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Ch.6 - Alkyl Halides; Nucleophilic Substitution
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
Chapter 6, Problem 59

A student adds NBS to a solution of 1-methylcyclohexene and irradiates the mixture with a sunlamp until all the NBS has reacted. After a careful distillation, the product mixture contains two major products of formula C7H11Br.
(a) Draw the resonance forms of the three possible allylic free radical intermediates.
(b) Rank these three intermediates from most stable to least stable.
(c) Draw the products obtained from each free-radical intermediate.

Verified step by step guidance
1
Step 1: Understand the reaction. NBS (N-Bromosuccinimide) is commonly used for allylic bromination in the presence of light or a radical initiator. The reaction proceeds via a free radical mechanism, targeting the allylic position of the alkene (the carbon adjacent to the double bond).
Step 2: Identify the allylic positions in 1-methylcyclohexene. The double bond in 1-methylcyclohexene creates two allylic positions: one on the carbon adjacent to the double bond in the ring and another on the methyl group attached to the ring. These positions are where the bromine atom can be added.
Step 3: Draw the resonance forms of the three possible allylic free radical intermediates. When the hydrogen atom is abstracted from the allylic positions, three possible allylic radicals can form. Use resonance structures to show the delocalization of the unpaired electron across the π-system. Ensure to draw all valid resonance forms for each radical.
Step 4: Rank the stability of the three allylic radicals. The stability of allylic radicals depends on the extent of resonance stabilization and the substituents attached. Generally, more substituted radicals (tertiary > secondary > primary) and those with greater resonance delocalization are more stable. Rank the radicals accordingly.
Step 5: Draw the products formed from each allylic radical. Each allylic radical can react with a bromine atom to form a brominated product. Show the structures of the products formed by adding bromine to the carbon atom where the radical was located in each case. These products will correspond to the two major products observed in the reaction.

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Key Concepts

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

Allylic Free Radicals

Allylic free radicals are reactive intermediates formed when a radical is located on a carbon atom adjacent to a double bond. These radicals are stabilized by resonance, allowing the unpaired electron to be delocalized over the π system of the alkene. Understanding the structure and stability of these radicals is crucial for predicting the products of reactions involving alkenes and radical initiators like NBS.
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Resonance Structures

Resonance structures are different ways of drawing the same molecule that illustrate the delocalization of electrons. In the case of allylic free radicals, resonance allows for the distribution of the radical character across multiple atoms, which can significantly affect the stability of the radical. Recognizing and drawing these structures is essential for analyzing the stability and reactivity of intermediates in organic reactions.
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Radical Stability

The stability of free radicals is influenced by factors such as hyperconjugation and the presence of electron-donating or withdrawing groups. Generally, tertiary radicals are more stable than secondary, which are more stable than primary radicals due to the degree of alkyl substitution. Ranking the stability of the allylic free radical intermediates formed in this reaction is key to predicting the major products resulting from their subsequent reactions.
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