$$n$$ - propyl alcohol and $$iso$$ - propyl alcohol gives different products on oxidation with $${K_2}C{r_2}{O_7}$$

Boiling point increases with increase in molecular mass and decreases with increase in branching.

The formation of ether from alcohol in the presence of base followed by alkylation is known as Williamson ether synthesis reaction.

TIPS/Formulae :
Conc. $$HCl,\,HBr$$   and conc. $$HCl + ZnC{l_2}$$   all are nucleophiles, thus convert alcohols to alkyl halides. However, conc. $${H_3}P{O_4}$$  is a good dehydrating agent which converts an alcohol to an alkene.

The organic liquid $$A$$ is $${C_2}{H_5}OH.$$
$$\left( {\text{i}} \right)$$ Ethyl alcohol is a colourless liquid with a characteristic pleasant smell, having boiling point $${78.1^ \circ }C.$$
\[\left( \text{ii} \right){{C}_{2}}{{H}_{5}}OH\xrightarrow{\text{conc}\text{.}\,{{\text{H}}_{2}}S{{O}_{4}}}\]      \[\underset{\begin{smallmatrix} \text{Decolourises }B{{r}_{2}}\,\text{water} \\ \text{and alkalin }KMn{{O}_{4}} \end{smallmatrix}}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,\]

Among the given compounds, hydroxybenzene $$(IV)$$ has least molar mass and therefore possess least boiling point. Among the three isomeric dihydroxybenzenes, 1,2 - dihydroxybenzene (I) forms intramolecular $$H$$ - bonding with the result it will not form intermolecular $$H$$ - bonding leading to lowest boiling point. On the other hand 1,3 - dihydroxybenzene (II) and 1, 4 - dihydroxybenzene (III) do not undergo chelation, hence they will involve extensive intermolecular $$H$$ - bonding leading to higher boiling point. Further intermolecular hydrogen bonding is stronger in the $$p$$ - isomer than the m - isomer hence former has highest b.p. Thus the decreasing order of boiling points is III > II > I >IV.

\[2{{C}_{2}}{{H}_{5}}OH\xrightarrow[\Delta ]{{{H}_{2}}S{{O}_{4}}}{{C}_{2}}{{H}_{4}}+{{H}_{2}}O\]

Acid-base reaction takes place.

$$O$$ - nitrophenol is insoluble in sodium hydrogen carbonate. While 2, 4, 6 - trinitrophenol, benzoic acid and benzene sulphonic acid are soluble in $$NaHC{O_3}.$$
Infact,
$${\text{Acid}} + NaHC{O_3} \to {\text{Salt + }}{{\text{H}}_2}C{O_3}$$
This reaction is possible in forward direction if acid is more acidic than $${H_2}C{O_3}.$$   $$o$$ - nitrophenol is less acidic than $${H_2}C{O_3}.$$   Hence, it is not soluble in sodium hydrogen carbonate.