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.
Dipole moment of phenol is smaller than that of methanol. Why?
Dipole moment depends upon the polarity of bonds. Higher the polarity of bonds in molecule, higher will be its dipole moment. In phenol carbon is $s p^2$ hybridised and due to this reason benzene ring is producing electron withdrawing effect.
On the other hand, carbon of methanol is $s p^3$ hybridised and produces electron releasing effect (+ $I$ effect). Thus, C - O bond in phenol is less polar than $\mathrm{C}-\mathrm{O}$ bond in methanol and therefore, the dipole moment of phenol is smaller than that of methanol.
Ethers can be prepared by Williamson synthesis in which an alkyl halide is reacted with sodium alkoxide. Di-tert -butyl ether can't be prepared by this method. Explain.
In order to prepare di-tert-butyl ether, sodium tert-butoxide must be reacted with tert-butyl bromide. Alkoxides are not only nucleophiles but they are strong base as well. They react with $3^{\circ}$ alkyl halides leading to the elimination reaction.
When tert-butyl-bromide reacts with sodium tert-butoxide instead of substitution, elimination takes place. As a result of this elimination reaction, Iso butylene is formed instead of di-tert butyl ether.
Why is the $\mathrm{C}-0-\mathrm{H}$ bond angle in alcohols slightly less than the tetrahedral angle whereas the $\mathrm{C}-0-\mathrm{C}$ bond angle in ether is slightly greater?
The bond angle in 
in alcohols is slightly less than tetrahedral angle $(109^\circ 28')$. It is due to the repulsion between the unshared electron pairs of oxygen. In alcohols, two lone pair of electrons are present. Therefore, there is comparatively more repulsion and less bond angle.
The $\mathrm{C}-\mathrm{O}-\mathrm{C}$ bond angle in ether is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky $(-R)$ groups.
