## Step-by-Step Solution ### 1. **Identify the Reaction Type** - The starting material is a terminal alkyne with a phenyl group (Ph). - The reagents are **H₂SO₄/H₂O**, which is an **acid-catalyzed hydration** condition. ### 2. **Markovnikov Addition** - In acid-catalyzed hydration of alkynes, water (H₂O) adds across the triple bond. - The reaction follows **Markovnikov's rule**: the OH group attaches to the more substituted carbon (closer to the phenyl group here). ### 3. **Enol Formation** - The initial product is an **enol** (a compound with a double bond and an OH group on the same carbon). ### 4. **Tautomerization** - Enols are unstable and rapidly **tautomerize** to form a **ketone** (keto-enol tautomerism). - For a terminal alkyne, the final product is a **methyl ketone** (acetophenone derivative). ### 5. **Determine the Major Product** - The correct structure will show a **carbonyl (C=O)** group on the carbon adjacent to the phenyl group. ### 6. **Match with Given Choices** - The correct answer is the structure that shows a **ketone** (C=O) next to the phenyl group (Ph). #### **Let's check the options:** - The fourth option (from the top) shows a ketone next to the Ph group: ![O=C-CH2-Ph] - This is the correct major product. --- ## **Final Answer** **The major product of the reaction is:** ``` O || Ph-C-CH3 ``` **This corresponds to the fourth structure from the top (not the one currently selected).** ### **Explanation** - Acid-catalyzed hydration of terminal alkynes leads to Markovnikov addition of water, forming a ketone via keto-enol tautomerism. - The product is **acetophenone**. --- **Correct answer:** **Fourth option from the top (structure with a ketone next to the Ph group).**