(a) p-nitrophenol is more acidic than o-nitrophenol.
p-nitrophenol has the most stable stereochemistry among o, m and p isomers of nitrophenol. Since the nitro group employs negative resonance (or mesomeric) effect there is decrease in the electron density at ortho and para positions.
The phenoxide ion after donating the proton can also show positive resonance (+R) effect and increasing the stability of the ion. This results into increase in the tendency to donate that proton.
In case of o-nitrophenol, it undergoes hydrogen bonding that involve six atom geometry (most stable form of hydrogen bonding) as shown in the figure given below.
The hydrogen bonding strongly reduces the tendency to donate that proton and in turn decreases the acidity of o-nitrophenol.
(b) Bond angle C ─ O ─ C in ethers is slightly higher than the tetrahedrahedral angle (109° 28’).
Both C ─ O ─ C and C ─ O ─ H undergo sp3 hybridization and according to Valence Bond Theory they form tetrahedral geometry. In case of C ─ O ─ H the lone pairs of oxygen push the C ─ O and C ─ H bonds closer due to lone-pair lone-pair repulsion.
In ethers due to presence of bulky alkyl groups (-R groups) on both sides the repulsion of lone pairs is interfered by molecular repulsion which in turn decreases the effect of lone pair repulsion and thus the bond angle is slightly higher for ethers.
(c) (CH3)3C ─ Br on reaction with NaOCH3 gives an alkene instead of an ether.
Formation of ether undergoes bimolecular nucleophilic substitution reaction. It implies simultaneous breaking and formation of bonds. Now this step involves the approach of alkaoxide as a nucleophile which experiences stearic hindrance from methyl groups in the compound. Also tert-butyl carbocation can be easily formed if the C ─ Br bond breakes easily.
The alkaoxide used in the reaction are strong bases which can easily break the C ─ Br bond to give the carbocation which then undergoes elimination reaction to give alkene.
Thus tert-butylbromide when treated with sodium ethoxide gives alkene.
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