Question

How will you convert

$1)$ Benzene into aniline

$2)$ $\text{Benzene into N,N-dimethylaniline}$

$3)$ $\text{Cl-(}C{H}_2{)}_4\text{-Cl into hexan-1,6-diamine}$

Answer 1

$A)$ Conversion of benzene into aniline
For the conversion of benzene into aniline, first we are doing the nitration of benzene which gives nitrobenzene via electrophilic substitution reaction then reduce the nitro group into amine by catalytic hydrogenation using $H_2/Pd$ in ethanol, as a result we get aniline as a product.

Involved reactions shown below-


$B)$ Conversion of benzene into $\text{N, N−dimethylaniline}$
For the conversion of benzene into $\text{N, N− dimethylaniline,}$
first we are doing the nitration of benzene which gives nitrobenzene via electrophilic substitution reaction then reduce the nitro group by catalytic hydrogenation, as a result we get aniline as a product. Aniline on reacting with two equivalents of iodomethane forms $\text{N, N−dimethylaniline.}$

Involved reactions are shown below-


$C)\text{Conversation of}Cl-(C{H}_2{)}_4-Cl\text{into hexan-1,6-diamine}$

For the conversion of $Cl-(C{H}_2{)}_4-Cl$ into $hexan-1,6-diamine,$ first of all we are doing the reaction of given reactant with ethanolic sodium cyanide, the $C{N}^{-}$ molecules replace both chlorine atoms via nucleophilic substitution reaction.

After that catalytic hydrogenation is done in presence of$H_2/Ni,$ in which both $-CN$ groups gets converted into $-C{H}_{2}N{H}_2$ and we get our desired product.

Involved reactions are shown below-