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.
Out of o-nitrophenol and o-cresol which is more acidic?
The acidic character of alcohols is due to the polar nature of $\mathrm{O}-\mathrm{H}$ bond. Higher the polarity of $\mathrm{O}-\mathrm{H}$ bond, more will be the acidic strength. Due to $-I$ and $-R$ effect of nitro group, electron density decreases in the $\mathrm{O}-\mathrm{H}$ bond of o-nitrophenol and thus polarity increases.
Whereas due to $+I$ effect of $-\mathrm{CH}_3$ group, electron density increases in the $\mathrm{O}-\mathrm{H}$ bond of o-cresol. Thus, $\mathrm{O}-\mathrm{H}$ bond of o-nitrophenol is weaker than $\mathrm{O}-\mathrm{H}$ bond of o-cresol and o-nitrophenol is more acidic than o-cresol.
When phenol is treated with bromine water, white precipitate is obtained. Give the structure and the name of the compound formed.
When phenol is treated with bromine water, white ppt. of 2, 4, 6 - tribromophenol is obtained.
Arrange the following compounds in increasing order of acidity and give a suitable explanation.
Phenol, o-nitrophenol, o-cresol
Nitro group shows $-I$ and $-R$ effect as follows
Due to this $-I$ and $-R$ effect of o-nitrophenol, it is a stronger acid than phenol. On the other hand, $-\mathrm{CH}_3$ group produces $+I$ effect $-I$ and $-R$ effect increases the acidic strength by increasing the polarity of -OH bond while $+I$ effect decreases the polarity due to increase in electron density on -OH bond. So, o-cresol is a weaker acid than phenol. Thus, the correct order is o-cresol < phenol
Alcohols react with active metals e.g., $\mathrm{Na}, \mathrm{K}$ etc., to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.
An electron releasing group produces $+I$ effect so it increases the electron density on oxygen and decreases the polarity of O - H bond. As the number of alkyl group increases from $1^{\circ}$ to $3^{\circ}$ alcohols, the electron density on the O - H bond increases. It will finally decrease the polarity and increase the strength of $\mathrm{O}-\mathrm{H}$ bond in going from $1^{\circ}$ to $3^{\circ}$ alcohols. Thus, acidic strength decreases in the following order
As we know that, sodium metal is basic in nature and alcohols are acidic in nature. Thus, reactivity of alcohol with sodium metal decreases with decrease in acidic strength. Therefore, the correct order is $1^{\circ}>2^{\circ}>3^{\circ}.$