Which of the following would not give 2-phenylbutane as the major product in a Friedel-Crafts alkylation reaction?

To solve this question, we need to determine which option does not produce 2-phenylbutane as the major product in a Friedel-Crafts alkylation reaction. Friedel-Crafts alkylation involves attaching an alkyl group to a benzene ring, forming a new carbon-carbon bond. The electrophile is typically a carbocation, which can rearrange to form a more stable carbocation. The target molecule, 2-phenylbutane, has the structure $$Ph-CH(CH_{3})CH_{2}CH_{3}$$, where the carbon attached to the benzene ring is secondary.

Now, let's evaluate each option step by step.

Option A: 1-butene + HF

1-butene is $$CH_{2}=CH-CH_{2}-CH_{3}$$. With HF, protonation occurs at the terminal carbon, forming a secondary carbocation: $$CH_{2}=CH-CH_{2}-CH_{3} + H+ \rightarrow CH_{3}-CH+-CH_{2}-CH_{3}$$. This carbocation attacks benzene in a Friedel-Crafts alkylation, yielding $$Ph-CH(CH_{3})CH_{2}CH_{3}$$, which is 2-phenylbutane. Thus, this option gives the desired product as the major product.

Option B: 2-butanol + H$$_2$$SO$$_4$$

2-butanol is $$CH_{3}-CH(OH)-CH_{2}-CH_{3}$$. With H$$_2$$SO$$_4$$, dehydration occurs, protonating the hydroxyl group and losing water to form a secondary carbocation: $$CH_{3}-CH(OH)-CH_{2}-CH_{3} + H+ \rightarrow CH_{3}-CH(OH_{2}+)-CH_{2}-CH_{3} \rightarrow CH_{3}-CH+-CH_{2}-CH_{3} + H_{2}O$$. This carbocation attacks benzene, forming $$Ph-CH(CH_{3})CH_{2}CH_{3}$$ (2-phenylbutane). No rearrangement occurs, as the secondary carbocation is stable. Hence, this option gives 2-phenylbutane as the major product.

Option C: Butanoyl chloride + AlCl$$_3$$ then Zn, HCl

Butanoyl chloride is $$CH_{3}CH_{2}CH_{2}C(=O)Cl$$. With AlCl$$_3$$, Friedel-Crafts acylation occurs, forming an acylium ion: $$CH_{3}CH_{2}CH_{2}C(=O)Cl + AlCl_{3} \rightarrow CH_{3}CH_{2}CH_{2}C{\equiv}O^+ + AlCl_{4}-$$. This electrophile attacks benzene, yielding the ketone $$Ph-C(=O)-CH_{2}CH_{2}CH_{3}$$ (phenyl propyl ketone). The subsequent step, "then Zn, HCl", refers to Clemmensen reduction, which reduces the carbonyl group to a methylene group: $$Ph-C(=O)-CH_{2}CH_{2}CH_{3} \rightarrow Ph-CH_2-CH_2CH_2CH_3$$ (1-phenylbutane). This is a straight-chain alkylbenzene, whereas 2-phenylbutane is branched. Therefore, this option does not give 2-phenylbutane as the major product.

Option D: Butyl chloride + AlCl$$_3$$

Butyl chloride typically refers to n-butyl chloride, $$CH_{3}CH_{2}CH_{2}CH_{2}Cl$$. With AlCl$$_3$$, it forms a primary carbocation: $$CH_{3}CH_{2}CH_{2}CH_{2}Cl + AlCl_{3} \rightarrow CH_{3}CH_{2}CH_{2}CH_{2}+ + AlCl_{4}-$$. Primary carbocations are unstable and rearrange via a hydride shift: $$CH_{3}CH_{2}CH_{2}CH_{2}+ \rightarrow CH_{3}CH_{2}CH+CH_{3}$$ (a more stable secondary carbocation). This carbocation attacks benzene, forming $$Ph-CH(CH_{3})CH_{2}CH_{3}$$ (2-phenylbutane). Due to rearrangement, this option gives 2-phenylbutane as the major product.

In summary, Options A, B, and D all yield 2-phenylbutane as the major product, while Option C yields 1-phenylbutane. Therefore, Option C does not give 2-phenylbutane as the major product.

Hence, the correct answer is Option C.