Compounds I and II can be distinguished by using reagent.
`{:("(I)",,"(II)"),("4-Hydroxy-4-methypent-2-enoic acid",,"5-Hydroxypent-2-ynoic acid "):}`

Explanation

To distinguish between the two compounds, 4-Hydroxy-4-methylpent-2-enoic acid (Compound I) and 5-Hydroxypent-2-ynoic acid (Compound II), we can use Lucas reagent, which is a solution of zinc chloride (ZnCl₂) in concentrated hydrochloric acid (HCl). Here’s a step-by-step solution:

### Step 1: Draw the Structures of the Compounds
- **Compound I (4-Hydroxy-4-methylpent-2-enoic acid)**:
- It has a total of 5 carbon atoms.
- The hydroxy group (OH) is attached to the 4th carbon, and a methyl group (CH₃) is also attached to the same carbon.
- The double bond is located between the 2nd and 3rd carbon atoms.

- **Compound II (5-Hydroxypent-2-ynoic acid)**:
- It also has 5 carbon atoms.
- The hydroxy group (OH) is attached to the 5th carbon.
- There is a triple bond between the 2nd and 3rd carbon atoms.

### Step 2: Identify the Type of Alcohol
- **Compound I** has a tertiary alcohol because the carbon bearing the hydroxy group is attached to three other carbon atoms.
- **Compound II** has a primary alcohol because the carbon bearing the hydroxy group is attached to only one other carbon atom.

### Step 3: React with Lucas Reagent
- **Lucas Reagent**: This reagent is used to differentiate between primary, secondary, and tertiary alcohols.
- Tertiary alcohols react quickly with Lucas reagent, resulting in the formation of an alkyl halide and immediate turbidity.
- Primary alcohols do not react significantly at room temperature with Lucas reagent.

### Step 4: Observations
- When Compound I (tertiary alcohol) is treated with Lucas reagent, it will show immediate turbidity due to the formation of an alkyl halide.
- When Compound II (primary alcohol) is treated with Lucas reagent, there will be no significant reaction at room temperature, and thus no turbidity will form.

### Conclusion
By using Lucas reagent, we can distinguish between Compound I and Compound II:
- **Compound I** will react and show turbidity (tertiary alcohol).
- **Compound II** will not react significantly and will remain clear (primary alcohol).