One of the heptanedione isomers below has a pka of approximately 9 rather than 20. Which isomer? Why? Draw its conjugate base and offer a full explanation of its acidit

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One of the heptanedione isomers below has a pka of approximately 9 rather than 20. Which isomer? Why? Draw its conjugate base and offer a full explanation of its acidity.
The image depicts four chemical structures of organic compounds featuring carbonyl groups. These are likely different isomers of diketones. Here's a detailed explanation of each structure:

1. **Structure 1:**
   - A line-bond structure with two ketone groups (C=O) separated by a three-carbon chain. 
   - The pattern is O=C-CH2-CH2-C=O-CH3.

2. **Structure 2:**
   - Similar to the first, with two ketone groups (C=O) but with a different positioning of the carbonyls, featuring a two-carbon spacer. 
   - The pattern is O=C-CH2-C=O-CH2-CH3.

3. **Structure 3:**
   - Features two carbonyl groups that are directly attached to a three-carbon chain and two methyl groups at each terminus.
   - The pattern is CH3-C=O-CH2-CH2-C=O-CH3.

4. **Structure 4:**
   - A linear arrangement with a carbonyl group on the second carbon from the right and another carbonyl on the fourth carbon from the left, creating a longer spacer of four carbons.
   - The pattern is CH3-C=O-CH2-CH2-CH2-C=O-CH3.

These structures illustrate the concept of isomerism in organic chemistry, where compounds have the same molecular formula but different structural arrangements. This can significantly influence the properties and reactivity of the compounds.
Transcribed Image Text:The image depicts four chemical structures of organic compounds featuring carbonyl groups. These are likely different isomers of diketones. Here's a detailed explanation of each structure: 1. **Structure 1:** - A line-bond structure with two ketone groups (C=O) separated by a three-carbon chain. - The pattern is O=C-CH2-CH2-C=O-CH3. 2. **Structure 2:** - Similar to the first, with two ketone groups (C=O) but with a different positioning of the carbonyls, featuring a two-carbon spacer. - The pattern is O=C-CH2-C=O-CH2-CH3. 3. **Structure 3:** - Features two carbonyl groups that are directly attached to a three-carbon chain and two methyl groups at each terminus. - The pattern is CH3-C=O-CH2-CH2-C=O-CH3. 4. **Structure 4:** - A linear arrangement with a carbonyl group on the second carbon from the right and another carbonyl on the fourth carbon from the left, creating a longer spacer of four carbons. - The pattern is CH3-C=O-CH2-CH2-CH2-C=O-CH3. These structures illustrate the concept of isomerism in organic chemistry, where compounds have the same molecular formula but different structural arrangements. This can significantly influence the properties and reactivity of the compounds.
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