot and acidic solution of $FeSO_4$

Column $I$	Column $II$
$A. H_2N-NH_3^+Cl^-$	$p. \text{Sodium fusion extract of the compound gives Prussian blue colour with } FeSO_4$
$B. HO-C_6H_4-CH(NH_3^+)COOH \text{ (with } I^- \text{ counterion)}$	$q. \text{Gives positive } FeCl_3 \text{ test}$
$C. HO-C_6H_4-NH_3^+Cl^-$	$r. \text{Gives white precipitate with } AgNO_3$
$D. (NO_2)_2C_6H_3-NH-NH_3^+Br^-$	$s. \text{Reacts with aldehydes to form the corresponding hydrazone derivative}$

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