\(\mathrm{C}_{6} \mathrm{H}_{6}+\angle \mathrm{\textrm{AlCl } _ { 3 }}{\longrightarrow} \mathrm{P}\) The major product formed in the above reaction is (a) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) CCC(C)C (c) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\)

First, we need to identify the electrophile that will be attacking the benzene ring in the reaction. In this case, it is AlCl3, which is a Lewis acid and can accept a lone pair of electrons.

We will look at each option individually to see how the electrophile might react with benzene and produce the given product. (a) C6H5-CH2-CH2-CH3 => The electrophile would need to transfer three carbons in a linear chain into the benzene ring for this product to form. However, AlCl3 cannot produce this electrophile. So, this option is not correct. (b) CCC(C)C => This option does not have a benzene ring, so this cannot be the product of the given reaction. (c) C6H5-CH=CH-CH3 => In this case, the electrophile would need to transfer a two-carbon chain with a double bond to the benzene ring. This can be achieved through a Friedel-Crafts alkylation reaction with AlCl3 and propene (CH2=CH-CH3). Therefore, this is a plausible product. (d) C6H5-CH2-CH=CH2 => Again, the electrophile needs to transfer a three-carbon chain with a double bond to the benzene ring. This can be achieved through a Friedel-Crafts alkylation reaction followed by a rearrangement. However, this reaction would be less selective compared to option (c) because it involves the rearrangement of an intermediate.

Based on the analysis, option (c) is the most plausible choice for the major product because it is formed through a direct Friedel-Crafts alkylation reaction without further rearrangement. Therefore, the major product of the given reaction is: (c) C6H5-CH=CH-CH3