To determine the stable carbocation formed in the given reaction, we need to consider the reaction conditions and the structure of the alkyl halide used:
1. The reaction shown involves a Friedel-Crafts alkylation, where an alkyl halide reacts with benzene in the presence of anhydrous AlCl3.
2. The key step is the formation of a carbocation from the alkyl halide, facilitated by AlCl3.
3. In the case of 1-chlorobutane, the initial carbocation formed would be a primary carbocation, which is not stable.
4. Carbocation rearrangement can occur to form a more stable carbocation. The primary carbocation rearranges to a secondary carbocation, which is more stable due to increased hyperconjugation and inductive effects.
5. Additionally, further rearrangement can lead to a tertiary carbocation, which is the most stable due to the greatest hyperconjugation and inductive stabilization.
In this reaction, the most stable carbocation is the one that can form the major product:
6. The rearranged tertiary carbocation formed is thus the most stable carbocation structure, aligning with the product formation in this Friedel-Crafts alkylation.
Conclusion: The correct answer is the carbocation represented in the third option, which is the stable tertiary carbocation.
