Match the compounds in Column I with their ch | vedclass

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(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 a

Vedclass · Accepted Answer

$A-r, s; B-p, q; C-p, q, r; D-p, s$

Vedclass · Answer

(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.

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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