In an attempt to prepare n−propylbenzene, a chemist alkylated benzene with 1−chloropropane and aluminum chloride. however, two isomeric hydrocarbons were obtained in a ratio of 2: 1, the desired n−propylbenzene being the minor component. what do you think was the major product? how did it arise?

2-Phenylpropane (Cumene)

Explanation:

Famous Friedel Craft Alkylation.

Aluminum Chloride grasped the 1-chloropropane forming an intermediate product composed of Aluminum tetrachloride and n-propylcation. It is well understood that primary carbocations are unstable and therefore undergo hydrogen shifting to attain stability. n-Propylcarbocation undergone hydrogen shifting, forming isopropylcarbocation which reacted with benzene forming 2-Phenylpropane as the major product and HCl as a byproduct.

AlCl3 + CH3CH2CH2Cl --> AlCl4-   +   CH3CH2CH2+

CH3CH2CH2+  ---> CH3CH(+)CH3

C6H6 + CH3CH(+)CH3 ---> C6H5CH(CH3)2   + H+

AlCl4- + H+ ---> HCl + AlCl3

From the given problem statement,he was attempting to prepare n−propylbenzene by alkylation benzene with 1−chloropropane and aluminum chloride,but 1-propyle benze was a major product in result.

50.23 i believe is correct but don't quote me on that.

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.