How will you bring about following conversions? (i) Chlorobenzene to phenol. (ii) iso-propyl bromide to n-propyl bromide (iii) Aniline to iodo benzene.

Chlorobenzene to phenol



C6H5Cl + Mg → C6H5MgCl

C6H5MgCl + H2O → C6H6
Benzene + H2SO4 ---> Benzene sulfonic acid + NaOH ---> Phenol
iso-propyl bromide to n-propyl bromide
conversion of 2-bromo propane to 1-bromo propane

1. dehalogenation reaction

the bromine attached to the propane chain can be removed by reacting the compound with alcoholic caustic potash, i.e. alc. KOH

CH3CH2CH2Br + alc. KOH -------> CH3CHCH2 + HBr

as the product we get propene and HBr

2. halogenation reaction

now on reaction of propene with HBr in presence of peroxide, will yield us 1 bromo propane.

HBr in presence of peroxide, follows anti-markonikoff's rule and hence forms 1 bromo propane
Aniline to iodo benzene.
when u react aniline +HNO2it gives benzene diazonium chlroide{C6H5N2+Cl-} than it react with CuI/HI it gives C6H5I.