Question

4. When benzene reacts with 1-chloro-2,2-dimethylpropane (neopentyl chloride) in the presence of aluminum chloride, the major product is 2-methyl-2-phenylbutane, not 2,2-dimethyl-1- phenylpropane (neopentylbenzene). Explain this result

Answer 1

Answer:

Explanation:

In this case, we have a Friedel-Craft reaction (see figure 1). Usually in this type of reaction, the benzene ring is bonded directly to the carbon that has the halide atom. But in this specific case, this is not happening.

To understand why we have to remember the reaction mechanism of this reaction. As first, step the Cl-C bond attacks the $AlCl_3$ and we form a new bond between the alkyl halide and the $AlCl_3$, finally the C-Cl bond is broken and a carbocation is formed. In this case, we will have a primary carbocation, a very unstable species. Therefore we will have a methyl shift to obtain a tertiary carbocation.

This tertiary carbocation will react with and continues the reaction. (See figure 2)

I hope it helps!