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.

Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $$\times$$ 4 matrix given in the ORS.

	Column I		Column II
(A)		(P)	sodium fusion extract of the compound gives Prussian blue colour with FeSO$$_4$$.
(B)		(Q)	gives positive FeCl$$_3$$ test.
(C)		(R)	gives white precipitate with AgNO$$_3$$.
(D)		(S)	reacts with aldehydes to form the corresponding hydrazone derivative.

$\mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CHCl}_3+\mathrm{KOH} \rightarrow$ Nitrogen containing compound $+\mathrm{KCl}+\mathrm{H}_2 \mathrm{O}$.

Nitrogen containing compound is :

$\mathrm{RCONH}_2$ is converted into $\mathrm{RNH}_2$ by means of Hofmann bromamide degradation.

In this reaction, RCONHBr is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.

How can the conversion of (i) to (ii) be brought about?