Chlorination of pentane gives a mixture of isomers having the molecular formula c5h11cl. the percentage of 1-chloropentane is 22%. assuming the secondary hydrogens in pentane are equally reactive to monochlorination, what is the percentage of 3-chloropentane in the mixture?

Hey there!:

Pentane is a 5 member hydrocarbon

1 chloropentane can be on either side and will still have thesame name because of the IUPAC rules .

To determine the relative amounts of products obtained from radical chlorination of an alkane, both probability (the number of hydrogens that can be abstracted that will lead to the formation of the particular product) and reactivity (the relative rate at which a particular hydrogen is abstracted) must be taken into account.

The precise ratios differ at different temperatures.

For 3- chloro pentane , There are two hydrogen only and the reactivity  is same for all  2° hydrogen

The 2°  hydrogen % should = 100-22 = 78% ( all remaining are 2° hydrogen only)

since all are equally recative 2° hydrogen  , proportionate distribution will be there.

% Yield  =   {( Total hydrogen which can give 3 choloropentane) /  (all hydrogen of 2° ) } * 78

             =  (2/6)* 78 =  26%  

Hope this helps!

it is called acid bonding

idk what the question is about