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Give step-by-step solution with explanation and final answer:=k y il $izsl : == gif] = HH ; = - P= ii E

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Give step-by-step solution with explanation and final answer:Uploaded Image=k y il $izsl : == gif] = HH ; = - P= ii E

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Asked by: Rahul
Created at: 2026-01-24 03:08:09

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Let's break down the problem step-by-step: --- # **Step-by-Step Solution** ## **Step 1: Understanding the Question** The question provides two \(^{13}C\) NMR spectra: - **Top spectrum:** For the starting material (contains a ketone at C17). - **Bottom spectrum:** For the product after reduction (C17 ketone is reduced to an alcohol). **Question:** If you chemically reduce the circled ketone (**C=O** at C17) to an alcohol (**OH**), which new peak would you expect to appear in the \(^{13}C\) NMR spectrum, and where (in ppm) would it show up? --- ## **Step 2: Analyzing the Key Changes** - **Starting material:** C17 has a **ketone** group. - **Reduced product:** C17 has a **secondary alcohol** group. --- ## **Step 3: Interpreting the NMR Peaks** ### **Ketone Carbon (C=O) Shift** - **Typical \(^{13}C\) NMR chemical shift for a ketone carbon:** Usually **~200–220 ppm**. ### **Alcohol Carbon (C–OH) Shift** - **Typical \(^{13}C\) NMR shift for a secondary alcohol carbon:** Usually **~60–80 ppm**. --- ## **Step 4: Locating the Relevant Peaks** Look at the provided spectra: - **Top spectrum:** The **peak at 220.9 ppm** is labeled as the ketone carbon (C=O). - **Bottom spectrum:** After reduction, this peak (220.9 ppm) **disappears**. Now, **look for a new peak** in the bottom spectrum, which should correspond to the new **C–OH** carbon. You can see a **new peak at 71.8 ppm** in the reduced product (bottom spectrum) that was not present in the starting material. --- ## **Step 5: Final Answer** ### **Explanation** - When you reduce the C17 ketone to an alcohol, the carbon that was previously a carbonyl (C=O, ~220 ppm) becomes a secondary alcohol carbon (C–OH, ~70 ppm). - The 220.9 ppm peak disappears and a new peak at **71.8 ppm** appears. --- ## **Final Answer** > **After reduction of the C17 ketone, you would expect the new \(^{13}C\) NMR peak for the C17 carbon to appear at approximately **71.8 ppm**.** --- ### **Summary Table** | Structure | C17 Peak (ppm) | |-------------------|-----------------| | Ketone (C=O) | 220.9 | | Alcohol (C–OH) | 71.8 | --- **The new peak at 71.8 ppm corresponds to the reduced C17 carbon (now an alcohol).**

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