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The most stable free radical among the following is (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\) (2) \(\mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{3}\) (3) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (a) \(3>1>4>2\) (b) \(1>3>2>4\) (c) \(3>1>2>4\) (d) \(3>2>1>4\)

Short Answer

Expert verified
The most stable radical configuration is option (c) \( 3 > 1 > 2 > 4 \).

Step by step solution

01

Analyze the options

In order to determine the stability of the free radicals provided, we need to evaluate the structure of each to determine how the nature of substituents and the possibility of resonance or hyperconjugation may influence their stability.
02

Evaluate the benzylic radicals

Examine the first and third options: (1) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\cdot \) and (3) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\cdot\). Both radicals are benzylic types, meaning they are adjacent to an aromatic benzene ring, which allows the radical electron to delocalize via resonance. Typically, benzylic radicals are more stable due to this resonance stabilization, especially if they are secondary or tertiary.
03

Evaluate the allylic radical

For option (3) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\cdot\), the radical is adjacent to both a benzene ring and a substituent group that provides even greater stabilization due to its secondary nature, making it particularly stable.
04

Compare benzylic and allylic stability

Compare radical (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\cdot\) with radical (3) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\cdot\). The latter is more stable because it benefits additionally from hyperconjugation due to the methyl group adjacent to the radical center making it a secondary radical.
05

Identify hyperconjugative and less stable radicals

Examine radicals (2) \(\mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{3}\cdot\), a secondary radical, and (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\cdot\), a primary radical. Secondary radicals are generally more stable than primary ones because they have more hyperconjugation and potential for electron distribution around the radical center.
06

Finalize the order of stability

Putting all the analysis together, the order of stability among the radicals is: \( 3 > 1 > 2 > 4 \). Thus, the most stable radical is (3), and the correct order is given by option (c) \( 3 > 1 > 2 > 4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Benzylic Radicals
Benzylic radicals are special types of radicals where the unpaired electron is located next to an aromatic benzene ring. This unique positioning allows the electron to spread out across the benzene ring.
This process is called resonance. The resonance effect makes benzylic radicals stable because the electron cloud can move through multiple conjugated bonds within the benzene ring.
  • Benzylic radicals are inherently stable.
  • Resonance within the benzene ring spreads the electron density.
The resonance stabilization provides additional stability compared to radicals that lack such interaction with an aromatic system.
Allylic Radicals
Allylic radicals are similar to benzylic radicals, yet instead of being next to a benzene ring, they are adjacent to a double bond. The unpaired electron can participate in resonance with this double bond system.
This allows for additional stability, much like in benzylic radicals. Allylic radicals often involve a carbon-carbon double bond and the associated unpaired electron.
  • They benefit from resonance due to the proximity to the double bond.
  • This creates a more delocalized electron structure.
In systems where both a double bond and a benzene ring are present, the radical can see a compounded stabilization effect, as displayed in certain compounds.
Resonance Stabilization
Resonance stabilization is a key concept for understanding the stability in both benzylic and allylic radicals. It describes how the distribution of an unpaired electron over multiple atoms creates a more stable radical. By allowing the unpaired electron to move across different bonds, the system reduces energy at the reactive site.
  • The more pathways available for resonance, the greater the stabilization.
  • Resonance contributes to decreased potential energy within a radical structure.
This effect is a significant factor in the radical stability ranking presented in the exercise, where resonance plays a crucial role.
Hyperconjugation
Hyperconjugation refers to a specific kind of stability in radicals linked to the overlap between a filled orbital in an adjacent single bond and the partly filled orbital of the radical. This kind of stabilization is mainly observed in secondary and tertiary radicals. Hyperconjugation increases the stability by allowing electron flow through these overlapping orbitals.
  • Occurs prominently in radicals with adjacent alkyl groups.
  • Secondary radicals benefit more from hyperconjugation than primary radicals.
It allows for increased distribution of electron density, making the radicals less reactive.
Radical Stability Comparison
When comparing the stability of different radicals, it's essential to consider how both resonance and hyperconjugation contribute to the stabilization of free radicals. Benzylic and allylic radicals tend to be more stable due to these factors. In the exercise given, options (3) and (1) exhibit high stability from resonance, while option (3) also benefits from hyperconjugation due to an adjacent methyl group.
  • Radicals with more resonance paths generally exhibit higher stability.
  • Secondary radicals are usually more stable than primary ones due to hyperconjugation.
Evaluating these effects in combination provides insights into determining the correct order of radical stability.

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Most popular questions from this chapter

Which of the following molecules has all the effects: inductive, mesomeric and Baker Nathan effect? (a) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}-\mathrm{CH}_{3}\)

Which of the following is correctly matched? (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{N}\left(\mathrm{CH}_{2}\right)_{3}^{\circ} \mathrm{OH} . \mathrm{E}_{2}\) reaction (b) CC(C)(C)Br E_reaction (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}, \mathrm{E}_{1}\) reaction (d) CC(C)(C)O \(\mathrm{E}_{1}\) cb reaction

The correct order of decreasing acidity of the acids given below is 1\. \(\mathrm{Cl}_{3} \mathrm{CCH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{COOH}\) 2\. \(\mathrm{H}_{3} \mathrm{CCH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{COOH}\) 3\. \(\mathrm{Cl}_{3} \mathrm{CCH}=\mathrm{CH}-\mathrm{COOH}\) 4\. \(\mathrm{H}_{3} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (a) \(1>3>2>4\) (b) \(3>1>2>4\) (c) \(3>4 \geq 1>2\) (d) \(3>1>4>2\)

Which one of the following pairs is correctly matched? (a) Saytzeff rule, least substituted alkene (b) Hoffmann rule, most substituted alkene (c) \(\mathrm{E}_{1}\) cb reaction, Hoffmann elimination (d) \(\mathrm{E}_{1}\) reaction, Hoffmann elimination

Which of the following statements is not correct (a) cyclopentadiene has an acidic hydrogen (b) squaric acid is highly acidic (c) aniline is more basic than ammonia (d) aniline is less basic than ammonia

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