Alcohol, phenol & ether MCQs for NEET — Chemistry Questions with Answers

In order for a ionic compound to be soluble in a solvent it has to be first get in between the solvent molecules, for this in case of  water molecules it has to break $O-H$ bonds.

(C)

     $\mathrm{M}+\left(\mathrm{x}+\mathrm{y}\right){\mathrm{NH}}_3\stackrel{ }{\to }{\left[\mathrm{M}{\left({\mathrm{NH}}_3\right)}_{\mathrm{x}}\right]}^{+}+{\left[\mathrm{e}{\left({\mathrm{NH}}_3\right)}_{\mathrm{y}}\right]}^{-}$

     Solutions are deep blue coloured due to the absorption of light in visible region by solvated electrons.

     It is paramagnetic due to the presence of unpaired electrons.

Concentrated sulfuric acid (conc. H2SO4) increases the acidity of phenol by protonating it, forming the phenoxonium ion. This makes phenol a stronger acid due to the stabilization of the phenoxide anion by delocalization of the negative charge over the benzene ring.

H₂O has the highest boiling point due to the presence of hydrogen bonding.

(b) Order of M.P. or B.P. or critical temperature : ${\mathrm{H}}_2\mathrm{O}>{\mathrm{H}}_2\mathrm{Te}>{\mathrm{H}}_2\mathrm{Se}>{\mathrm{H}}_2\mathrm{S}$

(d)       ${\mathrm{NH}}_3$ is a weak reducing agent than ${\mathrm{PH}}_3$, because $\mathrm{X}—\mathrm{H}$ bond strength

            decreases down the group. Due to absence of H-bonding, only weak van

            der waals force of attraction exists in ${\mathrm{PH}}_3$, it has lower critical

             temperature than ${\mathrm{NH}}_3$.

Anisole is produced by the reaction of phenol (C6H5OH) with sodium hydroxide (NaOH) and methyl iodide (CH3I). This is an example of the Williamson ether synthesis, where an alkoxide ion (from phenol and NaOH) reacts with an alkyl halide to form an ether.

Sodium phenoxide reacts with ethyl iodide in a nucleophilic substitution reaction to form phenetole (C₆H₅OC₂H₅). The phenoxide ion acts as a nucleophile, attacking the ethyl group and displacing the iodide ion.

(a) CH₃CH₂CH=CH₂$\underset{\mathrm{Anti}-\mathrm{Markownikoff}\text{'}\mathrm{s} \mathrm{rule}}{\overset{\mathrm{HBr}/{\mathrm{H}}_2{\mathrm{O}}_2}{\to }}$

      CH₃CH₂CH₂CH₂Br$\to$Bromo-butane (1$°$  product)

      CH₃CH₂CH₂CH₂Br$\underset{\mathrm{SN}2 \mathrm{reaction}}{\overset{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{ONa}}{\to }}$CH₃CH₂CH₂CH₂OC₂H₅  (williamson's synthesis)

Ethyl formate on reaction with excess of CH₃MgI, followed by hydrolysis gives a secondary alcohol i.e. isopropyl alcohol.