a) Interpretation: The structures of alkenes that would yield the alcohol shown on hydration are to be given. Among hydroboration-oxidation and oxymercuration-demercuration, which method could be used to prepare the alcohol also has to be stated. Concept introduction: The hydroboration reaction takes place with syn stereochemistry and results in a non-Markovnikov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the carbon with less number of alkyl substitutents. The oxymercuration-demercuration reaction takes place with anti stereochemistry and results in Markovnokov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the carbon with more number of alkyl substitutents. To give: The structures of alkenes that would yield the alcohol shown on hydration. To state: Among hydroboration-oxidation and oxymercuration-demercuration methods which method can be used to prepare the alcohol from alkenes.

Question

. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavage

Answer 1

Question

Chapter 8.SE, Problem 23VC

Interpretation Introduction

a)

Interpretation:

The structures of alkenes that would yield the alcohol shown on hydration are to be given. Among hydroboration-oxidation and oxymercuration-demercuration, which method could be used to prepare the alcohol also has to be stated.

Concept introduction:

The hydroboration reaction takes place with syn stereochemistry and results in a non-Markovnikov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the carbon with less number of alkyl substitutents.

The oxymercuration-demercuration reaction takes place with anti stereochemistry and results in Markovnokov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the carbon with more number of alkyl substitutents.

To give:

The structures of alkenes that would yield the alcohol shown on hydration.

To state:

Among hydroboration-oxidation and oxymercuration-demercuration methods which method can be used to prepare the alcohol from alkenes.

Interpretation Introduction

b)

Interpretation:

The structures of alkenes that would yield the alcohol shown on hydration are to be given. Among hydroboration-oxidation and oxymercuration-demercuration, the method which can be used to prepare the alcohol also has to be stated.

Concept introduction:

The hydroboration reaction takes place with syn stereochemistry and results in a non-Markovnikov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the carbon with less number of alkyl substitutents.

The oxymercuration-demercuration reaction takes place with anti stereochemistry and results in Markovnokov addition of water to the double bond in the alkene. The resulting product has the hydroxyl group on the carbon with more number of alkyl substitutents.

To give:

The structures of alkenes that would yield the alcohol shown on hydration.

To state:

Among hydroboration-oxidation and oxymercuration-demercuration, the method which can be used to prepare the alcohol.

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. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavage

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