Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

11. Radical Reactions

Free Radical Halogenation

Organic Chemistry

11. Radical Reactions

Free Radical Halogenation

Previous problemNext problem

9:47 minutes

Problem 67

Textbook Question

Because of the angle strain present in cyclopropanes, they tend to open up in the presence of nearby radicals. Show a mechanism for the following reaction that demonstrates this principle.

Verified step by step guidance

Begin by recognizing that the cyclopropane ring is highly strained due to its small bond angles. This strain makes it susceptible to ring-opening reactions, especially in the presence of radicals.

The presence of HBr and H2O2 suggests a radical mechanism. H2O2 can decompose to form hydroxyl radicals, which can initiate the radical chain reaction.

The hydroxyl radical abstracts a hydrogen atom from HBr, generating a bromine radical. This bromine radical is crucial for the next step of the mechanism.

The bromine radical attacks the cyclopropane ring, leading to the opening of the ring. This attack results in the formation of a new radical at the site where the ring opens.

The newly formed radical can then react with another molecule of HBr, abstracting a hydrogen atom and forming the final product with a bromine atom attached to the terminal carbon of the opened chain.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Angle Strain

Angle strain occurs when the bond angles in a molecule deviate from the ideal values, causing increased energy and instability. In cyclopropanes, the bond angles are approximately 60 degrees, which is significantly less than the ideal tetrahedral angle of 109.5 degrees. This strain makes cyclopropanes more reactive, particularly in the presence of radicals, as they seek to relieve this strain by breaking bonds.

Radical Mechanism

A radical mechanism involves the formation and reaction of species with unpaired electrons, known as radicals. These highly reactive intermediates can initiate chain reactions, such as the opening of strained rings like cyclopropanes. Understanding how radicals interact with cyclopropanes is crucial for illustrating the mechanism of the reaction, as they can lead to the cleavage of C-C bonds and the formation of new products.

Reaction Mechanism

A reaction mechanism is a step-by-step description of the process by which reactants transform into products. It outlines the sequence of bond-breaking and bond-forming events, including the involvement of intermediates like radicals. In the case of cyclopropanes, illustrating the mechanism helps to clarify how the angle strain facilitates the reaction and the specific steps taken during the bond cleavage and formation.

Watch next

Master The one reaction that alkanes will actually undergo. with a bite sized video explanation from Johnny

Start learning

Related Videos

Related Practice

Textbook Question

The radical fluorination of 2-methyl propane resulted in a 14:86 ratio of products.

(a) On the basis of this ratio, calculate the relative reactivity of 1° and 3° C―H bonds in the radical fluorination.

Textbook Question

The radical fluorination of 2-methyl propane resulted in a 14:86 ratio of products.

(b) From the relative reactivity, calculate the difference in energy between the transition states of the first propagation steps leading to a 1° and 3° radical.

Textbook Question

Show how free-radical halogenation might be used to synthesize the following compounds. In each case, explain why we expect to get a single major product.

(a) 1-chloro-2,2-dimethylpropane (neopentyl chloride)

(b) 2-bromo-2-methylbutane

Textbook Question

Rationalize the fact that reaction A results in an unequal mixture of products, but reaction B yields an equal mixture of two products.

Textbook Question

When (1R,3S)-1-tert-butyl-1,3-dimethylcyclopentane is halogenated, one stereoisomer is produced in excess.

(a) Predict the identity of the major stereoisomer

Textbook Question

When (1R,3S)-1-tert-butyl-1,3-dimethylcyclopentane is halogenated, one stereoisomer is produced in excess.

(b) explain why this reaction did not produce an equal mixture of stereoisomers.

Textbook Question

Predict the major products of the following alkane halogenation reactions. [The number of products shown ignores the formation of racemic mixtures.]

(b)

Textbook Question

Predict the major products of the following alkane halogenation reactions. [The number of products shown ignores the formation of racemic mixtures.]

(d)

Show more practices

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap