Let's solve the problem step by step: --- ## **Step 1: Analyze the Molecular Formula** **Given:** Molecular formula: **₄H₄O₃Let's calculate the degree of unsaturation\[ \text{Degree of Unsaturation} =frac{2C + 2 - H}{2} \] \[ =frac{24) +2 - }{2} \frac{ + 2 - }{2} \frac{}{2} 3 ] So, the molecule has **3 degrees of unsaturation** (rings and/or double bonds). --- ## **Step 2: Examine the ¹³C NMR Spectrum** There are **four distinct peaks**, indicating **four unique types of carbon**. **Chemical shift ranges (approximate):** 1. ~170 ppm — *Carboxyl or ester carbonyl carbon* 2. ~140 ppm — *Alkene or aromatic carbon* 3. ~120 ppm — *Alkene or aromatic carbon* 4. ~60 ppm — *Oxygenated sp³ carbon (like in an ester or ether)* --- ## **Step 3: Propose the Structure** Let's interpret the data: - **Three oxygens**: Suggests possible *ester* or *acid anhydride* or *lactone*. - **Three degrees of unsaturation**: Likely includes double bonds and/or a ring. - **NMR peaks**: One carbonyl (170 ppm), two unsaturated carbons (aromatic or alkene), one oxygenated sp³ carbon (60 ppm). ### **Possible Structure: Furan-2,5-dione (Maleic anhydride)** #### **Why?** - Maleic anhydride (C₄H₂O₃) is a common four-carbon ring with two carbonyls and an alkene; but our formula is **C₄H₄O₃** (2 more hydrogens). - Let's try **γ-butyrolactone** (a five-membered lactone ring): - C₄H₆O₂ (not matching). - Let's try **furan-2(5H)-one** (butenolide): - C₄H₄O₃, matches our formula. - Structure: a five-membered ring with one double bond and a lactone (cyclic ester). #### **Structure of 2(5H)-Furanone (Butenolide):** - **Four unique carbons:** 1. Carbonyl carbon (lactone, ~170 ppm) 2. Alkene carbon (conjugated to carbonyl, ~140 ppm) 3. Alkene carbon (conjugated to carbonyl, ~120 ppm) 4. Oxygenated sp³ carbon (~60 ppm) **This fits the NMR data and formula.** --- ## **Step 4: Draw the Structure** ### **Structure of 2(5H)-Furanone (Butenolide)** ``` O // C1—C2 | \ C4 C3 \ / O ``` Where: - C1 = carbonyl carbon (170 ppm) - C2 = alkene carbon (140 ppm) - C3 = alkene carbon (120 ppm) - C4 = oxygenated methylene (60 ppm) --- ## **Final Answer** ### **The structure that best fits the data is:** ### **2(5H)-Furanone (Butenolide)**  *Alt text: Structure of 2(5H)-furanone (butenolide), a five-membered ring with one double bond and a lactone group.* --- **Summary of Reasoning:** - The formula C₄H₄O₃ and the NMR peaks fit a γ-butyrolactone with a double bond (butenolide). - It has 3 degrees of unsaturation (1 ring + 2 double bonds). - The NMR peaks match: carbonyl (~170 ppm), alkenes (~140, ~120 ppm), and an oxygenated sp³ carbon (~60 ppm). --- **If you need the explicit Lewis structure or further confirmation, let me know!**