Practical Organic Chemistry

Amide ( $-\mathrm{CO}-\mathrm{NH}-$ ) bond is present. It is known as peptide bond It is a dipeptide.This gives ninhydrin test. The result is a purple color.Ninhydrin test is used to detect the presence of amines in a sample.The test works by reacting ninhydrin with the free amino group $\left(-\mathrm{NH}_2\right)$. The result of the test is a purple color.The above dipeptide react with ninhydrin, the alpha -amino group (present in the above structure) undergoes reaction to give purple color:The compound also reacts with $\mathrm{FeCl}_3$ to give violet color. It is due to the free - OH group. It is a characteristic reaction of phenols So, The compound $P$ reaction with ninhydrin gives purple color and it also reacts with $\mathrm{FeCl}_3$ to give violet color.$ P-2 $Ninhyduin test is possible with free $-\mathrm{NH}_2$ group. The above dipeptide on hydrolysis gives two amino acid molecules. So, after hydrolysis, ninhydrin test is possible. Because amino acids have free, $\mathrm{NH}_2$ group.But the compound does not give phthalein dye test.Phthalein dye test - It is used to identify the presence of phenolic functional groups. The compound has no free -OH group and hence it does not give phthalein dye test. For phthalein dye test, $-O H$ group must be attached directly to the momatic ring. So, The compound Q after complete hydrolysis give ninhydrin test and it does not give positive phthalein dye test.$ Q-5 $$ \text { Aniline hydrochboide. (Aniline + HCI) } $Aniline hydrochloride reacts with phenyl diazonium salt (benzene diazonium salt) and forms a dye ( $p$-aminoarobenzene) with yellow color.So, compound $R$ reaction with phenyl diazonium salt gives yellow dye.$ R-1 $It is a hydrazine. It reacts with glucose and give corresponding hydrazone.$ \text { It can be simply written as. } $So, S reaction with glucose give corresponding hydrazone.$S-3$$ \begin{aligned} &\text { Answer: Option (B) }\\ &P \rightarrow 2, Q \rightarrow 5, R \rightarrow 1, S \rightarrow 3 \end{aligned} $