Question

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

Answer 1

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

Complete step by step answer:
When benzene is treated with $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 - $ 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 - $ chloropropane) and the lewis acid leading to the formation of $n - $ propyl carbocation
The $n - $propyl carbocation being primary in nature has no practical stability so it undergoes $1,2$ proton shift to form more stable secondary isopropyl cation.
Therefore, isopropyl carbocation being more stable than $n - $ propyl action the forever has a greater population the latter so benzene later react more with isopropyl benzene than $n - $ propyl benzene
The overall reaction is
$2{C_6}{H_6} + 2C{H_3}{(C{H_2})_2}CL\xrightarrow{{an.ALC{L_3}}}{C_6}{H_5}CH{(C{H_3})_2} + {C_6}{H_5}C{H_2}C{H_2}C{H_3} + 2HCL$
The reaction mechanism is as follows:-
$1 - $ chloropropane reacts with $ALC{L_3}$ to give $1 - $ propyl carbocation & $ALC{L_4}^ - $ ion
$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 - $ propyl carbocation in $1^\circ $ carbocation and is unstable and undergoes $2,1 - $ hydride ion shift to form $9^\circ $carbocation ( $2 - $ propyl carbocation )
$C{H_3}C{H_2}{C^ + }{H_2}\xrightarrow[{suift}]{{2,1 - hydride}}C{H_3}{C^ + }HC{H_3}$
Benzene undergoes nucleophilic substitution with this $2 - $ propyl carbocation to give $2 - $ phenylpropane ( isopropylbenzene/ cumene).
${C_6}{H_6} + C{H_3}{C^ + } + C{H_3}\xrightarrow{{ALC{L_4}^ - }}{C_6}{H_5}CH{\left( {C{H_3}} \right)_2} + HCL$

Note:
Alkylation of benzene is an electrophilic substitution. In this reaction carbonium ion is formed as an intermediate. In case of $1 - $ 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.