List-I contains various reaction sequences and List-II contains different phenolic compounds. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List-I		List-II
(P)		(1)
(Q)		(2)
(R)		(3)
(S)		(4)
		(5)

The reagents mentioned in List-I are the classical name reactions of phenolic chemistry. Before matching, recall what each reaction sequence gives:

P  Dow’s process $$C_6H_5Cl \xrightarrow[\;300\;{\rm atm}\;]{NaOH,\;623\;{\rm K}}\;C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$$ → product is simple phenol.

Q  Kolbe-Schmitt reaction $$C_6H_5ONa \xrightarrow{CO_2,\;400\;{\rm K},\;4\!-\!7\;{\rm atm}} \;o\!-\!HO\!-\!C_6H_4COONa \xrightarrow{H^+} o\!-\!HO\!-\!C_6H_4COOH$$ → product is o-hydroxybenzoic acid (salicylic acid).

R  Reimer-Tiemann reaction $$C_6H_5OH \xrightarrow{CHCl_3/NaOH,\;343\;{\rm K}} o\!-\!HO\!-\!C_6H_4CHO$$ → product is o-hydroxybenzaldehyde (salicylaldehyde).

S  Nitration of phenol with conc. $$HNO_3/H_2SO_4$$ (333 K) $$C_6H_5OH \xrightarrow{conc\;HNO_3/conc\;H_2SO_4} 2,4,6\text{-trinitrophenol}$$ → product is picric acid.

Now read List-II (numbered phenolic compounds):

(1) 2,4,6-trinitrophenol (picric acid)
(2) …
(3) phenol
(4) o-hydroxybenzaldehyde (salicylaldehyde)
(5) o-hydroxybenzoic acid (salicylic acid)

Matching each entry:

P (Dow) → phenol → 3
Q (Kolbe-Schmitt) → o-hydroxybenzoic acid → 5
R (Reimer-Tiemann) → o-hydroxybenzaldehyde → 4
S (nitration) → picric acid → 1

Thus P-3, Q-5, R-4, S-1.

The option containing this combination is
Option C which is: P-3, Q-5, R-4, S-1.