What happens when C₆H₅-O-R is treated with HX

The correct option is RX and C₆H₅OH are formed.

Let us understand the step-by-step chemical reasoning behind this reaction:
The given compound is $C_6{H}_5-O-R$, which is an alkyl aryl ether (specifically, an anisole derivative or phenetole derivative depending on the alkyl group $R$).

When this ether is treated with a hydrogen halide ($HX$), the reaction proceeds as follows:

Step 1: Protonation of the ether
The oxygen atom of the ether has lone pairs of electrons. It accepts a proton ($H^{+}$) from the strong acid $HX$ to form a protonated ether (oxonium ion):
$C_6{H}_5-\stackrel{+}{O}\left(H\right)-R$ along with the halide ion $X^{-}$.

Step 2: Nucleophilic attack by halide ion ($X^{-}$)
The halide ion ($X^{-}$) acts as a nucleophile and must attack one of the carbons bonded to the oxygen atom to cleave the $C-O$ bond.
There are two possible sites for nucleophilic attack:
1. The $s{p}^3$ hybridized carbon of the alkyl group ($R$).
2. The $s{p}^2$ hybridized carbon of the phenyl ring ($C_6{H}_5-$).

Step 3: Bond resonance and bond strength comparison
In alkyl aryl ethers, the lone pair of electrons on the oxygen atom is in conjugation with the $\pi$-electrons of the benzene ring. This resonance imparts a partial double-bond character to the $C_6{H}_5-O$ bond:
$C_6{H}_5\&DoubleLongLeftRight;O-R$ resonance makes the $S{p}^2$ $C\left(\mathrm{phenyl}\right)-O$ bond much stronger and shorter than the $S{p}^3$ $C\left(\mathrm{alkyl}\right)-O$ bond.
Due to this high bond strength and partial double-bond character, the $C_6{H}_5-O$ bond is extremely difficult to break. Furthermore, $S{N}_2$ nucleophilic substitution cannot occur at the $s{p}^2$ hybridized phenyl carbon due to steric hindrance from the aromatic ring.

Conclusion:
Therefore, the nucleophile $X^{-}$ attacks the alkyl group ($R$), breaking the weaker $O-R$ bond.
This yields an alkyl halide ($RX$) and phenol ($C_6{H}_{5}OH$).
The overall reaction is:
$C_6{H}_5-O-R+HX\to C_6{H}_{5}OH+RX$