An organic compound whose structure is shown above is treated with...

The given organic compound is 1-methylcyclohexanol. In this structure, the carbon atom bonded to the hydroxyl ($-OH$) group is also bonded to three other carbon atoms (one from the methyl group and two from the cyclohexane ring). Therefore, it is a tertiary ($3^{\circ}$) alcohol.

The Lucas test is used to distinguish between primary, secondary, and tertiary alcohols. The Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$. The reaction proceeds via an $S_N1$ mechanism, forming an alkyl chloride which is insoluble and appears as a white turbidity.

The rate of reaction depends on the stability of the intermediate carbocation. Since tertiary carbocations are the most stable, tertiary alcohols react fastest.

Observations for the Lucas test at room temperature:
- Tertiary ($3^{\circ}$) alcohols: White turbidity appears immediately.
- Secondary ($2^{\circ}$) alcohols: White turbidity appears after 5 minutes.
- Primary ($1^{\circ}$) alcohols: No turbidity at room temperature (appears only upon heating).

Since the given compound is a tertiary alcohol, it will react with the Lucas reagent to produce white turbidity immediately.

Answer: White turbidity appears immediately.