Problem 201 - A Complete Resource Book in Chemistry for JEE Main

Question

The increasing order of stability of the following free radicals is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}\) (b) \(\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (c) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{C}_{6} \mathrm{H}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}<\left(\mathrm{CH}_{3}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \dot{\mathrm{C}}<\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\)

Step-by-Step Solution

Answer

Option (a) is the correct order of increasing stability.

Step 1: Identify Radical Types

Examine each option to determine the type of radicals involved. The radicals are either methyl, ethyl, phenyl, or triphenylcarbinyl.

Step 2: Assess Stability Factors

Recall that the stability of free radicals is often enhanced by hyperconjugation, resonance stabilization, and inductive effects. Radicals are more stable when resonance occurs, as seen with phenyl groups.

Step 3: Compare Options Based on Substituents

Identify resonance and hyperconjugation effects: phenyl groups provide resonance stabilization, while methyl groups provide hyperconjugation but less than phenyl. This implies radicals with phenyl groups are more stable.

Step 4: Evaluate Each Option's Order

Analyze each given option to judge the increasing stability: - In (a), radicals go from alkyl to phenyl, correctly increasing in stability. - In (b) and (c), phenyl radicals are placed less stably compared to alkyl, contradicting the expected order. - In (d), triphenyl radicals are misplaced; they should follow diphenyl.

Step 5: Conclude with the Correct Order

The correct order of stability considering resonance over charge delocalization and hyperconjugation is option (a): \(\left(\mathrm{CH}_3\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{CH}_3\right)_3 \dot{\mathrm{C}} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \dot{\mathrm{C}} \mathrm{H} < \left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \dot{\mathrm{C}}\).

Key Concepts

Hyperconjugation

Hyperconjugation is a fascinating concept that significantly contributes to the stability of free radicals. It's sometimes described as "no-bond resonance," which might sound strange, but it helps you understand what's happening at the molecular level. The idea is that electrons in a sigma bond (often from a hydrogen atom) next to an unoccupied p-orbital or a free radical can slightly move into that empty space.
This movement is like adding a stabilizing touch to the radical. It doesn't involve shifting entire bonds like in resonance, but just the slight overlap of electrons creates enough stability to matter.

Consider an example with a methyl radical (\( ext{CH}_3^\cdot\)). The carbon atom in this radical is looking for an electron, and it gets a little help from neighboring C-H bonds. This assistance is hyperconjugation in action. It stabilizes the radical by delocalizing the charge. This effect is why radicals like isopropyl and tert-butyl are more stable than methyl radicals.

• Stabilization occurs through slight electron delocalization.
• Only happens with adjacent sigma bonds and empty p-orbitals.
• Almost like a "soft” resonance effect without full bond shifts.

Understanding hyperconjugation can help you appreciate why certain radicals behave the way they do in organic reactions.

Resonance Stabilization

Resonance stabilization plays a crucial role in understanding the stability of free radicals, especially those containing aromatic groups like phenyls. When you have a radical on a benzene ring, the unpaired electron can be delocalized across the ring through resonance.
This means the electron doesn't just sit on one atom but instead spreads over the entire system, sharing and stabilizing the charge distribution.

The magic of resonance is that it allows multiple forms, called resonance forms, to exist simultaneously. This is why radicals containing phenyl groups are generally more stable than their alkyl counterparts. An easy way to visualize this is through the resonance structures that you can draw, showing the movement of the unpaired electron around the benzene ring.

• Resonance involves electron delocalization across multiple atoms.
• Phenyl radicals are significantly stabilized by this effect.
• Multiple resonance forms mean distributed charge and increased stability.

It's like spreading a load among several people rather than burdening just one, making it easier for each to manage. This principle is vital in predicting the stability and reactivity of aromatic radicals.

Inductive Effect

The inductive effect may not be as flashy as hyperconjugation or resonance, but it's nonetheless important in radical stability. This effect is all about how atoms or groups attached to a radical can influence its electron distribution based on their electronegativity.
Essentially, electronegative atoms pull electron density toward themselves through the sigma-bonds, known as a negative inductive effect.

Consider a chlorine atom bonded to a carbon radical. The chlorine would pull electrons towards itself, stabilizing the radical slightly by reducing electron deficiency. On the flip side, alkyl groups often push electrons away, having a positive inductive effect, helping to stabilize adjacent radicals by slightly increasing electron density towards them.

• Reflects electron donation or withdrawal through sigma bonds.
• Electronegative groups can stabilize radicals by pulling electron density.
• Alkyl groups can stabilize by donating electron density.

While the inductive effect is generally weaker compared to resonance and hyperconjugation, it still provides a nuanced level of stability that shouldn't be overlooked in determining free radical stability.