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The most stable free radical among the following is (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\) (2) \(\mathrm{CH}_{3}-\dot{\mathrm{CH}}-\mathrm{CH}_{3}\) (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 free radical arrangement is 3 > 1 > 2 > 4, so the answer is option (c).

Step by step solution

01

Understand the Concept of Stability in Free Radicals

Free radicals are atoms or molecules that contain an unpaired electron. The stability of free radicals depends on several factors such as resonance stabilization, hyperconjugation, and inductive effects. In general, the more resonance structures a free radical can form, the more stable it is.
02

Analyze the Radical Structures

For each given radical, identify the structural features that contribute to their stability:1. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\): This radical is benzyl free radical, having resonance stabilization with the phenyl ring.2. \(\mathrm{CH}_{3}-\dot{\mathrm{CH}}-\mathrm{CH}_{3}\): Isopropyl free radical, which is stabilized by hyperconjugation.3. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\): This is an allylic radical that is also stabilized by resonance with the phenyl ring, making it particularly stable due to extensive delocalization of the unpaired electron.4. \(\mathrm{CH}_{3} \mathrm{CH}_{2}\): The ethyl radical, which has limited hyperconjugation effects.
03

Compare Stability

Rank the radicals based on their stabilization: - The allylic radical \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\) (3) is the most stable due to extended resonance with both an alkene and a phenyl ring.- The benzyl radical \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\) (1) comes next due to its resonance stabilization with the phenyl ring.- The isopropyl radical \(\mathrm{CH}_{3}-\dot{\mathrm{CH}}-\mathrm{CH}_{3}\) (2) follows, being moderately stable from hyperconjugation.- The least stable is the ethyl radical \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (4), which is only stabilized by minimal hyperconjugation.
04

Select the Correct Option

Based on the stability ranking determined:- The order of stability is \(3>1>2>4\).- This corresponds with the option (c).

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Key Concepts

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

Resonance Stabilization
Free radicals prefer stability, and one way to achieve it is through resonance stabilization.
When a molecule has resonance, it can distribute the unpaired electron across multiple atoms.
This delocalization reduces the energy of the radical, making it more stable.
  • Phenyl rings: Considered excellent at providing resonance because they have cyclic structures which allow electron clouds to overlap.
  • Benzyl and allylic radicals: They are often stabilized by resonance with such structures.
An example is the benzyl radical, where resonance with the attached phenyl ring contributes significantly to its stability.
Hyperconjugation
This concept plays a vital role in the stability of radicals like the isopropyl radical.
Hyperconjugation involves the interaction of the radical's unpaired electron with adjacent sigma bonds.
  • Available Hydrogens: More adjacent C-H bonds mean more interactions, hence higher stability.
  • Alkyl groups: These groups with available C-H bonds provide a greater hyperconjugative effect.
In radicals, like isopropyl, the multiple C-H bonds enable hyperconjugation, distributing the radical character and increasing stability.
Inductive Effects
Inductive effects involve the influence of electronegative atoms or groups on the radical's structure.
These groups pull electron density towards themselves, helping stabilize the radical's charge distribution. However, for many free radicals, this effect is less pronounced compared to resonance and hyperconjugation.
  • Electronegative Atoms: Atoms like fluorine or chlorine can stabilize a radical by pulling electrons towards them.
  • Distance Matters: The effect weakens with increased distance from the radical center.
While not a dominant factor in radical stability, inductive effects can contribute to tuning the stability of specific radicals.
Benzyl Radical
A benzyl radical features a radical positioned right next to a phenyl ring.
It benefits greatly from resonance stabilization, which significantly improves its stability.
  • Phenyl Ring Interaction: The unpaired electron can resonate through the electrons in the ring.
  • Stability: A benzyl radical is more stable than many other free radicals due to this constant state of electron sharing.
The benzyl radical is often cited in exercises as an example of how resonance can significantly stabilize a molecule.
Allylic Radical
Allylic radicals are robust examples of stability through resonance.
Situated next to a double-bonded group, they allow the unpaired electron to be delocalized across adjacent pi bonds.
  • Adjacent Pi Bonds: Provide an opportunity for the radical to stabilize through resonance.
  • Phenyl Group Strength: When also adjacent to phenyl groups, like in some radicals analyzed, it provides even more resonance structures.
Due to these aspects, allylic radicals exhibit high stability, often more than radicals that solely benefit from hyperconjugation or inductive effects.

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Most popular questions from this chapter

Given below are the four compound which are stabilized either by inductive, mesomeric or hyperconjugation effects.. (a) CC(=O)C1=C(C)CCCC1 (b) BrC1=CCCCC1 (c) CC1=CCCCC1C (d) CC=CCl

The order of stability of the following carbocations: (I) \(\mathrm{CH}_{2}=\mathrm{CH}-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{2}\) \(\mathbf{[ 2 0 1 3}]\) (II) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{2}\) (III) is: [C+]c1ccccc1 (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}\) (b) \(\mathrm{III}>\mathrm{I}>\mathrm{II}\) (c) \(\mathrm{III}>\mathrm{II}>\mathrm{I}\) (d) \(\mathrm{II}>\mathrm{III}>\mathrm{I}\)

When a methyl radical is formed from \(\mathrm{CH}_{3} \mathrm{Cl}\), select the correct statement: (1) bond angle of \(109^{\circ} 28\) ' is retained (2) number of sigma bonds is three (3) carbon undergoes geometric change from tetrahedral to planar (4) hybridization changes \(\mathrm{sp}^{3}\) to \(\mathrm{sp}^{2}\) (a) 2,3 and 4 (b) 1,3 and 4 (c) 2 and 4 (d) 3 and 4

The correct order of increasing acid strength of the compound: (1) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (2) \(\mathrm{MeOCH}_{2} \mathrm{CO}_{2} \mathrm{H}\) (3) \(\mathrm{CF}_{3} \mathrm{CO}_{2} \mathrm{H}\) (4) \((\mathrm{Me})_{2}-\overline{\mathrm{CO}_{2}} \mathrm{H}\) (a) \(2<4<1<3\) (b) \(4<1<3<2\) (c) \(4<1<2<3\) (d) \(1<4<3<2\)

Which of the following order/statement (s) is/are correct? (a) In \(S_{N} 1\) reactions, \(\mathrm{AgNO}_{3}\) increases the rates of solvolysis (b) Soft bases have enhanced nucleophilicities towards \(\mathrm{S}_{\mathrm{N}} 2\) reactions. (c) \(\mathrm{PhSO}_{3}^{-}>\mathrm{Cl}_{3} \mathrm{CCOO}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{PhO}^{-}\)(leav- ing group ability) (d) HCOOH may lead a tertiary halide towards \(\mathrm{S}_{\mathrm{N}}^{2} .\)

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