In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom. Concept introduction: Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms. According to this rule stable and unstable molecule is mentioned below, The first compound is stable. Although double bond is at bridgehead, the system is not bridged But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Question

Want to see the full answer?

Answer 1

Question

Chapter 8, Problem 108P

Interpretation Introduction

Interpretation: In accordance with Bredt’s rule, it should be explained that why there cannot be a double bond at a bridgehead carbon atom.

Concept introduction:

Bredt’s rule: This rule states that a bridged bicyclic compound cannot have a double bond at the bridgehead position unless one of the rings contains at least 8 carbon atoms.

According to this rule stable and unstable molecule is mentioned below,

The first compound is stable. Although double bond is at bridgehead, the system is not bridged

But the second compound is the good example to illustrate the violation of this rule. This compound is unstable because double bond is at bridgehead and ring contains only 6 carbons.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)

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+

Li+ is a hard acid. With this in mind, which if the following compounds should be most soluble in water? Group of answer choices LiBr LiI LiF LiCl