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Science Chemistry Organic Chemistry: Principles And Mechanisms

It is to be identified which C − O bond in the given protonated epoxide you would expect to be longer and why. Concept introduction: Epoxides are three membered ring ethers. Epoxides react readily with nucleophiles under neutral and basic reaction conditions. The C − O bond in an epoxide partially breaks on protonation. As a result, both the carbon atoms develop a partial positive charge. The larger partial positive charge develops on the side of the ring with greater alkyl substitution. Thus, out of the two carbon atoms in the epoxide ring, the more substituted carbon atom carries the positive charge much better than the other carbon.

It is to be identified which C − O bond in the given protonated epoxide you would expect to be longer and why. Concept introduction: Epoxides are three membered ring ethers. Epoxides react readily with nucleophiles under neutral and basic reaction conditions. The C − O bond in an epoxide partially breaks on protonation. As a result, both the carbon atoms develop a partial positive charge. The larger partial positive charge develops on the side of the ring with greater alkyl substitution. Thus, out of the two carbon atoms in the epoxide ring, the more substituted carbon atom carries the positive charge much better than the other carbon.

Solution Summary: The author explains that the longer C- O bond in an epoxide is the one that involves the carbon that can better accommodate a positive charge.

Organic Chemistry: Principles And Mechanisms

Organic Chemistry: Principles And Mechanisms

2nd Edition

ISBN: 9780393630756

Author: KARTY, Joel

Publisher: W.w. Norton & Company,

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Chapter 10, Problem 10.54P

Interpretation Introduction

Interpretation:

It is to be identified which C−O bond in the given protonated epoxide you would expect to be longer and why.

Concept introduction:

Epoxides are three membered ring ethers. Epoxides react readily with nucleophiles under neutral and basic reaction conditions. The C−O bond in an epoxide partially breaks on protonation. As a result, both the carbon atoms develop a partial positive charge. The larger partial positive charge develops on the side of the ring with greater alkyl substitution. Thus, out of the two carbon atoms in the epoxide ring, the more substituted carbon atom carries the positive charge much better than the other carbon.

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Chapter 10, Problem 10.53P

Chapter 10, Problem 10.55P

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19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

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Chapter 10 Solutions

Organic Chemistry: Principles And Mechanisms

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