Suggest a reagent for the following conversion.
The given reactant is 
.
It is a secondary alcohol. Secondary alcohol 
gives ketone when oxidises by CrO$_3$ or pyridinium chlorochromate without carrying out oxidation at the double bond.
Out of 2-chloroethanol and ethanol which is more acidic and why?
2 -chloroethanol is more acidic than ethanol. Due to $-I$ effect (electron withdrawing group) of the Cl -atom electron density in $\mathrm{O}-\mathrm{H}$ bond decreases. So, $\mathrm{O}-\mathrm{H}$ bond of 2-chloroethanol becomes weaker than $\mathrm{O}-\mathrm{H}$ bond of ethanol. Thus, 2-chloroethanol is more acidic than ethanol.
Stronger acid due to $-I$ effect of Cl.
Suggest a reactant for conversion of ethanol to ethanal.
Ethanol can be oxidises into ethanal by using pyridinium chlorochromate. $\left(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{N}} \mathrm{HCl} \overline{\mathrm{C}} \mathrm{CrOO}_3\right)$ in $\mathrm{CH_2Cl_2}$.
$\mathrm{H}_3 \mathrm{C}-\underset{\text { Ethanol }}{\mathrm{CH}_2}-\mathrm{OH} \xrightarrow{\mathrm{PCC} / \mathrm{CH}_2 \mathrm{Cl}_2} \mathrm{\mathop {C{H_3}}\limits_{Ethanal} - CHO}$
Suggest a reagent for conversion of ethanol to ethanoic acid.
Ethanol can be converted into ethanoic acid by using acidified $\mathrm{KMnO}_4$ or $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$. Both $\mathrm{KMnO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ are strong oxidising agents.
$\underset{\text { Ethanol }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}} \xrightarrow{\substack{\text { Acidified } \\ \mathrm{KMnO}_4}} \underset{\text { Ethanoic acid }}{\mathrm{CH}_3 \mathrm{COOH}}$
Out of o-nitrophenol and p-nitrophenol, which is more volatile? Explain.
o-nitrophenol is more volatile than p-nitrophenol due to presence of intramolecular hydrogen bonding. In para nitrophenol intermolecular hydrogen bonding is present. This intermolecular hydrogen bonding causes the association of molecules.
