Let's solve the problem step by step: ### Step 1: Identify the Compound The structure shown is **propan-2-ol** (isopropanol), but the oxygen is shown with a negative charge and a hydrogen, so it is likely the **conjugate base of isopropanol**, or isopropoxide ion. ### Step 2: Identify the Different Types of Protons Let's label the protons: - **O–H proton:** The hydrogen attached to the oxygen atom. - **C–H protons:** The hydrogens attached to the carbon atoms (there are two types: methyl group hydrogens and a methine hydrogen). ### Step 3: Consider Acidity of Each Proton Acidity depends on how easily a proton can be removed, which is related to the stability of the conjugate base formed: - **O–H proton:** Removing this proton would result in a negative charge on the oxygen, forming an alkoxide ion (already shown here). - **C–H proton (methyl or methine):** Removing a proton from a carbon atom forms a carbanion, which is much less stable than an alkoxide ion. ### Step 4: Compare Acidity - **O–H proton:** Alcohol O–H protons are much more acidic than C–H protons because the resulting negative charge is better stabilized on oxygen (more electronegative). - **C–H protons:** These are far less acidic; the resulting carbanion is very unstable. ### Step 5: Conclusion and Final Answer **The most acidic proton in this molecule is the hydrogen attached to the oxygen atom (the O–H proton).** #### Correct Answer (with explanation): - **The most acidic proton is the one attached to the oxygen atom (O–H).** - **Reason:** Removing this proton forms an alkoxide ion, with the negative charge stabilized on the highly electronegative oxygen atom. In contrast, removing any of the C–H protons forms a carbanion, which is much less stable and thus much less acidic. --- **In summary:** The most acidic proton is the hydrogen atom attached to the oxygen atom (O–H group) in the molecule. If you were to "click" on the structure, **click on the hydrogen connected to the oxygen atom.**