1.

Benzene reacts with n-propyl bromide in presence of anhydrous AlCl_(3) to predominantly yield

Answer»

n-propylbenzene
ethylbenzene
isopropylbenzene
methylbenzene.

Solution :The initially formed less stable n - PROPYL carbocation `(1^(@))` REARRANGES to the more stable isopropyl carbocation `(2^(@))` before attacking the BENZENE ring to form isopropylbenzene, i.e.,
`CH_(3)CH_(2)CH-Broverset("Anhyd.")underset(AlCl_(3))rarr CH_(3)CH_(2)overset(+)CH_(2) overset("1,2-hydride")underset("shift")rarr CH_(3)-overset(+)CH-CH_(3)overset(C_(6)H_(6))underset(-H^(+))rarr underset("isopropyl benzene")(C_(6)H_(5)-CH(CH_(3))_(2))`


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