This is the fourth and final part of a multi-part module on Free Radical Halogenation.

View the first part here: Part 1: The Mechanism

View the second part here: Part 2: Regioselectivity

View the third part here: Part 3: Stereoselectivity

 

Question 1:

What principle accounts for the observed regioselectivity of radical bromination that is not observed for radical chlorination of alkenes?

1. Lechatlier’s Principle
2. Avagadro’s Lab
3. Hammond’s Postulate
4. Markovnikov Rule

 

Question 2:

Which product can NOT be prepared in high yield by radical halogenation of alkanes?

1. Iodoethane
2. 2-Bromo-2-methylheptane
3. Chlorocyclopentane
4. 2-Bromo-2,4,4,trimethylpentane

 

Question 3:

Predict the major product of the following reaction.

 

 

 

Question 4:

Describe the reaction conditions to produce the following product

1. NaBr
2. Br₂
3. H₂SO₄, Br₂
4. HBr

 

 

 

 

Answers and Explanations:

Question 1:  3

Difficulty Level: Easy

Explanation: Hammond’s Postulate describes the stability of the radical center is outweighed by the extreme exothermicity of radical chlorination (in contrast to bromination, which is endothermic), thus a mixture of chlorinated products is observed.

 

Question 2: 1

Difficulty Level: Medium

Explanation: Answers 2,3,4 all would start from alkanes containing secondary and tertiary carbon centers, which produce stable radical intermediates for halogenation.  Ethane contains two primary carbons coupled to the extreme endothermicity of iodination would yield very little product.

 

Question 3: B

Difficulty Level: Medium

Explanation:  The steric strain of the cyclopropane ring will drive the hemolytic cleavage of the 2,4 sigma bond with radical bromine to produce a radical intermediate on carbon 2.  Propagation of the radical bromine will result in the formation of product B.  Product C and D are not possible while the reagents to produce product A are not listed.

 

 

Question 4: 4

Difficulty Level: Hard

Explanation: Addition across the double bond with one equivalent of Br would more easily take place via an electrophilic addition using an equivalent of HBr, while radical bromination would produce dibromo-pentane.

 

 

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This is the second part of a multi-part module on Free Radical Halogenation.

View the first part here: Part 1: The Mechanism

View the second part here: Part 2: Regioselectivity

 

Radial intermediates (step 2a product) produce an sp2-like hybridization orbital with the lone electron in the vacant 2p orbital, therfore attack of the radical electron on the C-H bond can take place from either side of the molecule.  The result will always produce an RACEMIC MIXTURE (or equal amount) of the two enantiomers.

 

There are multiple examples of this reaction to review in the StudyOrgo.com Study Mode. When ready, test your knowledge in the StudyOrgo.com Quiz Mode.

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Stay tuned for Part 4: Practice Quiz

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This is the second part of a multi-part module on Free Radical Halogenation.

View the first part here: Part 1: The Mechanism

 

Radical bromination will always replace the C-H bond on the MOST subsituted carbon center because the stability of the radical intermediate is higher with increasing substituents on the carbon center.

(click on image to view larger)

This selectivity is the same, but a weaker consideration, for radical chlorination which obeys Hammond’s Postulate in that stability of the radical center is outweighed by the extreme exothermicity of radical chlorination (compared to bromination), thus a mixture of chlorinated products is observed.

 

There are multiple examples of this reaction to review in the StudyOrgo.com Study Mode. When ready, test your knowledge in the StudyOrgo.com Quiz Mode.

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Check out Part 3: Stereoslectivity – Determining stereochemistry of carbon centers

This is the first part of a multi-part module on Free Radical Halogenation.

 

There are THREE critical steps to free radical reactions – Memorize!

1) Initiation: The Br2 single bond is broken by high energy ligh (hv) to form radicals placing one electron on each atom.

2) Propagation: (Hint: One radical reacts with a sigle bond to form another radical, thus propigating the radical species to drive the reaction forward.

a) Radical Br abstracts one hydrogen from a C-H bond in propane to form radical propane and HBr.

b) Radical propane asbracts one Br from Br2 to form the bromoalkane and radical Br, thus restoring the reactants for another round as shown in step 2a.

3) Termination: Any two radicals combine to form a single bond.  These species will be in low abundance. Hint: Radicals are destroyed by combining two radicals to form a single bond.  This eliminates the radical necessary for radical alkane formation (green boxes) as shown in step 2a and ends the reaction.

 

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