The mechanism involves the following 5 steps: 1. Abstraction of a proton to form enolate anion 1; 2. Formation of a cyclopropanone intermediate 2 with expulsion of chloride ion; 3. Addition of hydroxide ion to form tetrahedral intermediate 3; 4. Collapse of the tetrahedral intermediate and breakage of the three-membered ring to form carbanion intermediate 4; 5. Proton transfer to form the rearranged carboxylic acid. For the following reaction, draw the reaction out on paper and then draw the structure of cyclopropanone intermediate 2 in the window. CO₂H & 8 base . You do not have to consider stereochemistry. . You do not have to explicitly draw H atoms. . Do not include lone pairs in your answer. They will not be considered in the grading. . Do not include counter-ions, e.g., Na", 1", in your answer.

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Chapter1: Chemical Foundations
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The mechanism involves the following 5 steps:
haloketone with base to yield a ring-contracted product.
1. Abstraction of a proton to form enolate anion 1;
2. Formation of a cyclopropanone intermediate 2 with expulsion of chloride ion;
3. Addition of hydroxide ion to form tetrahedral intermediate 3;
4. Collapse of the tetrahedral intermediate and breakage of the three-membered ring to form carbanion intermediate 4;
5. Proton transfer to form the rearranged carboxylic acid.
For the following reaction, draw the reaction out on paper and then draw the structure of cyclopropanone intermediate 2 in the window.
CO₂H
CI
8 5
base
. You do not have to consider stereochemistry.
. You do not have to explicitly draw H atoms.
. Do not include lone pairs in your answer. They will not be considered in the grading.
. Do not include counter-ions, e.g., Na, I, in your answer.
*****
O. Sn [F
Transcribed Image Text:R The mechanism involves the following 5 steps: haloketone with base to yield a ring-contracted product. 1. Abstraction of a proton to form enolate anion 1; 2. Formation of a cyclopropanone intermediate 2 with expulsion of chloride ion; 3. Addition of hydroxide ion to form tetrahedral intermediate 3; 4. Collapse of the tetrahedral intermediate and breakage of the three-membered ring to form carbanion intermediate 4; 5. Proton transfer to form the rearranged carboxylic acid. For the following reaction, draw the reaction out on paper and then draw the structure of cyclopropanone intermediate 2 in the window. CO₂H CI 8 5 base . You do not have to consider stereochemistry. . You do not have to explicitly draw H atoms. . Do not include lone pairs in your answer. They will not be considered in the grading. . Do not include counter-ions, e.g., Na, I, in your answer. ***** O. Sn [F
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