In this reaction, the reagent is Br2, but how come when the haloalkane is formed, there is only one bromine bonded. Since there are two bromine atoms in the reagent shouldn’t there be two in the new molecule???
If compound A had been an alkene (ie with a double bond) then it would have undergone a rapid addition reaction where both Br atoms are added on - one to each of the carbons of the double bond (which changes to a single bond). So there is only one product - a dibromoalkane.
But in this case the Br2 is reacted with an alkane which only contains single bonds which are much less reactive. So in the presence of UV light it only undergoes a very slow substitution reaction where a single Br is swapped for one of the H atoms. n this case there are two products - the one shown in your diagram and the other one is HBr formd by the unreacted Br joining onto the H atom that the other Br replaced from the alkane.
1 Like