Question

How do you convert the following:
(i) Prop-1-ene to 1-fluoropropane
(ii) Chlorobenzene to 2-chlorotoluene
(iii) Ethanol to propanenitrile
OR
Write the main products when
(i) n-butyl chloride is treated with alcoholic $KOH$
(ii) 2,4,6-trinitrochlorobenzene is subjected to hydrolysis
(iii) methyl chloride is treated with $AgCN$

Answer 1

$\left(i\right)$ Prop - 1 - ene to 1 - fluoropropane:

$C{H}_{3}CH=C{H}_2\to C{H}_{3}C{H}_{2}C{H}_{2}F$

Prop - 1 - ene 1 - fluoropropane

$C{H}_{3}CH=C{H}_2+H-Br\stackrel{Peroxide}{\to }C{H}_{3}C{H}_{2}C{H}_{2}Br$

First Prop - 1 - ene is converted to bromopropane by the addition of HBr in the presence of peroxide.

It follows Anti - Markowkov's rule:

$C{H}_{3}C{H}_{2}C{H}_{2}Br+AgF\to C{H}_{3}C{H}_{2}C{H}_{2}F+AgBr$

Swartz Reaction:

When an alkyl halide is treated with Fluoro compound $(AgF,Sb{F}_3,H{g}_2{F}_{2}etc..)$ alkyl flouride is formed.

$\left(ii\right)$ This is Friedel - crafts Alkylation Reaction.

MECHANISM:

$1)$ Generation of electrophile.

$2)$ Attack of benzene on electrophile.

$3)$ Regain of aromaticity.

$\left(iii\right)$ Ethanol to propane nitrite

$C{H}_{3}C{H}_{2}OH\to C{H}_{3}C{H}_{2}CN$

First ethanol is treated with $HBr$ to form bromoethane. Then bromoethane is treated with $KCN$ to form propane nitrate.

$OR$

$\left(i\right)$ n- butyl chloride is treated with alc. KOH.

Alcholic KOH acts as a base and attracts the proton. Elimination reaction takes place.

$\left(ii\right)$ 2,4,6-trinitro chlorobenzene is ubjected to hydrolysis.

$\left(iii\right)$ $C{H}_{3}Cl$ is treated with $AgCN$.

$AgCN$ being covalent in nature does not dissociate, so only nitrogen is available for the attack and wiryanide is formed.