Explain why is $O=\mathrm{C}=O$ non-polar while $\mathrm{R}-O-\mathrm{R}$ is polar?
$\mathrm{CO}_2$ is a linear molecule. The dipole moment of two $\mathrm{C}=\mathrm{O}$ bonds are equal and opposite and they cancel each other and hence the dipole moment of $\mathrm{CO}_2$ is zero and it is a non-polar molecule.
While for ethers, two dipoles are pointing in the same direction. These two dipoles do not cancel the effect of each other. Therefore, there is a finite resultant dipoles and hence $R-O-R$ is a polar molecule.
Why is the reactivity of all the three classes of alcohols with conc. HCl and $\mathrm{ZnCl}_2$ (Lucas reagent) different?
The reaction of alcohols with Lucas reagent (conc. HCl and $\mathrm{ZnCl}_2$ ) follow $\mathrm{S}_{\mathrm{N}}{ }^1$ mechanism. $\mathrm{S}_{\mathrm{N}}{ }^1$ mechanism depends upon the stability of carbocations (intermediate). More stable the intermediate carbocation, more reactive is the alcohol.
Tertiary carbocations are most stable among the three classes of carbocations and the order of the stability of carbocation is $3^{\circ}>2^{\circ}>1^{\circ}$. This order, inturn, reflects the order of reactivity of three classes of alcohols i.e., $3^{\circ}>2^{\circ}>1^{\circ}$.
Thus, as the stability of carbocations are different so the reactivity of all the three classes of alcohols with Lucas reagent is different.
Write steps to carry out the conversion of phenol to aspirin.
Aspirin can be prepared by the reaction of salicylic acid with acetic anhydride. Salicylic acid is prepared by the reaction of phenol with $\mathrm{CO}_2$ and NaOH.
This process is known as Kolbe's reaction. The product salicylic acid is used in the preparation of aspirin. After wards, when salicylic acid is treated with acetic anhydride then acetyl group replaces the hydrogen of - OH group i.e., acetylation occurs at $-$OH group of salicylic acid. Reaction is as follows
Nitration is an example of aromatic electrophilic substitution and its rate depends upon the group already present in the benzene ring. Out of benzene and phenol, which one is more easily nitrated and why?
Nitration of benzene and phenol is an electrophilic substitution reaction. During nitration $\stackrel{+}{\mathrm{N}} \mathrm{O}_2$ (nitronium ion) is produced as an intermediate as follows
This nitronium ion (electrophile) attacks on benzene or on phenol. Phenol is more easily nitrated than benzene as the presence of — OH group in phenol increases the electron density at ortho and para positions in benzene ring by $+R$ effect.
Since, the electron density is more in phenol than in benzene, therefore, phenol is more easily nitrated than benzene.
In Kolbe's reaction, instead of phenol, phenoxide ion is treated with carbon dioxide. Why?
In phenoxide ion, the ability to give lone pair of electrons to the benzene ring is more in comparison to phenols. Therefore, the reactivity of phenoxide ion towards electrophilic substitution reaction is more in comparison to phenols.
Thus, phenoxide ion being a stronger nucleophile reacts easily with $\mathrm{CO}_2$ (weak electrophile) than phenols in Kolbe's reaction.
