Write the starting substrate

The compound is 2-chloro-1-phenylbutane.

Its structure can be written as: $\mathrm{Ph-CH_2-CH(Cl)-CH_2-CH_3}$

Reaction with $EtOK/EtOH$

$EtOK/EtOH$ is a strong base, so the major reaction is dehydrohalogenation (E2 elimination).

The leaving group $Cl$ is on carbon-2. A $\beta$-hydrogen can be removed from either:

• carbon-1: $\mathrm{Ph-CH_2-}$
• carbon-3: $\mathrm{-CH_2-CH_3}$

So two alkenes are possible:

(i) Elimination between C1 and C2: $\mathrm{Ph-CH=CH-CH_2-CH_3}$ This is a double bond conjugated with the benzene ring.

(ii) Elimination between C2 and C3: $\mathrm{Ph-CH_2-CH=CH-CH_3}$ This is a non-conjugated alkene.

The conjugated alkene is more stable, so the major product $X$ is: $\boxed{\mathrm{Ph-CH=CH-CH_2-CH_3}}$ i.e. 1-phenylbut-1-ene.

Reaction of $X$ with $Hg(OAc)_2/H_2O$ followed by $NaBH_4$

This is oxymercuration-demercuration, which gives Markovnikov addition of water across the double bond without rearrangement.

For: $\mathrm{Ph-CH=CH-CH_2-CH_3}$

the double bond carbons are:

• left carbon: attached to $\mathrm{Ph}$ and $H$
• right carbon: attached to $\mathrm{H}$ and $\mathrm{CH_2CH_3}$

In oxymercuration, the reaction proceeds through a bridged mercurinium ion with greater positive character at the benzylic carbon, because that carbon is better stabilized by resonance with the phenyl ring.

Therefore, water attacks the benzylic carbon, and after demercuration the $OH$ group ends up there.

Thus product $Y$ is: $\boxed{\mathrm{Ph-CH(OH)-CH_2-CH_2-CH_3}}$

This is 1-phenylbutan-1-ol.

Final identification

Hence the major product $Y$ is: $\boxed{\mathrm{C}}$

Comparison with stored answer

Stored correct answer = C.

Our derived answer = C.

So they agree.