Let's break down the problem step by step! --- ## **Step 1: Identify the Functional Group and Reagent** - The molecule contains a benzene ring with two carbonyl groups in the **para** positions (a 1,4-benzoquinone structure), and an **ester** functional group attached. - The reagent is **SnCl₂** (tin(II) chloride). --- ## **Step 2: What Does SnCl₂ Do?** - **SnCl₂** is a **reducing agent**. - It is commonly used to reduce **quinones** (like 1,4-benzoquinone) to the corresponding **hydroquinone** (1,4-dihydroxybenzene). --- ## **Step 3: Predict the Product** - The two ketone groups on the benzene ring (the quinone) will each be **reduced to hydroxyl (OH) groups**. - The rest of the molecule (the ester and alkyl chain) will remain unchanged. --- ## **Step 4: Draw and Explain the Product** ### **Product Structure** - The benzene ring now has two **OH groups** at the 1 and 4 positions (para to each other). - The ester group and the alkyl chain remain the same. ### **Visual Description** - The product is **1,4-dihydroxybenzene** with a side chain containing an ester. --- ### **Summary Table** | Position on Ring | Group Before | Group After | |------------------|-------------|---------------| | 1 | =O | –OH | | 2 | H | H | | 3 | H | H | | 4 | =O | –OH | | Side Chain | Ester | Ester | --- #### **Annotated Product (Text)** ``` HO O \ // | / ----C6H4----OCH2C(=O)CH2CH(CH3)2 / HO ``` - The benzene ring now has two OH groups **para** to each other. --- ### **Key Takeaway** - **SnCl₂** reduces quinone (two C=O on benzene) to hydroquinone (two OH on benzene). - Ignore inorganic byproducts; focus just on the organic structure. --- **Let me know if you'd like a stepwise mechanism or a drawn structure!**