Problem 58 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

Rank, from the most stabilised to the least stabilised, the following free radicals according to their stabilisation energies. (i) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (ii) \(\dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{CH}_{3}\) (iii) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) (iv) \(\left(\mathrm{CH}_{2}=\mathrm{CH}-\dot{\mathrm{C}} \mathrm{H}_{2}\right)\) (a) \(\mathrm{iv}>\mathrm{iii}>\mathrm{ii}>\mathrm{i}\) (b) \(\mathrm{i}>\mathrm{iv}>\mathrm{iii}>\mathrm{ii}\) (c) \(\mathrm{iii}>\mathrm{iv}>\mathrm{i}>\mathrm{ii}\) (d) iii \(>\mathrm{iv}>\mathrm{ii}>\mathrm{i}\)

Step-by-Step Solution

Answer

The correct answer is (d) iii > iv > ii > i.

Step 1: Understanding Free Radical Stability

Free radicals are molecules that contain at least one unpaired electron and are often unstable. Their stability is influenced by factors such as the nature of the carbon atom holding the radical center, the presence of adjacent double bonds, and the inductive effects from other groups.

Step 2: Analyze Each Free Radical

(i) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) is a primary radical, lacking adjacent double bonds or significant stabilizing groups. (ii) \(\dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{CH}_{3}\) is also a primary radical with similar stability. (iii) \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\) has two methyl groups donating electron density through hyperconjugation and inductive effect, increasing radical stability. (iv) \(\left(\mathrm{CH}_{2}=\mathrm{CH}-\dot{\mathrm{C}} \mathrm{H}_{2}\right)\) is an allylic radical, highly stabilized by resonance with the adjacent double bond.

Step 3: Rank Based on Stability

The ranking of radical stability generally follows: allylic radicals (due to resonance) > tertiary radicals (due to hyperconjugation and inductive effects) > primary radicals. Therefore, the order from most stabilized to least is expected to be: iv > iii > ii and i.

Step 4: Choose the Correct Answer

Comparing the choices with our analysis, option (d) iii > iv > ii > i matches our calculated order of stabilization: tertiary radical > allylic radical > primary radical \((\dot{\mathrm{C}} \mathrm{H}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2})\). Therefore, (d) is the correct answer.

Key Concepts

Radical Stability

In organic chemistry, understanding radical stability is key to predicting reaction outcomes. Radicals are atoms or molecules that contain unpaired electrons and are typically unstable. The stability of a radical is determined by several factors: the type of carbon atom that carries the radical, nearby double bonds, and the influence of surrounding groups through effects like hyperconjugation and the inductive effect.
Primary radicals, such as

• \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) and \(\dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{CH}_{3}\),

do not have double bonds or additional stabilizing groups and are often less stable. Understanding these factors allows chemists to predict which radicals are more likely to participate in specific reactions.

Allylic Radical

An allylic radical is formed when the unpaired electron is adjacent to a double bond. This arrangement allows the radical to be stabilized by resonance, which occurs when electrons are delocalized over adjacent atoms, giving rise to multiple structures that depict the same molecule. The allylic radical is highly stabilized due to this delocalization.
The

• \(\left(\mathrm{CH}_{2}=\mathrm{CH}-\dot{\mathrm{C}} \mathrm{H}_{2}\right)\)

\ corresponds to an allylic radical. The electrons in the double bond can stabilize the unpaired electron through resonance, spreading the electron density over several atoms. This makes allylic radicals more stable compared to those that lack such delocalization.

Hyperconjugation

Hyperconjugation is a stabilizing interaction that occurs when electrons in a \(\sigma\)-bond, like those in carbon-hydrogen bonds, interact with an adjacent unoccupied or partially filled \(p\)-orbital. This interaction delocalizes the unpaired electron over several bonds, increasing radical stability.
In the case of

• \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\),

hyperconjugation plays a significant role. The two methyl groups provide electron density, which can interact with the radical center, stabilizing the unpaired electron. This effect is crucial in stabilizing more complex radicals, such as tertiary radicals, making them more stable than primary radicals.

Inductive Effect

The inductive effect refers to the transmission of charge through a chain of atoms, affecting the electron density at different points in a molecule. This effect can either push electrons towards or pull them away from a radical center, depending on the nature of the substituents.
In tertiary radicals, the presence of alkyl groups can donate electron density towards the radical center through the inductive effect. This can increase the stability of the radical by dispersing the electron density more broadly. For example, in radicals like

• \(\left(\mathrm{CH}_{3}\right)_{2} \dot{\mathrm{C}} \mathrm{H}\),

the inductive effect from the methyl groups enhances the stability of the radical. This understanding helps in predicting how radicals will behave in chemical reactions, making it easier to determine their relative stability.