Phenol does not react with heated $$Cu$$  at $${300^ \circ }C.$$

Halogens have both $$+R$$  and $$-I$$  effect, but the $$-I$$  effect predominates over the $$+ R$$ - effect. Therefore, $$m$$ - chlorophenol is most acidic due to electron withdrawing $$- Cl$$  group. Alcohols are less acidic than phenol.

TIPS/Formulae :
Riemer-Tiemann reaction involves electrophilic substitution on the highly reactive phenoxide ring.

Temperature is the effective factor for dehydration of alcohol by $$A{l_2}{O_3}$$  ( dehydrating reagent ).
\[R-C{{H}_{2}}-C{{H}_{2}}OH\xrightarrow[{{350}^{\circ }}-{{380}^{\circ }}C]{A{{l}_{2}}{{O}_{3}}}R-CH=C{{H}_{2}}+{{H}_{2}}O\]
While at $${220^ \circ } - {250^ \circ }C,$$    it forms ether.

Since $$1\,mole$$  of compound $$X$$ reacts with $$Na$$  to evolve $$1\,mole$$  of $${H_2}$$  gas, therefore the compound should have 2 active hydrogen atoms per mole which is possible only in option $$d.$$
$$C{H_2}OHC{H_2}C{H_2}C{H_2}OH + 2Na \to NaOC{H_2}C{H_2}C{H_2}C{H_2}ONa + 2H$$

(i) Since $$A$$ gives characteristic colour with aqueous $$FeC{l_3}$$  so it contains a phenolic group.
(ii) Since $$A$$ when treated with $$B{r_2}$$  forms $${C_7}{H_5}OB{r_3}$$   (ppt.) and considering the molecular formula of $$A,$$ it is most likely to be cresol.

(iii) Since $$A$$ on bromination forms tribromo derivative so, it is $$m$$ - cresol i.e.
The reactions are :

In such reactions, first write down the sequence of reactions and then proceed backward/forward from a known compound, reagent, or known fact (test).
\[A\xrightarrow[\left( ii \right)\,{{H}_{2}}O]{\left( i \right)\,C{{H}_{3}}MgI}B\xrightarrow{\text{Victor Meyer test}}\text{Blue colour}\]
Appearance of blue colour in Victor Meyer test indicates that $$B$$  is \[{{\text{2}}^{\circ }}\] alcohol and thus $$A$$  must be an aldehyde other than $$HCHO.$$   Thus among the given options, only option (C) is correct.

$$n$$ - Propyl alcohol on oxidation with potassium dichromate gives an aldehyde which on further oxidation gives an acid. Both aldehyde and acid contain the same number of $$C$$ atoms as the original alcohol.
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow[\frac{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}{{{H}_{2}}S{{O}_{4}}}]{{{25}^{\circ }}C}\]      \[C{{H}_{3}}C{{H}_{2}}CHO\xrightarrow{\frac{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}{{{H}_{2}}S{{O}_{4}}}}\]      \[C{{H}_{3}}C{{H}_{2}}COOH\]
Isopropyl alcohol on oxidation gives a ketone with the same number of $$C$$ atoms as the original alcohol.
\[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,OH \\ | \end{smallmatrix}}{\mathop{-CH}}}\,\xrightarrow{\frac{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}{{{H}_{2}}S{{O}_{4}}}}C{{H}_{3}}\overset{\begin{smallmatrix} O \\ \parallel \end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{3}}\]

Since the compound $$X\left( {{C_3}{H_6}O} \right)$$   gives iodoform test, it must have $$ - COC{H_3}$$   groups hence it should be $$C{H_3}COC{H_3}.$$   Therefore $$X$$ should be $$C{H_3}CH\left( {OH} \right)C{H_3}.$$

Ethylene oxide on treatment with Grignard reagent give additive product which undergo hydrolysis to give primary alcohol as final product