Identify the final major product [S] obtained in the following...

The starting material is acetophenone ($C_6H_5-CO-CH_3$).

Reaction with $LiAlH_4$ followed by acidic workup reduces the ketone to a secondary alcohol. Thus, $[P]$ is 1-phenylethanol ($C_6H_5-CH(OH)-CH_3$).

Heating $[P]$ with concentrated $H_2SO_4$ causes dehydration, forming an alkene. The major product is styrene ($C_6H_5-CH=CH_2$) due to conjugation with the benzene ring. Thus, $[Q]$ is styrene.

Hydroboration-oxidation of $[Q]$ using $B_2H_6$/THF followed by $H_2O_2/OH^-$ leads to anti-Markovnikov hydration of the alkene. This yields a primary alcohol, 2-phenylethanol ($C_6H_5-CH_2-CH_2-OH$). Thus, $[R]$ is 2-phenylethanol.

Finally, oxidation of $[R]$ with Pyridinium chlorochromate (PCC) selectively oxidizes the primary alcohol to an aldehyde without over-oxidation. The product is 2-phenylethanal ($C_6H_5-CH_2-CHO$).

Therefore, the final major product $[S]$ is 2-phenylethanal.

Answer: 2-Phenylethanal