Respuesta :
Answer:
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
Answer:
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.