Question

The percentage of n-propyl chloride obtained in the chlorination of propane is about 56%.
A) True
B) False

Answer 1

The answer to this question here is based on the reaction when propane is chlorinated and the products are the two isomeric forms and the answer is obtained by the calculation of their relative ratio based on the activity of hydrogen atoms.

Complete Solution :
We have studied in chemistry classes about the facts that deal with the chlorination, bromination etc., which is generally called as halogenations.
- Here, we shall now see the reaction of propane with chlorine that is chlorination of propane and the relative ratios of the product formed which gives us the required answer.
- The chlorination of propane is carried out in the present of light when treated with chlorine and is shown below,
$C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}\xrightarrow[h\upsilon ]{C{{l}_{2}}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl$

Here, the product formed is 1 – chloropropane and 2 – chloro propane.
- To see the percentage of the product formed can be obtained by their ratio that is the ratio of one degree hydrogen or primary hydrogen for 1-chloropropane multiplied by its activity to that of two degree hydrogen or secondary hydrogen multiplied by its activity of 2- chloro propane and is given as shown below:
$\dfrac{1-chloropropane}{2-chloropropane}=\dfrac{6\times 1}{2\times 3.8} = \dfrac{6}{7.6}$
Therefore, part of 1 – chloro propane formed is$\dfrac{6}{6+7.6}\times 100 = 44$
and that of 2-chloropropane is $\dfrac{7.6}{6+7.6}\times 100 = 55.8$
Therefore, the percentage of 1-chloropropane or simple n- propyl chloride is 44%
So, the correct answer is “Option B”.

Note: The main fact to be remembered is that the ratio of activities of one degree, two degree and three degree hydrogen in the chlorination reaction that is given by 1.0 : 3.8 : 5.0 and thus the questions based on this type can be answered easily.