Give preparation of haloalkanes from hydrocarbons.

Explanation

(i) From alkanes by free radical halogenation : Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono - and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any single compound is low.
`CH_(3)CH_(2)CH_(2)CH_(3)overset(Cl_(2)//Delta)underset(" or UV light ")rarr underset("2-Chlorobutane (55%)")(CH_(3)CHClCH_(2)CH_(3))+underset("1-Chlorobutane (45%)")(CH_(3)CH_(2)CH_(2)CH_(2)Cl)`
(ii) From alkenes : Alkyl halides can be prepared from alkenes by addition of hydrogen acids such as HCl, HBr and HI.
`underset("Ethene")(H_(2)C)=CH_(2)+HI to underset("Iodoethane")(CH_(3)CH_(2)I)`
In case of unsymmetrical alkenes, the addition takes place by Markovnikov.s rule.
`underset("Prop-1-ene")(CH_(3)CH=CH_(2))overset(+HI)rarr underset("(major)")underset("2-iodopropane")(CH_(3)CHICH_(3))+underset("(minor)")underset("1-iodopropane")(CH_(3)CH_(2)CH_(2)I)`
However, if the peroxide `(R_(2)O_(2))` is present, the reaction takes place contrary to Markovanikov.s rule. This is known as Kharasch effect.
`underset("Prop-1-ene")(CH_(3)-CH=CH_(2)+HBr)overset((C_(6)H_(5)CO)_(2)O_(2))rarr underset("1-Bromopropane")(CH_(3)CH_(2)CH_(2)Br)`
In the laboratory, addition of bromine in `C Cl_(4)` to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for detection of double bonds in a molecule. The addition results in the synthesis of (vic -dibromides which are colourless) vicinal dihalides.
`underset("Ethene")(H_(2)C)=CH_(2)+Br_(2)overset(C Cl_(4))rarr underset("1,2-dibromoethane")(underset("Br ")underset("| ")(CH_(2))-underset("Br ")underset("| ")(CH_(2)))`