Option 1
Step 1: Identify reaction (i). Phenol plus chloroform plus aqueous alkali is the classic \(Reimer\text{-}Tiemann\) formylation. Its job is to attach a \(-CHO\) group ortho to the phenolic \(-OH\).
Step 2: Reagent role. The active species produced from \(CHCl_3\) and \(NaOH\) is the electron-deficient carbene \(:CCl_2\). Meanwhile \(NaOH\) deprotonates phenol to phenoxide, whose ortho carbon is nucleophilic. Ortho attack followed by alkaline hydrolysis of the \(-CHCl_2\) intermediate delivers the aldehyde.
Step 3: Answer (i). Product \(=\) \(ortho\)-hydroxybenzaldehyde, i.e. salicylaldehyde.
Step 4: Identify reaction (ii). Heating phenol with zinc dust is a deoxygenation: the phenolic \(-OH\) is stripped off as \(ZnO\) and replaced by hydrogen.
Step 5: Answer (ii). Product \(=\) benzene, \(C_6H_6\).
Option 2
Step 1: Compare the two isomers. The key difference is whether the H-bond forms inside one molecule or between molecules.
Step 2: ortho isomer. Here \(-OH\) and \(-NO_2\) are neighbours, so a single molecule closes a six-membered chelate ring (intramolecular H-bond). Trapped this way, the molecule cannot bind neighbours well, so o-nitrophenol boils lower and distils in steam.
Step 3: para isomer. With the groups on opposite ends, only molecule-to-molecule bridging is possible (intermolecular H-bond). This ties molecules into an associated network, raising the boiling point and making it non-volatile in steam and more water-soluble.
Step 4: Steam distillation link. This difference is exactly why a mixture of o- and p-nitrophenol can be separated by steam distillation: only the ortho form comes over with the steam.
Step 5: Nitration outcome. Concentrated \(HNO_3\) exploits the powerful activation by \(-OH\) and substitutes three \(-NO_2\) groups at the 2, 4 and 6 positions, yielding picric acid, \(2,4,6\)-trinitrophenol.
\[\boxed{\text{Salicylaldehyde; benzene; picric acid}}\]