Treatment of compound O with KMnO4/H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2/NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH following by acidification, gave a compound T.
The compound T is
Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.
Match each of the compounds in Column I with its characteristic reaction(s) in Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | $$C{H_3}C{H_2}C{H_2}CN$$ | (P) | Reduction with $$Pd - C/{H_2}$$ |
| (B) | $$C{H_3}C{H_2}OCOC{H_3}$$ | (Q) | Reduction with $$SnC{l_2}/HCl$$ |
| (C) | $$C{H_3} - CH = CH - C{H_2}OH$$ | (R) | Development of foul smell on treatment with chloroform and alcoholic KOH |
| (D) | $$C{H_3}C{H_2}C{H_2}C{H_2}N{H_2}$$ | (S) | Reduction with diisobutylaluminium hydride (DIBAL-H) |
| (T) | Alkaline hydrolysis |
Statement 1 : Aniline on reaction with NaNO$$_2$$/HCl at 0$$^\circ$$C followed by coupling with $$\beta$$-naphthol gives a dark blue coloured precipitate.
Statement 2 : The colour of the compound formed in the reaction of aniline with NaNO$$_2$$/HCl at 0$$^\circ$$C followed by coupling with $$\beta$$-naphthol is due to the extended conjugation.