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Question: Why do we get isopropyl benzene on treating benzene with \(1 - \) chloropropane instead of \(n - \) ...

Why do we get isopropyl benzene on treating benzene with 1−1 - chloropropane instead of n−n - propyl benzene?

Explanation

Solution

Benzene here undergoes electrophilic substitution in the presence of anhydrous aluminium trichloride but 1∘1^\circ being unstable undergo hydride shift to forms 2∘2^\circ carbocation so we get isopropyl benzene.

Complete step by step answer:
When benzene is treated with 1−1 - chloropropane in presence of a lewis acid like aluminium trichloride, benzene undergoes friedel craft alkylation to give isopropyl benzene (also known as cummene) as a major product, and n−n - propyl benzene as minor product.
cummene) as a major product, and propyl benzene as a minor product.
The reason for the formation of isopropyl benzene as a major product over n-propyl benzene can be found on close inspection of the mechanism of the alkylation which is cited below
The first step involves the interaction of alkylating agent (here which is 1−1 - chloropropane) and the lewis acid leading to the formation of n−n - propyl carbocation
The n−n - propyl carbocation being primary in nature has no practical stability so it undergoes 1,21,2 proton shift to form more stable secondary isopropyl cation.
Therefore, isopropyl carbocation being more stable than n−n - propyl action the forever has a greater population the latter so benzene later react more with isopropyl benzene than n−n - propyl benzene
The overall reaction is
2C6H6+2CH3(CH2)2CL→an.ALCL3C6H5CH(CH3)2+C6H5CH2CH2CH3+2HCL2{C_6}{H_6} + 2C{H_3}{(C{H_2})_2}CL\xrightarrow{{an.ALC{L_3}}}{C_6}{H_5}CH{(C{H_3})_2} + {C_6}{H_5}C{H_2}C{H_2}C{H_3} + 2HCL
The reaction mechanism is as follows:-
1−1 - chloropropane reacts with ALCL3ALC{L_3} to give 1−1 - propyl carbocation & ALCL4−ALC{L_4}^ - ion
CH3(CH2)2CL+ALCL3→CH3CH2C+H2+ALCL4−C{H_3}{\left( {C{H_2}} \right)_2}CL + ALC{L_3} \to C{H_3}C{H_2}{C^ + }{H_2} + ALC{L_4}^ -
The 1−1 - propyl carbocation in 1∘1^\circ carbocation and is unstable and undergoes 2,1−2,1 - hydride ion shift to form 9∘9^\circ carbocation ( 2−2 - propyl carbocation )
CH3CH2C+H2→suift2,1−hydrideCH3C+HCH3C{H_3}C{H_2}{C^ + }{H_2}\xrightarrow[{suift}]{{2,1 - hydride}}C{H_3}{C^ + }HC{H_3}
Benzene undergoes nucleophilic substitution with this 2−2 - propyl carbocation to give 2−2 - phenylpropane ( isopropylbenzene/ cumene).
C6H6+CH3C++CH3→ALCL4−C6H5CH(CH3)2+HCL{C_6}{H_6} + C{H_3}{C^ + } + C{H_3}\xrightarrow{{ALC{L_4}^ - }}{C_6}{H_5}CH{\left( {C{H_3}} \right)_2} + HCL

Note:
Alkylation of benzene is an electrophilic substitution. In this reaction carbonium ion is formed as an intermediate. In case of 1−1 - chloropropane initially primary carbonium ion is obtained but due to hydride shifted converted into comparatively more stable secondary carbonium ion.
Thus isopropyl calion attacks benzene to form isopropyl benzene.